Issuu

Contents 1

Bivariate data analysis 1 1.01

Introduction to bivariate data

1.02

Describe data

1.03

Associations between categorical variables

1.04

Explanatory and response variables

Investigation: Scatterplots and body measurements

1.05

Associations between numerical variables

1.06

Correlation using r-values

Investigation: Statistical investigation process

Coefficient of determination

Chapter 1 review

1.07

Bivariate data analysis 2

2.01

Least-squares lines

2.02

Interpret slope and y-intercept

104

2.03

Residual plots

111

Investigation: You are the smartest generation

122

2.04

Predictions

129

2.05

Association and causation

139

2.06

Practical applications of bivariate data

146

Chapter 2 review

155

Time series analysis

160

3.01

Read and interpret time series graphs

162

3.02

Moving averages

178

3.03

Deseasonalise data

197

3.04

Time series data in tables

219

3.05

Make predictions using time series data

244

Investigation: Time series data

260

Further applications

264

Chapter 3 review

280

3.06

Contents mathspace.co

iii

Growth and decay in sequences

284

4.01

Sequences

286

4.02

Arithmetic sequences

295

4.03

Arithmetic sequences with tables and graphs

300

4.04

Application of arithmetic sequences

306

4.05

Geometric sequences

312

4.06

Geometric sequences with tables and graphs

317

4.07

Further applications of growth and decay

323

Investigation: Depreciation methods

328

Chapter 4 review

331

Earth geometry and time zones

334

5.01

Latitude and longitude

336

5.02

Distance between two places on Earth

348

Investigation: Geocaching

358

5.03

International time zones and time differences

362

5.04

Australian time zones and daylight savings

377

5.05

Longitude and time difference

389

5.06

Distance and time zone applications

395

Chapter 5 review

405

Loans, investments and annuities 1

408

6.01

Simple interest - Review

410

6.02

Compound interest using sequences

417

6.03

Compound interest using technology

430

6.04

Effective annual interest rate

440

6.05

Recurrence relations for present value annuities

449

6.06

Formula for present value annuities

458

6.07

Spreadsheet & TVM solver

465

6.08

Applications of present value annuities

473

Chapter 6 review

481

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Loans, investments and annuities 2 7.01

Recurrence relation

486

7.02

Future value annuity formula

491

Investigation: Annuities with spreadsheets

497

7.03

Practical applications of annuities

500

7.04

Perpetuity formula

507

Chapter 7 review

513

Graphs and networks

516

8.01

Introduction to graphs and networks

518

8.02

Network representations

528

8.03

Planar graph and Euler’s formula

534

8.04

Walks, trails and paths

538

Investigation: Algorithms for determining the shortest path

545

Eulerian and Hamiltonian graphs

548

Chapter 8 review

554

8.05

484

Networks and decision mathematics 1

558

9.01

Trees and spanning trees

560

9.02

Prim’s algorithm

569

9.03

Applications of minimum spanning trees

580

9.04

Project networks

591

9.05

Forward and backward scanning

603

9.06

Critical path

612

9.07

Float times

621

9.08

Applications of critical path analysis

631

Chapter 9 review

642

Contents mathspace.co

Networks and decision mathematics 2

646

10.01

Introduction to flow networks

648

10.02

Capacity of a cut

653

10.03

Maximum flow and minimum cut

659

10.04

Assignment and allocation

669

10.05

Optimisation by inspection

676

10.06

Introduction to the Hungarian algorithm

683

10.07

Maximisation with the Hungarian algorithm

691

10.08

Apply the Hungarian algorithm: Minimisation

698

10.09

Solve complex assignment problems

703

Chapter 10 review

711

Answers

716

Mathspace QLD 12 General Mathematics 2025 mathspace.co

1.01 Introduction to bivariate data After this lesson, you will be able to… • understand the meaning of bivariate data

Univariate versus bivariate statistics Univariate data involves the analysis of a single variable. This was studied in Unit 1. Univariate data Univariate data involves the analysis of a single variable. It primarily describes and analyses the distribution, central tendency (mean, median, mode), and spread (range, interquartile range, standard deviation) of that variable. This type of data is typically represented using visualisations such as dot plots, box plots, or histograms. Example: Comparing students’ test results, or their heights.

Students’ heights (cm) 160 160 165 165 165 165

155 160 165 170 175

165

Students’ heights (cm)

170 170 170 In Unit 3, the focus will be statistics with bivariate data. ‘Bi’ means two (think bicycle), so when comparing or ﬁnding an association between two sets of different variables, this is bivariate statistics.

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Bivariate data Bivariate data involves two variables and explores the relationship between them. It primarily identiﬁes associations, correlations, or trends between two variables, often represented using scatter plots or time series. This analysis is used to explore relationships and make predictions.

Students’ heights (cm)

Students’ arm spans (cm)

152

150

158

164

162

168

169

172

170

180

182

Students’ arm spans (cm)

Example: Determining the relationship between the hours studied and the marks scored. Analysing the association between litres of soft drink consumed per week and BMI for a set of people involves working with bivariate data as there are two variables (litres and BMI).

180 175 170 165 160 155 150 0 150 155 160 165 170 175 180

Students’ heights (cm)

Exploration Consider these scenarios: • A researcher records the average daily temperature in Brisbane over a month. • A high school surveys its students to compare the number of hours they spend studying with their average test scores. • A nutritionist tracks the weight of participants weekly over three months • A botanist measures the height of a plant each week and records the amount of water it received that week. 1. For each scenario, identify the variable(s) being collected. 2. Explain your reasoning for classifying each scenario as univariate or bivariate data. 3. Provide another real-world example of both univariate and bivariate data.

1.01 Introduction to bivariate data mathspace.co

Ex 1

Determine whether these are examples of univariate or bivariate data: a

A school collects data on the height of each student in two classes in order to compare the classes.

A political group collects data on the taxable income and the suburb of their local constituents in order to investigate if there is an association.

A factory inspects a batch of light bulbs to measure how long each bulb lasts.

John wishes to investigate whether there is an association between the amount of natural sunlight in a classroom and students’ exam results.

Determine if these statistical questions would result in univariate or bivariate data: a

What is the average rainfall in Cairns during the wet season?

Is there a relationship between the type of fertilizer used and the height of a tomato plant?

How does the number of hours a student spends on homework affect their grades in mathematics?

For each scenario, identify the variable(s) being collected and classify the data as univariate or bivariate: a

A sports scientist collects data on the number of hours athletes train each week and their performance scores in competitions.

A car manufacturer wants to know the distribution of fuel economy (litres per 100 km) for their new model.

A nutritionist investigates the relationship between the daily caloric intake and weight change in a group of adults over three months.

Examine these data tables. Determine whether each table represents univariate or bivariate data. Justify your answer. a

Table A: Height (cm)

160

165

170

Hours studied

Test score

162

158

Table B:

Examine this data table. Determine whether it represents univariate or bivariate data. Justify your answer. Temperature (°C)

Ice cream sales ($)

100

150

220

1.01 Introduction to bivariate data mathspace.co

1.02 Describe data After this lesson, you will be able to… • construct two-way frequency tables • determine row sums in two-way frequency tables • determine column sums in two-way frequency tables • determine percentages in two-way frequency tables

Two-way frequency tables and marginal totals Two-way frequency tables are powerful tools used to organise and display the relationship between two categorical variables. They provide a clear visual summary of how different categories of one variable relate to categories of another. Each category of one variable is listed along the top (columns), while the categories of the second variable are listed down the side (rows). The interior cells of the table show the frequency (or count) of observations for each combination of categories. To construct a two-way frequency table from raw data, identify the two categorical variables and their respective categories. Then, tally the occurrences for each combination of categories and place these counts into the corresponding cells. For example, consider a survey of 100 high school students about their preferred sport and their year level: Student 1: Year 10, Football Student 2: Year 11, Basketball Student 3: Year 10, Netball Student 4: Year 12, Football ... Student 98: Year 10, Basketball Student 99: Year 12, Netball Student 100: Year 11, Basketball After tallying the preferences for each combination, the raw data can be organised into a two-way frequency table. This table includes marginal totals for each row and column, which represent the total frequencies for each category. The cell at the bottom right represents the grand total, which is the total number of observations:

Football

Basketball

Netball

Total

Year 10

Year 11

Year 12

Total

100

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The marginal totals (row and column sums) provide a summary of the distribution of each variable independently. For instance, in the table above, 35 students preferred Football overall, and 30 students are in Year 10. The grand total of 100 students represents the total size of the surveyed group.

Exploration Consider these raw data collected from a survey of 50 individuals regarding their preferred mode of transport (Car, Bus, Train) and their age group (Under 30, 30 − 60, Over 60): • Under 30: Car, Bus, Car, Train, Car, Bus, Train, Car, Bus, Car • 30 − 60: Train, Car, Bus, Train, Car, Bus, Train, Car, Bus, Train, Car, Bus, Train, Car, Bus, Train, Car, Bus, Train, Car • Over 60: Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train, Bus, Train 1. What are the two categorical variables in this dataset? What are their categories? 2. Construct a two-way frequency table to summarise this data, including all row and column sums. 3. What does the grand total of your table represent in the context of this survey?

Example 1 A local library surveyed 150 patrons on their preferred book genre and their age group. The raw data is partially represented: • Children (0 − 12): 15 Mystery, 10 Fantasy, 5 Sci-Fi • Teens (13 − 18): 20 Mystery, 25 Fantasy, 15 Sci-Fi • Adults (19+): 30 Mystery, 15 Fantasy, 15 Sci-Fi Construct a two-way frequency table, including row and column sums, from this survey data.

Create a strategy Identify the two categorical variables and their categories. Tally the frequencies for each combination to ﬁll the interior cells. Then, calculate the sum of each row and column to determine the marginal totals, and ﬁnally, the grand total.

1.02 Describe data mathspace.co

Percentages in two-way tables While frequency tables show the raw counts, their true analytical power is unlocked by converting these counts into percentages. This allows for fair comparisons between groups of different sizes. These are the statistics of the passengers and crew who sailed on the Titanic on its fateful maiden voyage in 1912: First class

Second class

Third class

Crew

Total

Survived

202

118

178

212

710

Died

123

167

528

696

1514

Total:

325

285

706

908

2224

Although it is interesting to note that 202 first class passengers survived, it is more informative to consider the percentage of survivors within each class. This allows for a clearer comparison across groups of different sizes. To calculate this, use row percentages. To determine a row percentage, divide the value in the table by the total for that row, then multiply the result by 100%: Write the formula

Substitute the values

Evaluate First class

Second class

Third class

Crew

Total

Survived

28.5%

16.6%

25.1%

29.9%

100%

Died

8.1%

11.0%

34.9%

46.0%

100%

This percentage frequency table now provides more useful information than the raw data. The ﬁrst row gives the percentage breakdown of survivors by class. From this, it can be easily read that 46.0% of the people who died were crew members whereas only 8.1% of the people who died were in ﬁrst class. The percentage of people in each class that survived or died can also be calculated by determining column percentages. To determine the percentage divide the value in the table by the column total and multiply by 100%. The calculations are shown in the table: First class

Second class

Third class

Crew

Survived

62.2%

41.4%

25.2%

23.3%

Died

37.8%

58.6%

74.8%

76.7%

Total

100%

This percentage frequency table now provides more useful information. The first column shows the percentage breakdown of survivors and deaths in first class. From the raw data it may be observed that a similar number of first class (202) and third class (178) passengers survived. However this can be misleading. The percentage frequency table indicates that 62.2% of first class passengers survived whereas only 25.2% of third class passengers survived.

1.02 Describe data mathspace.co

Exploration Consider this two-way frequency table showing the results of a survey on pet ownership among urban and rural households: Cats

Dogs

None

Total

Urban

120

Rural

120

Total

120

240

1. Calculate the percentage of urban households that own dogs (using a row percentage). 2. Calculate the percentage of dog owners who live in rural areas (using a column percentage). 3. Which type of percentage (row or column) would be more appropriate to determine if location inﬂuences the choice between owning a cat or a dog? Explain your reasoning.

Example 2 Maria surveyed a group of people about the type of job they had. She recorded the data in this graph, which corresponds to the frequency table: 40

Number of people

35 30 25

Men

Women

15 10 5 0

None

Casual

Part-time

Full-time

Type of employment

None

Casual

Part-time

Full-time

Total

Men

Women

Total

175

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a Complete this two-way table displaying the row percentages, rounded to one decimal place: None

Casual

Part-time

Full-time

Total

Men

17.6%

29.4%

41.2%

100%

Women

22.2%

⬚

100%

27.8%

Create a strategy Determine the number of men or women having each job type and divide each by the total number of men or women surveyed respectively. Multiply the answers by 100%.

Apply the idea From the frequency table, there are 10 casual men employees, and the total number of men is 85. Determine the row percentage

Evaluate

There are 15 part-time women employees, and the total number of women surveyed is 90. Determine the row percentage

Evaluate

There are 30 full-time women employees, and the total number of women surveyed is 90. Determine the row percentage

Evaluate

So the row percentage table is given by: None

Casual

Part-time

Full-time

Total

Men

17.6%

11.8%

29.4%

41.2%

100%

Women

22.2%

27.8%

16.7%

33.3%

100%

Reﬂect and check To check the accuracy of the row percentages, verify that the percentages in each row sum to approximately 100% (allowing for rounding). Also, consider what each percentage means in context; for example, 11.8% for casual men means that 11.8% of all surveyed men are casual employees.

1.02 Describe data mathspace.co

b Complete this two-way table displaying the column percentage, rounded to one decimal place: None

Casual

Part-time

Full-time

28.6%

62.5%

Women

⬚

57.1%

37.5%

Total

100%

⬚

100%

Men

⬚

Create a strategy Determine the number of men or women having a particular job type in the missing column entries and divide each by the total number of men and women having the corresponding job type. Multiply the answers by 100%.

Apply the idea From the frequency table, there are 15 men in the ‘None’ category, and the total number of men and women in the same category is 35. Determine the column percentage

Evaluate

There are 35 full-time men, and the total number of men and women in the same category is 65. Determine the column percentage

Evaluate

There are 25 casual women employees, and the total number of men and women in the same category is 35. Determine the column percentage

Evaluate

From the frequency table, there are 30 full-time women employees, and the full-time column total is 65. Determine the column percentage

Evaluate

Verify the full-time men percentage by calculating the difference from 100%: Full-time men percentage = 100% − 46.2% Write the equation = 53.8%

Evaluate

So, the column percentage table is given by:

None

Casual

Part-time

Full-time

Men

42.9%

28.6%

62.5%

53.8%

Women

57.1%

71.4%

37.5%

46.2%

Total

100%

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A vet has a total of 28 pets visit their practice in a day.

Vaccinated

The pets are categorised based on whether they have been vaccinated and whether they have been microchipped as shown in the Venn diagram:

Microchipped

Construct a two-way frequency table including all marginal totals for this data: Microchipped

Not microchipped

Total

⬚

Vaccinated

⬚

Not vaccinated Total

⬚

Sophia asked some people in her community whether they were vegetarian or not. 29 responders said they were vegetarian, of which 8 were children. 14 children said they were not vegetarian, and 11 adults said they were not vegetarians: a

Construct a two-way frequency table including all marginal totals based on the results of Sophia’s survey:

Determine the proportion of responders that are vegetarian.

Children

⬚

Adults

Determine the proportion of adults that are vegetarian.

Total

⬚

This table shows the results of a survey on pet ownership and living location. Complete the two-way frequency table by ﬁlling in the missing cells (frequencies and marginal totals): Cats

Dogs

None

Total

Urban

Rural

⬚

120

⬚

Total 6

Not Vegetarian Total vegetarian

36 students were asked whether or not they were allergic to nuts and dairy. The two-way table is provided: Allergic to nuts

Not allergic to nuts

Allergic to dairy

Not allergic to dairy

Determine the number of students that are allergic to nuts.

Determine the number of students that are allergic to nuts or dairy, or both.

Determine the number of students that are allergic to at most one of the two things.

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Ex 2

A school surveyed 200 students about their mode of transport to school (Walk, Bike, Car) and their year level (Year 7, Year 12). The completed results are shown: Year 7

Year 12

Total

Walk

Bike

Car

100

Total

100

200

Calculate the percentage of students who walk to school.

Calculate the percentage of Year 7 students who bike to school.

Calculate the percentage of students who travel by car that are in Year 12.

Calculate the percentage of Year 12 students who travel by bike.

This table shows the distribution of students across different grade levels and their participation in after-school clubs: Sports club

Art club

Debate club

Total

Grade 9

Grade 10

Grade 11

Grade 12

Total

220

Construct a new two-way table displaying all row percentages. Round values to one decimal place. 9

A survey asked employees about their primary commute method (Car, Public transport, Bike) and their job satisfaction (Satisfied, Neutral, Dissatisfied). The results are shown in the table: Satisfied

Neutral

Dissatisﬁed

Total

Car

Public transport

Bike

Total

200

Construct a new two-way table displaying all column percentages. Round values to one decimal place.

1.02 Describe data mathspace.co

150 tennis players were asked whether they would support equal prize money for the women’s and men’s draw: a b

Determine how many more players are in support of equal prize money than against it.

Support

Do not support

Total

⬚

Calculate the percentage of the male ⬚ Total tennis players that support equal prize money. Round your answer to one decimal place if necessary.

⬚

Males Females

⬚

This table describes the departures of ﬂights out of an airport for the months of May and June: a

Construct a completed two-way frequency table including all marginal totals.

Determine the percentage of ﬂights in June that were delayed, rounded to one decimal place.

Departed on time

Delayed

May

144

June

116

Determine the fraction of the total number of ﬂights during the two months that departed on time in May.

Determine the percentage of ﬂights in June that were on time, rounded to one decimal place.

Two motels promoted packages for 1 night and 2 night stays. Results on the number of packages sold are recorded in the table: a

Determine the percentage of the stays that are two-night stays, rounded to two decimal places.

Determine the fraction of the two-night stays that were in Motel X.

Determine the percentage of the total accommodation that was in Motel X, rounded to two decimal places.

One-night stay

Two-night stays

Motel X

463

266

Motel Y

804

817

Complex familiar 13

Glen surveyed all of the students in Year 12 at his school and summarised the results in this table: Play netball

Do not play netball

Height ≥ 170 cm

138

Height < 170 cm

140

Identify the explanatory variable.

Construct a row percentage frequency table for this data. Round the values to the nearest percentage.

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170 people were surveyed about their music preference. The results have been recorded in the table: Music preference

Male

Female

Total

Rock and Roll

Classical

Pop

Rap

Country and Western

R and B

Punk

Total

170

Identify the explanatory variable for this data set.

Determine which of these 100% stacked column charts should be used to look for an association between the variables. A

Music preferences by gender 100 90 80 70 60 50 40 30 20 10 0

Rock & Roll

Classical

Pop

Rap

Country & Western Music Preference Male

Female

Music preferences by gender 100 90 80 70 60 50 40 30 20 10 0

R&B

Punk R&B Country & Western Rap Pop Classical Rock & Roll Male

Gender

Female

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Punk

Analyse if the stacked column chart suggests that there is an association between music preference and gender. Explain your answer.

Weekend workers at a fast food store were surveyed as to which pathway they followed at school (ATAR or Non-ATAR) and the type of phone (IPhone, Samsung or Other). The results are displayed in the column graph: Comparison of phones 70

50 40 30

20 0

iPhone

Samsung ATAR

Other

non-ATAR

Construct a completed frequency table, including marginal totals, for the given data:

ATAR non-ATAR

iPhone

Samsung

Other

Total

⬚

Identify which variable (School pathway or Type of phone) should be plotted on the horizontal x-axis.

Determine the statement that best describes the association between school pathway and type of phone: A

It appears that there is an association. Students following an non-ATAR pathway are more likely to have an iPhone.

There appears to be no association between the variables. The numbers in each category of phone are similar.

It appears that there is an association. Students following an ATAR pathway are more likely to have an iPhone.

It appears that there is an association. Students following an ATAR pathway are more likely to have a Samsung phone.

1.02 Describe data mathspace.co

A marketing ﬁrm surveyed 200 customers about their preferred method of receiving promotions (Email, SMS) and their age group (Under 25, 25 − 45, Over 45). The results are: Under 25

25-45

Over 45

Total

130

SMS

Total

200

To determine if there is an association between age group and preferred promotion method, which type of percentage (row or column) would be more appropriate? Explain your reasoning.

Based on your choice in part (a), construct the appropriate percentage table. Round values to one decimal place.

Analyse the percentages to discuss the association between age group and preferred promotion method.

A high school tracked its students’ participation in either the Arts or Sports program, and whether they achieved High Distinction (HD), Distinction (D), or Credit (C) in their end-of- year academic results. Out of 300 students: • 40 Arts students achieved HD. • 60 Arts students achieved D. • 50 Arts students achieved C. • 30 Sports students achieved HD. • 70 Sports students achieved D. • 50 Sports students achieved C.

Construct a two-way frequency table, including all marginal totals, for this data.

Calculate the percentage of Arts students who achieved HD. Round to one decimal place.

Calculate the percentage of students who achieved HD that were in the Sports program. Round to one decimal place.

Calculate the overall percentage of students who achieved a Distinction (D). Round to one decimal place.

Analyse if there is an association between participation in the Arts or Sports program and academic results. Justify your answer using relevant percentages.

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Complex unfamiliar 19

A study recorded the number of serious injuries among male and female athletes in two different sports over a year. The results are shown: Sport A injuries

Sport B injuries

Total injuries

Male injuries

Female injuries

Total

Based on the raw counts, a coach concludes that females have more injuries in Sport B than males. Construct both a row percentage table and a column percentage table for this data. Round values to one decimal place.

Analyse how these percentage tables provide a clearer picture of injury association compared to the raw counts, addressing the coach’s initial conclusion.

A local news report states: “Evidence shows that students who use social media for more than 3 hours a day have lower academic grades.” a

Propose how a two-way frequency table could be constructed to investigate this claim, including suitable variables, categories, and a brief description of how data might be collected.

Analyse how the percentages from such a table could support an association. Furthermore, discuss if this association necessarily implies a causal link, and suggest any potential confounding variables.

A company that sells two types of coffee blends, ‘Morning Brew’ and ‘Evening Roast’, collected data on customer preferences by age group: Under 30

30 − 50

Over 50

Total

Morning Brew

Evening Roast

Total

150

Construct a table of row percentages and a table of column percentages for this data. Round values to one decimal place.

Analyse how interpreting different percentage tables (row vs. column) can lead to different conclusions about customer preferences. Justify which percentage table is more appropriate for answering each of these questions: 1.

Which age group is most likely to prefer ‘Evening Roast’?

2. Which coffee blend is preferred by the 30 − 50 age group?

1.02 Describe data mathspace.co

1.03 Associations between categorical variables After this lesson, you will be able to… • use an appropriately percentaged two-way frequency table to identify patterns that suggest the presence of an association • understand an association in terms of differences observed in percentages across categories • interpret an association in the context of the data

Associations between categorical variables To determine if an association exists between the variables in the Titanic table, the question “Is survival rate dependent on the class of the passenger?” can be posed. To determine this, the explanatory variable must ﬁrst be identiﬁed. In this problem, the class of passenger is the explanatory variable. As the explanatory variable forms the heading of each column, the column percentage frequency table will best indicate any patterns.

Exploration Consider the data presented in the Titanic table: First class

Second class

Third class

Crew

Survived

62%

41%

25%

23%

Died

38%

59%

75%

77%

Total:

100%

1. Which variable is considered to influence the other: ‘passenger class’ or ‘survival status’? Explain the reasoning. 2. Based on the identified influencing variable, would a row or column percentage frequency table be more appropriate for identifying patterns? Justify the choice. 3. Examine the percentages in the provided table. What patterns are observed between passenger class and survival rates?

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Identifying associations using percentage frequency tables Two-way tables can be used to examine whether an association exists between two categorical variables by comparing percentages. When row or column percentages are calculated, it becomes easier to compare groups of different sizes and identify patterns. Interpreting percentage differences For example, when reading across the first row in a column percentage table, clear differences in the percentages of passengers who lived or died across passenger classes may be observed. If the survival rate increases with passenger class, this suggests an association between passenger class and survival. In this case, it appears that higher-class passengers were more likely to survive. Which percentage frequency table should be used? The type of percentage frequency table to use depends on the explanatory variable—the variable that is believed to influence the other. • If the explanatory variable is listed in the columns, use a column percentage frequency table. • If the explanatory variable is listed in the rows, use a row percentage frequency table. How to determine if an association exists? To identify whether an association exists, in a column percentage table, read across the rows: • If the percentages vary significantly, an association is suggested. • If the percentages are similar, no clear association exists. How to describe the association A good description of an association should include three parts: • State whether there is an association: “There appears to be an association between the variables” • Describe the nature of the association: “The class of passenger affects the survival rate of the passengers” • Support with an example: “The higher the class of passenger, the more likely the passenger was to survive.” Important note on association The term association describes a relationship between two variables. It is important to note that association does not imply causation. Just because two variables appear to be related does not mean that one causes the other to change. Additional statistical methods and analysis are needed to determine the strength and significance of any association.

1.03 Associations between categorical variables mathspace.co

Exploration Examine the two stacked column graphs provided. The ﬁrst displays the Titanic passenger data you have previously explored, and the second shows hypothetical data on household composition and money distribution. Death vs. Survival: Titanic 1912 100% 90% 80%

Died, 38 Died, 59

70% 60%

Died, 75

Died, 77

Survived, 25

Survived, 23

Third Class

Crew

50% 40% 30% 20%

Survived, 62

Survived, 41

10% 0%

First Class

Second Class

Household expenditure by domestic situation 100% 90% 80% 70% 60% 50% 40% 30% 20% 10%

Housing

Single Transportation

Married Food

Married with kids Insurance

Single with kids

Entertainment

Other

1. In the Titanic graph, how do the proportions of ‘Survived’ and ‘Died’ segments change across the different passenger classes? What does this suggest about an association? 2. In the money distribution graph, how do the proportions of the segments compare across the different household compositions? What does this suggest about an association? 3. Based on the observations, summarise the visual characteristics that indicate the presence of an association in a stacked column graph, versus the characteristics that indicate no clear association.

1.03 Associations between categorical variables mathspace.co

b Which of these 100% stacked column charts should be used to look for an association between the variables? Music preferences by gender 100% 80% 60% 40% 20% 0%

Rock & Classical Roll

Pop

Rap

Country & R&B Western

Punk

Music Preference Male

Female

Music preferences by gender 100 90 80 70 60 50 40 30 20 10 0

Punk R&B Country & Western Rap Pop Classical Rock & Roll Male

Female Gender

Create a strategy Use the fact that a correct chart should have the explanatory variable along the horizontal axis.

Apply the idea In part (a), gender was identiﬁed as the explanatory variable. So gender should be along the horizontal axis. The correct graph is Option B.

1.03 Associations between categorical variables mathspace.co

1.03 Practice questions Simple familiar 1

For each scenario, identify the explanatory and response variables: a

Does the amount of sleep a student gets affect their test scores?

Is there a relationship between a person’s age and their preference for a particular music genre?

Does the type of fertiliser used impact the yield of a crop?

A survey asked oﬃce workers about their preferred morning drink. The results were recorded in this percentage table: Coffee

Tea

Total

Day shift

60%

40%

100%

Night shift

85%

15%

100%

What percentage of day shift workers prefer tea?

Does there appear to be an association between the work shift and preferred drink? Explain the answer.

A total of 50 students were surveyed about their science subject choices.

Studies physics

The subjects are categorised based on whether they study physics and whether they study chemistry, as shown in the Venn diagram:

Studies chemistry

Complete this two-way table, including all marginal totals:

Studies physics Not physics Total

Studies chemistry

Not chemistry

Total

⬚

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⬚ ⬚

A travel agency surveyed people about their preferred holiday type. 35 respondents said they preferred a Beach holiday, of which 15 were under 40 years old. 10 people under 40 preferred a City Break, and 15 people aged 40 or over preferred a City Break. a

Complete this two-way table based on the survey results:

Determine what proportion of respondents prefer a City Break.

Determine what proportion of the ‘Under 40’ group prefer a Beach holiday.

40 and Over

City Break

⬚

200 council residents were surveyed on whether they supported a new recycling program: a

Complete this table, including marginal totals.

Determine how many more residents support the program than do not support it.

Calculate the percentage of ‘Under 50’ residents that support the program.

Support Under 50 50 and over

⬚

Do not support 30 20

A company tracks its website visits for the ﬁrst two quarters of the year by device type: a

Under 40

Beach

Calculate the percentage of website visits in Q2 that were from a mobile device, rounded to one decimal place.

Desktop

Mobile

250

150

200

400

Calculate the fraction of total visits during these two quarters that came from a desktop computer in Q1.

Calculate the percentage of Q1 visits that were from a desktop computer, rounded to one decimal place.

The sales ﬁgures for two smartphone models at two different electronics stores are recorded in this table: a

Calculate the percentage of total phone sales that are for Model Y, rounded to two decimal places.

Determine the fraction of Model Y sales that occurred at Store B.

Calculate the percentage of total sales that occurred at Store A, rounded to two decimal places.

Store A

Store B

Model X

300

250

Model Y

400

550

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A survey investigated if there was an association between a student’s year level and the industry of their part-time job. 50 students were chosen at random from each year group. a

Identify which variable, job industry or year level, represents the explanatory variable in this problem.

The results of the survey are recorded in this table: Retail

Hospitality

Other

Total

Year 11

Year 12

Total

Determine the values of: i c

d Ex 1

iii

Calculate the percentage of: i

Year 11 students who work in hospitality

Year 12 students who work in hospitality

Describe the association between year level and working in hospitality.

A coach surveyed students to see if there was a link between their height and whether they played basketball. The results are in this table: Plays basketball

Does not play

Total

Height ≥ 180 cm

100

Height < 180 cm

100

Total

135

200

Identify which variable is the explanatory variable.

To examine if there is an association between height and playing basketball, should a column or row percentage frequency table be used?

Construct a row percentage frequency table for this data.

Looking at the columns of the completed table, does there appear to be an association between height and playing basketball? Explain.

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Ex 2

A local community group surveyed residents about their pet ownership and whether they had a backyard. The results are in this table: Owns a pet

Does not own a pet

Has a backyard

No backyard

Identify the explanatory variable if the group wants to investigate if having a backyard inﬂuences pet ownership.

Determine whether a row or column percentage frequency table is more appropriate to investigate this association. Justify the answer.

Construct the appropriate percentage frequency table. Round to the nearest whole percent.

Describe any association observed from the table.

200 people were surveyed about their preferred device for watching movies. The results have been recorded in this table: Age group

Laptop

Phone

Total

Under 25

100

25 and over

100

Total

200

Identify the explanatory variable for this data set.

Identify which of these 100% stacked column charts should be used to look for an association between the variables. A

Age group by preferred movie device

Preferred movie device by age group

100%

80%

60%

40%

20%

TV Laptop Phone Preferred device 25 and over Under 25

TV Laptop Phone

Under 25 25 and over

Age group

Does this stacked column chart suggest that there is an association between age group and preferred device? Explain the answer.

1.03 Associations between categorical variables mathspace.co

A survey asked concert-goers about their ticket type (Seated or General Admission) and their preferred season for attending an outdoor concert (Summer or Winter). The results are displayed in this column graph:

Number of concert-goers

Preferred concert season by ticket type 100 90 80 70 60 50 40 30 20 10

Seated

General admission Age group

Summer

Winter

Construct a completed frequency table, including marginal totals, for the given data:

Identify which variable (Ticket type or Season) should be plotted on the horizontal axis (x-axis).

Select the statement that best describes the association between ticket type and season preference: A

There appears to be no association between the variables as the totals for each ticket type are the same.

It appears that there is an association. People with general admission tickets are more likely to prefer summer.

It appears that there is an association. People with seated tickets are more likely to prefer summer.

Complex familiar 13

Members of a gym were asked what kind of training they do. Each of them only did one kind of training. This table shows the results:

Cardio

Weight

Male

Female

Identify which variable is the explanatory variable.

Construct a row percentage frequency table for this data. Round the values to the nearest percentage.

Does there appear to be an association between the type of training and the gender of gym members? Explain the answer.

Does a person’s gender cause them to choose a certain type of training? Explain the answer.

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Ex 3

A study was conducted to see if there is an association between the type of part-time job a student has and their reported stress level. The results from the survey are displayed in the 100% stacked column chart shown. a

Identify the explanatory variable for this chart.

Does this chart suggest an association between job type and stress level? Explain the answer.

Can it be stated that working in hospitality causes a high chance of stress? Explain the answer.

Reported stress levels by part-time job type 100% 80% 60% 40% 20% 0%

High stress

The local library constructed divided bar graphs showing the last six months of book loans by category: Assume that the total number of loans was exactly the same for each month.

Retail Hospitality Other Type of part-time job

Determine in which month did subscribers borrow the most business books.

Identify which month was the best month for children’s books.

Identify the response variable for this data.

Does this divided bar graph suggest that there is an association between the variables? Explain the answer.

Medium suess

Low stress

Jun

D E

Jul

D E

Aug

Sep

D E

Oct Nov

D E

Children's Design Fiction Environmental Business

This stacked column chart shows the ﬁtness levels for various categories of smokers: 100%

Fitness of smokers

Below average fitness

80%

Normal fitness

60%

Above average fitness

40% 20% 0%

Very Heavy Heavy (1−5) (16−25)

Moderate (6−15)

Light Non−smoker (1−5) (1−5)

Type of part-time job a

Identify the response variable for this chart.

Does there appear to be an association between ﬁtness level and category of smoker?

If so, describe this association. 1.03 Associations between categorical variables mathspace.co

25 students were chosen at random from each year group at a particular school and asked these questions: • Question 1: What is your favourite genre? Cartoons, Reality TV, Movies or Documentaries? • Question 2: Which year group are you in? The results are displayed in this divided column graph: Student’s favourite TV genre

100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

Year 7

Year 8

Year 9

Reality TV

Cartoons

Year 10

Movies

Year 11

Year 12

Documentaries

Using the graph above or another method, describe the association between these variables: a

Cartoons and Year group

Genre and Year group

Documentaries and Year group

Complex unfamiliar 18

A city council is investigating the effectiveness of its public transport system. It surveyed 500 residents, asking about their primary mode of transport for commuting (Car, Bus, Train, Cycle) and their overall satisfaction with their commute (Satisfied, Neutral, Dissatisfied). The council then presented this statement: “Our data shows that a higher percentage of train commuters are satisfied compared to car commuters. Therefore, encouraging more residents to use trains will significantly increase overall commuter satisfaction in the city.”

Satisﬁed

Neutral

Dissatisﬁed

Car

110

Bus

Train

Cycle

Evaluate the council’s conclusion. Is it valid based on the provided data? Justify the answer by creating an appropriate percentage frequency table and interpreting the data.

What further data or analysis would be required to support a more robust recommendation for increasing overall commuter satisfaction?

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A local charity wants to understand if there’s a relationship between a donor’s age group and their preferred donation method (Online, Mail, Phone). They randomly selected 120 recent donors and collected this data. Mail

Phone

18 − 30 years

31 − 50 years

51 − 70 years

71+ years

Formulate an investigative question that the charity could ask to explore the relationship between donor age and donation method.

Which variable is the explanatory variable? Construct the appropriate percentage frequency table to investigate the association. Round to the nearest whole percent.

The charity considers launching an “Online Only” campaign for all donors under 50 years old. Evaluate this decision based on the data.

A tech company claims its new “Focus Mode” app signiﬁcantly improves employee productivity. They present this graph to support their claim: a

Online

Identify the explanatory variable presented in this graph. Explain why this is the correct choice for an explanatory variable in this context.

100%

Productivity level

80%

High productivity Moderate productivity Low productivity

60% 40% 20% 0%

Used Did not use focus mode focus mode App usage

Based on the graph, describe the association between app usage and productivity level. Does the graph support the company’s claim?

Can the company deﬁnitively state that “Focus Mode” causes increased productivity? Justify the answer by identifying potential limitations or other factors not accounted for in this graph.

A public health organisation is investigating the claim that “access to green spaces improves mental well-being.” They plan to collect data on residents’ proximity to parks (High, Medium, Low) and their self-reported mental well-being scores (High, Moderate, Low): a

If the organisation hypothesises that proximity to green spaces inﬂuences mental wellbeing, design a hypothetical percentage frequency table (including row/column totals) that would strongly suggest an association that supports their claim.

Sketch a 100% stacked column graph that visually represents the data from the table created in part (a). Label the axes and segments clearly.

Apart from collecting more data, what two qualitative explanations could be given to further support or refute the idea that proximity to green spaces causes improved mental well-being?

1.03 Associations between categorical variables mathspace.co

1.04 Explanatory and response variables After this lesson, you will be able to… • identify the explanatory variable • identify the response variable • construct a scatterplot to identify the association between two numerical variables • use a scatterplot to identify the association between two numerical variables

Explanatory and response variables The goal of bivariate data analysis is to see if two variables are associated in some way. The two variables that are studied in bivariate statistics are called the explanatory variable and the response variable. Explanatory Variable

Response Variable

The variable that is changed or controlled to see its effect on another variable, or that explains changes or trends in another variable. This is plotted on the horizontal or x-axis.

The variable that is measured or observed in response to changes in the explanatory variable. This is plotted on the vertical or y-axis.

Example: In a study of plant growth over time, the number of days (time) is the explanatory variable because it helps explain the change in the plant’s height, even though we don’t control time itself.

Example: For a tree growing at 5 cm per day, the height of the tree is the response variable as it changes based on the number of days.

A single coordinate point in a bivariate data set might be written in the form (x, y), and it would be understood that x is the explanatory variable and y is the response variable.

1.04 Explanatory and response variables mathspace.co

Exploration Use technology to construct a scatterplot for this data on hours studied and test scores: Hours studied

Test score

Based on the scatterplot displayed on your calculator, what can you observe about the direction and form of the association between hours studied and test score?

The form of a scatterplot refers to the type or shape of relationship that appears between two variables. This can be identiﬁed by the overall pattern of the data points. • A linear form occurs when the points lie roughly along a straight line. This suggests a constant rate of change between the two variables. • A non-linear form occurs when the points follow a clear curved pattern. Any distinct pattern that is not a straight line is considered non-linear. Linear

Non-Linear (Curved)

x 0

A scatterplot allows students to identify the association between two numerical variables. An association describes the relationship or pattern between the variables. When examining a scatterplot, look for the: • Direction: Is the association positive, negative, or neither? • Positive association: As the explanatory variable increases, the response variable also tends to increase (points rise from left to right). • Negative association: As the explanatory variable increases, the response variable tends to decrease (points fall from left to right). • No association: There is no clear pattern or relationship between the variables. • Form: Is the association linear or non-linear? • Linear association: The points tend to cluster around a straight line. • Non-linear association: The points tend to cluster around a curved pattern. 1.04 Explanatory and response variables mathspace.co

• Strength: How closely do the points follow the pattern? • Strong association: The points are tightly clustered around the line or curve. • Moderate association: The points show a general trend but are more spread out. • Weak association: The points show a very loose or scattered trend. The strength of a linear correlation relates to how closely the points resemble a straight line. • If the points lie exactly on a straight line, then there is said to be a perfect correlation. • If the points are scattered randomly, it can be said that there is no correlation. Most scatterplots will fall somewhere in between these two extremes and will display a weak, moderate or strong association. Perfect positive correlation

Perfect negative correlation

x 0

Strong positive correlation

Strong negative correlation

x 0

Weak negative correlation

Weak positive correlation

x 0

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No correlation

The strength of the correlation can be determined by how closely packed the points are to each other on the graph. Points that end up far outside the general cluster of points are known as outliers.

Test score 100 80 60 40 20

Hours studied 0

Many graphing calculators and statistical software can construct scatterplots and analyse associations. Students can input data into the calculator’s statistics function and generate a scatterplot. While the exact steps vary by calculator model, the general process involves: 1. Accessing the statistics menu. 2. Entering the explanatory variable data into one list and the response variable data into another. 3. Selecting the scatterplot option from the graphing menu. 4. Adjusting the viewing window to ﬁt the data points. 5. Interpreting the visual display to identify the association.

Example 3 A researcher collected data on the number of hours students spent studying for a mathematics test and their scores on the test. The data is shown in the table: Student

Hours studied

Test score

1.04 Explanatory and response variables mathspace.co

Construct a scatterplot to display the relationship between hours studied and test score.

Create a strategy Identify the explanatory and response variables. Then, draw a Cartesian plane, label the axes with appropriate scales, and plot each data point.

Apply the idea

Reﬂect and check

Hours studied is the explanatory variable (x-axis) as it inﬂuences the test score. Test score is the response variable ( y-axis) as it is affected by the hours studied.

The scatterplot visually represents the relationship, allowing for quick observation of patterns.

100

Test score

80 60 40 20 0

Hours studied

Example 4

Daily ice cream sales ($)

The scatterplot shows the relationship between daily ice cream sales and the maximum daily temperature. 400 350 300 250 200 150 100 50 0

Maximum daily temperature (°C) Describe the association between daily ice cream sales and the maximum daily temperature in terms of direction, form, and strength.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

1.04 Practice questions Simple familiar 1

Ex 1

Determine the explanatory and response variable in each given pair: a

Gender and favourite dog breed

Child’s eye colour and parent’s eye colour

Country of birth and language spoken at home

Amount of sleep and alertness level

Provide an example of a possible explanatory variable for these response variables: a

In a study on car accidents, the response variable was “type of injury”.

In a study on professional careers, the response variable was “income bracket”.

In a study on spending habits, the response variable was “allocation of income (health, food, rent, ...)”.

Consider these variables: • Ticket sales • Revenue from a show a

b 4

Determine if these statements are correct: i

Revenue from a show affects ticket sales.

Ticket sales affect revenue from a show.

Identify the explanatory variable and the response variable.

The graph shows the height of a ball after it is dropped off the side of a building: a

What is the response variable?

What is the explanatory variable?

Height

Time 5

A graph is to display the relationship between these variables: • Fitness level • Time spent exercising Identify which of these variables should be placed on the horizontal axis of the graph.

1.04 Explanatory and response variables mathspace.co

Determine the form of the association shown in each scatterplot. Is the relationship linear or non-linear or no relationship? a

y 9 8 7 6 5 4 3 2 1

2 3 4 5 6 7 8 9

9 8 7 6 5 4 3 2 1

x 0

y 9 8 7 6 5 4 3 2 1

9 8 7 6 5 4 3 2 1

x 0

2 3 4 5 6 7 8 9

Consider these three scatterplots: A

All of them show a positive relationship between two variables, but the strength of the correlation varies. Order the scatterplots (A, B, C) from weakest to strongest positive correlation. Justify the ordering by explaining how closely the data points cluster around a straight line in each plot. 56

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 2

For each set of variables, determine if one variable inﬂuences the other: a

Arm length (cm) and weight (kg)

Temperature (degrees Celsius) and distance from the equator (km)

Weekly income ($) and number of friends

Time of travel (minutes) and distance covered (m)

Age (years) and time spent watching television (hours)

Time spent working (hours) and wages earned ($)

A teacher wants to investigate the relationship between the number of hours spent watching TV and the results of a mathematics test for the students in her class. She collects the data, and is going to display her results in a scatterplot. Determine if each set of axes correctly displays the explanatory and response variables: a

Test score (%)

TV watched (hours)

100

TV watched (hours) 0 Ex 3

Test score (%)

100

Jack is tracking the growth of a new plant. He measures the height of the plant each week and records it in a table. Construct a scatterplot for these data in the table to display the plant’s growth over time:

Week (x)

Height in cm (y)

For each pair of variables: i

Determine if a relationship is likely between the two variables.

If applicable, determine whether there is a positive or negative correlation between the two variables.

Typing speed and dancing ability

Temperature and ice cream sales

Time spent studying and exam grade

Eye colour and IQ

The drama department is making costumes for their school’s upcoming musical. The amount of materials needed for costumes will be determined by the number of performers in the musical. Determine if these statements are true or false: a

The explanatory variable cannot be determined without further information.

The number of performers is the explanatory variable.

The time spent making costumes is the explanatory variable.

The amount of materials needed is the explanatory variable. 1.04 Explanatory and response variables mathspace.co

The scatterplot shows the relationship between air and sea temperature:

Sea temperature

Ex 4

Describe the association between air temperature and sea temperature in terms of direction, form, and strength.

35 30 25

Air temperature 14

The scatterplot shows the relationship between daily study time and test scores: a

Identify the response variable.

Identify the explanatory variable.

Describe the association between daily study time and test scores in terms of direction, form, and strength.

100 80

Test score

Ex 5

60 40 20 0

Daily study time (hours)

Complex familiar 15

The table shows the average hours of exercise per week and the resting heart rate (beats per minute) for a sample of individuals: Hours of exercise per week

Resting heart rate

Construct a scatterplot for these data.

Determine the correlation between the hours of exercise per week and resting heart rate based on the scatterplot.

Identify if this data set appears to contain any outliers. If so, identify the data point.

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After a mathematics exam, Lachlan commented that he left out Question 10 because it was not relevant to the topics that were assessed, but that the other questions were ﬁne. The teacher felt that Question 10 and Question 7 were assessing the same skills, so she decided to look at the 1 2 results of students across all 8 classes. She plotted the percentage who had gotten Question 10 correct against % that got the percentage who had gotten Question 7 correct: Q10 correct a

According to the graph, determine who is correct: Lachlan or the teacher.

Assume that Lachlan performed exceptionally well % that got Q7 correct in every other question. If a whole class of students had the same result as Lachlan, identify where this class’s results could appear.

The land size of a sample of properties and their selling price were graphed on the scatterplot:

Price (dollars)

1 200 000 1 000 000 800 000 600 000 400 000 200 000 0

1000 2000 3000 4000 5000 6000

Size (square metres) a

Which size range contains the most properties? A

4500 − 5500 m2

1500 − 2500 m2

2500 − 3500 m2

Another property that is not represented in the scatterplot sold for $704 000 and measured 3600 m2 in size. Determine if this is consistent with the relationship shown in the graph.

Complex unfamiliar 18

For each scenario, identify the response variable and the explanatory variable: a

A proposed area of research has the title “Can we use a person’s age to predict how much time they spend exercising each day?”

A group of students is investigating how effective different bacterial soaps are at killing germs.

A student observes a relationship between petrol prices and the day of the week.

1.04 Explanatory and response variables mathspace.co

Yvonne has a high income but does not like her job. Identify which dot is most likely to represent Yvonne.

Job satisfaction

2 3 1 0

Income 20

Consider the table of values that show four excerpts from a survey comparing the average daily screen time of teenagers and their average nightly sleep duration: Determine if the relationship would be expected to be strongly positive, strongly negative, or have no clear relationship.

Average sleep duration (hours)

8.5

7.0

6.0

5.5

The table lists the time taken to sprint 400 m by runners who all ran in different temperatures as part of a study:

Temperature (°C)

Time (s)

Construct a scatterplot to represent the data in the table.

Determine the number of runners tested in the study.

Determine the correlation between temperature and sprint time for the data.

Average daily screen time (hours)

Identify which data point represents an outlier.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

1.05 Associations between numerical variables After this lesson, you will be able to… • describe an association between two numerical variables • identify the direction (positive/negative) of an association between two numerical variables • recognise the form (linear/non-linear) of an association between two numerical variables • describe the strength (strong/moderate/weak) of an association between two numerical variables

Associations between numerical variables When analysing bivariate data, the relationship between an explanatory variable (x) and a response variable ( y) is examined. This relationship, also known as an association or correlation, is described by three key features observed in a scatterplot. • Form: The overall pattern of the data points. This can be linear (forms a straight line) or non-linear (forms a curve). • Direction: If the form is linear, the direction describes the trend. It can be positive (as x increases, y generally increases) or negative (as x increases, y generally decreases). • Strength: If the form is linear, this describes how closely the points follow a straight-line pattern. It can be described as strong, moderate, or weak. If there is no discernible pattern, there is no correlation. These scatterplots illustrate these different types of linear associations: 9

x 0

Strong positive correlation

x 0

Strong negative correlation

1.05 Associations between numerical variables mathspace.co

x 0

Weak positive correlation 9

Weak negative correlation

x 0

No correlation

x 0

Non-linear correlation

Data points that sit far from the main pattern of the data are called outliers.

Exploration Consider these scenarios where a strong correlation has been observed. Discuss these questions with a partner or as a class: • Scenario 1: A strong positive correlation is found between monthly ice cream sales and the number of drowning incidents. • Scenario 2: In a study of primary school students, a strong positive correlation is found between their shoe size and their reading ability. 1. For each scenario, does it seem plausible that the ﬁrst variable causes the change in the second? For example, does eating ice cream cause drowning? Does having bigger feet make you a better reader? Explain your initial thoughts. 2. If direct causation is unlikely, what other explanations could there be for the observed correlation? Can you think of a third factor (a confounding variable) that might be inﬂuencing both variables in each scenario? 3. Think of another real-world example where two variables might be correlated, but one does not cause the other. What is the likely explanation for the association?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

1.05 Practice questions Simple familiar 1

For each scatterplot, determine whether the correlation is positive, negative, or neither: a

x 0

Describe the association for each scatterplot by stating its form (linear or non-linear) and strength (strong, moderate, or weak): a

x 0

Ex 1

x 0

Mathspace QLD 12 General Mathematics 2025 mathspace.co

x 0

For each pair of variables, determine whether you would expect a positive, negative, or no correlation: a

A person’s height and their shoe size

The number of hours a phone is used and its remaining battery percentage

A person’s exam score and the colour of their hair

The daily temperature and sales of air conditioners

Scientists found that the longer a paragraph, the less information a person could retain from it. Is the correlation between the length of the paragraph and the information retained positive or negative?

The scatterplot shows the relationship between hours of sleep and reaction time for a group of people: Reaction time (s) 0.5 0.4 0.3

0.2

Hours of sleep 0

9 10

Describe the association between hours of sleep and reaction time.

One person slept for 5.5 hours and had a reaction time of 0.25 seconds. This is marked as point A. Does this point appear to be an outlier? Explain why or why not.

1.05 Associations between numerical variables mathspace.co

Ex 2

The table shows the time taken to ﬁnish a lap for several average speeds: Speed (km/h)

Time (min)

8.5

8.7

7.5

8.2

6.9

7.3

6.0

5.7

4.5

4.9

Construct a scatterplot for this data.

Describe the type of correlation between the variables in terms of direction and strength.

The table shows the marks of 10 students in two different tests: Test 1 score

Test 2 score

Construct a scatterplot for this data.

Describe the correlation between the two sets of scores in terms of its form, direction, and strength.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

The table shows the average IQ of people of different heights: Height (cm)

140

103

145

150

155

111

160

165

170

108

175

145

180

110

185

Construct a scatterplot to represent this data.

Describe the relationship between height and IQ based on this data.

Which data point appears to be an outlier?

Complex familiar 9

A database compares the income per capita of a country with the child mortality rate of that country. A sample of the data is shown. Income per capita ($)

Child mortality rate (per 1000)

1465

11 428

2621

32 468

If a scatterplot was created from the entire database, describe the association you would expect to see. Justify your answer.

1.05 Associations between numerical variables mathspace.co

The scatterplot shows the number of hours students spent watching TV the week before an exam, and their ﬁnal exam score. 100 90 80 70 60 50 40 30 20 10 0

Exam Score (%)

Hours of TV watched 2 4 6 8 10 12 14 16 18

A news report claims that “watching TV causes poor exam results”. Is this a valid conclusion to draw from the data? Explain why or why not, suggesting a possible confounding variable. 11

A researcher is studying the “bystander effect”, where the time taken for someone to offer help in an emergency is measured against the number of other people (passers-by) present. The results are shown in the table: Number of passers-by

Time to help (seconds)

Construct a scatterplot and describe the association (form, direction, and strength).

What does this correlation suggest about the bystander effect?

After an exam, a teacher plots the results of Question 7 versus Question 10 for each class. A student claimed Q10 was irrelevant, but the teacher felt they assessed the same skill:

Q10 correct (%) 80 60

Based on the general trend of the data, who is likely correct: the student or the teacher? Explain.

One class (Point A) had 95% correct on Q7 but only 40 15% correct on Q10. What does this outlier point 20 suggest about that particular class?

A Q7 correct (%) 0

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex unfamiliar 13

The scatterplot shows the relationship between anxiety level and performance on a task. Task Performance (%)

90 80 70 60 50 40 30 20 10

Anxiety Level 0

8 9

A student describes this as having ‘no correlation’ because it does not form a straight line. Explain why this description is incomplete. Provide a better description of the relationship shown. 14

A city council determines a strong negative correlation between its annual spending on public art and the number of potholes reported on its roads. A council member argues, “This data proves that spending money on art prevents potholes. To ﬁx our roads, we should increase the art budget.” Explain the ﬂaw in the council member’s reasoning. Propose an alternative explanation for the observed correlation.

The scatterplot shows data for two variables, x and y. The data is collected from two different groups, Group A (blue) and Group B (green): 9

8 7 6 5 4 3 2 1

x 0

Describe the correlation that exists within Group A only.

Describe the correlation that exists within Group B only.

If you ignored the groups and looked at all the data points together, what would the overall correlation appear to be? Explain why this could be misleading.

1.05 Associations between numerical variables mathspace.co

1.06 Correlation using r-values After this lesson, you will be able to… • calculate Pearson’s correlation coefficient, r, from raw data using technology • interpret Pearson’s correlation coefficient, r, to quantify the strength of a linear association

Correlation using r-values While describing an association with words like ‘strong’ or ‘weak’ is useful, a more precise method is to calculate a single number to represent the strength of a linear relationship. This number is called Pearson’s correlation coeﬃcient. Pearson’s correlation coeﬃcient, denoted by the letter r, is a value between −1 and 1 that measures the strength and direction of a linear association between two numerical variables. • A positive r-value indicates a positive linear association. • A negative r-value indicates a negative linear association. • An r-value close to 0 indicates a very weak or no linear association. • An r-value close to 1 or −1 indicates a strong linear association. This table provides a guide for interpreting the strength of the correlation based on the r-value. r-value

Strength of linear association

∣r∣ = 1

Perfect

0.75 ≤ ∣r∣ < 1

Strong

0.5 ≤ ∣r∣ < 0.75

Moderate

0.25 ≤ ∣r∣ < 0.5

Weak

0 ≤ ∣r∣ < 0.25

Very weak / No correlation

Note: ∣r∣ refers to the absolute value of r, ignoring the negative sign. The formula for calculating Pearson’s correlation coeﬃcient is complex, so it is almost always calculated using technology such as a spreadsheet program or statistical calculator. Typically, the two lists of data (for the explanatory and response variables) are entered, and the statistics function is used to calculate the value of r. It is critical to remember that correlation does not imply causation. Just because two variables are strongly correlated does not mean that a change in one variable causes a change in the other. A strong correlation might be observed for other reasons, such as: • Confounding variable: A third, unobserved variable is affecting both variables being studied. For example, the number of drownings and ice cream sales are correlated, but the confounding variable is hot weather. • Coincidence: The relationship is purely by chance and has no logical connection.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Example 1 The correlation coeﬃcients for three different sets of bivariate data are calculated as: • Set A: r = 0.75 • Set B: r = −0.9 • Set C: r = 0.3 Which dataset displays the strongest linear correlation?

Create a strategy The strength of a correlation is determined by the absolute value of the Pearson’s correlation coeﬃcient, ∣r∣. A value closer to 1 indicates a stronger correlation, regardless of whether the sign is positive or negative. Compare the absolute values of the three coeﬃcients.

Apply the idea Calculate the absolute value of each correlation coeﬃcient: • Set A: ∣0.75∣ = 0.75 • Set B: ∣−0.9∣ = 0.9

• Set C: ∣0.3∣ = 0.3

Comparing these values, the largest is 0.9. Therefore, Set B displays the strongest linear correlation.

Example 2 A health science student wants to investigate the relationship between hours of sleep and performance on a memory test. They collected this data:

Hours of sleep (x)

Memory test score ( y)

Use technology to calculate Pearson’s correlation coeﬃcient, r, for this data, rounded to two decimal places. Then, describe the association between the variables.

7.5

Create a strategy Enter the two lists of data into a statistical calculator or spreadsheet software. Use the function for calculating the Pearson’s correlation coeﬃcient. Once r is calculated, use its sign and value to interpret the direction and strength of the linear association.

Apply the idea Entering the data into technology gives a Pearson’s correlation coeﬃcient of: r ≈ 0.97 To describe the association, consider the direction and strength: • Direction: The r-value is positive. • Strength: An r-value of 0.97 is very close to 1, which indicates a strong linear association. Therefore, there is a strong, positive, linear association between hours of sleep and memory test score.

1.06 Correlation using r-values mathspace.co

1.06 Practice questions Simple familiar Ex 1

Three datasets have these correlation coeﬃcients: • Set A: r = 0.6 • Set B: r = −0.8 • Set C: r = 0.25 Which dataset has the strongest linear correlation?

For each of these scatterplots, choose the most appropriate correlation coeﬃcient (r) from the list: i

−0.9

−0.4

iii b

1 1

d 9

2 x

0.8

x 0

0.1

x 0

Determine whether these statements are true or false: a

If a causal relationship exists between two variables, they must be correlated.

If two variables are correlated, then one must cause the other.

A study on a new fertiliser ﬁnds a strong positive correlation between the amount of fertiliser used and crop yield. Is it reasonable to conclude that the fertiliser causes an increase in crop yield? Explain brieﬂy.

1.06 Correlation using r-values mathspace.co

A scatterplot shows a perfect negative linear correlation. What is the value of Pearson’s correlation coeﬃcient r?

If the correlation coeﬃcient between two variables is r = 0.05, what can be said about their linear relationship?

The scatterplot shows data for the height of a kicked ball over time: A student calculates Pearson’s correlation coeﬃcient and determines that r ≈ 0. They conclude there is no relationship between time and height. Explain why this conclusion is incorrect.

Height (m) 50 40 30 20 10

Time (s) 0

8 9

For each of these data examples, determine if there is likely a causal relationship between the variables: a

The number of hours spent training for a marathon and the runner’s ﬁnishing time

The sales of sunglasses and the number of shark attacks at a beach

Complex familiar 9

A study found the correlation coeﬃcient between a woman’s height and the probability of being turned down for a promotion was r = −0.90. Based on this r-value, describe the relationship between the variables.

Ex 2

A climate scientist investigates the relationship between the altitude of a city and its average maximum temperature. The data collected is shown in the table:

Temperature (°C)

940

627

Use technology to calculate the correlation coeﬃcient, r, for this data. Round your answer to two decimal places.

440

342

Describe the association between altitude and temperature.

302

Can the scientist conclude that a city’s altitude causes its average maximum temperature to change? Explain your answer.

525

250

775

896

116

Altitude (m)

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 3

A study found a strong positive correlation between the number of storks nesting in a village and the number of babies born in that village: a

Does this mean that storks cause babies to be born?

Suggest a likely confounding variable that could explain this correlation.

The scatterplot shows the blood glucose level and hormone level for a group of patients. The correlation coeﬃcient is r = −0.48: a

Based on the r-value, describe the linear association.

A researcher claims this data proves that high hormone levels do not cause low glucose levels. Is this claim correct based only on the r-value?

Glucose level

250 200 150 100 50

Hormone level 0

Complex unfamiliar 13

The manager of a car ﬂeet wants to estimate maintenance costs based on the distance driven. Data is collected and the correlation coeﬃcient is calculated to be r = 0.85: Cost ($100s)

500 450 400 350 300 250 200 150

Distance (1000s km) 0

Describe the association between distance driven and maintenance cost.

Is it reasonable to assume a causal relationship exists? Explain.

The manager notes four cars that were driven between 2000 and 10 000 km seem to have lower maintenance costs than expected. What might this suggest about the relationship for cars that are driven very little?

An analyst compares the number of libraries in a city to the city’s overall crime rate. They calculated a correlation of r = 0.92 and conclude that building more libraries is a highly effective way to reduce crime. Critique this conclusion. Identify a confounding variable and explain how it accounts for the strong positive correlation. 1.06 Correlation using r-values mathspace.co

Your Task: Investigate Surviving the Titanic The data The provided data ﬁle contains information about passengers on the Titanic. The ﬁrst few rows of the data table are shown below. The sample data is available in the provided downloadable file:

Notes about the data: Data name

Explanation

Survived

0 = No, 1 = Yes

Pclass

Passenger Class (1st = Upper, 2nd = Middle, 3rd = Lower)

Age

Age in years. Fractional if < 1, estimated if ends in 0.5

Sibsp

Number of siblings or spouses aboard

Parch

Number of parents or children aboard

Fare

Passenger fare in pounds sterling

Embarked

C = Cherbourg, Q = Queenstown, S = Southampton

Step 1: Identify a problem Select or create a suitable statistical question that captures the requirements of this investigation. For example: Is there an association between gender and survival? Is there an association between passenger class and survival? Step 2: Collect data Determine what data you will need to collect from the table to answer your question. Decide which is the explanatory variable and the response variable.

Investigation: Statistical investigation process mathspace.co

1.07 Coefficient of determination After this lesson, you will be able to… • calculate the coefficient of determination, R², from raw data using technology • use the coefficient of determination, R², to assess the strength of a linear association • assess the strength of a linear association in terms of the explained variation

Coefficient of determination While Pearson’s correlation coefficient, r, tells us about the strength and direction of a linear relationship, the coefficient of determination, denoted as R2, gives us more insight. It tells us the proportion of the variation in the response variable ( y) that can be explained by the variation in the explanatory variable (x). Calculating the Coefficient of Determination: The coefficient of determination is simply the square of the correlation coefficient.

R2 = r2 R2 r

The coeﬃcient of determination Pearson’s correlation coeﬃcient

Conversely, if you know R2, you can ﬁnd the magnitude of r by taking the square root. However, you must determine the sign (positive or negative) from the context of the problem or by looking at the trend of the scatterplot. A positive trend in the data indicates a positive r, while a negative trend indicates a negative r. Interpreting the Coeﬃcient of Determination: The value of R2 is always between 0 and 1. It is often expressed as a percentage to make it easier to understand. An R2 value of 0.75 means that 75% of the variation in the response variable can be explained by the variation in the explanatory variable. The remaining 25% of the variation is due to other factors. y (Response variable) Total variation

Explained variation

Unexplained variation (Residual)

x (Explanatory variable) 0

1.07 Coefficient of determination mathspace.co

Example 1 The correlation coeﬃcient for the relationship between hours spent studying and exam scores is r = 0.9: a Calculate the coeﬃcient of determination, R2.

Apply the idea To calculate the coeﬃcient of determination, square the correlation coeﬃcient. R2 = r2 = (0.9) = 0.81

Write the formula 2

Substitute the value of r Evaluate

b Interpret the value of R2 in the context of this problem.

Create a strategy

Apply the idea 2

To interpret the result, express R as a percentage and state what it means for the variables.

An R2 value of 0.81 means that 81% of the variation in exam scores can be explained by the variation in the hours spent studying. The remaining 19% is due to other factors.

Example 2 A study into the effect of car age on its value found a coeﬃcient of determination of R2 = 0.64: a Determine the value of the correlation coeﬃcient, r.

Create a strategy To determine the correlation coeﬃcient, take the square root of R2. The sign must also be determined.

Apply the idea Write the formula

Substitute the value of R2

Evaluate Logically, as a car’s age increases, its value decreases. This implies a negative association. Therefore, the negative value is chosen. r = −0.8 b Describe the linear association between car age and its value.

Apply the idea An r-value of −0.8 indicates a strong, negative, linear association between car age and its value.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

1.07 Practice questions Simple familiar Ex 1

The correlation coeﬃcient for a dataset is found to be r = 0.8. Calculate the coeﬃcient of determination, R2.

Ex 2

The coeﬃcient of determination for a dataset is R2 = 0.49. The scatterplot shows a negative association. What is the value of the correlation coeﬃcient, r?

Complete this table by calculating the missing values. Round your answers to two decimal places where necessary. Correlation coefficient (r)

Coefficient of determination (R2)

0.6

⬚

−0.9 ⬚ 4

⬚

0.25

A study on the relationship between hours of exercise per week and resting heart rate ﬁnds that R2 = 0.68. What percentage of the variation in resting heart rate can be explained by the variation in hours of exercise?

A research paper states that for their data on rainfall and crop growth, “85% of the variation in crop growth can be explained by the variation in rainfall.” What is the value of the coeﬃcient of determination, R2, for this data?

If R2 = 0.55 for a set of data, what percentage of the variation in the response variable is not explained by the variation in the explanatory variable?

Complex familiar Ex 3

A real estate agent collects data on the size of houses and their selling prices. The data is shown in the table: a

Use technology to calculate the coeﬃcient of determination, R2, for this data. Round your answer to three decimal places. Interpret this R2 value in the context of the problem.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Size (m2)

Price ($1000s)

120

550

150

620

180

700

210

780

240

850

The scatterplot shows the relationship between a car’s age and its resale value. The coeﬃcient of determination is R2 = 0.81: a

What is the value of the correlation coeﬃcient, r?

Based on the value of r, describe the association.

Value ($1000s) 50 40 30 20 10

Age (years) 0

8 9

An economist is studying factors that predict a person’s income. They create two models: • Model A: Predicts income based on years of education. It has R2 = 0.45. • Model B: Predicts income based on hours worked per week. It has R2 = 0.60. Which model is better at explaining the variation in income? Explain your reasoning.

Complex unfamiliar 10

A researcher is investigating the link between the price of a generic medicine and its effectiveness. They collect data and calculate a correlation coeﬃcient of r = −0.15: a

Calculate the coeﬃcient of determination, R2.

The researcher concludes: “Because R2 is so low, this proves there is no relationship between the price and effectiveness of the medicine.”

Explain the ﬂaw in this conclusion. What does the low R2 value actually signify? 11

A study examines the relationship between employees’ years of experience and their productivity. The initial data includes a senior manager who has been with the company for 30 years but now has low productivity due to being close to retirement. • With the manager included, R2 = 0.20 • With the manager excluded as an outlier, R2 = 0.70 Explain why the R2 value changes so signiﬁcantly when the outlier is removed. Which R2 value gives a more accurate understanding of the general trend for most employees?

1.07 Coefficient of determination mathspace.co

1 Chapter review SF

A survey of 150 gym members recorded their preferred workout time (Morning or Evening) and their main ﬁtness goal (Cardio or Weights): • 40 members who prefer morning workouts focus on Cardio. • 35 members who prefer morning workouts focus on Weights. • A total of 60 members focus on Cardio. Complete this two-way frequency table, including all marginal totals: Cardio

Weights

Total

Morning

Evening

⬚

Total

⬚

150

The table shows the results of a survey asking people their age group and preferred holiday destination: Beach

City

Adventure

Total

Under 30

100

30 and Over

Calculate the percentage of respondents ‘Under 30’ who prefer an ‘Adventure’ holiday. CF

A university surveyed 200 students to investigate an association between their faculty (Arts or Science) and their primary mode of transport (Public transport, Car, or Bicycle): Public transport

Car

Bicycle

Total

Arts

100

Science

100

Total

100

200

The university wants to know if faculty choice inﬂuences transport mode. Identify the explanatory variable.

Construct a row percentage frequency table to investigate the association. Round the percentages to the nearest whole number.

Comment on the association between faculty and mode of transport, using the percentages from your table to support your answer.

For each scenario, identify the explanatory and response variables: a

Investigating if the daily temperature affects the number of soft drinks sold.

A study looks at the relationship between a person’s age and their preferred social media platform.

A farmer tests whether the amount of rainfall inﬂuences the height of corn stalks.

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This scatterplot shows the relationship between the age of a second-hand car and its advertised price: Price ($1000s)

35 30 25 20 15 10 5

Age (years)

Describe the association between the variables in terms of direction, form, and strength. SF

A strong positive correlation is found between the number of firefighters attending a fire and the amount of damage caused by the fire. Does this mean that sending more firefighters causes more damage? Explain why or why not.

Match each correlation coeﬃcient, r, with the most appropriate scatterplot: A

r = 0.9

x 0

r = −0.7

x 0

A researcher ﬁnds a moderate positive correlation (r = 0.55) between the amount of coffee people drink per day and their reported levels of stress. A coffee company uses this to advertise that their coffee helps people handle stress: a

Explain the primary ﬂaw in the coffee company’s advertising claim.

Suggest a logical confounding variable that could explain this association.

A study into the relationship between the number of hours a student studies and their exam mark determines a correlation coeﬃcient of r = 0.85. Calculate the coeﬃcient of determination, R2. Round the answer to two decimal places.

Chapter 1 review mathspace.co

The relationship between a car’s weight and its fuel efficiency has a coefficient of determination of R2 = 0.64. The association is negative. Determine the value of the correlation coefficient, r.

For a set of bivariate data, the coeﬃcient of determination is R2 = 0.45. Interpret this value by completing the sentence:

45% of the variation in the ⬚ variable can be explained by the variation in the ⬚ variable.

A sports scientist investigates the relationship between an athlete’s daily protein intake (in grams) and their muscle mass gain (in kg) over a 3 month period. The data is recorded: Protein Intake (g)

Muscle Gain (kg)

100

1.2

120

1.5

140

2.1

160

2.4

180

2.9

200

3.2

Construct a scatterplot of the data with Protein Intake as the explanatory variable.

Use technology to calculate the correlation coeﬃcient, r, and the coeﬃcient of determination, R2. Round your answers to three decimal places.

Interpret the coeﬃcient of determination in the context of this study.

A school analyses the pass rates for two subjects, Maths and English, at two different campuses, North and South. The data is as follows: North Campus

Pass

Fail

Maths

English

South Campus Pass

Fail

Maths

English

For each campus, calculate the pass rate for Maths and the pass rate for English. At both individual campuses, which subject has a higher pass rate?

Now, combine the data for both campuses into a single table. Calculate the overall pass rate for Maths and for English across the entire school.

Based on the overall results, which subject appears to have the higher pass rate? Explain how this result seems to contradict the ﬁndings from the individual campuses (this is known as Simpson’s Paradox).

A technology blog reviews new laptops. For the relationship between a laptop’s battery life (in hours) and its price (in dollars), they report a correlation coeﬃcient of r = 0.2. The blog concludes: “Our analysis shows there is almost no relationship between what you pay and how long the battery lasts, so price shouldn’t be a factor if you want a long battery life.”

Mathspace QLD 12 General Mathematics 2025 mathspace.co

2.01 Least-squares lines After this lesson, you will be able to… • model a linear relationship by using technology to fit a least-squares line to data in the form of y = mx + c • understand and use the formulas to determine the equation of a leastsquares line • identify the components of the least-squares line equation, including slope, y-intercept, correlation coefficient, and means and standard deviations of xand y-values

Determine a least-squares line with technology When analysing a scatterplot, it is often possible to see a trend in the data. If the trend appears to be linear, a single straight line can be drawn that best represents the overall pattern of the data. This line is often called a ‘line of best ﬁt’. Ice creams sold vs Hours of sunshine 100

Ice creams sold

90 80 70 60 50 40 30 20 0

x 2

Hours of sunshine The most accurate line of best ﬁt is called the least-squares line. Least-squares line The unique line that minimises the sum of the squares of the vertical distances from each data point to the line. It is also known as the least- squares regression line or line of best ﬁt.

The equation of the least-squares line is given in the form y = mx + c, where m represents the gradient (slope) and c represents the y-intercept. Note that different technologies may present this equation as y = ax + b, so it is important to identify which variable represents the gradient and which represents the y-intercept. The gradient is always represented by the letter in front of the x-variable. 96

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Calculate a least-squares line from summary statistics If the original data is not available, the equation of the least-squares line can still be calculated using the given summary statistics. To calculate the least-squares line, ﬁve key summary statistics are needed: • The correlation coeﬃcient (r) • The mean of the x-values • The mean of the y-values • The sample standard deviation of the x-values (sx) • The sample standard deviation of the y-values (sy ) The gradient (m) and y-intercept (c) are calculated using the formulas:

It is essential to calculate the gradient (m) ﬁrst, as this value is needed to calculate the y-intercept (c).

Example 2 A researcher studying the link between the age of high school students and the average number of hours they sleep per night provides these summary statistics: • • sy (stand. dev. sleep) = 0.5 hours (mean age) = 16.5 years • • r (correlation coeff.) = −0.85 (mean sleep) = 7.8 hours • sx (stand. dev. age) = 1.2 years Calculate the equation of the least-squares line that models this relationship. Round the gradient and y-intercept to two decimal places.

Create a strategy First, calculate the gradient (m) using the formula: m = r × the y-intercept (c) using the formula, c = written in the form y = mx + c.

−

. Then, use this gradient to calculate

. Once both values are known, the equation can be

Apply the idea Calculate the gradient (m): Write the formula

Substitute the known values

Evaluate and round to two decimal places

Calculate the y-intercept (c): c=

–

Write the formula

c ≈ 7.8 − (− 0.35) × 16.5

Substitute the known values

c ≈ 13.58

Evaluate and round to two decimal places

The equation of the least-squares line is y = −0.35x + 13.58

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A personal trainer records the number of weeks a client has been training and the time it takes them to run 1 kilometre: Weeks training (x)

Time (minutes, y)

Use technology to determine the equation of the least-squares line. Round values to one decimal place. 4

A coach tracks the number of practice sessions a basketball player has completed and their free-throw shooting percentage. The data is recorded in the table: Practice sessions (x)

Shooting percentage (%, y)

Use technology to determine the equation of the least-squares line. Round values to one decimal place. 5

A gardener records the number of days since planting and the height of a plant in centimetres. The data is recorded in the table: Days since planting (x)

Plant height (cm, y)

Use technology to determine the equation of the least-squares line. Round values to one decimal place. 6

A music teacher notes the number of hours a student practices and their score on a sightreading test. The data is recorded in the table: Practice hours (x)

Test score (%, y)

Use technology to determine the equation of the least-squares line. Round values to one decimal place. Ex 2

For a set of bivariate data, these summary statistics were calculated: • = 25 • = 80 • sx = 4.5 • sy = 12.1 • r = 0.92

100

Calculate the gradient, m, of the least-squares line, rounded to two decimal places.

Calculate the y-intercept, c, of the least-squares line, rounded to two decimal places.

Determine the equation of the least-squares line.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A study on the relationship between daily screen time and test scores produced these summary statistics: • r = − 0.75 • (screen time) = 4.2 hours • (test score) = 106.4 • sx = 1.1 hours • sy = 2.8 Calculate the equation of the least-squares line. Round values to two decimal places.

The summary statistics for a dataset comparing years of experience to weekly income (in thousands of dollars) are provided: • r = 0.88 • = 10.5 years • = 7 (thousands) • sx = 3.2 years • sy = 2.0 (thousands) Calculate the equation of the least-squares line. Round values to two decimal places.

A study of the relationship between hours spent exercising per week and resting heart rate produced these summary statistics: • r = − 0.85 • = 5.20 • = 72.40 • sx = 1.80 • sy = 8.50 Calculate the equation of the least-squares line. Round values to two decimal places.

Research investigating the link between years of education and annual income provided these summary statistics: • r = 0.76 • = 14.50 • = 58.20 • sx = 2.30 • sy = 15.70 Calculate the equation of the least-squares line. Round values to two decimal places.

An analysis of daily hours of sleep and concentration test scores reported these summary statistics: • r = 0.65 • = 7.10 • = 82.60 • sx = 1.20 • sy = 10.30 Calculate the equation of the least-squares line. Round values to two decimal places.

2.01 Least-squares lines mathspace.co

101

The age and diameter of several trees of the same species were recorded: Age (years, x)

Diameter (cm, y)

Use technology to determine the equation of the least-squares line. Round values to one decimal place.

Identify the gradient of the line.

Identify the y-intercept of the line.

Complex familiar 14

102

A gardener records the height of a plant over several weeks: Week (x)

Height (cm, y)

5.5

8.0

11.5

13.0

16.5

18.0

Use technology to calculate the ﬁve key summary statistics for this data. Round values to two decimal places.

Using the statistics from part (a), calculate the equation of the least-squares line.

Data was collected to investigate the relationship between the distance a person lives from a city centre and their average weekly transport cost: Distance (km, x)

Cost ($, y)

Use technology to calculate the summary statistics for this data. Round values to two decimal places.

Using the statistics from part (a), calculate the equation of the least-squares line.

Determine the equation of the least-squares line using technology and compare this to the least-squares line using the summary statistics by commenting on any difference caused by rounding.

The scatterplot shows the relationship between the number of customers in a restaurant and the wait time for a meal: a

Extract the data from the scatterplot and enter into a table of values.

Determine the equation of the least-squares line for this data. Round values to two decimal places.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

28 Waiting time 26 24 22 20 18 16 14 12 10 8 6 4 2 Number of customers 0 5 10 15 20 25 30 35 40 45

The scatterplot shows the relationship between hours spent studying and test scores, along with the leastsquares line of best ﬁt, y = 2x + 5:

Identify the coordinates of two points that lie exactly on the line of best ﬁt.

Use the two points from part (a) to calculate the gradient of the line.

Test score

25 20 15

Compare the gradient calculated in part (b) with the gradient given in the equation.

10 5

Hours studying 0

10 12 14

Complex unfamiliar 18

The equation of a least-squares line is y = 1.5x + 10. The summary statistics for the data used to create this line are: • = 20 • sx = 5 • sy = 12 Calculate r, the correlation coeﬃcient, rounded to two decimal places.

Two friends, Alex and Ben, track their performance in a video game. They record the hours played each week and the number of games they win. Their summary statistics are shown: Alex

Ben

2.1

2.5

4.2

3.5

0.9

0.85

Identify which person’s model has a steeper gradient. 20

A dataset has one signiﬁcant outlier. If this outlier is removed, describe the effect this would likely have on the equation of the least-squares line when calculated using the summary statistics formulas.

2.01 Least-squares lines mathspace.co

103

2.02 Interpret slope and y-intercept After this lesson, you will be able to… • interpret the y-intercept of a fitted line • interpret the slope (gradient) of a fitted line • recognise the meaning of the y-intercept in context • understand the meaning of the slope (gradient) in context

Interpret slope and y-intercept Once the least-squares line has been calculated, the equation can be used to describe the relationship between two variables in context.

Exploration Consider this scatterplot which shows the relationship between daily temperature and ice cream sales for a local shop: Temperature vs. Sales 1600 1400

Sales ($)

1200 1000 800 600 400 200 0

Temperature (°C) The equation of the least-squares line is y = 50x + 20: 1. Looking at the graph, what happens to the sales for every 1°C increase in temperature? How does the number 50 in the equation relate to this? 2. According to the line, what would the sales be if the temperature was 0°C? Where is this shown on the graph? How does the number 20 in the equation relate to this? 3. Does it make sense for an ice cream shop to make sales when the temperature is 0°C? Why or why not?

104

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The y-intercept (c) is the predicted value of the response variable ( y) when the explanatory variable (x) is zero. One interpretation of the y-variable: When the explanatory variable is 0, the predicted value of the response variable is the value of c. It is important to note that this interpretation is only meaningful if a value of x = 0 makes sense in the context. For example, a $65 call-out fee for an electrician at 0 hours worked is a practical interpretation. However, predicting the height of a person with a shoe size of 0 is not a practical interpretation. The slope (m), also called the gradient, describes the rate of change. It tells us how much the response variable is predicted to change for every one-unit increase in the explanatory variable. To interpret the slope: For every one unit increase in the explanatory variable, the response variable is predicted to increase or decrease by the value of m. Hours studied vs. Test score

Test score

10 8 6 4 2 0

Hours studied A positive slope means as the explanatory variable increases, the response variable is predicted to increase. Age of car vs. Value

Value ($)

10 8 6 4 2 0

Age of car (years) A negative slope means as the explanatory variable increases, the response variable is predicted to decrease.

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Example 1 The cost of hiring an electrician is modelled by the least-squares line: Cost = 90 × Hours + 65 a Interpret the slope (gradient) in the context of this scenario.

Create a strategy

Apply the idea

Use the value of the slope to identify by how much the response variable changes for every one unit increase in the explanatory variable.

The explanatory variable is ‘hours’ and the response variable is ‘cost’. The slope (m) is 90. For every one hour increase in the hours worked, the cost is predicted to increase by $90.

b Interpret the y-intercept in the context of this scenario.

Create a strategy

Apply the idea

Use the y-intercept of the equation to predict the value of the response variable when the explanatory variable is zero.

The y-intercept (c) is 65. When the hours worked is 0, the predicted cost is $65. This can be interpreted as a ﬁxed call-out fee before any work begins.

Example 2 The resale value of a mobile phone is modelled by the equation, Value = −250 × Age + 1200, where age is in years and value is in dollars: a Interpret the gradient (slope) in this context.

Create a strategy Use the value of the slope to identify by how much the response variable changes for every one unit increase in the explanatory variable.

Apply the idea The slope (m) is −250. For every one year increase in the phone’s age, the resale value is predicted to decrease by $250.

Reﬂect and check A positive gradient implies an overall increasing trend, and a negative gradient implies a decreasing trend.

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Ex 2

A personal trainer determined the relationship between weeks of training (x) and the time to run 1 km ( y) is y = −0.5x + 9.5: a

Interpret the gradient in the context of this scenario.

Interpret the y-intercept in the context of this scenario.

For a set of bivariate data where x = age (months) and y = height (cm), these summary statistics were calculated: • = 18 • = 75 • sx = 3.2 • sy = 8.5 • r = 0.95

Determine the equation of the least-squares line. Round values to two decimal places.

Interpret the gradient in the context of this scenario.

Interpret the y-intercept in the context of this scenario.

A study on the relationship between daily screen time and test scores produced these summary statistics: • r = −0.80 • (screen time) = 5.5 hours • (test score) = 78.2 • sx = 1.5 hours • sy = 6.2

Calculate the equation of the least-squares line. Round values to two decimal places.

Interpret the gradient in the context of the study.

Interpret the y-intercept in the context of the study.

The summary statistics for a dataset comparing years of experience to weekly income are provided: • r = 0.88 • (experience) = 10.5 years • = $1.20 (thousands) • sx = 3.2 years • sy = $0.55 (thousands)

108

Calculate the equation of the least-squares line. Round values to two decimal places.

Interpret the slope in the context of this data.

Interpret the y-intercept. Is this interpretation practical?

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Complex familiar 7

A gardener records the height of a plant over several weeks: Week (x)

Height (cm, y)

4.0

6.5

9.0

10.5

13.5

15.0

Determine the equation of the least-squares line. Round values to two decimal places.

Interpret the slope in the context of the plant’s growth.

Interpret the y-intercept. Does this value make sense in this context?

The scatterplot shows the relationship between the daily maximum temperature and the number of swimmers at a local pool: Number of swimmers 100 80 60 40 20

Temperature (°C) 0

12 14 16 18 20 22 24 26 28

Determine the equation of the least-squares line for this data. Round values to two decimal places.

Interpret the gradient in the context of the swimming pool.

Interpret the y-intercept. Is this interpretation practical?

The relationship between hours spent studying (x) and test scores ( y) is modelled by the line y = 5.2x + 45.5: a

A student increases their study time by 2 hours per week. What is the predicted change in their test score?

Another model for a different subject is y = 3.8x + 52.1. Which subject gives a greater return (in marks) for each hour of study?

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Complex unfamiliar 10

A least-squares line for car value ( y, in dollars) versus age (x, in years) is y = −1500x + c. The mean age of the cars in the dataset is = 6 years and the mean value is = $12 000: a

Recalling that the least-squares line must pass through the point ( , ), determine the value of the y-intercept, c.

Interpret the value of the y-intercept determined in part (a).

Two students, Sarah and Mike, track their performance in a biology class. They record the hours they study each week (x) and their score on the weekly quiz ( y). Their summary statistics are shown: Sarah

Mike

2.1

2.5

4.2

3.5

0.9

0.85

Determine the least-squares equation for both Sarah and Mike. Round values to two decimal places.

Whose quiz score is predicted to improve more for each additional hour of study? Justify your answer by interpreting the gradients.

Who has the better predicted score with no study? Justify your answer by interpreting the y-intercepts.

A researcher creates a model for house prices ( y, in thousands of dollars) based on the ﬂoor area (x, in square metres). The model is y = 2.5x − 50: a

Interpret the slope of this line.

Interpret the y-intercept. Explain why this interpretation is not practical.

The scatterplot shows the relationship between the number of weeks since planting and the height of a plant, along with the least-squares line of best fit, y = 1.5x + 10: a

Identify the coordinates of two points that lie exactly on the line of best ﬁt.

Use the two points from part (a) to calculate the gradient of the line.

Compare the gradient calculated in part (b) with the gradient given in the equation.

Plant height (cm)

35 30 25 20 15 10 5

Weeks 0

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10 12 14

2.03 Residual plots After this lesson, you will be able to… • construct a residual plot • use a residual plot to assess the appropriateness of fitting a linear model to data • understand how a residual plot indicates the appropriateness of a linear model

Residuals The correlation coeﬃcient, r, describes the strength and direction of a relationship between two variables. An r-value of 0.8 or − 0.8 indicates a strong relationship. Whereas, an r-value of 0.45 or − 0.45 indicates a weak relationship. A positive r value indicates the relationship is positive, alternatively a negative r value indicates a negative relationship. The purpose of least-squares regression line is to be able to make statistical predictions. However, there are some requirements that need to be met in terms of the data. One of those requirements is for the data to be linear. The two scatterplots shown illustrate this idea. Although both have a strong correlation and similar least-squares lines, the patterns differ. In the left scatterplot, the data points are scattered randomly above and below the line, which is typical when a linear model is appropriate. In contrast, the right scatterplot shows a clear pattern: the points start below the line, then rise above it, and return below. This systematic pattern suggests the linear model does not capture the true relationship.

Profit

Sales

Sales 0

Linear data

Non-linear data

Residual plots help identify whether a linear model is appropriate by revealing such non-random patterns.

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Residual The vertical distance from an actual data point to the least-squares regression line.

A calculator determines the least-squares regression line by minimising the sum of the squares of the residuals to choose the optimal coefficients for the line of best fit. These scatterplots show how the residuals are short when the line of best fit is chosen appropriately, and longer for a line that is a poor fit to the data. Appropriately fitted least-squares line: y

28 26 24 22 20 18 16 14 12 10 8 6 4 2

1.9 0.50 0.20 2.61

6.07

0.20 2.64 1.68

3.04

Inappropriately ﬁtted least-squares line: y

28 26 24 22 20 18 16 14 12 10 8 6 4 2

10.12 6.6 4.07

2.02 2.51 4.56 9.61

4.46

8.03

11.09 x

Interactive exploration Discover this concept in action online

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mathspace.co

To calculate a residual for a data value: Residual = Actual value − Predicted value =y− where the predicted value, , is obtained from the equation of the least-squares regression line. A positive residual means the actual data point is above the least-squares regression line and a negative residual means the raw data point is below the line. In a real-life example, if the purpose of the regression line was to predict a students exam score, a positive residual would indicate the student scored better than predicted.

Example 1 The table shows the sets of data (x, y) and the predicted -values based on a least-squares regression line: x

22.7

22.3

24.2

21.8

21.5

25.2

23.4

21.6

19.8

⬚

Residuals

Complete the table by calculating the residuals.

Create a strategy Use the formula: Residual = y −

Apply the idea Solving for the ﬁrst column: Residual = y –

Write the formula

= 22.7 − 25.2

Substitute the values

= −2.5

Evaluate

Repeat this process for the remaining values in the table: x

22.7

22.3

24.2

21.8

21.5

25.2

23.4

21.6

19.8

18.0

−2.5

−1.1

2.6

2.0

3.5

Residuals

Reflect and check Check the signs of the residuals. For the first data point, the actual value (22.7) is less than the predicted value (25.2), so the residual must be negative. For the third data point, the actual value (24.2) is greater than the predicted value (21.6), so the residual must be positive. This confirms the calculations are reasonable.

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Consider this scatterplot and its corresponding residual plot: 100 90 80 70 60 50 40 30 20 10 0

y 2 1 0 −1 −2

x 10

x 5

10 15 20 25 30 35

The original data has a strong correlation (0.994) and appears to follow a linear distribution. However, the residual plot shows a clear pattern, suggesting that using a linear regression model with the original data for making predictions may not be appropriate.

Example 2 This table shows a company’s monthly sales, y (in thousands) and advertising budget, x. The equation y = 5x + 12 is being used to model the data: a Complete the table of residuals: x

Residual ⬚

⬚

⬚ ⬚

⬚

Create a strategy To calculate the predicted value, , substitute each value of x into the equation of the least-squares line. To calculate the residual value, use the formula: Residual = y −

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Apply the idea Solving for the ﬁrst row: y = 5x + 12

Write the equation

= 5 × 1 + 12

Substitute x = 1

= 17

Evaluate

Residual = y –

Write the formula

= 22 – 17

Subtract the actual and predicted values

Evaluate

Repeat this process for the remaining values in the table: x

−3

−4

−3

−1

Residual

b Plot the residuals on the scatterplot.

Create a strategy

Apply the idea

Plot each value of x along with its residual value.

The points have the coordinates: (1, 5), (2, 3), (4, 1), (6, −3), (9, −4), (12, −3), (14, −1), (17, 2). The residual plot is shown: 5 4 3 2 1 0 −1 −2 −3 −4 −5

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Residual

x 2 4 6 8 10 12 14 16 18

2.03 Practice questions Simple familiar Ex 1

These tables show sets of data (x, y) and the predicted values, , based on a least-squares regression line. Complete the tables by calculating the residuals: a

105

110

112

118

120

106.5

108.0

113.0

117.5

121.0

⬚

Residuals b

37.7

37.2

21.1

27.1

28.9

30.4

31.9

33.4

34.9

⬚

Residuals c

38.5

55.1

46.2

40.1

50.5

54.0

47.3

⬚

− 10.5

− 12.0

− 15.8

− 17.2

− 20.0

− 11.0

− 11.5

− 16.0

− 17.5

− 19.5

⬚

2.5

3.1

4.5

5.0

6.2

For the given data and predicted values, calculate the corresponding residuals: a

x y

⬚

12.8

⬚

Residual

0.6

− 30.3

⬚

− 29.7

⬚

0.2

⬚

− 1.4

−8

11.4

⬚

11.1

⬚

⬚ 13

6.5

− 21.5

⬚

− 28.5

26 x

⬚

45.5

48.2

50.0

53.8

56.1

46.0

47.5

51.2

52.9

55.5

⬚

−8

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Residual − 33.3

Residuals 118

Residuals 2

Residuals d

− 27.4

− 1.1

Ex 2

= 2.5x + 10. For the data point (8, 32),

For a set of data, the least-squares line is given by calculate the residual.

If a residual is a positive value, is the actual data point above or below the least-squares regression line?

If a measured point in a dataset is below the least-squares regression line, will the corresponding residual be positive or negative?

The time taken to complete a task y (in minutes) after x practice sessions is recorded. The equation y = −3x + 50 is being used to model the data:

Practice sessions (x)

Time (minutes) ( y)

Complete a table of residuals for this data.

Plot the residuals on a scatterplot.

Comment on the appropriateness of using a linear model to ﬁt the data.

The number of ice creams, y, sold at a beach kiosk is recorded against the maximum daily temperature, x (in °C). The equation y = 5x − 80 is used to model the data: Temperature (°C) (x)

Ice creams sold ( y)

110

Complete a table of residuals for this data.

Plot the residuals on a scatterplot.

Comment on the appropriateness of using a linear model to ﬁt the data.

Identify which of these residual plots suggest that a linear model is appropriate for the data. Justify your choice. A

Residual

2 0 −2

2 x 2

−2

Residual

Residual x 2

Residual 2

x 2

0 −2

x 2

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For the data in the table, the residuals are calculated and shown: x

23.5

−1.5

26.0

1.0

28.5

−0.5

31.0

1.0

Residual

Calculate the sum of the residuals.

Why is the sum of the residuals for a least-squares line of best ﬁt always expected to be zero (or very close to it)?

The residual plot for a set of data is shown: 5 4 3 2 1 0 −1 −2 −3 −4 −5

Residual

x 1

2 3 4 5 6 7 8 9 10

Draw the scatterplot showing the original data using the least-squares line y = 1.25x + 6

Complex unfamiliar 14

The table shows a set of bivariate data: x

Determine, using numerical and graphical analysis, if a linear model is a good ﬁt for the data. 15

A dataset has a strong linear ﬁt. A new point is added that is an outlier, lying far below the regression line. Describe what the residual for this point will look like and how this single point might affect the conclusion about the model’s suitability when analysing the residual plot.

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Calculate the least-squares line Calculating the model While technology can determine the line of best ﬁt instantly, understanding the calculation helps see how the model is created. Calculate the slope (m) and y-intercept (c) by hand using a set of formulas and ﬁve key summary statistics. The formulas to determine the slope (m) and y-intercept (c) of a least-squares line, y = mx + c, are:

where: • r = Pearson’s correlation coeﬃcient • sy = Standard deviation of the y-values (response variable) • sx = Standard deviation of the x-values (explanatory variable) • = Mean of the y-values Let’s calculate the equation for the Flynn effect data. The summary statistics calculated from the data are: r ≈ 0.998, sy ≈ 7.37, sx ≈ 22.82,

= 109.25,

= 1977.25

Step 1: Calculate the slope

Step 2: Calculate the y-intercept c = 109.25 − (0.322 × 1977.25) ≈ −527.43 The equation of the least-squares line is approximately y = 0.322x − 527.43. Using the summary statistics provided in the worked example, conﬁrm the calculations for the slope and y-intercept.

Investigate 1.

How closely do you expect the technology’s equation to match the one calculated by hand? Why might there be a small difference?

For the “Similarities” data from Activity 1, the summary statistics are: r ≈ 0.999, sy ≈ 10.45, sx ≈ 22.82, = 114.125, and = 1977.25. Calculate the equation of the least-squares line for this data.

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Model and assess the fit Making predictions with technology Now that you understand the calculation, we can use technology to speed up the process. We will generate the line of best fit automatically and then assess whether our linear model is actually appropriate for the data. What is a residual? A residual is the vertical distance between an actual data point and the point predicted by the line of best fit. It measures the error of our prediction for that specific point. Residual = Actual y-value − Predicted y-value A positive residual means the actual data was greater than predicted, while a negative residual means the actual data was less than predicted. Interpretation of the residual plot will determine whether a linear model is appropriate for the data. This is a scatterplot of the residuals against the explanatory variable (x-values). A good fit: If the linear model is appropriate, the residual plot will show a random scatter of points around the zero line. There should be no obvious pattern. 1.00 0.75

Residuals

0.50 0.25 0.00 −0.25

−0.50 −0.75 −1.00

Explanatory variable A poor ﬁt: If the linear model is not appropriate, the residual plot will show a clear pattern (like a curve or U-shape). This tells you that a straight line is not the best way to model this data. 2.00

Residuals

1.50 1.00 0.50 0.00 −0.50

−1.00

Explanatory variable

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Return to the scatterplot you created in Activity 1. Use your spreadsheet software to add a line of best fit. Display the equation on the graph. Determine the equation of the line of best fit, rounding values to two decimal places. Interpret the slope of your line of best fit in the context of the Flynn effect. (e.g., “For every one year that passes, the average IQ score increases/decreases by...”) Interpret the y-intercept of your line of best fit. Explain if this value is meaningful in the real world.

Year

Average IQ

Predicted IQ

Residual

1947

100

95.61

4.39

1972

107

103.61

3.39

1989

113

109.05

3.95

2001

117

112.89

4.11

In your spreadsheet, create three new columns: ‘Predicted IQ’, ‘Residual’, and ‘Year’. Use the equation from your line of best ﬁt to calculate the predicted IQ for each year in the data. Then, calculate the residual for each point (Actual IQ − Predicted IQ)

Create a new scatterplot with ‘Year’ on the horizontal axis and ‘Residual’ on the vertical axis. This is your residual plot. Analyse your residual plot. Does it show a random scatter or a clear pattern? Based on this, is a linear model appropriate for the Flynn effect data? Justify your answer.

Investigate 1.

Use your equation to predict the average IQ in the year you were born and in the current year.

Use your equation to calculate the year in which the average IQ is predicted to reach 130 (classiﬁed as “very superior”). Comment on the reliability of this prediction.

If a residual plot for a different set of data showed a distinct U-shape, what would this tell you about the relationship between the variables, and what would your next step be?

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2.04 Predictions After this lesson, you will be able to… • distinguish between interpolation and extrapolation • use the equation of the least-squares line to make predictions • recognise and explain the potential dangers of extrapolation

Predictions The least-squares line is a linear representation of the general trend of data. Once the least-squares line has been determined, it can be used as a model to predict the likely value of the response variable based on a given value of the explanatory variable, or vice versa. • To make a prediction using the least-squares line, substitute the x-value into the linear model to determine the predicted value y. • To make predictions using a graph with the line of best ﬁt, locate the given x-value on the graph, move vertically to the line of best ﬁt, then move horizontally to the y-axis to read off the predicted y-value. Both of these methods can be used to predict an x-value for a given y-value in a similar way. This example shows a scatterplot, with the least-squares line for the relationship between daily ice cream sales and the maximum daily temperature. It can be predicted using the least-squares line that for a 30°C day, there will be approximately 242 ice cream sales. S

400

Ice cream sales

300

200

100

T 0

Temperature (°C)

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Example 1 A bivariate dataset has a line of best ﬁt with equation y = −8.71x + 6.79 Predict the value of y when x = 3.49

Create a strategy Substitute the value of x into the equation and solve to predict the y-value.

Apply the idea y = −8.71x + 6.79

Write the formula

= −8.71(3.49) + 6.79

Substitute x = 3.49

= −30.3979 + 6.79

Evaluate the multiplication

= −23.6079

Evaluate

Example 2 This scatterplot shows the relationship between arm span and height in centimetres: Arm span vs. Height 200

Height (cm)

180

160

140

160

180

200

220

Arm span (cm) Use the least-squares line to predict the arm span of a person if they have a height of 182 cm.

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Interpolation and extrapolation An important consideration when making predictions is to recognise if the prediction is within the range of the original data. A prediction made within this range is referred to as an interpolation. A prediction made outside this range is referred to as an extrapolation. This diagram shown illustrates these terms: Interpolation

Extrapolation

• Interpolation refers to making a prediction using a value that lies within the range of the data used to develop the model. • Extrapolation refers to making a prediction using a value that lies outside the range of the data used to develop the model. These scatterplots shown are annotated to show examples of interpolation and extrapolation from the line of best ﬁt: y

Interpolations

Extrapolations

To judge the reliability of the prediction, a few factors are considered: • The strength of correlation • Whether the predicted value used interpolation or extrapolation. Reliability of predictions: • If interpolation is used the correlation is strong, then the prediction will be reliable. • If using interpolation, and the correlation is moderate or weak, then the prediction is less reliable as the data values are more widely scattered, so there will be greater uncertainty in the prediction than for data with a strong correlation. • If using extrapolation, the prediction is always unreliable, especially if the correlation is weak or when extrapolating far beyond the range of available data. This is because no similar measurements have been made in this region, and the linear relationship may no longer be valid. 132

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Example 3 Research on the number of cigarettes smoked during pregnancy and the birth weights of the newborn babies was conducted: Average number of cigarettes per day (x) Birth weight in kilograms ( y)

46.3 13.0 21.4 25.0

8.6

36.5

1.0

17.9

10.6 13.4 37.3 18.5

2.2

3.7

2.4

4.1

3.1

3.5

3.6

3.0

2.8

3.4

2.3

3.2

a Using technology, calculate the correlation coeﬃcient between the average number of cigarettes per day and birth weight. Round your answer to three decimal places.

Create a strategy

Apply the idea

Use the linear regression function on a calculator.

Using the statistics mode, enter each x-value along with its y-value into a data table on a calculator then determine the linear regression. The correlation coeﬃcient (r) is −0.957

b Use technology to determine an equation for the least-squares line of y on x. Round all values to two decimal places.

Create a strategy

Apply the idea

Use the linear regression function on a calculator.

Using the statistics mode, enter each x-value along with its y-value into a data table on a calculator then determine the linear regression. y = −0.04x + 3.92

c Use the equation to calculate the predicted birth weight of a newborn whose mother smoked on average 5 cigarettes per day. Round your answer to two decimal places.

Create a strategy Substitute the given x-value into the equation and evaluate to determine the predicted y-value.

Apply the idea y = −0.04x + 3.92

Write the equation

= −0.04 × 5 + 3.92

Substitute x = 5

= 3.72

Evaluate

Therefore, the predicted birth weight would be 3.72 kg.

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d Comment on the validity of the prediction in part (c).

Create a strategy Use the strength of the correlation and check whether the prediction used interpolation or extrapolation to comment on reliability.

Apply the idea The correlation coeﬃcient is −0.957 indicating a very strong relationship, which suggests a reliable prediction. The given x-value of 5 cigarettes per day sits within the range of the given data, therefore the prediction uses interpolation. Since interpolation is used and the correlation is very strong, the prediction is deemed reliable.

Reﬂect and check If the original dataset has a small number of data points, it can reduce the reliability of the predictions. For the purpose of this course, the focus will be on the strength of the correlation and whether interpolation or extrapolation was used.

e Calculate the predicted birth weight of a newborn whose mother smoked on average 55 cigarettes per day. Comment on the reliability of this prediction.

Create a strategy Substitute the given x-value into the equation to calculate the predicted y-value. To comment on the reliability, determine if the prediction is an interpolation or extrapolation, and consider the strength of the correlation.

Apply the idea y = −0.04x + 3.92

Write the equation

= −0.04 × 55 + 3.92

Substitute x = 55

= 1.72

Evaluate

Therefore, the predicted birth weight would be 1.72 kg. The range of the original data for cigarettes per day is [1.0, 46.3]. The value of x = 55 lies outside this range. Therefore, the prediction uses extrapolation.

Reﬂect and check Any prediction based on extrapolation is considered unreliable. This is because the linear relationship observed in the original data may not continue in the same way for values far outside the measured range.

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This scatterplot displays the market value of a car model based on its age: 35 000

Value ($)

30 000 25 000 20 000 15 000 10 000 5000

Age (years) 0 1

2 3 4 5 6 7 8 9

A car of this model is currently valued at $15 000. Using the regression line, estimate its age. 7

For each dataset, determine if the prediction is an interpolation or an extrapolation: a

A prediction for the y-value when x = 5 is made from the dataset: x

A prediction for the y-value when x = 33 is made from the dataset: x

For this dataset, predict the value of y for the given x-value, and comment on whether the prediction is reliable, referring to both the strength of correlation and whether interpolation or extrapolation is used. Correlation coeﬃcient: r = 0.93, x = 15, line of best ﬁt: y = 0.84x + 2.66 x

Complex familiar 9

For this dataset, predict the value of y for the given x-value, and comment on whether the prediction is reliable. Correlation coeﬃcient: r = −1, x = 1, line of best ﬁt: y = −3.06x + 93.51 x

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Ex 3

A sample of families were interviewed about their annual family income and their average monthly expenditure. The results are given in the table:

Income (x)

Expenditure ( y)

66 000

1100

75 000

1700

65 000

1400

73 000

1300

54 000

600

90 000

1800

87 000

1100

87 000

1500

94 000

1800

96 000

2200

Determine the equation for the least-squares line. Round values to three decimal places.

Predict the monthly expenditure for a family whose annual income is $80 000.

Comment on the reliability of the prediction made in part (b). The correlation coeﬃcient is r ≈ 0.79

The scores from a practice spelling test and the ﬁnal spelling test were recorded as percentages for 8 students. The correlation coeﬃcient between the two sets of scores is r ≈ 0.99: Practice (x)

56.3

79.0

59.4

77.0

71.6

64.4

61.6

68.2

Real ( y)

48.9

69.3

51.9

66.0

62.5

56.1

54.5

59.2

Predict the real spelling test result of a student who scored 60% in their practice test. Round the answer to two decimal places.

Explain the validity of this prediction.

The data in the table compares the number of marketing emails sent per month to the proportion of subscribers who open them. The data point (1, 1.28) is an outlier, the correlation coeﬃcient is r ≈ 0.79: a

Determine the equation for the least-squares line with the outlier removed. Round values to two decimal places. Predict the proportion of emails opened if they are sent 20 times a month. Round your answer to two decimal places. Describe the reliability of the prediction in part (b).

Frequency (x)

Proportion ( y)

0.41

0.59

1.28

0.47

0.26

0.62

0.75

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2.05 Association and causation After this lesson, you will be able to… • recognise that an observed association between two variables does not necessarily mean there is a causal relationship between them • explain that an observed association between two variables does not necessarily mean there is a causal relationship between them • identify possible non-causal explanations for an association, including coincidence • communicate possible non-causal explanations for an association, including the influence of another variable

Association and causation A least-squares line is used to identify trends and make predictions. For example, a trend may show that as a café’s advertising budget increases, its number of customers also increases. However, it is important to analyse what this connection between the variables means. To understand this connection, the deﬁnitions of association and causation must be explored.

Exploration Consider this graph, which shows data collected from a sample of primary school children: 1.

Describe the association shown in the graph between a child’s shoe size and their reading test score. Does a change in one variable appear to cause the change in the other? For example, does having bigger feet cause a child to become a better reader? Identify a possible reason that both variables are increasing together. (This is known as a confounding or lurking variable).

Reading test score (%) 80 60 40 20

Child shoe size (Youth) 0

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Association

Causation

A connection or a pattern between two variables. When one variable changes, the other seems to change in a predictable way.

A change in one variable directly causes a change in another. One thing makes the other happen.

Example: There is an association between students who bring lunch from home and their test scores. The data might show that students with a packed lunch tend to have higher scores.

Example: Increasing hours of study before a test can cause test performance to improve.

When analysing bivariate data, whether it is numerical or categorical data, it is important to note that an association between two variables does not necessarily imply causation. One reason why an association may not be a causation is if an observed pattern happens purely by coincidence or chance. For example, data might show an association between the price of bread in Brisbane and the amount of rainfall in London over the last ten years. This is almost certainly a coincidence, as one does not inﬂuence the other. However, more often, there are other factors, such as a third variable that is causing both things being observed to change. For example, data shows that when ice cream sales increase, sunburns also tend to increase. However, it’s known that eating ice cream does not cause sunburns. Other factors, such as hotter weather, may be involved.

Example 1 A local council observes that suburbs with more public parks have a larger life expectancy. They are asked to explain this association: a Does this mean building parks causes people to live longer? Explain your answer.

Create a strategy Consider if there is a direct cause-and-effect relationship.

Apply the idea No, this does not necessarily mean that building parks causes people to live longer. This is an example of an association, but not necessarily causation.

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b Identify a possible non-causal explanation for this association.

Create a strategy Consider a possible third variable that may inﬂuence both of these variables.

Apply the idea A possible third factor that could explain the association between having more parks in a suburb and higher life expectancy is socioeconomic status. Often, suburbs with more parks may also be wealthier areas. People in wealthier areas often have better access to healthcare, healthier food, less physically demanding jobs, and safer living conditions. These factors can lead to higher life expectancy, not necessarily the number of parks. So while parks and life expectancy are associated, it may be the higher socioeconomic status that is inﬂuencing both.

Example 2 A factory manager notices that on days with more workplace accidents, the company spends more money on coffee for the staff room. The manager concludes that buying coffee is causing accidents and suggests cutting the coffee budget. Is this a valid conclusion? Explain your reasoning.

Create a strategy Consider if there is a direct cause-and-effect relationship. If not, think about what other factors could be involved.

Apply the idea This is not a valid conclusion. The manager has observed an association between coffee spending and accidents, but this does not prove causation. A likely third factor may be the nature of the workday. For example, longer or more demanding workdays could cause workers to be more tired, leading to more accidents. These same diﬃcult days might also lead to management buying more coffee to boost employee morale and energy. The coffee is not causing the accidents; the demanding workday is likely inﬂuencing both.

Reﬂect and check Making business decisions based on a misunderstanding of data can lead to poor outcomes. In this case, cutting the coffee budget would likely lower morale without making the workplace any safer.

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Determine whether each statement is true or false: a

If two variables have a strong association, it always means one variable causes the other to change.

A third variable can be the real reason for an observed association.

Causation is a special, direct type of association.

If a scatterplot shows a clear linear pattern, it must be causation.

For each scenario, identify if it describes a clear causation or an association: a

The more hours that are worked at a casual job, the more money is earned.

As a person’s height increases, their shoe size increases.

The more kilometres a car drives, the more petrol it uses.

When a shop sells more ice cream, the neighbouring store sells more sunglasses.

Match each scenario to the correct non-causal explanation. i

Coincidence

Data shows that the number of bees in a town is associated with the number of babies born.

For the last decade, the average price of bananas in Australia has shown an association with the number of games won by the Brisbane Lions.

Third variable

A study shows that people who own expensive gym equipment tend to have better overall health. What is the most likely third factor that explains this association?

Ex 1

Exercising on expensive equipment is more effective.

People with more disposable income can afford both expensive gym gear and other aspects of a healthy lifestyle, like nutritious food or more leisure time.

People who own gym equipment live in areas with better air quality.

The instruction manuals for expensive equipment are better written.

A school observes that students who bring their own lunch tend to get higher test scores. They are asked to explain this association: a

Does this mean bringing lunch from home causes students to get higher test scores? Explain why or why not.

Identify a possible non-causal explanation for this association.

A city council shows that suburbs with more bicycle lanes have fewer cases of heart disease. They are asked to explain this association: a

Does this mean building more bicycle lanes directly reduces heart disease? Explain why or why not.

Identify a possible non-causal explanation for this association.

A university notices that students who attend more tutorials tend to report higher levels of happiness. They are asked to explain this association: a

Does this mean attending tutorials causes students to feel happier? Explain why or why not.

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A researcher observes that people who own electric cars tend to eat more plant-based meals. They are asked to explain this association: a

Does this mean owning an electric car causes someone to eat more plant-based food? Explain why or why not.

Identify a possible non-causal explanation for this association.

Complex familiar 10

A retail store manager observes that when the store sells more umbrellas, it also has more “wet ﬂoor” signs placed out. The manager concludes that putting out “wet ﬂoor” signs causes people to buy umbrellas. Is this a valid conclusion? Explain your reasoning.

An analyst for a city council determines that suburbs with a higher number of public libraries also have a higher average level of education among residents. Does this mean building more libraries would directly cause the education level of the community to rise? Explain a more likely reason for this association.

A health researcher notices that people who own a dog tend to have lower blood pressure than people who do not own a dog. Is it correct to conclude that owning a dog causes a person’s blood pressure to decrease? Explain a possible non-causal reason for this association.

A car dealership analyses its sales records. This scatterplot shows a car’s age and its average resale value: 40 000

Resale value ($)

Ex 2

35 000 30 000 25 000 20 000 15 000 10 000 5000

Age of car (Years)

144

Describe the association shown in the graph.

A car dealer states: “This proves that a car getting older is the only thing that causes its value to drop.” Comment on the validity of this sentence.

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2.06 Practical applications of bivariate data After this lesson, you will be able to… • identify associations between two variables in practical problems • analyse associations between two variables in practical problems • describe associations between two variables in practical problems • solve practical problems involving associations between two variables

Practical applications of bivariate data The skills of identifying, describing, and analysing associations between two variables can be applied to solve practical problems using real-world data. This involves numerical and graphical analysis, followed by an interpretation of the analysis in the practical context of the data. A review of the key ideas and vocabulary for analysing bivariate data is provided: Bivariate data

Least-squares line

Data consisting of two variables recorded for each individual, item, or event.

A linear line that best represents the trend in a scatterplot, also known as the line of best ﬁt.

Association A relationship or connection between two variables.

Correlation coeﬃcient (r)

Scatterplot

A number between −1 and 1 that measures the strength and direction of a linear association.

A graph used to display the relationship between two numerical variables.

Causation A relationship where a change in one variable directly causes a change in another.

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The equation for a least-squares line can be used to make predictions.

Response variable = Slope × Explanatory variable + y-intercept Response variable The variable being measured or predicted (plotted on the vertical axis). Slope The rate of change. It describes how much the response variable is predicted to change for every one-unit change in the explanatory variable. Explanatory variable The variable that is changed or observed, which may explain changes in the response variable (plotted on the horizontal axis). y-intercept The predicted value of the response variable when the explanatory variable is equal to zero.

Example 1

Hot chocolates sold

A local cafe owner tracks the daily temperature and the number of hot chocolates sold. The data is shown in the scatterplot, along with the least-squares regression line and correlation coeﬃcient: Temperature vs. Hot chocolates sold Sales = −2.5 (Temp) + 68 60 r = −0.92 50 40 30 20 10 0 5.0

7.5

10.0 12.5 15.0 17.5 20.0 22.5 25.0

Temperature (°C) a Describe the association between temperature and hot chocolate sales.

Create a strategy To describe the association, comment on form, direction, and strength. Use the scatterplot to see the form and direction, and use the correlation coeﬃcient (r) to determine the strength.

Apply the idea Form: The points on the scatterplot roughly follow a straight line, so the association is linear. Direction: As the temperature increases, the number of sales decreases. This is a negative association. Strength: The correlation coeﬃcient is r = −0.92. Since this value is very close to −1, the association is strong. Overall, there is a strong, negative, linear association between the daily temperature and the number of hot chocolates sold.

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b Interpret the slope of the least-squares line in the context of the problem.

Create a strategy

Apply the idea

The slope of the line is the coeﬃcient of the independent variable (Temperature). It represents the predicted change in the dependent variable (Sales) for every one unit change in the independent variable.

The equation of the line is Sales = −2.5 (Temp) + 68 The slope is equal to −2.5 This means that for every 1°C increase in temperature, the number of hot chocolates sold is predicted to decrease by 2.5

c Predict the number of hot chocolates sold on a day when the temperature is 10°C.

Create a strategy Substitute the given temperature value into the equation of the least-squares line to calculate the predicted number of sales.

Apply the idea Sales = −2.5 (Temp) + 68

Write the equation of the least-squares line

= −2.5(10) + 68

Substitute Temp = 10

= −25 + 68

Evaluate the product

= 43

Evaluate

On a 10°C day, the cafe is predicted to sell 43 hot chocolates.

Reﬂect and check This is considered a reliable prediction because the correlation coeﬃcient indicates a strong association and the prediction is made within the range of the original data (interpolation).

d Does this association imply causation? Explain.

Create a strategy Consider if the change in temperature is the direct cause of the change in sales, or if there might be other related factors.

Apply the idea While there is a strong association, it does not necessarily imply causation. It is very likely that lower temperatures cause people to want warm drinks, so in this case, a causal link is plausible. However, certainty is not possible without more information. For example, rainy weather might also cause an increase in sales, and rainy days tend to be colder. The temperature itself might not be the only cause.

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Example 2 A school surveyed 100 students about their mode of transport to school and whether they arrived on time or late. The results are shown: Transport mode

On time

Late

Total

Walk

Bus

Car

Total

100

a Identify the explanatory variable.

Create a strategy

Apply the idea

Identify which variable is likely to inﬂuence changes in the other.

The mode of transport used is likely to inﬂuence whether a student is late or on time. Therefore, the mode of transport is the explanatory variable.

b A row percentage frequency table has been started to analyse the association between the variables. Complete the table. Transport mode

On time

Late

Total

Walk

93%

Bus

⬚

Car

⬚

100% ⬚

Create a strategy Complete the table by using this formula: Percentage =

× 100

Apply the idea The ﬁrst cell in the table is calculated by: Walk and on time =

× 100 ≈ 93%

The remaining values in the row percentage frequency table can be calculated similarly as shown: Transport mode

On time

Late

Total

Walk

93%

100%

Bus

62.5%

37.5%

100%

Car

93%

100%

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2.06 Practice questions Simple familiar 1

Match each term to its correct deﬁnition. a

Bivariate data

A line that best represents the trend in a scatterplot.

Association

Scatterplot

A number between − 1 and 1 measuring the strength and direction of a linear relationship.

Least-squares line

iii

A relationship where a change in one variable directly causes a change in another.

Correlation coeﬃcient (r)

A relationship or connection between two variables.

Data that has two variables, collected from the same source.

Causation

A graph used to display the relationship between two numerical variables.

For each scenario, identify the two variables and classify them as numerical or categorical: a

Comparing the colours of cars from different manufacturers (e.g., Ford, Toyota, Hyundai).

Investigating the relationship between the number of hours a student sleeps and their score on a test.

Analysing the average daily screen time for people in different age groups (e.g., child, teenager, adult).

For each correlation coeﬃcient (r) value, describe the strength and direction of the linear association: a

r = − 0.95

r = 0.3

r = 0.8

r = − 0.6

The predicted score on a ﬁnal exam based on the hours a student studies is given by the least-squares regression equation: Score = 4.5 × Hours + 55 a

Determine the slope of the line.

Determine the y-intercept.

Using the equation, Score = 4.5 × Hours + 55, predict the exam score for a student who studies for 8 hours.

For each pair of variables, determine which is the explanatory variable and which is the response variable:

Number of cats a person owns vs. amount of money spent on pet food each week

Score on a maths test vs. number of hours a student studies

Final exam grade vs. number of missed school days

Is the observed association likely to be causal? Explain: a

People who carry lighters are more likely to get lung cancer.

Schools with more playgrounds tend to have more active students.

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Complex familiar Ex 1

A study was conducted to analyse the relationship between the age of a second-hand car (in years) and its value (in dollars). The data is shown in this scatterplot: Age vs. Value 35 000

Value = −2800 (Age) + 35 000 r = −0.96

Value ($)

30 000 25 000 20 000 15 000 10 000 5000

Age (years)

Ex 2

Describe the association between the age of the car and its value by commenting on the form, direction, and strength. The correlation coeﬃcient is r = − 0.96.

The equation of the least-squares regression line is Value = − 2800 × Age + 35 000. Interpret the slope in the context of this problem.

Use the equation to predict the value of a 6-year-old car.

Does this association imply causation? Explain.

A survey of 120 students looked at whether students ate breakfast and whether they reported high or low concentration during morning classes: High concentration

Low concentration

Total

Ate breakfast

Skipped breakfast

Total

120

Identify the response variable.

Construct a frequency table.

Describe the association between breakfast habits and concentration levels.

A study investigated the relationship between the number of hours teenagers spend on screens per day and the number of hours they sleep at night. The correlation coeﬃcient is r = − 0.85 and the equation of the least-squares line is: Sleep (hrs) = − 0.6 × Screen time (hrs)+9.2 a

Describe the association by commenting on form, direction, and strength.

Interpret the slope in the context of the problem.

Predict the average sleep duration for a student who spends 5 hours on screens per day.

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Complex unfamiliar 15

A group of students recorded the number of hours they studied for an exam and their corresponding exam scores. The data recorded is shown: Hours studied

1.5

3.5

4.5

2.5

Exams score (%)

Provide a complete analysis and description of the relationship between hours studied and exam score. 16

A teacher recorded where 100 students studied for a test and whether they passed or failed. The results are shown: Study location

Passed

Failed

Total

Library

Home

Total

100

Analyse the data and describe the relationship between study location and test results. 17

A used car dealership collected data on the age of second-hand cars (in years) and the resale values (in dollars). The data collected is shown: Age ( years) Resale values ($)

28 000 24 000 20 000 26 000 22 000 18 000 15 000 14 000 23 000 19 500

Write a complete analysis and description of the relationship between the age of the car and its resale value. Include comments on the form, direction, strength, and any potential outliers or clusters.

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2 Chapter review SF

This table shows data collected on the daily temperature and the number of customers at a local swimming pool: Temperature (°C, x)

Customers ( y)

110

145

180

210

255

290

Use technology to determine the equation of the least-squares line that models this data, in the form y = mx + c. Round the gradient and y-intercept to two decimal places. SF

The value of a delivery van over time is modelled by the least-squares line: Value = − 3500 × Age + 40 000 where value is in dollars and age is in years:

Interpret the slope (gradient) of the line in the context of this scenario.

Interpret the y-intercept in the context of this scenario.

A study investigated the link between the number of hours a phone is charged and its battery percentage. The least-squares line is: Percentage = 25 × Hours + 5 The data was collected for charging times between 0.5 and 3.5 hours: a

Predict the battery percentage after charging for 2.5 hours.

Is the prediction in part (i) an example of interpolation or extrapolation?

If the battery percentage is at 30%, how many hours would it have been charged according to this model?

For a bivariate dataset, these summary statistics were calculated: • = 15, = 50 • sx = 3, sy = 10 • r = 0.8 Calculate the equation of the least-squares line. Round values to one decimal place.

A linear model is used to predict a student’s test score based on their attendance. For a student with 90% attendance, the model predicts a score of 85. The student’s actual score was 82: a

Calculate the residual for this data point.

Was the actual data point above or below the least-squares line?

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This graph shows the association between the air temperature, x, and the number of vehicles parked at a train station, y:

Number of vehicles

300 250

Least-squares fit y = 25x − 610 r = 0.986 R2 = 0.972

200 150 100 50 0

Air temperature (°C) a

Identify Pearson’s correlation coeﬃcient and use it to describe the strength of the association between x and y.

It is suggested that the time of day, t, could be a confounding variable in this situation. Deﬁne confounding variable.

Explain why t could be a confounding variable in this situation.

(QCAA 2025 Paper 1 Q20) 15

Hiroki believes that more fish are caught on warmer days. Jiro believes that the number of fish caught in a day is more dependent on the number of people fishing. Bivariate datasets for six days are shown: Temperature, t (° C)

Number of fish caught, f

530

400

320

220

180

120

Number of people fishing, p

Number of fish caught, f

530

400

320

220

180

120

Calculate the correlation coefficient for each dataset and use the results to identify the explanatory variable for the stronger linear association. Use the least-squares line equation for the stronger linear association to predict the number of fish caught on a 25°C day when 50 people are fishing. (QCAA 2023 Paper 2 Q4)

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3.01 Read and interpret time series graphs After this lesson, you will be able to… • construct a time series graph from a given set of data • identify and describe the underlying trend in a time series graph • determine if seasonality is present

Time series data A time series graph shows how a variable changes over time. Time (the explanatory variable) is represented in the x-axis and the response variable (such as temperature, rainfall, population) is represented in the y-axis. A time series graph is a bit like a scatterplot; however, the points are connected in sequential order by lines. It is measured at regular time intervals such as monthly, weekly, yearly, etc.

Exploration

Average temperature (°C)

Before learning the formal terms, take a moment to explore a time series graph. This graph shows the average monthly temperature in a city over 3 years: 35 30 25 20 15 10 0

Time (months) 1. Determine the general direction of the temperature over the entire 3-year period. Is it generally increasing, decreasing, or staying about the same? 2. Is there a repeating pattern in the temperature data? If so, how many months does it take for one full pattern to complete before it starts again? 3. Is there any point on the graph that seems unusually high or low and does not ﬁt the general pattern? What could be a possible real-world reason for this?

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Time series data can be broken into 3 main components: Trend The long-term direction of the data over time. It can be upwards or positive, downwards or negative, or ﬂat. Example: Electricity usage showing a long-term upward trend over a year for a high-rise building. These graphs show whether the overall response variable is increasing (positive trend), decreasing (negative trend), or stable (not increasing or decreasing).

This graph shows an increasing trend. Note the ‘peaks’ and ‘troughs’ are steadily getting higher.

This graph shows a decreasing trend. Note the ‘peaks’ and ‘troughs’ are steadily getting lower.

This graph shows a stable trend. Note the ‘peaks’ and ‘troughs’ are at a similar level.

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Seasonality A pattern in the data that repeats at regular intervals of time due to seasonal effects, for instance, monthly sales peaks in the month of December due to the festival season in a private retail shop. Example: Electricity usage showing seasonal increase during summer due to the increased usage of air conditioners.

By following the rise and fall, or cyclical nature of the data, the number of seasons or time periods that constitute one cycle can be determined. To determine the number of seasons in a cycle, count the number of points from peak to peak or trough to trough.

This is a peak

60 50

This graph shows how to count the number of points in a cycle.

20 10

Start 0

This is a trough

2 4 6 8 10 12 14 16

Note: To be seasonal, each cycle must have the same number of points. If there is a repeating pattern, but the number of data points is different for each cycle, then it is described as cyclical rather than seasonal. In the example, each cycle has 4-points, so it is described as seasonal with 4 seasons per cycle. Irregular ﬂuctuations Unpredictable changes usually caused by events that are unfortunate or unexpected such as natural disasters or inﬂation in the economy or tensions between countries. Example: Irregular or sudden spike in electricity usage during a heat wave due to increased use of air conditioners within a country.

70 60 50 40 30 20

Outlier

10 0

164

Outliers - Once trend and seasonality have been observed, outliers can be determined. Outliers are data points that do not follow the pattern of the rest of the data.

2 4 6 8 10 12 14 16

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Example 1 Consider this time series graph:

y 30 25 20 15 10 5 x

a Describe the underlying trend of the data.

Create a strategy Check whether the peaks and troughs are increasing or decreasing.

Apply the idea The overall trend of the data needs to be considered. The peaks and troughs are generally increasing, so the underlying trend of the time series graph is increasing.

Reﬂect and check When identifying trends in time series data, it is important to focus on the overall direction over time rather than short-term ﬂuctuations within the dataset. b Determine the number of points per cycle.

Create a strategy Count the points in a cycle from peak to peak.

Apply the idea Consider one cycle of the data.

y 30

By counting the points from the ﬁrst peak, and stopping before the next peak, it can be seen that there are 3-points per cycle.

25 20

15 10

2 3

x 0

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Example 2 Consider this set of time series data: Time period Data

1 5

2 10

3 18

4 1

5 11

6 20

7 7

8 13

9 22

10 10

11 15

12 25

a Plot this data on a graph.

Create a strategy Consider each pair of values in the table as a set of coordinates: (Time period, Data)

Apply the idea

Reﬂect and check

Sketch the graph of the data with the time period on the x-axis and the data on the y-axis.

Time series data is usually joined together using line segments.

Data 20 15 10 5

Time period 0

b Determine the number of seasons (points per cycle) in this data.

Create a strategy Count the points in each cycle from trough to trough.

Apply the idea For time series, a more general deﬁnition of seasons is used. This involves looking at the number of data points it takes for the data to start a similar-looking cycle again.

Data 20

3 15

Consider one cycle of the data. It can be noticed that the graph repeats a similar pattern.

Counting can start from the ﬁrst trough and stop before the next trough. It can be seen on the graph that there are 3 seasons in each cycle of this data.

Time period 0

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3.01 Practice questions Simple familiar Consider this graph:

Identify the explanatory variable.

Identify the underlying trend in the data as increasing, decreasing, or constant.

Temperature

30 25 20 15 10 5

. a. m . 12 p. m . 2 p. m .

a. m 8

a. m 6

Time Consider this graph:

168

Arctic Sea ice extent (Area of ocean with at least 15% sea ice)

National Snow and Ice Data Center, Boulder CO

Extent (millions of square kilometres)

12 10 8 6 4 0

2007 2005 Record Low 1979-2000 Average Jul

Aug

Sep

Oct

Nov

Which year had more ice: 2005 or 2007?

What can be concluded about the amount of sea ice as time passes?

Which month is the warmest in the Arctic?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

This graph shows energy usage in two countries over several years: y 225 000 200 000 175 000 150 000 125 000 0 2004 2005 2006 2007 2008 2009 2010

Australia

x 2011

2012 2013

United Kingdom

Determine whether each of these statements is true or false:

Australia used more energy in 2012 than it did in 2004.

The United Kingdom used more energy in 2011 than in 2006.

The United Kingdom is cutting back on its energy use overall.

From 2004 to 2012, the United Kingdom’s usage decreased by more than Australia’s usage increased.

The graph does not give us any information about how the type of energy is being used.

The graph does not give us any information about energy use per capita.

From the graph, it can be assumed that the population of the United Kingdom is declining.

More information would be needed to comment on the energy usage per capita, and hence draw a conclusion about how well either country is doing about reducing primary energy usage.

This graph shows the change in Apple Inc.’s share price along with major releases of its products: Apple (AAPL) $140 weekly closing prices $120 $100 $80 $60 $40

First iPod ($399)

iPod Touch ($299)

iPod Shuffle ($99) iPod Video iTunes store ($299) launches iPod Photo ($499)

$20

iPod Mini ($249)

0 2001

2002

2003

2004

iPhone ($499)

Redesigned iPod Nano iPod Shuffle ($199) ($79) 2005

2006

2007

Using the graph, determine whether each of these is true or false: a

Apple’s share price always increased.

The release of Apple products promoted a general trend upwards in the share price.

Apple’s share price increased by the same amount each year. 3.01 Read and interpret time series graphs mathspace.co

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This graph shows the change in Australian stock prices (in green) and US stock prices (in blue) over several years: US and Australia stock prices 8000

Index

Index Index

1600

7000 6000

1400

S & P500 (RHS)

1200

5000 4000

A SX200 (LHS)

1000

3000

800

2000

600

Oct. 2002 2003 2004 2005 2006 2007 2008 2009

1800

Identify the underlying trends of the US and Australian stock prices from October 2002 to October 2006 as increasing, decreasing, or constant.

During October of which year does the trend ﬁrst change direction?

Using the graph, list the months when the rainfall in London was less than the rainfall in Berlin:

Amount of rainfall (mm)

Monthly rainfall (mm) in major cities 70 65 60 55 50 45 40 35 30 0

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Berlin

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London

Consider this graph:

Number of responses

Connection responses 4000 3000 2000 1000 0

April 21

April 28

May 05

May 12

May 19

April 2024 - May 2024

Identify the underlying trend as increasing, decreasing, or constant.

On what date does an outlier appear to occur?

Count the approximate number of days in one cycle.

Consider this time series graph: Company sales 2002 - 2016 (millions) Sales (millions) 60 50 40 30 20 10

Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Identify the underlying trend as increasing, decreasing, or constant.

Determine the number of seasons per cycle.

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171

Ex 1

For each of these time series graphs: i

Identify the underlying trend of the data as increasing, decreasing, or constant.

Determine the number of points per cycle.

Data

Time period 0

Time period

Data

30 10

Time period

172

2 2

8 10 12 14 16

160 Data 140 120 100 80 60 40 20 −20 −40 −60 −80 −100

Data

Time period 4

10 12

Time period 0

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Consider this time series graph: 14 12 10 8 6 4

Nov 19

Jul 19

Mar 19

Nov 18

Jul 18

Mar 18

Nov 17

Jul 17

Mar 17

Nov 16

Jul 16

Mar 16

Nov 15

Jul 15

2 Mar 15

Visitors to King’s Park ('000s)

Days

Identify the underlying trend as increasing, decreasing, or constant.

Determine the number of seasons per cycle.

Identify the month in which the cycle peaks.

Identify the month that is usually the lowest season and determine if there was an exception to this.

Complex familiar 11

The local police station records the number of speeding ﬁnes issued each quarter. This table has the data for each quarter from 2012 to 2014: Period

Time

Number of speeding ﬁnes

Percentage of yearly mean

March 2012

83.33%

June 2012

116.67%

September 2012

150%

December 2012

50%

March 2013

81.36%

June 2013

118.64%

September 2013

142.37%

December 2013

57.63%

March 2014

89.51%

June 2014

109.09%

September 2014

142.66%

December 2014

58.74%

Construct a time series graph for the number of ﬁnes recorded each quarter against the period.

Identify the underlying trend of the data as increasing, decreasing, or constant. 3.01 Read and interpret time series graphs mathspace.co

173

From the beginning of 2012, the number of new houses built in the suburb of Woodford was recorded and ﬁgures are released every 4 months.

Time period

This table shows the data from the beginning of 2012 to the end of 2015:

August 2012

December 2012

105 99

April 2013

Determine if the data appears to be seasonal.

August 2013

Identify the underlying trend of the data as increasing, decreasing, or constant.

December 2013

104

April 2014

Determine the number of seasons per cycle.

August 2014

December 2014

103

April 2015

August 2015

December 2015

For each table of time series data: i

Construct a time series graph for the data.

Determine the number of seasons per cycle.

iii

Identify the underlying trend of the data as increasing, decreasing, or constant.

Identify the time period that appears to be an outlier, if there is one.

Time period

Data

Time period

Data

100

135

110

105

174

103

Construct a time series graph for the data.

April 2012

a c

Ex 2

Houses built

Month Watch time (hours)

Month

Number of bags (’000s)

Jan ’17

Dec ’05

9.5

Apr ’17

Jun ’06

9.1

Jul ’17

Dec ’06

9.7

Oct ’17

Jun ’07

9.5

Jan ’18

Dec ’07

9.9

Apr ’18

Jun ’08

8.2

Jul ’18

Dec ’08

8.5

Oct ’18

Jun ’09

8.3

Jan ’19

Dec ’09

8.5

Apr ’19

Jun ’10

Jul ’19

Dec ’10

8.1

Oct ’19

Jun ’11

7.7

Dec ’11

7.3

Jun ’12

7.5

Dec ’12

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Consider this time series graph, which shows the average hours worked per week recorded for each month: a

Identify the underlying trend for the data as increasing, decreasing, or constant.

Determine the number of seasons per cycle.

Identify the number of the month in which there is an outlier.

Hours 70 60 50 40 30 20 10

Month 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

A beachside cafe tracks its proﬁt each quarter over 3 years. The data is recorded in this table: Quarter

Proﬁt (in thousands of dollars)

Q1 2021 (Summer)

Q2 2021 (Autumn)

Q3 2021 (Winter)

Q4 2021 (Spring)

Q1 2022 (Summer)

Q2 2022 (Autumn)

Q3 2022 (Winter)

Q4 2022 (Spring)

Q1 2023 (Summer)

Q2 2023 (Autumn)

Q3 2023 (Winter)

Q4 2023 (Spring)

Construct a time series plot for the data.

Describe the underlying trend.

Explain the seasonal pattern in the context of the problem.

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This graph shows the number of US dollars that one Australian dollar could buy over a 5-year period:

1.05

0.95

0.85

0.75

0.65

2012

2013

2014

2015

Identify the underlying trend as increasing, decreasing, or constant.

During the 5-year period, the ﬁrst irregular and sustained ﬂuctuation occurred in what year?

Compare the 5-year graph to this 10-year graph:

1.15

1.00

0.85

0.70

0.55

2007

2008

2009

2010

2011

2012

2013

During what year did the value of the Australian dollar crash?

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2014

2015

The time series graph shows the monthly ice cream sales for a shop over 3 years:

Ice cream sales

1200 1000 800 600 400 200 0

r r r c c c n p n p n p Ma r-Ju -Se t-De -Ma r-Ju -Se t-De n-Ma r-Ju -Se t-De n n n n p p p c c A Ju Oc Ja A Ju O Ja A Ju O 2006 2007 2008

The shop was forced to close for two weeks in May of the second year for emergency renovations. Describe how this event would likely appear on the time series graph as an irregular ﬂuctuation.

Complex unfamiliar 18

This table shows the proﬁt, in thousands of dollars, for a particular company for the years 2010 to 2016: Year

2010

2011

2012

2013

2014

2015

2016

Proﬁt (in thousands of dollars)

Construct a time series graph for the data.

Identify the underlying trend of the data as increasing, decreasing, or constant.

Determine the percentage growth in proﬁt from 2010 to 2016. Round your answer to the nearest percent.

The quarterly visitor numbers, in thousands, for a tropical resort in North Queensland are recorded in the table for the years 2021 to 2023: Year

Q1 (Jan-Mar)

Q2 (Apr-Jun)

Q3 (Jul-Sep)

Q4 (Oct-Dec)

2021

2022

2023

Describe the seasonal pattern observed in the data.

Provide a possible real-world explanation for this seasonal pattern. Justify your reasoning.

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3.02 Moving averages After this lesson, you will be able to… • understand the purpose of smoothing time series data to reveal underlying trends • calculate a simple moving average and moving median for an odd number of data points • interpret a smoothed time series graph to analyse the underlying trend • uses spreadsheets to calculate moving averages

Moving average Exploration Consider this time series data, which shows the number of ice creams sold at a shop over 7 days: Day (t)

Sales (y)

1. Calculate the average (mean) of the sales for the ﬁrst 3 days (Day 1, 2, and 3). 2. Which day (1, 2, or 3) is the middle time period for these 3 days? 3. Calculate the average of the sales for Day 2, 3, and 4. Which day is the middle time period for these days? 4. How does using the average of 3 days compare to the original daily sales data? Does it look smoother or more jagged?

A moving average is a method used to ‘smooth out’ time-series data by averaging a set number of consecutive points. This reduces short-term ﬂuctuations, making long-term trends easier to identify. Smooth out means removing the random spikes and processing the data to make the trend clearer. To apply this method, the moving averages of the response variable, y, and plot them against time, t. A linear regression line can then be ﬁtted to the smoothed data to highlight the trend. 178

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The most common method is to smooth over an odd number of points, for example, 3 or 5: 3-point moving average

5-point moving average

Average of each group of 3 consecutive values (odd moving average)

Average of each group of 5 consecutive values (odd moving average)

Example: For crop yield data recorded over a 3-season cycle, a 3-point moving average helps smooth the variation between each season and highlight long-term trends in productivity.

Example: For weekly retail sales data with a 5-week repeating business cycle, a 5-point moving average helps smooth variations and highlight underlying consumer trends.

Odd moving averages are used for cycles with an odd number of data points. This is one of the most common types of moving averages, as it includes a central value—eliminating the need for graphical adjustment to align with the time axis. They are named 3MA (3-point moving average), 5MA (5-point moving average), 7MA (7-point moving average), etc. For a 3MA, the average of response variables for 3 data points (For example: t = 1, t = 2 and t = 3) is calculated, and the result is plotted against the middle point (t = 2). The ﬁrst and last points do not have values on each side, so they are left out. For a 5MA, the average of the response variables for 5 data points (For example: t = 1, t = 2, t = 3, t = 4 and t = 5) is calculated, and the result is plotted against the middle point (t = 3). The ﬁrst two and the last two points do not have values on each side, so they are left out. This graph shows a time series with both a 3-point and a 5-point moving average applied: 160 140 120 100 80 60 40 20 0

Raw data

3MA

5MA

Generally, a moving average with more points creates a smoother line. However, the primary goal of smoothing is often to remove seasonality. To do this most effectively, the number of points in the moving average should match the length of the seasonal cycle. In the graph shown, the cycle length is 3-points. Therefore, the 3-point moving average is the most appropriate for deseasonalising this speciﬁc data set, as it effectively removes the 3-point cycle. By examining the raw data graph and counting the data points from one peak to the next, it becomes clear that there are 3-points per cycle. If there are 3-points per cycle, then a 3MA will be the best moving average to use to smooth or deseasonalise the data - this is because a 3MA will account for each of the three seasons present in the data. The moving average used should always match the number of seasons per cycle in the original data. 3.02 Moving averages mathspace.co

179

Example 1 Calculate the missing values of x and y in the table. Time period

Raw data

3MA

5MA

119

106.7

107.6

131

107.33

102.4

105

109.67

109.33

114.8

130

112.67

109.6

115

116.67

116.8

105

120.33

121.2

141

120.33

116.6

115

121

107

Create a strategy Calculate x and y using the average formula.

Apply the idea To calculate x, the 3-point moving average is calculated for the data centred at time period 3. This uses the raw data from time periods 2, 3, and 4:

To calculate y, the 5-point moving average is calculated for the data centred at time period 5. This uses the raw data from time periods 3, 4, 5, 6, and 7:

Reﬂect and check Blank cells appear at the beginning and end of the moving average columns. This is because a 3-point moving average requires a data point on either side of the centre point, and a 5-point moving average requires two data points on either side.

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Example 3 This table contains time series data: Time period

Raw data

3-moving average

5-moving average

113.8

109

112.1

113.4

107.4

97.7

99.7

52.6

86.4

93.8

107

85.3

91.6

96.3

101.8

90.7

89.5

95.6

81.4

88.2

46.5

80.9

86.7

100.5

78.6

a Calculate the values of a and b in the table. Round your answer to two decimal places.

Create a strategy Use three data points to calculate a and ﬁve data points for b, then use average formula.

Apply the idea The three data points used to calculate a are 113.4, 99.7, and 52.6. Write the formula

Substitute the data points

Evaluate the addition

Evaluate

The five data points used to calculate b are 109, 113.4, 99.7, 52.6, and 107. Write the formula

Substitute the data points

Evaluate the addition

Evaluate

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Moving median Moving median is a smoothing technique used in time series analysis to reduce short-term fluctuations and highlight long-term trends. Unlike the moving average, which uses the mean of values, this method takes the median of a fixed number of consecutive data points. Because the median is less affected by extreme values, moving median smoothing is particularly useful when the data contains outliers or sharp spikes. It is common to carry out the smoothing over an odd number of points such as 3-point moving median, 5-point moving median, etc. For 3-point moving median, the median of the data values is calculated for t = 1, t = 2 and t = 3 and the result will be plotted against the middle point (t = 2). The first and last points do not have values on each side, so they are left out. Since the median is the middle value when the numbers are arranged in order, in 3-median smoothing, the median is always the second value in the ordered set. For 5-point moving median, the median of five data values (For example: t = 1, t = 2, t = 3, t = 4 and t = 5) is calculated and the result is plotted against the middle point (t = 3). The first two and the last two points do not have values on each side, so they are left out. Since the median is the middle value when the numbers are arranged in order, in 5-median smoothing, the median is always the third value in the ordered set. Consider the time series shown: For 3-point moving median, the y values for t = 1, t = 2 and t = 3 are 2, 1 and 3 respectively. When arranged in order (1, 2, 3), the median will be 2 which will be the smoothed median value for t = 2.

y 4 3 2 1

t 0

For the next point, centred at t = 3 the y values for t = 2, t = 3, and t = 4 are 1, 3, and 4. When arranged in order (1, 3, 4), the median is 3 which will be the smoothed median value for t = 3. This can be continued until the last data point. The 3-point moving median line is the blue line. A similar process is used for 5-point moving median which uses 5 consecutive data points at a time.

Moving median smoothing can be done directly on the graph without needing to know the exact values of each data point.

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Example 4 This table contains time series data: Time period

Raw data

3-moving median

5-moving median

113.8

−

109

−

113.4

⬚

99.7

52.6

107

96.3

⬚

−

⬚ ⬚

⬚ −

a Complete the table by determining the moving median values.

Create a strategy Use three consecutive data points to calculate 3-point moving median values and ﬁve consecutive data points for 5-point moving median values.

Apply the idea The first three data points are 113.8, 109, and 113.4. When arranged in order the median will be 113.4. The next three data points are 109, 113.4, and 99.7. When arranged in order the median will be 109. And so on. The first five data points are 113.8, 109, 113.4, 99.7, and 52.6. When arranged in order the median will be 109. The completed table is:

186

Time period

Raw data

3-moving median

5-moving median

113.8

−

109

113.4

−

113.4

109

99.7

107

52.6

99.7

107

96.3

−

96.3

−

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Use of spreadsheet with time series data Spreadsheets are a powerful tool for calculating moving averages and medians and also graphing both the original and smoothed data sets.

Exploration Consider this spreadsheet showing sales data:

The formula in cell C3 calculates the 3-point moving average centred at Day 2. 1. What values are being averaged by the formula in cell C3? 2. If this formula is copied from C3 and pasted into cell C4, what will the new formula be? 3. Why is this automatic change (called relative referencing) useful when calculating a moving average for a long time series?

Spreadsheets use formulas to perform calculations on data entered into cells. All formulas must begin with an equals (=) sign. To calculate the moving average, =AVERAGE(), function is very useful. It calculates the average of a range of cells. For example, =AVERAGE(B2:B4) calculates the average of the values in cells B2, B3, and B4. To calculate the moving median, =MEDIAN(), function is very useful. It calculates the median of a range of cells. For example, =MEDIAN(B2:B4) calculates the median of the values in cells B2, B3, and B4. When writing formulas that will be copied to other cells, it is important to understand cell referencing: • A relative cell reference (e.g., A1) changes when a formula is copied to another cell. If you copy a formula from C1 that references A1, and paste it into C2, the reference will automatically change to A2. • An absolute cell reference (e.g., $A$1) remains constant no matter where it is copied. The dollar signs lock the column and row. This is useful when you want all calculations to refer back to one speciﬁc cell, like a cycle average.

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Example 5 Consider the time series data shown on the spreadsheet: A

Time period

Raw data

3-moving average

3-moving median

5-moving average

108.4

98.2

⬚

99.9

109.5

111.2

91.2

⬚

a Write the formula for cell C3 that can be used to calculate the 3-point moving average for time period 2.

Create a strategy Write the formula using the AVERAGE function and the cells that contain the values.

Apply the idea

Reﬂect and check

3-point moving average is calculated using the values in cells B2, B3 and B4.

The 3-point moving average can also be calculated using the formula: =(B2+B3+B4)/3

So, the formula is =AVERAGE(B2:B4).

b Write the formula for cell D3 that can be used to calculate the 3-point moving median for time period 2.

Create a strategy Write the formula using the MEDIAN function and the cells that contain the values.

Apply the idea 3-point moving median is calculated using the values in cells B2, B3 and B4. So, the formula is =MEDIAN(B2:B4).

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A cafe records the number of coffees sold each day for a week: 45, 51, 48, 55, 110, 58, 62

Ex 1

Calculate the ﬁrst possible 3-point moving average.

Calculate the ﬁrst possible 5-point moving average.

For each table containing time series data: i

Calculate the values of a, b, c, and d, rounded to one decimal place.

Determine which moving average best smooths the data.

Time period

Raw data

3-moving average

5-moving average

113.8

109

112.1

113.4

107.4

97.7

99.7

52.6

86.4

93.8

107

85.3

91.6

96.3

101.8

90.7

89.5

95.6

81.4

88.2

46.5

80.9

86.7

100.5

78.6

Calculate the values of a, b, and c, rounded to one decimal place. Time period

Raw data

3-moving average

5-moving average

43.8

46.5

41.2

33.4

42.5

45.8

40.5

39.4

32.8

40.3

41.8

42.2

39.6

39.9

38.7

32.3

39.6

41.1

41.3

39.5

44.8

39.2

31.5

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Consider the time series graphs drawn, along with two sets of moving averages. For each graph, explain which moving average is most appropriate for this data:

Data

Ex 2

15 5 0

10 5

Time period data c

MA3

540

Data

480

MA3

8 9

MA5

0 1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

Time period

data

192

440

500 460

Ex 3

Time period

520

420

data

MA5

MA3

MA5

data

MA7

MA5

This table contains time series data showing a company’s quarterly proﬁts in thousands of dollars: Quarter

Profit ($000s)

3-point MA

5-point MA

150

165

156.3

154

158.8

170

163.3

166

165.3

160.2

160

161.3

Calculate the missing values for a, b, and c. Round your answers to one decimal place where necessary.

Determine which moving average best smooths the data.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 4

This table contains time series data: Time period

Raw data

13.8

10.9

13.4

19.7

12.6

10.7

9.3

3-moving median

5-moving median

⬚ ⬚

⬚

Complete the table by determining the moving median values. 8

Use the time series given to determine the smoothed value for: Data 4 3 2 1

t 0

t = 4 using: i

3-point moving median

5-point moving median

t = 7 using: i

3-point moving median

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Ex 5

These data are entered into a spreadsheet: A

Year

Million ($)

3MA

5MA

4CMA

19.0

18.3

19.8

19.3

20.7

26.2

24.1

28.7

26.4

27.3

34.0

25.4

28.6

28.3

30.6

29.8

25.7

35.4

31.8

35.7

36.2

36.0

46.7

35.8

39.9

42.7

42.0

42.3

39.3

47.2

44.3

44.0

47.6

46.3

55.3

Write a spreadsheet formula that could be typed into cell C3 to calculate the 3-point moving average for the year 2003.

Write a spreadsheet formula that could be typed into cell D4 to calculate the 5-point moving average for the year 2004.

If the formulas were copied down their respective columns, what formula would be in: i

Cell C5?

Cell D6?

Complex familiar 10

194

This table shows the monthly rainfall, in millimetres, for a town. A data entry error in April resulted in an outlier: Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Rainfall (mm)

210

Calculate the 3-point moving average centred on April.

Calculate the 3-point moving median centred on April.

Explain which smoothed value (the mean or the median) is a better representation of the rainfall trend for that period, and why.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Consider the time series data displayed in the table: Calculate the values of a, b, c, and d, rounded to one decimal place

Time period

Raw data

Use the calculator to determine, rounded to four decimal places:

3.9

13.0

3-point moving average

The correlation coeﬃcient for the raw data

The correlation coeﬃcient for the 3-point moving average data

31.3

3.8

13.9

Is the correlation coeﬃcient stronger for the raw data or the 3-point moving average data?

6.6

14.9

34.4

6.1

16.4

8.6

15.6

37.3

10.3

18.1

11.8

39.5

Time period

Raw data

29.7

14.9

9.2

31.1

18.9

34.9

18.9

18.4

34.5

24.6

37.3

24.6

34.5

17.3

39.5

Consider the time series data displayed in the table: a

Calculate the values of a and b.

Use the calculator to determine, rounded to four decimal places:

The correlation coeﬃcient for the raw data

The correlation coeﬃcient for 5-point moving median data

Is the correlation coeﬃcient stronger for the raw data or the 5-point moving median data?

5-point moving median

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Complex unfamiliar 13

This time series graph shows the maximum daily temperatures, in degrees Celsius, in a city over a period of 9 consecutive days: 28

Temperature

26 24 22 20 18 16

Day

Determine the smoothed temperature for: a

Day 4 using: i

5-point moving median

7-point moving median

Day 6 using: i

3-point moving median

This time series graph shows the number of births in a local hospital over a period of 12 months: 28

Number of births

26 24 22 20 18 16 0

Month 1

Draw the 5-point moving median smoothed plot for the time series.

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Calculating the seasonal index using the average percentage method: 1. For each cycle (e.g., year), calculate the cycle average. 2. For each individual data point, express it as a percentage or proportion of its cycle’s average. This is the seasonal proportion:

3. For each season (e.g., March, Quarter 1), calculate the average of all the seasonal proportions for that season. This average is the seasonal index:

Example 1 The local police station records the number of speeding ﬁnes issued each quarter. The table has the data for each quarter from 2012 to 2014: Time period

Data

Percentage of yearly mean

March 2012

105

106.06%

June 2012

September 2012

101

102.02%

December 2012

100%

March 2013

101

June 2013

89.01%

September 2013

102.95%

December 2013

99.73%

March 2014

108.2%

June 2014

89.62%

September 2014

102.73%

December 2014

a For 2012, 2013, and 2014, calculate the mean number of speeding tickets issued in each time period. Round your answers to two decimal places.

198

Year

2012

2013

2014

Mean

⬚

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Create a strategy To calculate the mean for each year, add the number of speeding tickets for each quarter of that year, then divide by the number of data values.

Apply the idea Calculate the mean of the data for that year

Evaluate and round Calculate the mean of the data for that year

Evaluate and round

Year

2012

2013

2014

Mean

99.00

93.25

91.50

b Use the answers from part (a) to calculate the value of x. Round your answer to two decimal places.

Create a strategy To calculate the percentage of the yearly mean, the time period’s data value is divided by the yearly mean, then multiplied by 100%.

Apply the idea x is for June 2012 which had 91 ﬁnes. From part (a), the mean for 2012 was 99. Divide the value by the mean and multiply by 100%

Evaluate and round

c Use the answers from part (a) to calculate the value of y. Round your answer to two decimal places.

Apply the idea y is for March 2013 which had 101 ﬁnes. From part (a), the mean for 2013 was 93.25. Divide the value by the mean and multiply by 100%

Evaluate and round

3.03 Deseasonalise data mathspace.co

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Deseasonalised raw data The seasonal index is used when predicting from time series data. The data is ﬁrst smoothed either by a moving average or by deseasonalising. The value is then calculated using the equation of the least- squares regression line from the smoothed data. Finally, the seasonal index is used to adjust the predicted value to take the particular season into consideration. For example, a predicted value for December would be adjusted to be higher, whereas a predicted value for March would be adjusted lower. Deseasonalising data is also called making seasonal adjustments. The seasonal indices are used to smooth or deseasonalise data in a similar way that a moving average or moving median is used. Both methods smooth the data as shown in these graphs: 140

120 100 80 60 40 20 0

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Raw data

Deseasonalised Data

7MA

Note that the 7-point moving average line is smoother than the deseasonalised data line, but both methods are used in the real world to assist with predicting from time series data. Deseasonalising is more appropriate when you want to analyse or forecast the underlying trend in time series data without the effects of regular seasonal ﬂuctuations. Deseasonalised data formula:

Note that the seasonal index should be a decimal when using this formula. The raw data value can be calculated by rearranging the formula: Raw value = Deseasonalised data × Seasonal index

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Example 2 Three times a year, Neil records the growth of his bean plant (starting with a new plant every year). The data provided is from the beginning of 2012 to the end of 2015: Time period

Growth (in cm)

Proportion of yearly mean

April 2012

95.6

0.99

August 2012

106.7

December 2012

87.8

0.91

April 2013

0.99

August 2013

101.2

1.1

December 2013

84.1

0.91

April 2014

86.3

1.01

August 2014

93.6

1.09

December 2014

77.3

0.9

April 2015

76.1

0.99

August 2015

83.4

December 2015

71.8

0.93

a Calculate the value of a in the table. Round your answer to two decimal places.

Create a strategy Divide the corresponding data value by the mean growth for 2012.

Apply the idea First, the mean for 2012 is calculated: Calculate the average

Evaluate

The growth for August 2012 was 106.7. This value is divided by the mean: a = 106.7 ÷ 96.7 = 1.10

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Divide 106.7 by 96.7 Evaluate

b Calculate the value of b in the table. Round your answer to two decimal places.

Create a strategy Divide the corresponding data value by the mean growth for 2015.

Apply the idea First, the mean for 2015 is calculated: Calculate the average

Evaluate

The growth for August 2015 was 83.4. This value is divided by the mean: b = 83.4 ÷ 77.1

Divide 83.4 by 77.1

= 1.08

Evaluate

c If the mean for 2013 is 92.2, calculate the value of c. Round your answer to two decimal places.

Create a strategy Multiply the mean for 2013 by the proportion for April 2013.

Apply the idea In the previous parts, the growth was divided by the mean to get the proportion:

Now, to calculate the growth, the opposite operation is required: multiply the mean for 2013 by the proportion for April 2013: Growth = Mean × Proportion The proportion for April 2013 was 0.99, and the mean for 2013 is given as 92.2: c = 92.2 × 0.99 = 91.28

Multiply the mean by the proportion Evaluate

d Calculate the seasonal component for April. Round your answer to two decimal places.

Create a strategy Average the proportions for April.

Apply the idea Average the proportions for April

Evaluate and round

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From the beginning of 2016, the number of new houses built in the suburb of Woodford was recorded and ﬁgures are released every four months. The table contains the data from the beginning of 2016 to the end of 2019: a

Ex 1

2016

iii

2018

2017

Calculate the value of each of these values, rounded to two decimal places: i

iii

A local grocery store records the number of lemons sold in the ﬁrst quarters of 2016, 2017, 2018 and 2019. The table shows the data for each ﬁrst quarter up to 2019: a

208

Calculate the mean number of houses built in each of these time periods, rounded to two decimal places:

Calculate the mean number of lemons sold in each year’s ﬁrst quarter, rounded to two decimal places:

Houses built

Percentage of yearly mean

April 2016

60%

August 2016

December 2016

30%

April 2017

August 2017

180%

December 2017

37.5%

April 2018

84.91%

August 2018

141.51%

December 2018

April 2019

77.59%

August 2019

144.83%

December 2019

77.59%

Time period

Data

Percentage of quarterly mean

January 2016

71.43%

February 2016

March 2016

128.57%

January 2017

50.7%

Time period

2016

2017

February 2017

iii

2018

2019

March 2017

147.89%

January 2018

138.46%

February 2018

56.04%

March 2018

105.49%

January 2019

105.41%

February 2019

137.84%

March 2019

Calculate the value of each of these values, rounded to two decimal places: i

iii

Calculate the seasonal index of each month, rounded to four decimal places: i

January

iii

March

February

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This table shows the number of new gaming apps released each quarter, from the beginning of 2016 through to the end of 2018: a

Calculate the seasonal component for these quarters, rounded to two decimal places:

Proportion of yearly mean

March 2016

0.77

June 2016

0.81

September 2016

0.97

December 2016

1.45

First quarter

Third quarter

March 2017

0.82

iii

Fourth quarter

June 2017

0.8

September 2017

0.96

December 2017

1.42

March 2018

0.81

June 2018

0.74

September 2018

0.98

December 2018

103

1.47

Time period

Data

Percentage of yearly mean

April 2016

60%

August 2016

210%

Deseasonalise the data for March 2018.

Deseasonalise the data for December 2017.

Consider the data provided for each of these time periods from 2016 to 2019: a

Data

Time period

Calculate the seasonal index for the April season, rounded to two decimal places.

Deseasonalise the data for April 2018.

December 2016

30%

Calculate the seasonal index for the August season.

April 2017

83%

Deseasonalise the data for August 2016.

August 2017

180%

December 2017

38%

April 2018

85%

August 2018

142%

December 2018

74%

April 2019

78%

August 2019

145%

December 2019

78%

Calculate the seasonal index for the December season.

Deseasonalise the data for December 2017.

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Ex 2

Consider the data provided for each of these time periods from 2016 to 2018: Round all answers to two decimal places. a

Calculate the value of: i

108.2

1.12

June 2016

80.9

0.84

September 2016

102.4

December 2016

94.6

0.98

Calculate the seasonal component for the third quarter.

March 2017

95.3

1.1

June 2017

68.7

Deseasonalise the data for September 2017.

September 2017

93.5

1.08

December 2017

88.1

1.02

March 2018

91.3

1.06

June 2018

0.74

September 2018

87.5

1.02

December 2018

81.5

0.95

Calculate the seasonal index for Winter for the data: Summer

Autumn

Winter

Spring

1.20

0.92

⬚

1.48

A company sells central heating systems. The table shows the quarterly seasonal indices for sales in the last three quarters of a year. One year represents one cycle:

Seasonal index

210

March 2016

If the mean for 2018 is 86.075, calculate the value of c.

Seasonal index

Proportion of yearly mean

Data

Time period

Quarter 1

Quarter 2

Quarter 3

Quarter 4

⬚

1.30

1.45

0.58

Calculate the seasonal index for Quarter 1.

The deseasonalised sales for Quarter 1 in 2018 were $2 500 000. Calculate the actual sales in dollars.

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This table shows the number of complaints for neighbourhood disturbance received by the Springﬁeld Council during weekdays for 3 weeks: Day

Complaints

Proportion of weekly mean

Week One

−

Monday

1656

1.47

Tuesday

254

Wednesday

1705

1.51

Thursday

878

0.78

Friday

1155

1.02

Week Two

−

Monday

1979

1.52

Tuesday

0.54

Wednesday

1774

1.36

Thursday

794

0.61

Friday

1259

0.97

−

Monday

1433

0.9

Tuesday

917

0.57

Wednesday

1609

1.01

Thursday

1413

Friday

2609

1.63

Week Three

Round all answers to two decimal places. a

Calculate the value of: i

If the mean for Week 2 is 1302.20, calculate the value of c.

Calculate the seasonal index for Wednesday.

Deseasonalise the data for Thursday in Week 2.

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To provide information to traﬃc engineers, the number of vehicles using an intersection each day between 6:00 a.m. and 9:30 a.m. was recorded over three weeks. The data is tabulated: Week 1

Vehicles

Percentage of weekly mean

Monday

836

77%

Tuesday

1898

175%

Wednesday

1648

Thursday

1339

124%

Friday

640

59%

Saturday

980

90%

Sunday

241

22%

Monday

1321

94%

Tuesday

1989

142%

Wednesday

129%

Thursday

2064

147%

Friday

648

46%

Saturday

1301

93%

Sunday

692

49%

Monday

1503

85%

Tuesday

2635

148%

Wednesday

2532

143%

Thursday

2292

Friday

1136

64%

Saturday

1738

98%

Sunday

586

33%

Calculate the value of these values, rounded to the nearest percentage: i

212

Day

If the mean for Week 2 is 1402.57, calculate the value of b. Round your answer to two decimal places.

Calculate the seasonal index for Monday. Round your answer to two decimal places.

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The table contains the seasonal indices for the monthly sales of spring water in a particular supermarket. The full year represents one cycle:

Month

Seasonal index

January

1.00 ⬚

Calculate the seasonal index for February.

February

If the actual sales ﬁgure for June was $108 000, calculate the deseasonalised ﬁgure.

March

1.00

If the deseasonalised ﬁgure for December is $110 000, calculate the actual sales for December.

April

1.00

May

0.95

June

0.80

July

0.75

August

0.95

September

0.95

October

1.10

November

1.10

December

1.15

Month

Seasonal index

Jan

1.23

The seasonal indices for the ﬁrst 11 months of the year for sales in a sporting equipment store are shown in the table. The full year represents one cycle: a

Calculate the seasonal index for December.

Feb

0.99

In May, the store sold $210 000 worth of sporting equipment. Calculate the deseasonalised value of the May sales.

Mar

1.12

Apr

1.11

In February, the deseasonalised value of sales was $200 000. Calculate the actual sales ﬁgures for February.

May

0.89

Jun

0.96

Jul

0.86

Aug

0.73

Sep

0.77

Oct

0.96

Nov

1.10

Dec

⬚

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Ex 3

A bus company records the number of passengers using public transport on a particular route each day over a 3-week period. The spreadsheet with the recorded data is shown below. Note the table in columns I and J has been displayed on top of column F and G in the image:

214

Which cell should contain this information: i

The cycle mean for Week 1

The seasonal index for Wednesday

iii

The deseasonalised score for Monday of Week 3

John enters the following incorrect formula into cell E11 to calculate the cycle mean for Week 2: =AVERAGE(D11:D18). Write the correct formula.

What formula must be entered into cell F18 to calculate the percentage of cycle mean for Monday of Week 3?

What formula must be entered into cell J2 to calculate the seasonal index for Monday?

What formula must be entered into cell G18 to deseasonalise the data for Monday Week 3?

One method to check that the calculation of the seasonal indices is correct is to check that the sum of the indices is 700. What formula could be used to do this?

Another method to check that the calculation of the seasonal indices is correct is to make sure the mean is equal to 100%. What formula could be used to do this?

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Complex familiar 16

Determine whether the raw data is higher or lower than average for these seasonal indices: i

The seasonal index is greater than 1.0

The seasonal index is less than 1.0

iii

The seasonal index is greater than 100%

The seasonal index is less than 100%

Explain how to deseasonalise raw data, using the seasonal index.

Explain how to convert a deseasonalised ﬁgure to the raw data ﬁgure, using the seasonal index.

On a time series graph there are ‘peaks’ and ‘troughs’. Explain the meaning of these terms in relation to the seasonal index.

The number of customers for an online food delivery service is recorded for periods of 4 months. The seasonal indices calculated as a percentage are displayed in the given table: Months

Jan - Apr

May - Aug

Sept - Dec

Seasonal index

38.36

173.11

88.54

Which is the least popular period to order food online?

Which is the most popular period to order food online?

In a certain year the number of customers from September to December was 1024. Deseasonalise this number. Round your answer to nearest whole number.

The deseasonalised ﬁgure from May to August in a certain year was 1432. What was the raw data for this period in that year? Round your answer to nearest whole number.

A local cafe stays open for 5 days a week, and records how many cappuccinos they sell on Friday, Saturday, and Sunday for four weeks, as shown in the table: a

Calculate the mean number of cappuccinos sold during each week, rounded to two decimal places: i

Week 1

Week 2

iii

Week 3

Week 4

Week

Day

Data

Percentage of weekly mean

Week 1

Friday

107.14%

Saturday

Sunday

90%

Friday

37.89%

Saturday

Sunday

148.42%

Friday

58.54%

Saturday

43.9%

Sunday

197.56%

Friday

39.56%

Saturday

75.82%

Sunday

Week 2

Calculate these values, rounded to two decimal places: i

iii

Calculate the seasonal index for each day, rounded to four decimal places: i

Friday

iii

Sunday

Saturday

What does the seasonal index for Friday tell us about coffee sales on Friday?

Week 3

Week 4

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The table displays the quarterly newspaper sales (in thousands of dollars) of a corner store. The table also contains the associated seasonal indices for the respective quarters:

Sales Seasonal Index

Quarter 1

Quarter 2

Quarter 3

Quarter 4

⬚

1058

1852

1650

0.8

1.4

⬚

0.9

To what value must the seasonal indices sum?

Determine the seasonal index for Quarter 4.

Which quarter is peak season for newspaper sales? Explain your answer.

Calculate the deseasonalised sales ﬁgure for Quarter 2.

Calculate the deseasonalised sales ﬁgure for Quarter 3.

The deseasonalised sales ﬁgure for Quarter 1 is 1238. Calculate the actual sales ﬁgure for Quarter 1.

A pizza shop that is open seven days a week has these seasonal indices:

Seasonal index

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

0.4

0.3

0.5

0.8

⬚

1.9

1.2

The index for Friday has not been recorded. Calculate the missing index. 21

216

The number of waiters employed by a restaurant chain in each quarter of one year, along with the quarterly seasonal indices are displayed in this table: Quarter 1

Quarter 2

Quarter 3

Quarter 4

Number of waiters

197

144

170

Seasonal index

1.31

⬚

0.57

1.08

Calculate the seasonal index for the second quarter.

Explain the meaning of seasonal index for Quarter 1 this context.

Deseasonalise the data for each quarter. Round your answers to the nearest whole number.

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A taxi company records the weekly number of taxis ordered to go to the airport over a 4-month period. The data is displayed in the spreadsheet:

Week 2 in April corresponds to which time value (t)?

Which week does the time value t = 7 correspond to?

What formula must be entered into cell G11 to calculate the cycle mean for March?

What formula must be entered to calculate the percentage of cycle mean for time period t = 10?

A mistake has been made and the raw data ﬁgure for t = 5 should be 822. Will the cycle mean for this cycle now be higher or lower?

The seasonal indices are calculated and shown in this table:

Explain how to check that a mistake has not been made. g

Which week represents a ‘peak’ season in each cycle?

Are the taxi orders in Week 1 of each cycle higher or lower than average?

The raw data taxi orders for t = 18 are recorded as 1610 orders. Deseasonalise this ﬁgure, rounding your answer to two decimal places.

The deseasonalised ﬁgure for t = 20 is 1321 . Calculate the number of taxis ordered in this time period.

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Complex unfamiliar 23

An online bookstore records its quarterly sales, in thousands of dollars, over a 3-year period. The raw data is shown in the table: Year

Quarter

Time (t)

Sales ($000s)

Year 1

10.2

13.5

11.8

14.0

11.5

15.1

13.5

16.0

12.8

16.8

15.0

18.1

Year 2

Year 3

After analysis, these seasonal indices were calculated:

Seasonal index

Quarter 1

Quarter 2

Quarter 3

Quarter 4

0.85

1.10

0.95

1.10

A least-squares regression line fitted to the deseasonalised data is given by the equation: Deseasonalised sales = 0.5 × t + 11.5 Round all answers to three significant figures where appropriate:

218

Calculate the deseasonalised sales data for Year 3.

Use the regression equation to forecast the deseasonalised sales for Quarter 1 of Year 4 (t = 13).

Using the appropriate seasonal index, adjust the forecast from part (b) to predict the actual sales for Quarter 1 of Year 4.

Explain the meaning of the seasonal index for Quarter 2 in the context of the bookstore’s sales.

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3.04 Time series data in tables After this lesson, you will be able to… • solve practical problems involving time series data • analyse practical problems involving time series data • describe practical problems involving time series data

Calculate and interpret time series data In time series analysis, it is often helpful to present data in tabular form. This makes patterns easier to identify and calculations more straightforward. To observe the underlying trend clearly, without being distracted by sharp rises and falls, smoothing techniques are used.

Exploration Consider these quarterly sales data for an ice cream shop: Quarter (t)

Sales ($ thousands)

A simple line graph of this data would illustrate the seasonal ﬂuctuations. To see the underlying trend, the data can be smoothed by calculating a moving average. Instructions: 1. Plot the raw sales data on a simple line graph. 2. Calculate the 5-point moving average for the data at t = 3 and t = 4. 3. Observe how the smoothed points compare to the raw data points. Guiding questions: 1. What is the main difference between the graph of the raw data and the smoothed moving average points? 2. How does calculating the moving average help in identifying the underlying trend of the sales data? 3. What is the purpose of smoothing data in time series analysis?

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There are three key methods used to smooth data: 1. Moving average: A moving average smooths the data by calculating the mean of a fixed number of consecutive values. This helps reduce random fluctuations (peaks and troughs), making it easier to identify the long-term trend in the data. 2. Moving median: A moving median smooths the data by calculating the median of a fixed number of consecutive values. It is particularly useful when the data contains outliers or sharp spikes, as the median is less affected by extreme values than the mean. 3. Deseasonalising using seasonal indices: Some data varies in a predictable seasonal pattern— like sales increasing in December or ice cream sales dropping in winter. To remove these seasonal effects and focus on the trend, seasonal indices are used to adjust the data. This process is called deseasonalising. Dividing each data point by its corresponding seasonal index strips away seasonal influence and reveals the underlying behaviour more clearly. The key formulas used in deseasonalising data are:

where: • Seasonal index: The value which indicates how a particular season compares to the average. An index above 1 (or 100%) is an above-average season, while an index below 1 (or 100%) is a below- average season. • Value for a speciﬁc season: The raw data point for the season being analysed. • Average value for a cycle: The mean of all data points within one full cycle (e.g., the mean of 4 quarters for yearly data). The sum of all seasonal indices in a cycle must equal the number of seasons (e.g., 4 for quarterly data, 12 for monthly data).

where: • Deseasonalised value: This is the data point with the seasonal component removed, which helps to reveal the underlying trend. • Original value: This is the raw data point. • Seasonal index: The seasonal index for that particular season, expressed as a decimal (e.g., 110% is 1.10).

Example 1 Determine the missing seasonal index in the table: Quarter

Jan

Apr

July

Oct

Seasonal Index

92.34%

124.12%

⬚

98.45%

Create a strategy The sum of the seasonal indices for the four quarters must be 4 × 100% = 400%. Let the missing index be x, write an equation where the sum of the four indices equals 400% and solve for x.

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Apply the idea 92.34% + 124.12% + x + 98.45% = 400% 314.91% + x = 400% x = 85.09%

he sum of the indices for the four seasons T is 400% Collect like terms Subtract 314.91% from both sides

Quarter

Jan

Apr

July

Oct

Seasonal Index

92.34%

124.12%

85.09%

98.45%

Reﬂect and check For seasonal indices written as percentages, the sum will always be k × 100%, where k is the number of seasons. For seasonal indices written as decimals, the sum will be k. Therefore, this calculation could also begin by setting the average of the indices equal to 100%.

Example 2 The table shows some time series data where t represents time. t

120

160

180

250

210

⬚

163

134

142

a Calculate the 3-point moving average at t = 4.

Create a strategy The 3-point moving average at t = 4 is the mean of the values at t = 3, t = 4, and t = 5.

Apply the idea Write the equation

Evaluate and round to two decimal places

b The 5-point moving average at t = 4 is 191.4. Calculate the missing data value for t = 6.

Create a strategy The 5-point centred moving average at t = 4 is the mean of 5 data points from t = 2 to t = 6. Use the formula for the mean to solve for the missing value.

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Apply the idea Write the formula

Substitute the values, with x as the missing value

Simplify the numerator

Multiply both sides by 5

Subtract 800 from both sides

The missing value is 157.

Example 3 A company specialising in repairing clothes dryers records its number of call-outs. Determine the values of a, b, c, and d in the table:

Year

Data number (n)

Quarter

Call-outs

Quarterly mean

Percentage of quarterly mean

Deseasonalised data (D)

2017

Jan

491.25

25.45%

427

Apr

654

133.13%

499

Jul

841

171.20%

Oct

345

70.23%

477

Jan

159

30.80%

543

Apr

672

130.17%

513

Jul

867

167.94%

518

Oct

367

71.09%

507

Jan

176

601

Apr

723

130.04%

551

Jul

904

162.59%

541

Oct

421

75.72%

582

2018

2019

556

The cycle mean here is the average number of call-outs per quarter for that year. This column may sometimes be labelled “Quarterly mean”, “Yearly mean” or simply the “Cycle mean”. Calculate a, b, c and d.

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Create a strategy For a, the cycle mean is the mean of the 4 seasons in the cycle. Sum the data points and divide by 4. For b, the percentage of the quarterly mean is the raw data score divided by the quarterly mean, multiplied by 100% (to show it as a percentage). For c, deseasonalised data is the raw data score divided by the seasonal index (with the index written as a decimal). Since this is a July value, the seasonal index for July must be calculated first. This is done by determining the mean of the three July “percentage of quarterly mean” values. For d, three methods can be used: using the quarterly mean, the percentage of the quarterly mean, or the deseasonalised figure. Any of these can be used to form an equation to solve for d.

Apply the idea Calculation for a: Calculate the mean of the four data points for the year

Evaluate

Calculation for b: Divide the raw value by the yearly mean and multiply by 100%

Evaluate and round to two decimal places

Calculation for c: First, calculate the seasonal index for July: Average the percentages for the July season Evaluate and round to two decimal places Convert the percentage to a decimal Now, calculate c: Write the formula

Substitute the raw data and seasonal index

Evaluate and round to the nearest whole number

Note: The raw data are given as whole numbers, so the deseasonalised data should also be rounded to the nearest whole number. Calculation for d: To determine d, we can use one of three methods. Any method will give the same result. Method 1: Using the quarterly mean Set up the equation for the quarterly mean

Simplify the numerator and multiply both sides by 4

Subtract 1840 from both sides

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Method 2: Using the percentage of the quarterly mean Set up the equation for the percentage of the mean

Divide both sides by 100%

Multiply both sides by 491.25

Evaluate and round to the nearest whole number

Method 3: Using the deseasonalised ﬁgure First, calculate the seasonal index for January: verage the percentages for A the January season Evaluate and round to two decimal places Convert the percentage to a decimal Now, use the deseasonalised value to determine d: Write the formula

Substitute the known values

Multiply both sides by 0.293

Evaluate and round to the nearest whole number The answer using any of the 3 methods is d = 125.

Reflect and check The longest method is Method 3, which uses the deseasonalised figure. This is because it requires calculating the seasonal index for January first, which needs the value of b to be determined beforehand. Methods 1 and 2 can be used directly with the given information.

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Example 4 The quarterly power bills of a suburban household are recorded in the table: Year

Month

Time (t)

Bill ($)

Quarterly mean ($)

Percentage of quarterly mean

Deseasonalised data ($)

2016

Jan

456.24

348.37

130.97%

341.37

Apr

95.40%

347.23

Jul

300.43

86.24%

354.12

Oct

304.45

87.39%

354.85

Jan

477.05

132.85%

356.94

Apr

343.77

95.73%

359.16

Jul

305.98

85.21%

360.66

Oct

309.54

86.20%

360.78

Jan

494.22

134.49%

369.79

Apr

352.56

368.34

Jul

310.65

84.54%

366.17

Oct

312.43

85.02%

364.15

Jan

510.45

136.28%

381.94

Apr

358.76

95.78%

Jul

312.25

83.37%

368.05

Oct

316.76

84.57%

369.20

2017

2018

2019

367.47

374.56

The seasonal indices are: Month

Jan

Apr

Jul

Oct

Seasonal index

95.716%

84.839%

84.797%

a Calculate the values of A, B, C, D and E. Round your answer to the nearest cent for dollar amounts and to two decimal places for percentages.

Create a strategy For A, use the quarterly mean for 2016 to construct and solve an equation. For B, ﬁnd the mean of the bill payments in 2017. For C, the percentage of the quarterly mean for a particular quarter is the bill amount for that quarter divided by the corresponding quarterly mean, multiplied by 100. For D, to deseasonalise data, divide the original value by the relevant seasonal index. For E, the sum of the seasonal indices is equal to the number of seasons per cycle, which is 4 or 400%.

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Example 5 Data on the number of cartons of chocolate milk sold at the school canteen is collected every day over a 6-week time period. The seasonal indices for each day are calculated in order to deseasonalise the data. a Calculate the missing value in the table of seasonal indices. Monday

Tuesday

Wednesday

Thursday

Friday

102.21%

97.54%

114.65%

88.98%

⬚%

Create a strategy The sum of the indices for the 5-day cycle is 5 × 100% = 500%. Add the known indices and subtract the total from 500%.

Apply the idea Sum of known indices = 102.21 + 97.54 + 114.65 + 88.98 = 403.38

Sum the given values Evaluate the sum

Friday Index = 500 − 403.38

Subtract the sum from 500%

= 96.62%

Evaluate

Monday

Tuesday

Wednesday

Thursday

Friday

102.21%

97.54%

114.65%

88.98%

96.62%

b Which day is the most popular day for buying chocolate milk?

Create a strategy

Apply the idea

Look for the highest seasonal index in the completed table.

Wednesday, with an index of 114.65%, is the most popular day.

c Will the deseasonalised data for Wednesday be higher or lower than the original raw data for Wednesday?

Create a strategy To deseasonalise data, divide the raw value by the seasonal index. Consider whether the seasonal index for Wednesday is greater or less than 1 (or 100%).

Apply the idea The seasonal index for Wednesday is 114.65%, which is 1.1465 as a decimal. Since the raw data is divided by a number greater than 1, the deseasonalised value for Wednesday will be lower than the original data.

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Forecasting with time series data Once time series data has been deseasonalised, the underlying trend can be modelled with a least- squares regression line. This trend line can then be used to make future predictions, or forecasts. Forecasting is a two-step process: 1. Model the trend: A least-squares regression line (e.g., y = a + bt) is ﬁtted to the deseasonalised data, where t is the time period. 2. Reseasonalise the prediction: The trend line is used to predict a future deseasonalised value. This value is then multiplied by the relevant seasonal index to re-introduce the seasonal component, giving the ﬁnal forecast. The formula for this process is:

Forecasted value = Predicted deseasonalised value × Seasonal index Forecasted value The ﬁnal prediction for a future time period, including the expected seasonal effect. Predicted deseasonalised value The value obtained from substituting a future time period (t) into the least-squares regression line equation. Seasonal index The index for the season being forecast, expressed as a decimal. Predictions made within the time frame of the original data are called interpolation. Predictions made beyond the time frame are called extrapolation. Extrapolations should be viewed with caution as they assume the identiﬁed trend will continue into the future.

Example 6 The quarterly power bills of a suburban household have been deseasonalised. A least-squares regression line is ﬁtted to the deseasonalised data versus the time period, t. The equation for the trend line is determined to be: Deseasonalised bill = 3.4t + 338 The seasonal indices for each quarter are: Quarter

Jan (Q1)

Apr (Q2)

Jul (Q3)

Oct (Q4)

Seasonal index

1.346

0.957

0.848

a Use the trend line to predict the deseasonalised power bill for the ﬁrst quarter of the next year, t = 17.

Create a strategy Substitute t = 17 into the given least-squares regression line equation.

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Apply the idea Deseasonalised bill = 3.4t + 338

Write the formula

= 3.4(17) + 338

Substitute t = 17

= 57.8 + 338

Evaluate the product

= 395.80

Evaluate

The predicted deseasonalised bill for t = 17 is $395.80.

b Calculate the forecast for the power bill for t = 17, which is in the ﬁrst quarter (Jan). Round to the nearest cent.

Create a strategy Multiply the predicted deseasonalised value from part (a) by the seasonal index for the ﬁrst quarter.

Apply the idea Forecasted value = Predicted deseasonalised value × Seasonal index

Write the formula

= 395.80 × 1.346

Substitute the values

= 532.7508

Evaluate

= 532.75 Round to the nearest cent The forecasted power bill for the ﬁrst quarter of the next year is $532.75.

Reﬂect and check This prediction is an extrapolation because the time period t = 17 is outside the range of the original data. This forecast relies on the assumption that the increasing trend in power bills will continue at the same rate.

Example 7 A surf shop records its sales data over three years. The deseasonalised sales data is shown in this table, where t is the quarter number and sales are in thousands of dollars: Time (t)

Deseasonalised sales ($000s)

The seasonal indices for the shop are: Quarter

Summer (Q1)

Autumn (Q2)

Winter (Q3)

Spring (Q4)

Seasonal index

1.40

0.85

0.75

1.00

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a Determine the equation of the least-squares regression line for the deseasonalised sales data. Let deseasonalised sales be y and time be t in the form y = a + bt. Write the values of a and b rounded to two decimal places.

Create a strategy Enter the time data (t = 1, 2, ..., 12) as the independent variable (x-list) on a calculator and the deseasonalised sales data as the dependent variable (y-list). Use the calculator’s statistical function to perform a linear regression analysis to determine the y-intercept (a) and the gradient (b).

Apply the idea Using a calculator, the values for the linear regression equation are: • y-intercept (a): 48.32 • Gradient (b): 1.10 The equation of the least-squares regression line is: y = 48.32 + 1.10t

b Use the trend line to predict the deseasonalised sales for the second quarter of the fourth year, t = 14.

Create a strategy Substitute t = 14 into the equation found in part (a).

Apply the idea y = 48.32 + 1.10t

Write the equation

= 48.32 + 1.10(14)

Substitute t = 14

= 48.32 + 15.4

Evaluate the product

= 63.72

Evaluate

The predicted deseasonalised sales are $63.72 thousand, or $63 720.

c Calculate the ﬁnal sales forecast for t = 14.

Create a strategy The time period t = 14 corresponds to the second quarter of the fourth year, which is Autumn (Quarter 2). Multiply the predicted deseasonalised value by the seasonal index for Autumn.

Apply the idea Forecasted sales = Deseasonalised prediction × Seasonal index for Q2

Write the formula

= 63.72 × 0.85

Substitute the values

= 54.162

Evaluate

The final sales forecast is $54.16 thousand (rounded to two decimal places), or $54 160.

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Ex 4

This table contains time series data showing a company’s quarterly proﬁts in thousands of dollars: Year

Quarter

Time (t)

Subscribers (000s)

Yearly mean

Percentage of yearly mean

Deseasonalised data

2020

120

140.00

85.71%

139.05

180

128.57%

142.54

150

107.14%

142.64

110

78.57%

133.72

130

85.25%

150.64

127.87%

154.42

160

104.92%

152.15

125

81.97%

151.95

145

86.57%

168.02

210

125.37%

166.30

175

104.48%

166.41

140

170.19

160

87.67%

185.40

225

123.29%

178.18

190

104.11%

155

84.93%

188.43

2021

2022

2023

152.50

182.50

The seasonal indices for the service are: Quarter

Seasonal index

86.30%

105.16%

82.26%

Calculate the values of A, B, C, D, and E. Round your answer to two decimal places where necessary. a

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Ex 5

Data on the number of cartons of chocolate milk sold at the school canteen is collected every day over a 6-week time period. The seasonal indices for each day are calculated in order to deseasonalise the data: a

Monday

Tuesday

Wednesday

Thursday

Friday

Seasonal index

102.21%

97.54%

114.65%

88.98%

⬚

Which day is the most popular day for buying chocolate milk?

Will deseasonalised data for Wednesday be higher or lower than the original raw data for Wednesday?

The number of chocolate milk cartons sold on Thursday of Week 2 was 57. Calculate the deseasonalised score for Thursday of Week 2, rounding your answer to the nearest whole number.

Data on the number of cloudy days is collected quarterly over a number of years. The seasonal indices are calculated in order to deseasonalise the data: Calculate the missing value in the table of seasonal indices: Season

Spring

Summer

Autumn

Winter

Seasonal index

0.8

⬚

1.3

1.4

Will deseasonalised data for Spring be higher or lower than the original raw data for Spring?

Will deseasonalised data for Autumn be higher or lower than the original raw data for Autumn?

The number of cloudy days in Winter 2018 was 65. Calculate the deseasonalised score for Winter 2018. Round your answer to the nearest whole number.

Data on the number of customers at a local cafe was collected daily over a number of weeks. The seasonal indices are calculated in order to deseasonalise the data: a

236

Day

Calculate the missing value in the table of seasonal indices:

Calculate the missing value in the table of seasonal indices: Day

Mon

Tue

Wed

Thu

Fri

Seasonal index

67%

78%

82%

⬚

139%

Which days of the week have raw scores that are higher than the average in a cycle?

Which days of the week have raw scores that are lower than the average in a cycle?

If the number of customers for Wednesday of Week 3 was 84, calculate the deseasonalised score. Round your answer to the nearest whole number.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A university records the number of students cycling to the library each day and records the data in this table:

Week 1

Week 2

Week 3

Day

Number

Weekly mean

Percentage of weekly mean

Deseasonalised data

Mon

103.57

63.72

100

Tue

65.66

Wed

68.55

100

Thu

62.76

103

Fri

246.21

104

Sat

111

107.17

107

Sun

85.93

107

Mon

66.57

103

Tue

65.59

100

Wed

70.49

101

Thu

64.62

105

Fri

250

244.76

102

Sat

107

104.76

103

Sun

83.22

102

Mon

66.96

100

Tue

101

Wed

73.04

101

Thu

62.09

Fri

244

247.54

Sat

100.43

Sun

80.14

98.57

The table also shows the seasonal index for the days:

Day

Mon

Tue

Wed

Thu

Fri

Sat

Sun

Seasonal index

0.66

0.67

0.71

2.46

1.04

0.83

Calculate the value of: i

iii

Does the value of the percentage of weekly mean for Wednesday in Week 1 imply that the number of students cycling that day is above or below the average for that week?

Without using your calculator, determine the average of the percentage of weekly means for Sunday.

Using the deseasonalised number column, determine whether the trend is decreasing, increasing or remaining stable.

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Complex familiar 13

A cafe owner in a suburban supermarket recorded the number of customers they had for each quarter over four years in this table:

2016

2017

2018

2019

Month

No. of customers

Yearly mean

Percentage of yearly mean

Deseasonalised number

Jan

1687

1293.75

130.396

1284

Apr

1218

94.145

1256

July

886

68.483

Oct

1384

106.976

1318

Jan

1789

130.825

1361

Apr

1327

97.038

1369

July

905

66.179

1358

Oct

1449

105.960

1380

Jan

2325

132.971

1769

Apr

1745

99.800

1800

July

63.597

1669

Oct

1812

103.632

1726

Jan

2565

131.471

1952

Apr

1890

1949

July

1333

68.324

2000

Oct

2016

103.332

1920

1748.5

1951

The table also shows the seasonal index for the quarters: Jan

Apr

July

Oct

Seasonal index

131.42

96.964

104.975

Calculate the value of:

Using deseasonalised number column, determine whether the trend is increasing, decreasing, or remaining stable.

Use the seasonal index for the January quarter to explain why an accountant might recommend employing more staff during that period.

240

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iii

A pizza delivery company records the number of daily deliveries over a period of 3 weeks and records the data in this table:

Week 1

Week 2

Week 3

Day

Time (t)

Number of orders

Weekly mean

Percentage of weekly mean

Deseasonalised data

Mon

69.1

50.62%

44.25

Tue

53.51%

69.22

Wed

60.74%

71.41

Thu

75.21%

71.72

Fri

104

150.41%

72.46

Sat

114

164.88%

72.51

Sun

100

144.63%

73.86

Mon

52.93%

50.57

Tue

78.57

Wed

62.19

79.91

Thu

75.43%

78.62

Fri

114

150.85%

79.43

Sat

124

164.08%

78.87

Sun

105

138.94%

77.55

Mon

120

133.76%

151.70

Tue

51.27%

86.05

Wed

53.50%

81.61

Thu

66.88%

82.75

Fri

116

129.80%

80.82

Sat

128

142.68%

81.42

Sun

110

122.61%

75.6

89.7

The seasonal indices are shown in the table:

Day

Mon

Tue

Wed

Thu

Fri

Sat

Sun

Seasonal index

79.10%

58.81%

72.50%

143.52%

157.21%

135.39%

Calculate the values of: i

iii

Construct a time series graph for the number of orders against t.

There appears to be an outlier in the seasonal trend. Which value of t contains the outlier?

On what day of the week does the outlier occur?

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Ex 6

After analysing several years of quarterly data, a car dealership developed a least-squares regression line to model the trend in its deseasonalised sales figures. The equation for the trend line is: Deseasonalised sales = 2500t + 120 000 where t is the quarter number (t = 1 for Q1 of Year 1). The seasonal indices for each quarter are shown in the table:

Ex 7

Quarter

Seasonal index

0.85

1.05

0.90

1.20

Use the trend line to predict the deseasonalised sales for the third quarter of Year 5 (t = 19).

Calculate the ﬁnal sales forecast for Q3 of Year 5.

Is this forecast an example of interpolation or extrapolation? Explain your reasoning.

A city zoo records the number of visitors each season over three years. The deseasonalised visitor numbers (in thousands) are shown in the table. Time (t)

Deseasonalised visitors (′000s)

22.5

23.1

23.9

24.5

25.0

25.8

26.2

27.0

Time (t)

Deseasonalised visitors (′000s)

27.5

28.1

28.9

29.5

The seasonal indices are:

242

Quarter

Summer (Q1)

Autumn (Q2)

Winter (Q3)

Spring (Q4)

Seasonal index

1.50

0.90

0.60

1.00

Determine the equation of the least-squares regression line for the deseasonalised visitor data. Let deseasonalised visitors be y and time be t. Write the values in the equation, rounded to two decimal places.

Interpret the slope of the regression line in the context of this problem.

Use the trend line to predict the deseasonalised number of visitors for the Winter of Year 4 (t = 15).

Calculate the ﬁnal forecast for the number of visitors in the Winter of Year 4.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex unfamiliar 17

A shop owner collects data on the number of thefts per quarter and uses a spreadsheet to calculate a number of moving averages: Year

Quarter

Time (t)

Data

19.33

27.67

24.75

25.25

27.33

25.75

21.33

21.50

23.00

22.00

22.67

22.50

3 pt moving avg

4 pt CMA

22.67

23.13

23.33

23.50

25.00

24.33

24.50

23.00

24.88

25.00

25.13

26.67

25.50

25.67

Calculate the value of: i

B, given the value for time t = 17 is the centre of the 3-point average.

iii

D, given that the 3-point moving average centred at t = 18 is 31.

Determine the purpose of calculating a centred moving average for time series data.

Explain why a 4-point centred moving average is the most appropriate choice for this particular data set.

Based on the ‘4 pt CMA’ column, describe the underlying trend in thefts over the ﬁrst three years (t = 3 to t = 11).

Sketch the graph of the original data and the 4-point centred moving average (4 pt CMA) on the same set of axes.

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3.05 Make predictions using time series data After this lesson, you will be able to… • fit a least-squares line to model long-term trends in time series data • use a least-squares line to make predictions for time series data • understand how to model long-term trends in time series data using a leastsquares line

Make predictions using time series data The purpose of smoothing time series data, either by using a moving average, moving median or by deseasonalising, is to remove the cyclical ‘peaks’ and ‘troughs’ to reveal the underlying trend. Often, the smoothed data will appear to follow a linear trend. When this is the case, a least-squares regression line can be ﬁtted to the smoothed data. This uses the same calculation process as determining a line of best ﬁt for bivariate data. The equation of this trend line can then be used to make future predictions. Forecasting Making predictions about future values based on a time series model.

Exploration A line of best ﬁt can be drawn by eye to model a linear trend. However, this method can be subjective. An objective method, like calculating the least-squares regression line, ensures consistency.

Consider this scatterplot showing deseasonalised sales data:

1. Using a ruler, draw a line of best ﬁt that you think best represents the trend in the data. 2. Choose two points on your line and calculate its equation.

50 40

20 10 0

3. How might your line and equation differ from a classmate’s? 4. Why would a standardised method for determining the line of best ﬁt be useful for statisticians?

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Once a predicted value from the underlying trend line is determined, it must be adjusted to account for seasonality. This process is often called reseasonalising the data. If the prediction is for a ‘peak’ season, the value should be adjusted upwards. If it is for a ‘trough’ season, it should be adjusted downwards. This adjustment is done by multiplying the predicted value from the regression line by the appropriate seasonal index. Step

Action

Smooth the data using the appropriate moving average or by deseasonalising the data.

Give each time period a number, e.g., Mon Week 1 becomes t = 1, Tues Week 1 becomes t = 2 etc.

Calculate the equation of the least-squares regression line using the time period in list 1 and the smoothed data in list 2 of the calculator. Round numbers to four decimal places unless instructed otherwise.

Substitute a future time value into the equation of the least-squares regression line to obtain a predicted score.

Multiply the predicted score by the appropriate seasonal index to factor back in the seasonality of the data.

If asked to comment on the reliability of the prediction, consider whether the future time value is close to the data. Generally predicting within one cycle of the existing data is considered reliable.

Note: • Moving average, moving median or deseasonalised data can be used to determine the regression line. Read the question carefully. • Correlation has little meaning with time series data as the smoothed line will almost always have a strong, linear correlation for an underlying linear trend. • Predictions for future time periods are a form of extrapolation. Their reliability is assessed by how far the prediction is from the original data, often described as being ‘within one cycle’. • Showing working is essential. Always state the value of t being used, the equation of the regression line, and the seasonal index applied in the calculation. • Time is always the explanatory variable (plotted on the horizontal axis). For example, if a cycle is four quarters and the last data point is for December 2019, any prediction for 2020 (up to four quarters ahead) is considered reliable. Predictions for 2021 and beyond would be considered unreliable.

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Example 1 An aluminium window company records quarterly ﬁgures on the number of windows manufactured. For planning purposes, the company management wishes to predict the number of windows that will be manufactured in March 2020: Month/Yr

Time (t)

Raw data

Proportion of yearly mean

Deseasonalised data

Mar 2017

471

0.9598

505.81

Jun 2017

480

0.9781

485.01

Sep 2017

492

1.0025

485.43

Dec 2017

520

1.0596

487.97

Mar 2018

427

0.9143

458.56

Jun 2018

463

0.9914

467.84

Sep 2018

484

1.0364

477.54

Dec 2018

494

1.0578

463.57

Mar 2019

425

0.9190

456.41

Jun 2019

462

0.9995

466.83

Sep 2019

463

1.0016

456.82

Dec 2019

499

1.0795

468.26

Create a strategy Look at the table carefully. It is possible to see that steps 1 and 2 have been completed already. • Step 1: The data has been smoothed. The deseasonalised data can be ﬁtted to a linear regression model. • Step 2: The time periods have been allocated a number (t = 1, t = 2, t = 3 etc.). Do step 3. • Step 3: Calculate the least-squares regression line. Using a calculator, enter the time values in list 1 and the deseasonalised data in list 2, and then ﬁt a linear regression model. To complete the remaining steps, use the linear regression model to make a prediction for March 2020, and then adjust this value to include seasonality.

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Apply the idea Write the equation y = −3.2519t + 494.4746 for the least-squares regression line. To predict for March 2020, the t value must ﬁrst be determined. March 2019 was t = 9 so one cycle or four more time periods ahead means March 2020 will be t = 13. This value is now substituted into the regression line to make the deseasonalised prediction. y = −3.2519t + 494.4746

Write the least-squares equation

= −3.2519 × 13 + 494.4746

Substitute t = 13

= 452.1999

Evaluate

Now, it is necessary to adjust for seasonality. The value just determined is the deseasonalised prediction for March 2020, taken from the trend line of the deseasonalised values. However, the point of interest is the more realistic value that includes the seasonal trend from the actual data. So, the predicted value must be multiplied by the appropriate seasonal index. In this case it is necessary to multiply by the seasonal index for March. The seasonal indices for this data are: Month

March

June

September

December

Seasonal index

0.9310

0.9897

1.0135

1.0656

So, the predicted value for March 2020 is 452.1999 × 0.9310 = 420.9981. In other words, the window company predicts that 421 windows will be manufactured in March 2020.

Reflect and check The original data, as shown in the graph, contains peaks and troughs that reflect seasonality. The deseasonalised values are used to fit a least-squares line, which is then used to make a deseasonalised prediction. To include seasonality, this prediction is adjusted using the seasonal index. Deseasonalised prediction

600 500 400 300

Seasonalised prediction

200 100 0

Raw data

Deseasonalised data

Least-squares

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Example 2 This data shows the sales of air conditioners at a leading retailer over four quarters of 3 consecutive years: Time period

Time (t)

Number of air conditioners sold

Proportion of yearly mean

March year 1

1042

0.8529

June year 1

486

0.3978

Sept year 1

613

0.5017

Dec year 1

2746

2.2476

March year 2

1160

0.8183

June year 2

609

0.4296

Sept year 2

1139

0.8035

Dec year 2

2762

1.9485

March year 3

1795

0.9638

June year 3

1181

0.6341

Sept year 3

1094

0.5874

Dec year 3

3380

1.8148

a Calculate the seasonal component for the quarters ending in March, June, September and December, rounding to four decimal places if necessary. Month

March

June

September

December

Seasonal index

⬚

Create a strategy Determine the seasonal component, also known as the seasonal index, for a particular quarter by calculating the average of the yearly mean proportions for that quarter.

Apply the idea The seasonal component for the quarter ending in March will be equal to = 0.8783. Follow the same procedure for the quarters ending in June, September, and December.

248

Month

March

June

September

December

Seasonal index

0.8783

0.4872

0.6309

2.0036

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b The data is smoothed using a 5-point moving average as shown in this table: Time period

Time (t)

Number of air conditioners sold

March year 1

1042

June year 1

486

Sept year 1

613

1209.4

Dec year 1

2746

1122.8

March year 2

1160

1253.4

June year 2

609

Sept year 2

1139

1493

Dec year 2

2762

1497.2

March year 3

1795

1594.2 ⬚

June year 3

1181

Sept year 3

1094

Dec year 3

3380

5MA

⬚

Calculate the missing values.

Create a strategy Use the values in the column “Number of air conditioners sold” along with the formula for calculating the 5-point moving average to determine the missing values.

Apply the idea For June year 2: Use the 5MA formula

Evaluate the addition

Evaluate

Follow the same procedure for June year 3 and complete the table. Time period

Time (t)

Number of air conditioners sold

5MA

March year 1

1042

June year 1

486

Sept year 1

613

1209.4

Dec year 1

2746

1122.8

March year 2

1160

1253.4

June year 2

609

1683.2

Sept year 2

1139

1493

Dec year 2

2762

1497.2

March year 3

1795

1594.2

June year 3

1181

2042.4

Sept year 3

1094

Dec year 3

3380

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b Calculate the value of X in the table. Round the value to two decimal places if necessary.

Create a strategy

Apply the idea

Divide the raw ﬁgure by the seasonal index.

Write the equation

Evaluate

c Calculate the value of Y in the table. Round the value to two decimal places if necessary.

Create a strategy Multiply the deseasonalised ﬁgure by the seasonal component.

Apply the idea Y = 2431.09 × 0.9251 Write the equation = 2249

Evaluate

d Determine the equation of the least-squares regression line for the deseasonalised data, where t = 1 represents Friday of Week 1, giving the equation in the form y = at + b with the coeﬃcients, rounded to two decimal places.

Create a strategy

Apply the idea

Enter the t-values in the ﬁrst list or column of the calculator, and the deseasonalised values in the second list or column.

The value of the gradient is a = 49.88 and the value of the intercept is b = 2010.84. Substituting these values in the equation of the least-squares regression, the resulting equation is:

It is possible to then ﬁt a linear regression to the data using a calculator. This will give the gradient and intercept of the regression line.

y = 49.88t + 2010.84

e Predict the sales for Friday of the 6th week, giving the answer in dollars and rounded to two decimal places.

Create a strategy Use the regression line equation determined in part (d) and the relevant seasonal index to make the prediction.

Apply the idea Just before the 6th Friday, 5 cycles will have been completed, and each cycle is 3 days long. So, the 6th Friday will be when t = 16. Substitute this t-value into the regression line equation, then multiply by the seasonal index to reverse the effect of deseasonalisation. Prediction = (49.88 × 16 + 2010.84) × 0.9690 = $2721.84

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Substitute t = 16 Evaluate

3.05 Practice questions Simple familiar 1

A least-squares regression line is ﬁtted to some seasonally adjusted data, given by: y = 3.5t + 22.8 Use the regression line to predict the deseasonalised value for time period 20.

A least-squares regression line is ﬁtted to some seasonally adjusted data for time periods 1 to 15, given by: y = 1.3t + 45.7

Use the regression line to predict the deseasonalised value for time period 16.

Use the regression line to predict the deseasonalised value for time period 60.

Which prediction is more reliable?

The monthly average cost of a hotel room in Sydney in 2000, is shown in this table: Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Hotel price, P ($)

250

240

235

237

239

230

228

237

332

Let January 2000 be t = 1. Which month appears to be an outlier?

Remove the outlier and determine the equation of the least-squares regression line for the remaining data. Round all values to four decimal places.

Predict the average cost of a hotel room in Sydney in November 2000 (t = 11).

This data has a 5-point cyclical pattern and is collected and seasonally adjusted for time periods 1 to 14. A least-squares regression line is ﬁtted to the seasonally adjusted data, which appears linear, and is given by: y = 2.4378t + 66.2925

254

Calculate the predicted deseasonalised value for time period 15, rounded to four decimal places.

If the seasonal index for this period was 77%, calculate the true predicted value rounded to four decimal places.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex familiar Ex 1

The number of customers served at a shopping centre cafe are recorded quarterly over a period of 4 years and the results are entered into this table:

2016

2017

2018

2019

Month

Time (t)

Customer numbers

Jan

Apr

Cycle mean

Percentage of cycle mean

Deseasonalised number

1687

130.396%

1284

1218

94.145%

1256

Jul

886

68.483%

1329

Oct

1384

106.976%

1318

Jan

1789

130.823%

1361

Apr

1327

97.038%

1369

Jul

905

66.179%

1358

Oct

1449

105.960%

1380

Jan

2325

132.971%

1769

Apr

1745

99.800%

1800

Jul

1112

63.597%

1669

Oct

1821

103.632%

1726

Jan

2565

131.471%

1952

Apr

1890

96.873%

1949

Jul

1333

68.324%

2000

Oct

2016

103.332%

1920

1293.75

1367.5

1951

The seasonal indices are: Month

Jan

Apr

Jul

Oct

Seasonal index

131.42%

96.964%

66.646%

104.975%

Use your calculator to determine the equation of the least-squares regression line for the deseasonalised data, in terms of t. Round all values to three decimal places.

Use the regression line from the deseasonalised data and the seasonal index for April to predict the number of customers for April 2020. Round your answer to the nearest whole number.

3.05 Make predictions using time series data mathspace.co

255

The trend for the quarterly sales of electric cars in a region is modelled by the least-squares line: Deseasonalised sales = 150t + 2000 where t is the quarter number. The seasonal indices are:

Ex 2

Quarter

Seasonal index

0.80

1.10

1.20

0.90

Predict the actual number of electric cars sold in the second quarter of the fourth year (t = 14).

If the actual sales in that quarter were 10% higher than the forecast, how many cars were actually sold?

The quarterly online sales (in thousands of dollars) for a clothing retailer are recorded over 3 consecutive years:

Year 1

Year 2

Year 3

Time (t)

Sales ($000s)

Proportion of yearly mean

March

210

0.9417

June

185

0.8305

Sept

240

1.0769

215.8

Dec

255

1.1444

224.2

March

190

0.8520

June

195

0.8744

230.0

Sept

260

1.1659

240.2

Dec

250

1.1211

March

295

1.1217

255.4

June

255

0.9696

264.4

Sept

245

0.9316

Dec

260

0.9886

5MA

Calculate the seasonal index, rounded to four decimal places for each quarter: i

256

Quarter

March

June

iii

September

December

The data is smoothed using a 5-point moving average (5MA) as shown in the table. Calculate the missing values A and B.

Use your calculator to determine the equation of the least-squares regression line that fits the 5MA data, in terms of t. Round values to four decimal places.

Predict the sales revenue for the third quarter (Q3) of Year 4, rounding to the nearest dollar.

Determine the reliability of your prediction.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 3

The petrol price cycle at a local service station is monitored. The results over two weeks are given in this table:

Week 1

Week 2

Day

Time (t)

Price (cents)

Deseasonalised data

Mon

102.97

Tue

85.2

98.19

Wed

105.13

Thu

104.7

103.10

Fri

132.5

Sat

113.9

103.15

Sun

105.4

103.44

Mon

114.5

119.10

Tue

108.1

124.58

Wed

93.2

116.65

Thu

120.8

118.96

Fri

140.6

114.00

Sat

118.91

Sun

120.8

118.56

The seasonal indices are:

Day

Mon

Tue

Wed

Thu

Fri

Sat

Sun

Seasonal index

0.9614

0.8677

0.7990

1.0155

1.2333

1.1042

1.0189

Which is the best day of the cycle to purchase petrol?

Calculate the missing values X and Y, rounded to two decimal places.

Using your calculator, determine the equation of least-squares regression line for the deseasonalised data, in terms of t. Round all values to two decimal places.

Predict the price of petrol for Thursday in the third week.

Determine the reliability of your prediction.

A company’s quarterly proﬁts (in thousands of dollars) are recorded for three years: Year

200

180

250

220

210

190

260

230

220

200

270

240

Calculate the seasonal indices for all four quarters. Round the values to two decimal places.

Deseasonalise the data for Year 3.

A least-squares regression line ﬁtted to the full set of deseasonalised data is y = 4.55t + 195.33. Forecast the actual proﬁt for Quarter 1 of Year 4 (t = 13). 3.05 Make predictions using time series data mathspace.co

257

Complex unfamiliar 10

After analysing several years of quarterly sales data, a retail company developed a leastsquares regression line to model the trend in its deseasonalised sales ﬁgures. The equation for the trend line is: Deseasonalised sales = 2500t + 120 000 where t is the quarter number (t = 1 for Q1 of Year 1). The seasonal indices for each quarter are shown in the table:

Quarter

Seasonal index

0.85

1.05

0.90

1.20

Use the trend line to predict the deseasonalised sales for the third quarter of Year 5 (t = 19).

Calculate the ﬁnal sales forecast for Q3 of Year 5.

Is this forecast an example of interpolation or extrapolation? Explain your reasoning.

A national park records the number of visitors each season over three years. The deseasonalised visitor numbers (in thousands) are shown in the table: Time (t)

Deseasonalised 22.5 23.1 23.9 24.5 25.0 25.8 26.2 27.0 27.5 28.1 28.9 29.5 visitors (′000s) The seasonal indices are:

258

Quarter

Summer (Q1)

Autumn (Q2)

Winter (Q3)

Spring (Q4)

Seasonal index

1.50

0.90

0.60

1.00

Interpret the slope of the regression line in the context of this problem.

Use the trend line to predict the deseasonalised number of visitors for the Winter of Year 4 (t = 15).

Calculate the ﬁnal forecast for the number of visitors in the Winter of Year 4.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

This data shows the sales of air conditioners at a leading retailer over 4 quarters from 2017 to 2019:

2017

2018

2019

Month

Time (t)

No. of air conditioners sold

Proportion of yearly mean

March

332

0.3054

June

320

0.2943

Sept

966

0.8885

Dec

2731

2.5118

March

987

0.6340

June

926

0.5948

Sept

1117

0.7175

Dec

3197

2.0536

March

1216

0.6910

June

939

0.5336

Sept

1414

0.8035

Dec

3470

1.9719

Deseasonalise the data and use a least-squares regression line to predict the whole number of air conditioners sold in the quarter ending December 2020. 13

A nursery records the demand of herb seedlings by making note of the sales during periods of the year. The results are given in this table: t

Time period

Seedlings sold

Deseasonalised data

Apr 2017

1636

1959.05

Aug 2017

1472

1957.19

Dec 2017

2977

2107.16

Apr 2018

2027

2427.25

Aug 2018

1730

2300.23

Dec 2018

3463

2451.16

Apr 2019

2342

2804.45

Aug 2019

2219

2950.41

Dec 2019

3641

2577.15

The seasonal indices are: Month

April

August

December

Seasonal index

0.8351

0.7521

1.4128

Use a least-squares regression line to predict the whole number of seedlings sold in August 2020.

3.05 Make predictions using time series data mathspace.co

259

Collecting the Data For the context of this investigation, the analysis is set in late February 2016. Therefore, “next week” refers to the period from Monday, 29th February to Sunday, 6th March 2016. Historical data was sourced from Perth’s ‘FuelWatch’ website for a single Caltex petrol station in Balcatta. The unleaded petrol (ULP) prices for each day in February 2016 are shown below. Feb Date

ULP Price (cents)

Feb Date

ULP Price (cents)

Mon 1/02/2016

102.9

Mon 15/02/2016

98.9

2/02/2016

125.9

16/02/2016

119.9

3/02/2016

121.9

17/02/2016

115.9

4/02/2016

118.9

18/02/2016

111.9

5/02/2016

113.5

19/02/2016

108.5

6/02/2016

109.5

20/02/2016

105.9

7/02/2016

106.5

21/02/2016

98.9

Mon 8/02/2016

104.5

Mon 22/02/2016

98.9

9/02/2016

123.9

23/02/2016

117.9

10/02/2016

117.9

24/02/2016

114.9

11/02/2016

112.9

25/02/2016

111.9

12/02/2016

109.9

26/02/2016

107.9

13/02/2016

106.9

27/02/2016

99.9

14/02/2016

101.9

28/02/2016

96.9

Price

140 120 100 80 60 40 20

Time 0

Step 1: Visualise the Data Enter the data into a spreadsheet. Use the software to create a time series plot of the data. Your plot should look like the one below. What key features do you observe? Note the length of the repeating cycle (seasonality) and the overall trend.

Investigation: Time series data mathspace.co

261

Price

130 125 120 115 110 105 100 95 90

Time 0

ULP Price

Deseasonalised Data

Step 2: Deseasonalise the Data To analyse the trend more formally, we first remove the seasonal component. Follow these calculations using your spreadsheet: 1. Calculate the weekly means. 2. Calculate the seasonal index for each day of the week. 3. Divide each data point by its corresponding seasonal index to get the deseasonalised data. Plot the original and deseasonalised data on the same axes to see the smoothing effect.

Step 3: Model the Trend and Forecast Use the spreadsheet to fit a least-squares regression line to the deseasonalised data. The resulting equation should be approximately: y = −0.3278t + 114.9397 Use this equation to predict the deseasonalised price for Monday, 29th February 2016 (where t = 29). Finally, re-seasonalise your forecast by multiplying it by the seasonal index for Monday to get the final predicted price. Step 4: Interpret and Communicate the Result The final predicted price for Monday, 29th February 2016, is 96.95 cents/litre. This forecast’s reliability is high because predicting for day 29 is an example of interpolation (since it’s close to the original data range). For context, the actual price on that day was 93.9 cents/litre. Write a concluding sentence that directly answers the initial statistical question.

262

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3.06 Further applications After this lesson, you will be able to… • solve practical problems involving the analysis of time series data

Analyse and forecast with time series data Further applications of time series analysis require combining several techniques to solve practical problems. This involves a multi-step process to move from raw, seasonal data to making a final, realistic prediction. The goal is to first remove the seasonal effects to identify the underlying trend, and then re-apply the seasonal effect to a future forecast. The general process for analysing seasonal time series data and making predictions can be summarised in these stages: 1. Calculate seasonal indices The first step is to quantify the seasonal effect. This is done by calculating a seasonal index for each season (e.g., each quarter or month). A seasonal index is a measure of how a particular season’s value compares to the average value for a cycle. The sum of the seasonal indices must equal the number of seasons in a cycle. 2. Deseasonalise the data To reveal the underlying trend in the data, the seasonal variations are removed. This process is called deseasonalising. The deseasonalised data is calculated by dividing each raw data value by its corresponding seasonal index.

3. Fit a trend line and make a forecast With the seasonal component removed, a least-squares regression line can be fitted to the deseasonalised data points. This line models the underlying trend over time. This trend line is then used to forecast a deseasonalised value for a future time period by substituting the appropriate time value (t) into the regression equation. 4. Reseasonalise the forecast The forecast obtained from the trend line is a smoothed, deseasonalised value. To make a realistic prediction for a specific future season, the seasonal component must be put back in. This is done by multiplying the deseasonalised forecast by the seasonal index for that specific season. Predicted figure = Deseasonalised forecast × Seasonal index 5. Assess the reliability Finally, it is important to comment on the reliability of the prediction. Predictions made for time periods close to the original dataset are generally more reliable than those made far into the future. A common guideline is that predictions made within one cycle of the end of the data are considered reasonably reliable. 264

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Example 1 A beachside ice cream shop records its sales, in thousands of dollars, every 4 months, with seasons ending in January, May, and September. The sales data, together with some calculations, are shown in this table: Year

Data number (d)

Season

Sales ($’000)

Yearly mean

Percent of yearly mean

Deseasonalised figure

2017

Jan

50.33

127.2%

2017

May

50.33

103.3%

2017

Sept

50.33

2018

Jan

144.0%

2018

May

90.0%

2018

Sept

66.0%

2019

Jan

125.1%

2019

May

93.1%

2019

Sept

81.8%

a Calculate the value A.

Create a strategy Use the yearly mean for 2018 to create an equation containing A.

Apply the idea Equate the mean to the average of the data

Simplify the numerator

Multiply both sides by 3

Subtract 78 from both sides

b Calculate the value B, rounded to two decimal places.

Create a strategy Calculate the average sales in 2019.

Apply the idea Determine the average

Evaluate

3.06 Further applications mathspace.co

265

c Calculate the value C, express the answer as a percentage, rounded to one decimal place.

Create a strategy The ‘Percent of yearly mean’ is calculated by dividing the actual data value by the yearly mean and multiplying by 100%.

Apply the idea C is the percentage for the September 2017 season. The sales for September 2017 were 35, in thousands, and the yearly mean for 2017 was 50.33. Divide the data by the mean and multiply by 100%

Evaluate and round

d If the seasonal index for May is 0.9545, calculate the value D, rounded to the nearest whole number.

Create a strategy Use the formula: Deseasonalised data =

Apply the idea The sales ﬁgure for May 2018 was 45. Divide the raw data by the seasonal index

Evaluate

Example 2 A cat boarding kennel records its number of boarders every 4 months, ending at January, May, and September. The seasonal indices are shown in the table: Trimester

Jan

May

Sept

Seasonal index

⬚

0.9545

0.7245

Complete the table by calculating the seasonal index for January.

Create a strategy The sum of the seasonal indices is equal to the number of seasons in a cycle, which is 3.

266

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Apply the idea Let E be the seasonal index for January. E + 0.9545 + 0.7245 = 3

Write the equation

E + 1.679 = 3

Evaluate the addition

E = 3 − 1.679

Subtract 1.679 from both sides

E = 1.321

Evaluate

Trimester

Jan

May

Sept

Seasonal index

1.321

0.9545

0.7245

Example 3 A jet ski rental business records its number of rentals every 4 months, with seasons ending in January, May, and September. The data, together with some calculations, are shown in the table: Year

Data number (d)

Season

Number of rentals

Yearly mean

Percent of yearly mean

Deseasonalised figure

2017

Jan

50.33

127.2%

2017

May

50.33

103.3%

2017

Sept

50.33

69.5%

2018

Jan

144.0%

2018

May

90.0%

2018

Sept

66.0%

2019

Jan

62.33

125.1%

2019

May

62.33

93.1%

2019

Sept

62.33

81.8%

a The equation of the least-squares line for the deseasonalised number of rentals against the data number is determined to be y = 2.0333d + 44.0556. Predict the number of jet ski rentals for September 2020. Round your answer to the nearest whole number.

Create a strategy Use the given equation to calculate a deseasonalised forecast, and then multiply the result by the September seasonal index to reseasonalise it.

3.06 Further applications mathspace.co

267

a Calculate the seasonal indices for each quarter, rounded to four decimal places.

Create a strategy First, calculate the average sales for each year (cycle mean). Then, for each quarter, divide the sales ﬁgure by its corresponding yearly average to get the sales as a percentage of the mean. Finally, average these percentages for each quarter (e.g., average all Q1 percentages) to determine the seasonal index.

Apply the idea Step 1: Calculate the yearly averages.

Step 2: Express each sales ﬁgure as a proportion of its yearly mean. Year

1 2 3

Step 3: Average the percentages for each quarter to get the seasonal indices.

Reﬂect and check Check that the seasonal indices sum to the number of seasons (4): 0.9431 + 0.6948 + 1.2827 + 1.0793 = 3.9999 ≈ 4 The sum is correct, accounting for rounding.

3.06 Further applications mathspace.co

269

b Deseasonalise the sales ﬁgure for Quarter 3 of Year 2.

Create a strategy Divide the raw sales ﬁgure by the corresponding seasonal index calculated in part (a).

Apply the idea The raw sales for Q3 Year 2 is $125 000. The seasonal index for Q3 is 1.2827.

The deseasonalised sales ﬁgure is $97 451, rounded to the nearest dollar.

c A least-squares regression line is ﬁtted to the deseasonalised data, with the equation Deseasonalised sales = 2.5t + 85.1 where t is the time period (Q1 Year 1 is t = 1). Predict the sales for Quarter 2 of Year 4.

Create a strategy First, determine the time period t for Q2 of Year 4. Use the regression equation to get a deseasonalised forecast. Then, reseasonalise this forecast by multiplying it by the seasonal index for Quarter 2.

Apply the idea Step 1: Determine the value of t. Year 3 ends at t = 12. Therefore, Q1 Year 4 is t = 13, and Q2 Year 4 is t = 14. Step 2: Calculate the deseasonalised forecast. Deseasonalised sales = 2.5(14) + 85.1

Write the equation

= 35 + 85.1

Evaluate the multiplication

= 120.1

Evaluate

Step 3: Reseasonalise the forecast using the Q2 seasonal index (0.6948). Predicted sales = Deseasonalised forecast × Seasonal index = 120.1 × 0.6948

Substitute the values

= 83.445

Evaluate

The predicted sales for Quarter 2 of Year 4 are $83 445.

270

Write the equation

Mathspace QLD 12 General Mathematics 2025 mathspace.co

3.06 Practice questions Simple familiar 1

A hairdressing salon at a local shopping centre records the number of customers per day and the results are shown in this graph:

120 100 80 60 40 20 0

Mon Tue Wed Thu Fri Mon Tue Wed Thu Fri Mon Tue Wed Thu Fri Mon Tue Wed Thu Fri

Number of customers

140

Week 1

Week 2

Week 3

Week 4

Week

Comment on the trend and seasonality of the data. 2

The graph represents the sales of sausages at a Bunnings store in the four weeks leading up to Christmas: 800

Number of sausages

700 600 500 400 300 200 100

Days Fri

Sat Sun

Fri

Sat Sun

272

Fri

Sat Sun

Fri

Which day tends to be the busiest for sausage sellers?

What type of moving average would best smooth the data?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Sat Sun

The table shows some time series data where t represents time: t

Calculate the 5-point moving average at t = 3.

Calculate the 5-point moving median at t = 3.

An ice cream store manager recorded the number of ice creams sold on Friday, Saturday and Sunday over a 4-week period. The given table shows the data for the third and fourth weeks:

Week 3

Week 4

Day

Sales

Fri

145

Sat

492

Sun

604

Fri

180

Sat

530

Sun

675

Calculate the 3-point moving average for Sunday of Week 3. Round your answer to two decimal places.

The seasonal indices for Friday and Sunday are shown in the second table: Friday

Saturday

Sunday

36.11%

⬚

143.70%

Calculate the seasonal index for Saturday. c

Complete the third table by calculating the deseasonalised ﬁgures for Week 4. Day

Raw data

Fri

180

Sat

530

Sun

675

Deseasonalised data ⬚ ⬚ ⬚

Round your answers to the nearest whole number.

3.06 Further applications mathspace.co

273

Complex familiar Ex 1

A cat boarding kennel records its number of boarders every 4 months, ending in January, May and September. The data for the number of cats, some calculations and the seasonal indices are shown:

2017

2018

2019

Time, t

Trimester

No. of boarders

Yearly mean

Percentage of yearly mean

Deseasonalised figure

Jan

55.00

127.3%

May

100.0%

Sept

Jan

130.8%

May

96.2%

Sept

73.1%

Jan

124.4%

May

95.1%

Sept

80.5%

52.00

The seasonal indices are:

Ex 2

Trimester

Jan

May

Sept

Seasonal index

0.9750

0.7550

Calculate the value A.

Calculate the value B, rounded to two decimal places.

Calculate the value C as a percentage, rounded to one decimal place.

Given the seasonal index for May, calculate the value D, rounded to the nearest whole number.

Determine E, the seasonal index for January, rounded to three decimal places.

The number of children coming to visit Santa Claus on weekdays at a suburban shopping centre is recorded over a 4-week period leading up to Christmas. The seasonal indices are shown in the table. Complete the table by calculating the seasonal index for Wednesday.

274

Day (in Week 4)

Mon

Tue

Wed

Thu

Fri

Seasonal index

40.22%

78.09%

⬚

134.05%

162.60%

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 3

A ballet company performed 4 times a week for 4 weeks at the Perth Concert Hall which is able to seat 1731 patrons. To break-even, the attendance must be more than 60% of maximum capacity. The attendance at the daily performances are shown in the table: Performance day Week 1

Week 2

Week 3

Week 4

Attendance number (A)

Fri

1087

Sat (matinee)

844

Sat (evening)

1731

1179.4

141.5%

Sun

1232

100.7%

Fri

1003

1299.8

Sat (matinee)

802

1169.2

69.5%

Sat (evening)

1731

113.4

150.1%

Sun

1078

1061.4

93.5%

Fri

953

1239

Sat (matinee)

743

1083.6

68.5%

Sat (evening)

1690

1047.8

155.8%

Sun

954

987.2

87.9%

Fri

899

1154.6

Sat (matinee)

650

Sat (evening)

1580

158.2%

Sun

865

86.6%

5MA

Weekly mean

Percentage of weekly mean

1223.5

88.8% 69.0%

1153.5

1085

998.5

87.0%

87.8%

90.0% 65.1%

Calculate the value of: i

Performance number (n)

The equation of the regression line from the 5MA calculated from the raw attendance data for the 4 weeks is: A = −15.75n + 1254.5

Given that the seasonal index for Friday is 88.4%, predict the attendance figure for Friday of Week 6. Round your answer to the nearest whole number. c

Assess the reliability of your prediction.

3.06 Further applications mathspace.co

275

Complex unfamiliar 9

An oﬃce manager recorded the number of coffee pods used daily by staff over a 4-week period. The data and seasonal indices are recorded in the tables:

Week 1

Week 2

Week 3

Week 4

Weekday

Day, d

Number of coffee pods

Mon

134

Tues

154

Wed

143

Thurs

165

Fri

175

Mon

137

Tues

158

Wed

145

Thurs

169

Fri

179

Mon

140

Tues

160

Wed

147

Thurs

171

Fri

180

Mon

145

Tues

163

Wed

151

Thurs

176

Fri

186

Day

Monday

Tuesday

Wednesday

Thursday

Friday

Seasonal index

1.9053

2.1771

2.0043

2.3303

2.4580

Deseasonalise the data and use a least-squares regression line to predict the actual number of coffee pods used for Tuesday of Week 6. Round your ﬁnal answer to the nearest whole number.

3.06 Further applications mathspace.co

277

This data shows the sales of heaters at a leading retailer over 4 quarters from 2022 to 2024:

2022

2023

2024

Month

Time (t)

No. of heaters sold

Seasonal index

March

260

0.276

June

1260

1.319

Sept

1200

1.240

Dec

160

0.163

March

340

0.276

June

1620

1.319

Sept

1520

1.240

Dec

200

0.163

March

420

0.276

June

1980

1.319

Sept

1840

1.240

Dec

240

0.163

Use the seasonal indices provided to deseasonalise the data. Then, use a least-squares regression line to predict the whole number of heaters that will be sold in the quarter ending December 2025. 11

The electricity bills for a household are recorded over 2 years and are shown in this table:

2018

2019

Month

Time (t)

Billed amount

Proportion of yearly mean

Feb

122

0.7114

Apr

242

1.4111

Jun

110

0.6414

Aug

156

0.9096

Oct

159

0.9271

Dec

240

1.3994

Feb

0.5287

Apr

169

1.3538

Jun

0.6729

Aug

127

1.0174

Oct

124

0.9933

Dec

179

1.4339

Deseasonalise the data, determine the least-squares regression line that ﬁts the deseasonalised data, and use the model to predict the electricity bill amount for February 2021.

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3 Chapter review CF

This time series graph shows the number of active monthly subscribers for a new streaming service over its first two years of operation: 90 Subscribers (in thousands) 80 70 60 50 40 30 20 10 0

Month 12

Describe the underlying trend in the number of subscribers.

Does the data appear to be seasonal? If so, what is the approximate length of the cycle?

In which month does an outlier or irregular fluctuation appear to occur? Provide a possible reason for this fluctuation.

The number of coffees sold by a cafe each day over a 7-day period is recorded as: 85, 92, 88, 95, 250, 105, 110

Calculate the 3-point moving average for the number of coffees sold, centred on the third day.

The large number of sales on the ﬁfth day was due to a half-price promotion and is considered an outlier. To better represent the underlying trend, would a moving mean or a moving median be more appropriate for smoothing the data around this day? Justify your answer.

The quarterly seasonal indices for the sales of a retail clothing store are shown in the table:

Seasonal index

280

Quarter 1

Quarter 2

Quarter 3

Quarter 4

0.85

0.70

1.15

Assuming a cycle length of 4 quarters, calculate the seasonal index for Quarter 4.

In Quarter 3, the store’s actual sales were $287 500. Calculate the deseasonalised sales ﬁgure for this quarter.

Interpret the meaning of the seasonal index for Quarter 2.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

This table shows the quarterly number of visitors (in thousands) to a national park over a three-year period: Year

Quarter 1

Quarter 2

Quarter 3

Quarter 4

2021

2022

2023

Calculate the seasonal index for each quarter. Show workings for at least one quarter. Round indices to three decimal places.

Deseasonalise the visitor numbers for Quarter 2 of 2023. Round the answer to the nearest whole number.

A spreadsheet is used to analyse the monthly sales data for a business. The raw data is in column B, a 3-point moving average is in column C, and a 5-point moving average is in column D. A

Month

Sales

3-point MA

5-point MA

4523

7812

5572

⬚

6689

5577

6673

6287

6108

6657

6368

5875

6267

⬚

6714

6269

6282

6429

6702

6450

6352

6379

6420

6457

6180

6420

6395

6701

4678

Write the spreadsheet formula that should be entered into cell C3 to calculate the ﬁrst value of the 3-point moving average.

What formula would be found in cell D6 if the formula in cell D4 was correctly created and ﬁlled down?

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The table shows Darwin’s actual rainfall (mm) each season for two years. 2022

2023

Autumn

410

390

Winter

Spring

205

150

Summer

1135

1100

Calculate the seasonal index for each season in Darwin.

This table shows Hobart’s actual rainfall (mm) each season for 2023 and the long-term seasonal indices. Autumn

Winter

Spring

Summer

2023 rainfall (mm)

130

145

155

132

Seasonal index

0.92

1.02

1.12

0.94

Deseasonalise the Hobart rainfall data to identify the 2023 season with the highest seasonally adjusted rainfall. (QCAA 2024 Paper 1 Q25) 9

Buffalo ﬂy bites cause skin wounds on cattle. The table shows the average number of skin wounds per animal in a herd for two years. Autumn

Winter

Spring

Summer

2021

285

195

460

2022

276

170

392

Deseasonalise the data. (QCAA 2023 Paper 2 Q2)

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4.01 Sequences After this lesson, you will be able to… • understand the meaning of recursion in sequences • identify an arithmetic sequence • use recursion to generate an arithmetic sequence

Sequences Sequences

Term

An ordered collection of numbers where a rule or formula is followed is called a sequence. It is similar to a set, but may contain repetition and the order of the numbers is important.

The elements of a sequence are called terms. To refer generally to the nth term, the notation tn is used. The n is a variable which represents the position of a term in the sequence.

Example: − 3, 5, 13, 21, … and 1, 10, 100, 1000, … are two interesting mathematical sequences.

Example: If n = 1 then tn is t1, and if n = 6 then tn is t6. For any term, tn, using this notation, the term before tn is tn − 1 and the term after tn is tn + 1.

If the sequence ends, it is known as a finite sequence. If the sequence continues indefinitely, it is said to be infinite. For the sequence 2, 3, 5, 7, 11, … (the sequence of prime numbers) the first term is 2, the second term is 3, and term 3 is 5. This can be shown with the following notation, t1 = 2, t2 = 3, t3 = 5, and so on. The “rule” here is to only include numbers that are divisible by 1 and the number itself or a number with exactly two factors in order. Term 6 of this sequence is 13, t6 = 13. A sequence does not have to follow a simple add or multiply pattern, it just needs a clear definition for what comes next. For instance, a sequence can be generated by writing the digits of π : 3, 1, 4, 1, 5, 9, … The digits of π form a sequence because each digit can be systematically determined, even if there’s no simple formula for the nth digit.

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For the sequence 3, 5, 7, 9, …, the explicit rule is tn = 2n + 1. To determine the 5th term, substitute n = 5 directly into the formula: t5 = 2 × 5 + 1 = 11. This rule can be applied to determine a much later term, such as t43 : t43 = 2 × 43 + 1 = 87. If a question asks for a ‘rule’ or ‘formula’ for a sequence, it usually means the explicit rule. If a question asks for a ‘recurrence relation’ or ‘recursive rule’, then the recurrence format is required.

Example 3 Using this explicit rule, determine the ﬁrst 5 terms of the sequence in order starting with n = 1. tn = n2 + 6

Create a strategy The explicit rule allows for determining each term given its position n. Substitute n = 1, 2, 3, 4, 5 into the explicit rule.

Apply the idea Using the explicit rule: t1 = 12 + 6 = 7 t2 = 22 + 6 = 10 t3 = 32 + 6 = 15 t4 = 42 + 6 = 22 t5 = 52 + 6 = 31 The ﬁrst 5 terms of the sequence are 7, 10, 15, 22, 31.

Example 4 Determine a recurrence relation and an explicit rule to describe the nth term of the sequence 5, 9, 13, 17, 21, …

Create a strategy For the recurrence relation, identify the ﬁrst term and the rule that connects each term to the previous one. In this sequence, each term is determined by adding 4 to the previous term, and the ﬁrst term is 5. For the explicit rule, determine a formula that connects the term’s position, n, directly to its value, tn. This can be done by observing the pattern in a table of values.

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4.01 Practice questions Simple familiar Ex 1

If tn describes the nth term of a sequence, determine the indicated term: a b

t4 in the sequence: 200, 20, 2, 0.2, 0.02, …

2t2 − t4 in the sequence: 9, 12, 15, 18, 21, …

In the sequence 5, − 5, 7, − 7, 9, − 9, …, determine n if tn = 7.

For each sequence:

Describe the pattern in words.

Determine the next number in the sequence.

− 1, 1, 3, 5, 7, …

c 5

292

64, 32, 16, 8, 4, 2, …

Determine whether these are recurrence relations: a

Ex 2

t3 in the sequence: 3, − 6, 5, − 8, 7, …

= n + 1 tn = tn − 1 + 9

Rn = (3Rn − 1 )5 + Rn − 2

Vn = 3 (n − 1)

Consider the sequence given in the table: a

Determine the explicit rule for tn in terms of n.

Use the explicit rule to determine t20.

…

Write the recursive rule, letting tn be the nth term, for each description: a

The next term is calculated by adding 5 to the previous term

The next term is calculated by adding the two previous terms

The next term is calculated by multiplying the previous term by negative 1 and then adding 7

The next term is calculated by subtracting one and a half from the previous term

For each recursive rule, determine the ﬁrst 5 terms of the sequences in order: a

tn = 3tn − 1, t1 = 1

bn = −bn − 1 + 3, b1 = 0

an = an − 1 + 6, a1 = − 8

Consider the sequence which has a ﬁrst term of 11 and a second term of 22, and subsequent terms are calculated by adding the two previous terms: a

Write the recursive rule for the sequence, letting tn be the nth term.

Determine the ﬁrst ﬁve terms of the sequence.

Determine whether these are explicit relations: a

bn = bn − 1 bn − 2

an = 8an − 1 + a1

Mathspace QLD 12 General Mathematics 2025 mathspace.co

dn = n2 + 3n + 8

cn = 3n2

Ex 3

For each explicit rule, determine the ﬁrst 5 terms of the sequences in order starting with n = 1: a

Ex 4

bn = 5n − 2

sn = n2 + 2

For each sequence: i

Determine the recursive rule that describes tn in terms of tn − 1, and the initial term t1.

Determine the explicit rule that describes tn in terms of n.

A sequence starts with a ﬁrst term of 1300 and each subsequent term increases by 2.5% of the previous term.

A sequence starts with a ﬁrst term of 44 and each term is 77 more than the previous term.

For the explicit rule tn = 0.7n − 5 which starts at n = 1, determine the sum of the ﬁrst 50 terms.

Complex familiar 13

Consider the sequence deﬁned by an = 2an − 1 + 3, a1 = 12. Calculate the sum of the ﬁrst 25 terms.

For each pair of sequences, compare the ﬁrst 10 terms of each sequence on a graph to determine which sequence has a greater value at n = 10: a

tn + 1 = tn + 22, t1 = 12

An + 1 = An + 25, A1 = 12

Ex 5

tn + 1 = 2.5 tn, t1 = 5 An + 1 = 3.5An, A1 = 5

For the explicit rule tn = (0.2n)2 + 5n which starts at n = 1, determine the sum of the terms from the 20th to the 30th term inclusive.

Consider the sequence 3, 7, 11, 15, 19, 23, …

If tn is the value of the nth term in the sequence, determine t10.

Determine the ﬁrst term greater than 100.

If Sn is the sum of the ﬁrst n terms, determine S10.

Starting from n = 1, determine the minimum number of terms for the sum to ﬁrst exceed 500.

Consider the sequence 1, 2, 4, 8, 16, … a

Determine t13.

Determine the ﬁrst term greater than 10 000.

If Sn is the sum of the ﬁrst n terms, determine S20.

Starting from n = 1, determine the greatest number of terms such that the sum is still less than 10 000 000.

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4.02 Arithmetic sequences After this lesson, you will be able to… • use the rule for the nth term of an arithmetic sequence • calculate the nth term of an arithmetic sequence given the first term, term number, and common difference • identify the components of the arithmetic sequence rule: tn, t₁, n, and d

Arithmetic sequences Arithmetic sequences A sequence in which each term changes from the previous term by adding or subtracting a constant amount is called an arithmetic sequence. The constant by which the terms are changing is referred to as the common difference, which can be calculated by subtracting any term from its succeeding term (d = tn + 1 − tn).

The sequence − 3, 5, 13, 21, … is an arithmetic sequence with a common difference of 8. On the other hand, the sequence 1, 10, 100, 1000, … is not arithmetic because the difference between consecutive terms is not constant. The ﬁrst term is denoted by t1 and the common difference by the letter d. Any arithmetic sequence can be written as the recurrence relation: tn = tn − 1 + d, with a starting term of t1 For example, the recurrence relation tn = tn − 1 + 2, t1 = 5, starts with 5 and 2 is added to determine each subsequent term, generating the sequence 5, 7, 9, 11, 13, 15, … To determine a speciﬁc term, such as the 100th term, without listing the entire sequence, an explicit formula is needed. The pattern for this formula can be seen in a table. For the sequence − 3, 5, 13, 21, …, the starting term is t1 = − 3 and the common difference is d = 8. n

−3

−3 + 8

−3 + 2 × 8

−3 + 3 × 8

…

− 3 + (n − 1) × 8

Pattern

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295

b Determine t9, the 9th term in the sequence.

Create a strategy Substitute n = 9 into the given formula.

Apply the idea tn = 15 + 5(n − 1)

Write the formula

t9 = 15 + 5 × (9 − 1)

Substitute n = 9

= 15 + 5 × 8

Evaluate the subtraction inside the brackets

= 15 + 40

Evaluate the multiplication

= 55

Evaluate the addition

Reﬂect and check Since the ﬁrst term and the common difference are both positive, all terms in the sequence will be positive and increasing. Therefore, the 9th term must be a positive value greater than the 8th term.

Example 3 In an arithmetic sequence, t5 = 15 and t20 = 45. Determine the explicit rule for this sequence.

Create a strategy Set up two equations using the explicit rule tn = t1 + (n − 1) d and the information given. Then, solve these equations simultaneously to determine t1 and d.

Apply the idea For t5 = 15, substitute n = 5: 15 = t1 + (5 − 1) d ⟹ 15 = t1 + 4d

45 = t1 + (20 − 1) d ⟹ 45 = t1 + 19d

For t20 = 45, substitute n = 20:

Subtract equation 1 from equation 2 to eliminate t1 and solve for d: 45 − 15 = (t1 + 19d) − (t1 + 4 d) 30 = 15d d=2 Now, substitute d = 2 into equation 1 to determine t1: 15 = t1 + 4(2) 15 = t1 + 8 t1 = 7 The explicit rule is tn = 7 + 2(n − 1).

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297

Ex 1

Determine if these sequences are arithmetic sequences: a

4, 0, − 4, − 8, …

1, 4, 9, 16, 25, …

3, 6, 12, 24, …

5, 7, 5, 7, …

Identify the common difference for these arithmetic sequences: a

330, 280, 230, 180, …

8, ⬚, 16, 20, ⬚

tn + 1 = tn + 5, t1 = 4

⬚, 0, ⬚, 10, ⬚

Determine the ﬁrst 3 terms of the sequences using these recursive deﬁnitions: a

tn = tn − 1 − 4, t1 = − 12

Write a recursive rule for tn + 1 in terms of tn and an initial condition for t1 for a sequence where each term is obtained by increasing the previous term by 35, and the ﬁrst term is 60.

The nth term of a sequence is deﬁned by tn = 5n + 8: a

Ex 2

Determine the missing terms in these arithmetic sequences: a

5, 1, − 3, − 7, …

Determine the ﬁrst four terms.

Determine d, the common difference.

The nth term of a sequence is defined by: tn = 11 + (n − 1) × 10. Determine the first four terms of the sequence. Consider the first three terms of the arithmetic sequence: 10, 3, − 4 … a

Determine the common difference.

Determine the 9th term in the sequence.

Determine the rule for tn.

An arithmetic sequence is deﬁned by the rule tn = 4 + 5 (n − 1): a

Identify t1, the ﬁrst term, and d, the common difference.

Calculate the 9th term, t9.

For the sequence 12, 15, 18, 21, …, determine the 16th term.

Complex familiar 13

The ﬁrst term of an arithmetic sequence is 5. The third term is 17. Calculate the common difference, d.

Consider the arithmetic sequence with terms t3 = 14 and t12 = 59: a

Ex 3

Calculate d, the common difference.

Determine the ﬁrst term, t1.

An arithmetic sequence contains the terms t2 = 9 and t5 = 27: a

Determine the ﬁrst term and the common difference.

Calculate t30.

The ﬁrst three terms of an arithmetic sequence are 46, 41, 36, … a

Determine the range of values of n for which the terms in the sequence are positive.

Determine the last positive term in the sequence.

In an arithmetic sequence, tn = − 530. Solve for n given t1 = 28 and d = − 18. 4.02 Arithmetic sequences mathspace.co

299

Ex 4

Calculate the value of x such that these three terms form successive terms in an arithmetic sequence:

x + 4, 4x + 4, 10x − 20

2x − 2, 5x − 6, x + 4

Complex unfamiliar 19

In an arithmetic sequence, the 6th term is − 11 and the 8th term is − 17. Calculate the 14th term.

The nth term of the sequence

The terms x + 6, 5x, and 5x + 10 are consecutive terms of an arithmetic sequence:

, … is

Calculate the value of x.

Determine the ﬁrst 5 terms of the sequence.

. Calculate the value of n.

4.03 Arithmetic sequences with tables and graphs After this lesson, you will be able to… • use the rule for the nth term of an arithmetic sequence to calculate terms • understand how to represent arithmetic sequences using tables • construct graphs to display arithmetic sequences

Arithmetic sequences in tables and graphs The rule for the nth term of an arithmetic sequence is given by:

tn = t1 + (n − 1) d tn

is the nth term of the sequence

is the ﬁrst term of the sequence

n is the term number, which must be a positive integer d

is the common difference

If the rule tn = t1 + (n − 1) d is expanded, it takes the form tn = dn + k (where k = t1 − d). This is the form of a linear equation, y = mx + c, where the gradient (m) is the common difference (d). • If the common difference is positive (d > 0), the terms increase. This models linear growth. • If the common difference is negative (d < 0), the terms decrease. This models linear decay.

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Because of this linear relationship, when the terms of an arithmetic sequence are plotted on a graph, the points will form a straight line. Since the term number, n, must be a positive integer, arithmetic sequences are used to model discrete situations. tn 12 10 8 6 4 2

n 0

This graph shows an example of linear growth. The ﬁrst term is t1 = 5 and the common difference is d = 2. The sequence can also be represented in a table: n

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 The nth term of an arithmetic sequence is given by the equation tn = 12 + 4(n − 1): a Complete the table of values. n

⬚

Create a strategy

Apply the idea

Substitute the corresponding values of n into the equation.

The completed table is: n

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301

b Plot the points in the table on a graph.

Apply the idea tn

56 48 40 32 24 16 8

n 0

Example 2 The plotted points represent terms in an arithmetic sequence: tn

14 12 10 8 6 4 2

n 0

Write a simpliﬁed expression for the general nth term of the sequence, tn.

Create a strategy

Apply the idea

From the graph, identify the ﬁrst term t1 and calculate the common difference d. Substitute these into the formula tn = t1 + (n − 1) d.

From the graph, the ﬁrst term is t1 = 1.

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The common difference is d = t2 − t1 = 5 − 1 = 4. tn = t1 + (n − 1) d

Write the formula

= 1 + (n − 1) × 4

Substitute t1 = 1, d = 4

= 1 + 4n − 4

Expand the brackets

= 4n − 3

Combine like terms

Ex 1

For the recursive rule tn = tn − 1 + 2, t1 = 5: a

Determine the next four terms of the sequence, from t2 to t5.

Plot the ﬁrst ﬁve terms on a graph.

For the arithmetic sequence given by tn = 10 + 3 (n − 1): a

Create a table of values for tn using n = 1, 2, 3, 4.

Determine the common difference.

For the sequence given by the recursive rule tn + 1 = tn + 3, t1 = 5, determine the explicit rule for tn in terms of n.

An arithmetic sequence is deﬁned by tn = 40 − 3(n − 1). Is this a model of linear growth or decay? Justify your answer.

An arithmetic sequence has a ﬁrst term of − 8 and a common difference of 5:

Determine the rule for the nth term, tn.

Create a table of values for n = 1, 2, 3, 4.

For each sequence given by a recursive rule: i

Determine the value of d.

Determine the explicit rule for tn in terms of n.

iii

Create a table of values for tn using n = 1, 2, 3, 4, and 10.

Plot the points from the table on a number plane.

Determine whether the points on the graph lie in a straight line.

tn + 1 = tn + 3, t1 = 5

tn + 1 = tn − 4, t1 = 50

tn + 1 = tn + 4, t1 = − 46

tn + 1 = tn + 2.5, t1 = 5

Complex familiar Ex 2

For each graph: i

Create a table of values for the given points.

Determine d.

iii

Determine the general rule for tn.

Calculate the indicated term in the sequence.

Calculate the 14th term.

Calculate the 18th term.

tn 14

9 6

n 0

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−3

n 1

Calculate the 12th term.

Calculate the 11th term.

0 10 tn 9 8 7 6 5 4 3 2 1 0

−2

n tn

−4 −6 −8 −10

n 1

−12

For the arithmetic sequence given by tn = 12 + 3 (n − 1), a straight line is drawn connecting the points of the sequence. Determine the equation of this line in the form y = mx + c.

A local gym offers a new member deal. The initial joining fee is $50, and the membership costs $15 per week. The total cost for a member after n weeks forms an arithmetic sequence:

Determine the rule for Cn, the total cost after n weeks.

A person has a budget of $500. Determine the maximum number of full weeks they can be a member.

For the sequence given by the recursive rule tn + 1 = tn + 4, t1 = − 1, determine the ﬁrst position n, where the term becomes greater than or equal to 500.

Complex unfamiliar 14

For the sequence given by the recursive rule tn + 1 = tn − 4, t1 = 2, determine the ﬁrst position n, where the sequence becomes less than or equal to − 400.

An arithmetic sequence has a 5th term of 22 and a 12th term of 57:

Determine the rule for the nth term, tn.

Calculate the 30th term.

The plotted points are reﬂected across the horizontal axis to form three new points. If these new points represent consecutive terms of an arithmetic sequence, write the general rule for An, the nth term in this new sequence.

14 tn 13 12 11 10 9 8 7 6 5 4 3 2 1 0

n 1

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4.04 Application of arithmetic sequences After this lesson, you will be able to… • model practical situations involving linear growth or decay using arithmetic sequences • define an initial value (t0) and a common difference (d) from a real-world scenario • apply the rule tn = t0 + nd to solve problems involving simple interest and straight-line depreciation • calculate the future value of an item or the time required for it to reach a specific value in a linear model

Applications of arithmetic sequences Arithmetic sequences can model many real-life situations involving linear growth or decay. This includes simple interest calculations, straight-line depreciation, and regular payment plans. Any time a quantity changes by a constant amount over a set time period, it can be modelled as an arithmetic sequence. Depreciation is when an item loses value over time. In straight-line depreciation, the value reduces by the same amount each period, which is a model of linear decay. When modelling real-world scenarios, it is often more intuitive to start with an initial value at time zero, denoted as t0. This represents the value before the ﬁrst change occurs (e.g., the purchase price of an item before it depreciates, or the principal of a loan before interest is added). The standard formula starting from t1 is: tn = t1 + (n − 1) d When starting from an initial value t0, the explicit formula becomes:

tn = t0 + nd tn

is the value after n time periods

is the initial value (at n = 0)

is the number of time periods after the start

is the common difference per time period

Using t0 is often simpler, as the term number, n, directly matches the number of time periods that have passed.

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Example 1 A taxi service charges a $1.50 pick-up fee and $1.95 per kilometre of travel: a Calculate the total charge for a 10 km journey.

Create a strategy

Apply the idea Total charge = 1.50 + 1.95 × 10

Add the pick-up fee to the total cost for the kilometres travelled.

= 1.50 + 19.50 = $21.00

b Determine a recurrence relation for Tn, the price of an n km trip.

Create a strategy

Apply the idea

The initial condition, T0, is the cost for 0 km, which is the pick-up fee. The common difference, d, is the cost per kilometre.

The initial condition is the pick-up fee: T0 = 1.50 The cost increases by $1.95 for each kilometre, so d = 1.95 The recurrence relation is: Tn = Tn − 1 + 1.95, where T0 = 1.50

Example 2 A sum of $3000 is invested in an account that pays simple interest at a rate of 4% per annum. The value of the investment, Vn, forms an arithmetic sequence: a Calculate the amount of interest earned each year.

Apply the idea Interest per year = 0.04 × 3000 = $120 The amount of interest earned each year is $120. This is the common difference, d.

b Determine a recurrence relation for Vn, the value of the investment after n years.

Apply the idea The initial investment is $3000, so V0 = 3000. The common difference is d = 120. The recurrence relation is: Vn = Vn − 1 + 120, where V0 = 3000

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c Determine the value of the investment after 8 years.

Create a strategy Use the explicit rule Vn = V0 + nd to calculate the value of V8.

Apply the idea Vn = V0 + nd

Write the formula

V8 = 3000 + 8 × 120

Substitute n = 8, V0 = 3000, and d = 120

= 3000 + 960

Evaluate the product

= $3960

Evaluate

The value of the investment after 8 years is $3960.

Example 3 A car is purchased for $1800. The value of the car depreciates by a constant amount of $50 each year: a Determine an explicit rule for the value of the car, Vn, after n years.

Create a strategy

Apply the idea

Identify the initial value, V0, and the common difference, d. Use the rule Vn = V0 + nd.

The initial value is V0 = 1800. Depreciation means the value decreases, so the common difference is d = − 50. The explicit rule is Vn = 1800 + n(− 50), which simpliﬁes to Vn = 1800 − 50n.

b Determine the number of years after which the car will be worth half its purchase price.

Create a strategy Set Vn equal to half the purchase price and solve the rule from the previous part for n.

Apply the idea Half the purchase price is

= 900. Let V = 900. Set the value of the car to 900

Rearrange to isolate the term with n

Simplify the right-hand side

Divide both sides by 50

Evaluate

After 18 years, the car will be worth half its purchase price.

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A motorbike’s value is modelled by the recurrence relation Vn + 1 = Vn − 1200, V0 = 15 000, where Vn is the value after n years: a

Identify the purchase price of the motorbike.

Determine an explicit rule for Vn.

Calculate the value of the motorbike after 9 years.

The graph shows the depreciation of a car’s value over 4 years: a

Identify the initial value of the car.

Calculate the annual depreciation of the car.

Determine the value of the car after 4 years.

Value ($)

36 000 32 000 28 000 24 000 20 000 16 000 12 000 8 000 4 000

Age (years) 0

A TV depreciates at a constant rate each year. Its value after n years is shown in the graph: a

Identify the initial value of the TV.

Determine a recursive rule, Vn, that gives the value of the TV after n years.

n 1

For a ﬁbre-optic cable service, Christa pays a one-off amount of $200 for installation costs and then a monthly fee of $30: a

310

160 Vn 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0

Complete the table of values for the total cost (T ) of Christa’s service over n months. n

⬚

Determine the increase between consecutive terms in the sequence.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex familiar Ex 2

Ex 3

A sum of $5000 is invested in an account that pays simple interest at a rate of 3.5% per annum: a

Calculate the amount of interest earned each year.

Determine the value of the investment after 10 years.

A ball starts rolling down a slope. It has rolled 23 cm after the ﬁrst second, 44 cm after the second, 65 cm after the third, and so on: a

Explain why the ball’s distance can be modelled by an arithmetic sequence.

Determine how far the ball will have rolled after 12 seconds.

A car bought at the beginning of 2009 is worth $1200 at the beginning of 2015. The value of the car has depreciated by a constant amount of $40 each year since it was purchased: a

Calculate the purchase price of the car in 2009.

Determine an explicit rule for the value of the car after n years from purchase.

Jack is learning to drive. His ﬁrst lesson is 28 minutes long, and each subsequent lesson is 4 minutes longer than the lesson before: a

Determine a recursive rule that describes the length of the next lesson, tn + 1, based on the previous one, tn.

Calculate the length of his 14th lesson.

Each day, Namika withdraws $5 from her bank account to spend on lunch. Before she withdraws this amount on January 1, she has $3000 in her bank account: a

Calculate the amount left in her bank account after she withdraws $5 on January 7.

Determine a recursive rule for tn, the amount in her account at the end of day n, using an initial condition for t0.

Complex unfamiliar 16

A telecommunications company sells 1700 mobile phones in the ﬁrst month of its operation. The company plans to increase its sales by 150 mobile phones each month. Calculate how many phones the company plans to sell in the last month of the 4th year of its operation.

When a new school ﬁrst opened, t1 students started. Each year, the number of students increases by a constant amount, d. At the beginning of its 7th year, it had 361 students. At the end of the 11th year, the school had 536 students. At the end of the nth year, the school will reach its capacity of 921 students. Determine the value of n.

For a sprint training exercise, a number of balls are placed in a straight line. The ﬁrst ball is 2 metres from the start, and there is a 3 metre distance between each of the remaining consecutive balls. An athlete must run from the start, collect the nearest ball, run back to the start, and deposit the ball in a box. She repeats this process until all balls have been collected: a

Determine an expression for the total distance the athlete runs to collect and deposit the kth ball.

Calculate the distance she runs to collect and deposit the 12th ball. 4.04 Application of arithmetic sequences mathspace.co

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4.05 Geometric sequences After this lesson, you will be able to… • identify a sequence as geometric by finding a common ratio (r) • generate the terms of a geometric sequence using a recurrence relation • determine the explicit rule (tn = t₁ r n − 1 ) for a geometric sequence • use the explicit rule to solve for an unknown term (tn), the first term (t1), or the common ratio (r)

Geometric sequences Geometric sequences A sequence where each term is calculated by multiplying the previous term by a constant factor is called a geometric sequence. This constant factor is called the common ratio (r), which can be calculated by dividing any term by its preceding term

For example, the sequence 4, 8, 16, 32, … is geometric with a ﬁrst term t1 = 4 and a common ratio r = 2. The sequence 100, − 50, 25, − 12.5, … is geometric with t1 = 100 and r =

A geometric sequence can be described with a recurrence relation, which deﬁnes a term based on the one before it. tn = r × tn − 1, with a starting term of t1 To determine a speciﬁc term without listing all previous terms, an explicit formula is used. The pattern can be seen in a table. For the sequence 5, 10, 20, 40, …, the sequence has t1 = 5 and r = 2. n

Pattern

5 × 20

5 × 21

5 × 22

5 × 2n − 1

This pattern leads to the general explicit rule for any geometric sequence:

tn = t1 r n − 1 tn is the nth term t1 is the ﬁrst term r

is the common ratio

n is the term number

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Example 1 For the geometric sequence 4, 12, 36, …, calculate the next two terms.

Create a strategy Determine the common ratio by dividing the second term by the ﬁrst. Then, multiply the last known term by the common ratio to determine the next terms.

Apply the idea The common ratio is r =

= 3. 4th term = 36 × 3 = 108 5th term = 108 × 3 = 324

The next two terms are 108 and 324.

Example 2 In a geometric sequence, t4 = 32 and t6 = 128. Determine the possible value(s) for the common ratio, r.

Create a strategy To get from the 4th term to the 6th term, the common ratio is multiplied twice. Use this relationship to form an equation and solve for r.

Apply the idea The relationship between the terms is t6 = t4 × r2. Write the formula

Substitute the values

Divide both sides by 32

Simplify the fraction

Take the square root of both sides

Evaluate The common ratio could be 2 or − 2.

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4.05 Practice questions Simple familiar Ex 1

Calculate the next two terms for these sequences: a

Ex 2

2, 6, 18, …

12, − 48, 192, …

− 1, 8, − 64, …

− 6, 9,

Determine t5

,…

Consider the ﬁrst four terms of these geometric sequences: i

Evaluate

a c

iii

Evaluate

8, 16, 32, 64, …

− 4, − 8, − 16, − 32, …

2, − 6, 18, − 54, …

− 64, − 16, − 4, − 1, …

Evaluate

Identify the common ratio between the terms of these sequences: a

9, 36, 144, 576, …

2, − 16, 128, − 1024, …

Determine the ﬁrst four terms for a geometric sequence where the ﬁrst term is 7 and the common ratio is − 2.

Determine the ﬁrst four terms in a geometric sequence where the ﬁrst term is 700 000 and the common ratio is 1.04.

Determine the ﬁrst 4 terms in the sequence deﬁned by tn = 4tn − 1, t1 = 0.5.

For the sequence 9, 45, 225, 1125, …, determine the recursive rule for tn in terms of tn − 1, including the initial term t1.

For the sequence 1, 3, 9, 27, … a

Determine the general rule for tn.

Calculate the 15th term of the sequence.

Explain how the common ratio of a geometric sequence can be calculated.

If the ﬁrst term of a sequence is 27 and the common ratio is , calculate the 10th term.

For the sequence − 0.3, − 1.5, − 7.5, − 37.5, …, determine the formula for the nth term.

Complex familiar 12

A geometric sequence is defined by the recursive rule tn = 2tn − 1. If the fifth term, t5, is 32, determine the first term, t1.

The ﬁrst term of a geometric sequence is 7, and the fourth term is 875: a

Determine the common ratio, r, of this sequence.

Determine the recursive rule for tn + 1 in terms of tn, including the initial term t1.

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4.06 Geometric sequences with tables and graphs After this lesson, you will be able to… • represent a geometric sequence using a table of values and a graph • describe the shape of the graph of a geometric sequence • classify a geometric sequence as representing exponential growth or decay based on the value of its common ratio • determine the rule for a geometric sequence from its graph

Geometric sequences with tables and graphs The explicit rule for a geometric sequence, tn = t1 r n − 1, is an exponential function. Unlike arithmetic sequences, which have a linear graph, the points of a geometric sequence follow an exponential curve. Given a rule, a table of values can be generated. For the rule tn = 12 × (1.5)n − 1, the ﬁrst few terms are: n

40.5

The shape of the graph depends on the value of the common ratio, r. • If r > 1, the terms increase at an accelerating rate. This models exponential growth. • If 0 < r < 1, the terms decrease towards zero. This models exponential decay. • If r is negative, the terms alternate in sign, creating a ‘zig-zag’ pattern on the graph. These graphs compare the sequence tn = 12 × (1.5)n − 1 (growth) with tn = 12 × (− 1.5)n − 1 (zig-zag). 140 120 100 80 60 40 20 0 −20 −40 −60 −80 −100 −120 −140

Value of nth term

n 1

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Interactive exploration Discover this concept in action online

mathspace.co

Example 1 The nth term of a geometric sequence is given by the equation tn = 2 × 3n − 1: a Complete a table of values for the ﬁrst four terms of the sequence.

Apply the idea Substitute n = 1, 2, 3, and 4 into the rule: n

b Plot the points from the table on a graph.

Apply the idea 55 tn 50 45 40 35 30 25 20 15 10 5 0

n 1

c If the points on the graph were joined, would they form a straight line or a curve?

Apply the idea The points do not form a straight line. They lie on an exponential curve, which is characteristic of a geometric sequence. Therefore, the points lie on a curve.

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4.06 Practice questions Simple familiar 1

The plotted points represent terms in a geometric sequence. Explain the shape of the graph.

tn 6 4 2

−2 −4

Consider this sequence plot: a

Identify the ﬁrst ﬁve terms of the sequence.

Is the sequence arithmetic or geometric? Explain your answer.

11 10 9 8 7 6 5 4 3 2 1

n 0

Ex 1

The table of values represents terms in a geometric sequence. Determine the general rule.

For the recursive rule tn = 2tn − 1, t1 = 5, plot the ﬁrst four terms on a graph.

Given the nth term of these geometric sequences:

320

Complete the table of values.

Determine the common ratio.

iii

Plot the points in the table that correspond to n = 1, n = 2, n = 3 and n = 4 on a Cartesian plane.

Identify whether the joined points would form a straight line, a zig-zag pattern, a parabola, or an exponential curve.

tn = 3 × 2n − 1

tn = 6 × (− 2)n − 1

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For the sequence given by the recursive rule tn + 1 = 3tn, t1 = 5, determine the explicit rule for tn.

A geometric sequence is deﬁned by tn = 100 × (0.8)n − 1. Is this a model of exponential growth or decay? Justify your answer.

A geometric sequence has a ﬁrst term of 90 000 and a common ratio of 1.11. Determine the 6th term, rounded to three decimal places.

The value of Beirut Bank shares is decreasing by 5% each day. At the beginning of today’s trading, the shares are worth $42.54. Determine a recursive rule for Vn + 1 in terms of Vn, which deﬁnes the value of the shares at the end of day n, and an initial condition for V0.

In a laboratory, an antibiotic is tested on a sample of 5 bacteria. The bacteria are observed to double each hour. Determine the equation linking the number of bacteria (B) and number of hours passed (h).

Complex familiar Ex 2

For the graph shown: a

Determine the general rule for tn.

Determine the 8th term.

tn 125 100 75 50 25

n 0

Consider the sequence: − 54, − 18, − 6, − 2, … a

Determine the general rule for tn.

Determine t9, the 9th term of the sequence.

The ﬁrst term of a geometric sequence is 6, and the third term is 96: a

Determine the possible values of the common ratio, r.

Determine the recursive rule for tn that uses the positive common ratio.

The third term of a geometric sequence is 7500, and the seventh term is 12: a

Determine the possible values of the common ratio, r.

Determine the ﬁrst term of this sequence.

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Complex unfamiliar 15

Two friends, Ali and Ben, are tracking their investments. Ali’s investment value after n years is given by the rule An = 250 × (1.1)n − 1. Ben’s investment value is represented by the ﬁrst four terms plotted on the graph:

400

Determine the explicit rule for the value of Ben’s investment, Bn, after n years.

200

Whose investment is worth more after 5 years? Show your calculations.

100

Using a calculator, determine the ﬁrst year in which Ben’s investment value will exceed Ali’s investment value.

The graph shows the ﬁrst three terms of a geometric sequence tn: a

A new sequence, An, is created by squaring each term of tn. Is An a geometric sequence? Justify your answer by determining its common ratio. Another new sequence, Bn, is created by adding 10 to each term of tn. Is Bn a geometric sequence? Justify your answer.

Value ($)

450 350 300 250 150 50

n (Year)

35 30 25 20 15 10 5

n 0

A pendulum is released and its maximum displacement from the centre is recorded on each swing. The graph shows these displacements, where a positive value is a swing to the right and a negative value is a swing to the left:

Determine the recursive rule that models the pendulum’s displacement, Dn, on the nth swing.

−20

On which swing will the absolute displacement of the pendulum ﬁrst be less than 5 cm?

−60

Displacement (cm)

40 20 0

n (Swing) 1

−40 −80

The population of a town is 8000 and is growing at a rate of 5% per year. This growth can be modelled by the rule tn = 8000 × (1.05)n − 1, where tn is the population in the n-th year. Determine the ﬁrst year in which the population will exceed 12 000.

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4.07 Further applications of growth and decay After this lesson, you will be able to… • model practical situations involving a constant percentage change (geometric growth/decay) using geometric sequences • calculate the common ratio (r) from a given percentage increase or decrease • apply the rule tn = t0 × r n to solve problems involving population growth and diminishing-value depreciation • calculate the future value of an item or estimate the time required for it to reach a specific value in a geometric model

Model geometric growth Geometric growth occurs when a quantity is repeatedly multiplied by a common ratio, r, that is greater than 1 (r > 1). This models situations where a quantity increases by a constant percentage or factor over time. Common applications include population growth and compound interest. To calculate the common ratio for a percentage increase, convert the percentage to a decimal and add it to 1. For example, a growth of 5% per year means the value is 100% + 5% = 105% of the previous year’s value. Therefore, the common ratio is r = 1.05. When modelling applications, it is often useful to use an initial value, t0, and the explicit rule tn = t0 × r n. In this form, n directly corresponds to the number of time periods after the start.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 Suppose an amount of $1 is saved on the ﬁrst day of a month, $2 on the second day, $4 on the third day, and so on, with the amount saved doubling each day. Calculate the amount that will be put aside for savings on the 17th day of the month.

Create a strategy This is a geometric sequence where the term number corresponds to the day. Use the formula tn = t1 r n − 1.

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4.07 Practice questions Simple familiar Ex 1

Ex 2

Calculate the amount that will be put aside for savings on the 12th day.

Calculate the amount that will be put aside for savings on the 30th day.

A bacteria colony doubles in size every hour. If there are 1000 bacteria now, calculate the number of bacteria that will be present after 4 hours.

Computer equipment bought for $8000 depreciates by 25% annually:

Ex 3

An initial saving of $1 is made on the ﬁrst day of a month, with $2 saved on the second day, $4 on the third day, and so on, with the amount saved doubling each day:

Calculate its value at the end of the 1st, 2nd, and 3rd years.

Calculate the percentage of its original value that remains after 3 years. Round to two decimal places.

The government plans to decrease its spending on a programme from an initial $11 million by 6.85% each year: a

Determine the common ratio for the sequence of the government’s spending.

Calculate the government’s spending in the 6th year, rounding to the nearest dollar.

The population of some native bees is declining at a rate of 10% per year: a

Determine an explicit rule for the population Pn after n years. Let P0 be the starting population.

If there are 23 700 bees in a hive now, calculate the number of bees that will be present in 6 years.

The average annual rate of inﬂation is 2.6%. A loaf of bread costs $3.65 in 2015: a

Calculate the cost of the bread in 2016.

At this rate, calculate the cost of the bread in 2018.

A ball dropped from a height of 21 metres will bounce back off the ground to 50% of the height of the previous bounce. Calculate the height of the ﬁfth bounce, rounded to two decimal places.

In an image, the height of a waterfall is 35 mm. After a zoom function is applied once, the height is 66.5 mm. After a second application, its height is 126.35 mm. Determine the factor by which the image is enlarged each time the zoom is applied.

This year, 640 people are expected to enter the workforce as registered nurses. This number is expected to increase by 2% each year. Calculate the number of nurses expected to enter the workforce 6 years from now. Round your answer to the nearest whole number.

Luigi starts his career with a monthly wage of $5400. At the beginning of each year, he receives a raise of 2.2% on his previous year’s wage. Calculate his monthly wage in his third year of service.

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Complex familiar 11

A factory machine bought for $50 000 loses 15% of its value each year: a

Determine an explicit rule to represent the machine’s value, Vn, after n years.

Determine the number of full years after which the value will ﬁrst fall below $20 000.

Average annual salaries are expected to increase by 5% each year. If the average annual salary this year is $49 000: a

Calculate the expected average annual salary in 4 years’ time.

Calculate the expected average annual salary in the 4th year of the new plan.

A phone battery loses 20% of its charge every hour. It is initially at 100%: a

Determine an explicit rule for the battery charge, Bn, after n hours.

Determine the number of full hours after which the battery percentage will ﬁrst drop below 40%.

Initially, there are 14 bacteria in a body. The population quadruples every four hours: a

Calculate the number of bacteria that will be present after 24 hours.

An antibiotic is applied after 24 hours, which kills half of the bacteria every four hours. Calculate the number of bacteria that will be left 24 hours after applying the antibiotic. Round to the nearest whole number.

Complex unfamiliar 15

An online business sold 200 pairs of shoes in its ﬁrst quarter. In the second quarter, it sold 800 pairs, and in the third, 3200 pairs. Assume this rate of growth continues: a

Determine an equation for the number of sales, Sn, in the nth quarter.

Use the formula to predict the number of sales they will have in the 9th quarter (2.25 years after starting).

On the release date, movie ticket sales were $275 000. For the first 10 days, sales increased by 5% per day. Calculate the value of ticket sales on the 10th day.

A gym trainer provides two progression plans based on Monday’s workout: MONDAY: 9 rope skips. • PLAN A (BEGINNER): Each day, increase the number of skips by

of the ﬁrst day’s total.

• PLAN B (INTERMEDIATE): Each day, increase the number of skips by day’s total.

of the previous

Calculate the number of rope skips that must be completed on Wednesday for each plan. Round to the nearest whole number where necessary.

Which level of training plan is more realistic for long-term, sustainable improvement? Justify your answer.

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Part A: Straight-line depreciation Model the value of the car using the straight-line method. • Determine a recursion relation for the value of the car Vn. • Determine the explicit rule for the value of the car, Vn, after n years. • Complete a table of values for the car’s value for the ﬁrst 8 years of ownership (i.e., for n = 0, 1, …, 8). Part B: Reducing balance depreciation Model the value of the car using the reducing balance method. • Calculate the common ratio, r, for the 15% depreciation rate. • Determine a recursion relation for the value of the car, Vn. • Determine the explicit rule for the value of the car, Vn, after n years. • Complete a table of values for the car’s value for the ﬁrst 8 years of ownership. Round the values to the nearest cent. 35 000

Value ($)

30 000 25 000 20 000 15 000 10 000 5000 0

Straight-line

Years

Reducing balance

Part C: Comparing the models Using either graph paper or spreadsheet software, plot the values from both depreciation tables onto a single set of axes. Use n (Years) for the horizontal axis and Value ($) for the vertical axis.

Investigate 1.

Describe the shape of the graph for each depreciation method. (e.g., linear, exponential curve).

What is the ‘scrap value’ (the value of the car after 8 years) under each model?

Use your explicit rules to calculate the value of the car after 4 years for both methods.

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4 Chapter review SF

Consider the sequence 10, 4, − 2, − 8, − 14, …. If tn represents the nth term, identify the value of t5.

A sequence is deﬁned by the recursive rule tn = tn − 1 + 2tn − 2, with initial terms t1 = 1 and t2 = 3. Determine the ﬁrst four terms of the sequence.

An arithmetic sequence is given by 5, 9, 13, 17, …. Identify the ﬁrst term, t1, and calculate the common difference, d.

An arithmetic sequence has a ﬁrst term of 50 and a common difference of − 4. Determine the explicit rule for tn.

An arithmetic sequence has its fourth term equal to 11 and its ninth term equal to 26: a

Determine the rule for the nth term, tn.

Determine the ﬁrst term in the sequence to exceed a value of 50.

The terms k − 1, 2k + 3, and 5k − 2 are three consecutive terms of an arithmetic sequence. By forming an equation, calculate the value of k.

This table shows the ﬁrst four terms of an arithmetic sequence. Determine the recursive rule for this sequence: n

−5

−2

Sarah starts a savings plan. In the ﬁrst week, she saves $20. Each subsequent week, she saves $5 more than the previous week. Calculate the amount she will save in the 10th week.

A geometric sequence is given by 100, 50, 25, 12.5, …. Identify the ﬁrst term, t1, and calculate the common ratio, r.

A geometric sequence has a second term of 40 and a ﬁfth term of 5: a

Determine the rule for the nth term, tn.

Determine the ﬁrst term in the sequence that has a value of less than 1.

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5.01 Latitude and longitude After this lesson, you will be able to… • understand the meaning of angles of latitude and longitude (in decimal degrees, and degrees and minutes) in relation to the equator and the prime meridian respectively • locate positions on Earth’s surface given latitude and longitude, for example, using a globe, map, GPS and other digital technologies • state latitude and longitude for positions on Earth’s surface, for example, investigating a map of Australia and locating boundary positions for Aboriginal peoples’ and Torres Strait Islander peoples’ language groups, Australian landmarks or local land boundaries

Latitude and longitude To ﬁnd the exact location of any place on Earth, a special coordinate system called latitude and longitude is used. This system works like an invisible grid wrapped around the globe. These coordinates are essential for navigation and are used in tools like GPS and digital mapping systems, such as Google Maps. They are represented as a coordinate pair (latitude, longitude) such as (27.5°S, 154.3°E). Latitude

Longitude

Latitude lines are horizontal circles whose radius decreases with distance from the equator towards the poles. In coordinate systems (like GPS), latitude is the ﬁrst number in an ordered pair. The equator is the starting point for measuring latitude and is marked as 0° latitude. They are parallel to each other and never meet, not even at the poles.

Longitude lines are vertical lines on a map and meet at the poles, which can also be called meridians. In coordinate systems like GPS, longitude is the second number in the ordered pair. The prime meridian is the starting point for measuring longitude. It runs through Greenwich, England and is marked as 0° longitude.

Latitude is measured from 0° at the equator to +90° at the North Pole and −90° at the South Pole. They are always written with a degree value between 0 and 90. Example: Cairns, QLD is located at about 17° South, meaning it is 17 degrees south of the equator.

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Longitudes are measured between 0° and 180° in the direction from east or west of the prime meridian. Example: Brisbane, QLD is located at about 153° East, meaning it is 153 degrees east of the prime meridian.

By combining a latitude and longitude coordinates, such as (33.8523°S, 151.2108°E) for the Sydney Harbour Bridge, any point on Earth can be pinpointed accurately on maps, globes, or digital devices. These coordinates enable accurate positioning on globes, maps, and digital technologies, supporting applications like mapping Australian landmarks or delineating boundaries, such as those of Aboriginal and Torres Strait Islander language groups.

Exploration Using an online map or globe (e.g. Google Earth or a physical globe), ﬁnd and record the coordinates of these cities in both decimal degrees and degrees/minutes: City

Latitude

Longitude

Sydney

⬚

Tokyo Nairobi London

⬚

Example 1 Identify which of these are examples of great circles on Earth. Explain your reasoning for each: a The equator

Create a strategy Recall the deﬁnition of a great circle. A great circle is the largest circle that can be drawn on a sphere, and its centre is the same as the centre of the Earth.

Apply the idea The equator is a great circle. It is the line of latitude at 0° and is the only line of latitude that is a great circle.

b The Tropic of Capricorn (a line of latitude at 23.5°S)

Create a strategy Consider if this line of latitude has its centre at the centre of the Earth.

Apply the idea The Tropic of Capricorn is not a great circle. It is a line of latitude that is south of the equator, so its centre is not at the Earth’s centre. It is an example of a small circle.

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Create a strategy Using the map, identify the number of degrees corresponding to the vertical and horizontal lines respectively for Darwin.

Apply the idea Longitude refers to the vertical lines on the map. Darwin has a longitude corresponding to 130°E. Latitude refers to the horizontal lines on the map. Darwin has a latitude corresponding somewhere between 10°S and 15°S. Therefore, the correct answer is Option A.

Example 4 Convert between decimal degrees and degrees/minutes format: a Convert the latitude 27.5°S to degrees and minutes.

Create a strategy The whole number part is the degrees part. To convert the decimal part to minutes, multiply it by 60.

Apply the idea The degrees part is 27°. The decimal part is 0.5. Minutes = 0.5 × 60 = 30′ So, 27.5°S is equal to 27°30′S.

b Convert the longitude 153°2′E to decimal degrees.

Create a strategy The value before the degree symbol is the full degrees. To convert the minutes part to decimal degrees, divide it by 60.

Apply the idea The degrees part is 153°. The minutes part is 2′.

So, 153°2′E is approximately equal to 153.03°E.

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Ex 2

This world map shows the lines of longitude and latitude. Several locations are marked on the map: 90°

75°

90°

75°

60°

45° 30° 15° 0°

60°

45° 30°

B 150° 120°

15°

90°

30°

60°

0°

30°

60°

90°

120°

15° 30°

45°

45° 60°

60°

J 75°

0°

30°

Ex 3

15°

150° 180°

90°

Identify the point that is located on the prime meridian.

Determine the point which has coordinates of (60°S, 60°W).

Determine the coordinates of point I.

How many degrees north is point I from point B?

How many degrees east is point I from point F?

Identify the location that has the same latitude as C.

Identify the location that has the same longitude as F.

75°

Determine the latitude and longitude, rounded to the nearest degree, of each of the labelled locations on this map: 110° E

15° S

115° E

120° E 125° E

30° S

150° E 155° E 10° S 15° S

La Grange

20° S

25° S

140° E 145° E

Darwin

20° S

Alice Springs Brisbane Menzies Perth

25° S

30° S Adelaide Melbourne

35° S

Sydney Canberra

35° S

125° E 130° E 135° E 140° E

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This table shows the latitudes and longitudes of several countries: Determine which of these cities is closest to: a

South Pole

Equator

This table shows the latitudes and longitudes of several countries: Determine which of these cities is closest to:

Ex 4

Nazareth

33°N

35°E

Bilbao

43°N

3°W

Swakopmund

23°S

15°E

Nelson

41°S

173°E

Whakatane

38°S

177°E

Port Douglas

16°S

145°E

City

Latitude

Longitude

Zagreb

46°N

16°E

Comallo

41°S

70°W

Line of latitude at 10°S

Eden

37°N

80°W

San José

10°N

84°W

Nagele

5°N

40°E

Lobamba

26°S

31°E

City

Latitude

Longitude

Bangkok

14°N

101°E

This table shows the latitudes and longitudes of several cities: a

Line of longitude at 80°E

Paris

49°N

2°E

Line of longitude at 175°W

Wellington

41°S

175°E

Suva

18°S

178°E

Perm

58°N

56°E

Salem

37°N

80°W

Amsterdam is located at coordinates (52° N, 5° E). Determine whether these locations are further south than Amsterdam: a

Beijing: (40° N, 116° E)

London: (51° N, 0° W)

Dublin: (53°N, 6°W)

Copenhagen: (55° N, 12° E)

Convert these latitudes from decimal degrees to degrees and minutes: 35.5°S

51.25°N

Convert these longitudes from degrees and minutes to decimal degrees: a

Longitude

Line of latitude at 40°N

a 14

Latitude

Determine which of these cities is closest to:

City

145°30′ E

112°15′ W

Consider the city of Pyongyang at latitude 39°N and longitude 126°E. Determine the latitude and longitude of: a

20° north of Pyongyang

20° south of Pyongyang

20° east of Pyongyang

20° west of Pyongyang

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Consider the city of Melbourne at latitude 38°S and longitude 145°E. Determine the latitude and longitude of:

30° north of Melbourne

30° south of Melbourne

30° east of Melbourne

30° west of Melbourne

Consider the city of Aloﬁ at latitude 19°S and longitude 170°W. Determine the latitude and longitude of:

Ex 5

15° north of Aloﬁ

15° south of Aloﬁ

15° east of Aloﬁ

15° west of Aloﬁ

Consider the city of Springbok at latitude 30°S and longitude 18°E. Determine the latitude and longitude of: a

39° north of Springbok

39° south of Springbok

39° east of Springbok

39° west of Springbok

Determine the coordinates of town A for each of these locations: a

Town A is 8° south of town B and town B’s position is (20°S, 155°E).

Town A is 6° west of town B and town B’s position is (21°N, 140°E).

Town A is 7° south and 10° west of town B and town B’s position is (25°N, 154°E).

The location of city A is (35°N, 173°W) and the location of city B is (44°N, 160°W): a

Calculate the difference in latitude between the pair of cities.

Determine which city is further south.

Calculate the difference in longitude between city A at (23°S, 22°W) and city B at (17°S, 72°E).

Complex familiar 22

Convert the coordinates (44°26′N, 26°06′E) to decimal degrees, rounded to two decimal places, and identify the city located at these coordinates.

A ship starts at a position of (15°45′S, 160°30′E). It travels 10 degrees due north. Determine its new position in degrees and minutes.

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Patricia is going to travel to various destinations, and begins her journey at point D shown on the map: 75°

90°

90° A

60°

75° 60°

45° 30° 15° 150° 120° 90° 60° 30° 0°

B 0° 30°

60°

45°

30°

15°

30°

30° 45° 60°

45° 60° 75°

90°

75°

Identify the coordinates of her starting position, D.

Her change in latitude and longitude after each part of her journey is represented in the table given. Complete the table to determine the coordinates of her starting and ﬁnishing positions after each part of her journey:

15° 0°

90° 120° 150° 180°

Part of journey

Starting position

Change in latitude

Change in longitude

Finishing position

(45° N, 75° E)

9° N

7° W

⬚

12° S

16° E

⬚ ⬚

Calculate the differences in longitude and latitude between her position at the end of the second part of her trip, and point A.

The expression “travel to the ends of the Earth” suggests travelling to the farthest places. Determine the coordinates of a point on Earth that is the greatest possible distance from the point with coordinates (67°S, 99°E).

Complex unfamiliar 26

Using a map of Australia, identify and locate these three Aboriginal language groups: the Warlpiri (Northern Territory), Noongar (Western Australia), and Yidinji (Queensland). For each language group: a

Provide the precise latitude and longitude of a named central location within their traditional lands (e.g., a town or cultural site).

Describe one speciﬁc cultural practice or historical event tied to the language group’s connection to their land, referencing a veriﬁable source or example.

Analyse how one specific geographic feature (e.g., a river, mountain range, or desert) defines or influences the extent of their traditional territory, including its role in their cultural practices.

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5.02 Distance between two places on Earth After this lesson, you will be able to… • calculate angular distance and distance between two places on Earth on the same meridian • calculate angular distance and distance between two places on Earth on the same parallel of latitude • solve practical problems involving latitude, longitude, angular distance and distance

Angular distance Interactive exploration Discover this concept in action online

mathspace.co

Angular distance The angle formed by two points on a sphere, such as Earth, with respect to its centre, measured in degrees. Lines of longitude, running north-south, measure angular distance relative to the prime meridian, while lines of latitude, running east-west, measure angular distance relative to the equator. These coordinates provide a system for locating points on Earth’s surface.

For points with the same longitude, the angular distance between two points sharing the same longitude depends on their latitudinal positions. Two cases arise: • Both points on the same side of the equator: The angular distance is the difference between the latitudes. • Points on opposite sides of the equator: The angular distance is the sum of the values of the latitudes. For points with the same latitude, the angular distance between two points sharing the same latitude depends on their longitudinal separation. Two cases are considered: • Both points in the same longitudinal hemisphere: The angular distance is the difference between the longitudes. • Points in opposite longitudinal hemispheres: The angular distance is the sum of the values of the longitudes. If this results in a reﬂex angle (greater than 180°), subtract it from 360° to ﬁnd the smaller, relative angle.

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These calculations are based on a simpliﬁed spherical model of the Earth and are valid for points that lie along the same longitude or latitude, where the geometry can be represented in a twodimensional plane. It is important to note that angular distance is always expressed as a positive value, representing the absolute difference between two positions. These diagrams show each case of angular distance between two points on the same longitude. Similar diagrams can be constructed for points with the same latitude. Case 1: A B

The angular

distance between A and B is the angle between these two lines.

Both positions lie on the same side of the equator, the angular distance between the two positions will be the difference between the latitude of A and the latitude of B.

Both positions lie on opposite sides of the equator, the angular distance between the two positions will be the sum of the latitude of A and the latitude of B.

Case 2: A The angular

distance between A and B is the sum of the two angles.

Example 1 Determine the angular distance between Bali (8°S, 115°E) and Mandurah (33°S, 115°E), rounded to the nearest degree.

Create a strategy The positions both lie on the same longitude, and are both on the same side of the equator, the angular distance can be calculated by subtracting the latitudes from one another.

Apply the idea

Reﬂect and check

Angular distance = 33° − 8° Subtract the values

It shows the angular separation in degrees between the two places on Earth — not the actual distance in kilometres.

= 25°

Evaluate

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Substituting the appropriate values into the arc length formula gives:

This can be simpliﬁed by evaluating the terms which do not change. Since, simpliﬁed formula is:

= 111.2, the

D = 111.2 cos θ × Angular distance where D is the distance in kilometres, θ is the latitude of the small circle and the angular distance is the angle between the two points. It is important to note that this is not the shortest distance between the two points, that would be the great circle distance between the two points. This is the distance along the line of latitude between the two points.

Example 3 Ned wants to ﬂy from Newcastle (32.4°S, 151.8°E) to Perth (32.4°S, 115.5°E) and then on to Denpasar (8.6°S, 115.5°E): a Determine the distance between Newcastle and Perth travelling along the parallel.

Create a strategy The formula for distance along a parallel of latitude is used: D = 111.2 cos θ × Angular distance

Apply the idea • The latitude angle for this parallel is θ = 32.4°. • The angular distance is the difference of the longitude coordinates of both cities since they are both east of the prime meridian, Angular distance = 151.8 − 115.5 = 36.3°. D = 111.2 cos θ × Angular distance

Write the formula

= 111.2 cos (32.4°) × 36.3°

Substitute the values

= 3408 km

Evaluate

b Determine the distance between Perth and Denpasar, rounded to the nearest kilometre.

Create a strategy The formula for distance along a meridian is used: D = 111.2 × Angular distance

Apply the idea The angular distance is the difference of the latitude coordinates of both cities since they are both south of the equator, Angular distance = 32.4 − 8.6 = 23.8°. D = 111.2 × Angular distance

352

Write the formula

= 111.2 × 23.8

Substitute the values

= 2647 km

Evaluate

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Consider this diagram of Earth: 90°N

75°N 60°N 45°N 30°N

135°120° W W 105°90° W W 75° 60° 45° 30° 15° W W W W W

15°N Equator 15° E 15°S 30°S

45°S 60°S 90°S

Ex 1

Ex 2

Two locations on Earth have the coordinates (30°S, 70°E) and (70°S, 70°E). Do they lie along the same meridian or the same parallel?

Do (10°S, 70°E) and (10°S, 30°E) lie along the same longitude or the same latitude?

Does a parallel form a great circle or small circle?

Does a meridian form half of a small circle or half of a great circle?

Determine the angular distance between: a

Bali (8°S, 115°E) and Mandurah (33°S, 115°E)

Points on Earth with coordinates of (32°N, 55°E) and (29°N, 55°E)

The coordinates of two cities A and B are (32°S, 84°E) and (22°S, 84°E). a

Determine whether each is a great circle that passes through cities A and B: i

354

75°S

10°S

84°E

iii

32°S

54°S

Determine the angular distance between the two cities.

Determine the shortest distance between the two cities to the nearest kilometre, given that the radius of the Earth is 6371 km.

The angular distance between points A and B on the Earth’s surface is 148°: a

The radius of the Earth is 6371 km to the nearest kilometre. Determine the radius of the Earth to the nearest hundred kilometres.

Calculate the arc length AB to the nearest kilometre.

A sailor began a journey from position (27°N, 146°E) and continued along the same line of longitude until he arrived at an island with a latitude of 18°N: a

Determine the angular distance he covered.

Determine the distance he covered to the nearest kilometre, given that the radius of the Earth is 6371 km.

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Consider the two points with coordinates of (39°N, 121°E) and (24°S, 121°E). Using the approximation that the radius of the Earth is 6371 km, determine the distance between them to the nearest kilometre.

Determine the distance between these points on the surface of the Earth. Assume the radius of the Earth is 6371 km. Round your answers to the nearest kilometre: a

A (3°N, 42°E) and B (4°N, 42°E)

A (4°N, 75°W) and B (36°N, 75°W)

A (25°S, 62°E) and B (20°S, 62°E)

A (11°S, 132°W) and B (36°N, 132°W)

The Arctic Circle is at a latitude of 66.5°N. Given that the Earth has a radius of approximately 6371 km, what is the shortest distance, to the nearest kilometre, from any point on the Arctic Circle to: a

The equator

The Antarctic Circle, which is at a latitude of 66.5°S

Consider the two locations A (25°N, 70°E) and B (25°N, 150°E): a

Do the two locations A and B lie on the same meridian or parallel?

Does a parallel form a great circle or small circle?

How far east of point A is point B? Give your answer in degrees.

Determine the distance between the two locations along the parallel, rounded to the nearest kilometre.

Calculate the distance between these points along their common parallel. Round your answers to the nearest kilometre: a

A (85°S, 25°W) and B (85°S, 60°W)

A (15°N, 35°W) and B (15°N, 70°W)

A (55°S, 45°E) and B (55°S, 165°W)

A (65°N, 55°E) and B (65°N, 150°W)

A (66°15′ S, 78°36′ E) and B (66°15′ S, 32°12′ E)

A (54°39′ N, 25°42′ W) and B (54°39′ N, 34°6′ E)

Consider the points A (40°S, 40°E) and B (40°S, 75°E). Determine the distance between the two positions along their common parallel, rounded to the nearest kilometre.

Consider the points A (20°S, 40°W) and B (20°S, 85°W): a

Determine the angular distance between the two points.

Determine the distance between the two positions along their common parallel, rounded to the nearest kilometre.

Consider the point A (40°N, 10°W): a

Calculate the radius of the circle which forms the (40°N) parallel. Assume the Earth’s radius equals 6371 km. Round your answer to two decimal places.

Determine the distance that point A is from the prime meridian along the parallel, rounded to the nearest kilometre.

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Complex familiar Ex 3

Lucas wants to ﬂy from Brisbane (27.5°S, 153.0°E) to Geraldton (27.5°S, 114.6°E) and then on to Jakarta (6.2°S, 114.6°E): a

Determine the distance between Brisbane and Geraldton travelling along the parallel. Give the answer rounded to the nearest kilometre.

Determine the distance between Geraldton and Jakarta, rounded to the nearest kilometre.

Determine an estimate for the total ﬂight time for the journey if the plane travels at a speed of 810 km/h. Use the distances found in parts (a) and (b) and give your answer in hours, rounded to two decimal places.

Luke is travelling between Nanjing (32°N, 118°E) and Port Hedland (20°S, 118°E). Determine the time it will take to ﬂy between the two cities if the plane travels at a speed of 610 km/h. Give your answer in hours rounded to two decimal places.

Ben wants to know how long it will take him to travel between Sydney, Australia (34°S, 151°E) and Cape Town, South Africa (34°S, 18°E). Determine the time it will take to ﬂy between the two cities if the plane travels at a speed of 670 km/h travelling along the same parallel of latitude. Round to two decimal places.

Consider the two points with coordinates of (9°S, 58°W) and (8°N, 58°W). Assume the radius of the Earth is 6371 km. Determine the time it will take to ﬂy between the two points if the plane travels at a speed of 640 km/h. Give your answer in hours, rounded to two decimal places.

A hiker is orienteering and travels due south for 277.5 km. How many degrees south has its coordinate position changed? Rounded to one decimal place.

The coordinates of two cities A and B are (43°N, 30°W) and (37°S, 30°W) respectively:

Identify which city, A or B, is closer to the prime meridian, or if they are the same distance from it.

Determine the shortest distance between the two cities to the nearest kilometre, using 6400 km as the radius of the Earth.

A more accurate radius of the Earth is 6371 km. Determine the shortest distance between the two cities using this radius to the nearest kilometre.

Calculate the difference between these two measurements of the shortest distance between the cities.

The coordinates of the Devil’s Sea are (58°N, 170°E). A line is drawn from this position through the centre of the Earth to a point B on the opposite side. Determine the shortest distance along the Earth’s surface between the Devil’s Sea and point B, given that the radius of the Earth is 6371 km.

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Calculate the shortest distance, D, for each of the pairs of points. Round your answers to the nearest kilometre: a

A (20°N, 60°E) and B (20°S, 60°E)

A (66°35′ S, 42°40′ W) and B (12°5′ S, 42°40′ W)

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A (42°15′ N, 36°55′ W) and B (30°5′ S, 36°55′ W)

Determine the circumference of a great circle around a sphere with a diameter of 4 km. Round your answer to two decimal places.

Determine the circumference of a small circle with a radius of 2 m that lies on a sphere with a radius of 3 m. Round your answer to two decimal places.

If Mercury has a radius of 2878 km, determine the circumference of a great circle that lies on its surface. Round your answer to one decimal place.

Brisbane is located on the 27.5°S parallel and Melbourne is located on the 37.8°S parallel. Assume 6371 km is the Earth’s radius: a

Determine the distance a ﬂight from Melbourne would travel non-stop around the globe on the 37.8°S parallel before landing back in Melbourne, rounded to the nearest kilometre.

If a plane travels at 390 km/h, how long will a similar trip departing from Brisbane take? Round your answer to two decimal places.

Complex unfamiliar 28

Two travellers, Ava and Leo, are flying from Darwin (12.4°S, 130.9°E) to Suva (18.1°S, 178.4°E), but take different routes. • Ava flies from Darwin directly east along the parallel to the longitude of Suva, then south along the meridian to Suva. • Leo flies from Darwin directly south along the meridian to the latitude of Suva, then east along the parallel to Suva. Both planes travel at the same constant speed of 900 km/h. Who has the shorter total flight time? Justify your answer.

A pilot is planning a flight from Santiago, Chile (33.5°S, 70.7°W) to Johannesburg, South Africa (26.2°S, 28.0°E). A navigation system estimates the great-circle distance between the two cities to be approximately 6850 km, which would represent the shortest possible path over the surface of a spherical Earth. However, due to airspace restrictions and strong prevailing winds, the airline instead follows a two-leg journey: first flying due east along the 33.5°S, then turning north to Johannesburg along a meridian. Determine by how much the chosen route significantly increases the total distance compared to the great-circle route. Comment on how much time is lost if the plane cruises at 900 km/h.

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Common types of geocache hunts include: • Traditional cache is the most common type and consists of a container with a logbook. The exact coordinates where the cache is located are given. • Multi-cache consists of one or more stages, each of them containing the coordinates for the next one; the final stage contains a physical container with the logbook. • Offset cache is a type of multi-cache in which the initial coordinates are for a location containing information that encodes the final cache coordinates. An example would be to direct the finder to a plaque where the digits of a date on the plaque correspond to the coordinates of the final cache. • Mystery/puzzle caches require one to discover information or solve a puzzle to find the cache. • Moving/travelling caches are found at a listed set of coordinates. The finder hides the cache in a different location, and updates the listing, essentially becoming the hider, and the next finder continues the cycle.

Let’s first find a few caches in your area. This will give you some ideas about what makes a great cache and how the system works. A great place to start is Geocaching Australia (geocaching.com.au). You will need to create an account. Once you have an account you can search for caches near you. Symbols on the pins represent the type of cache located. Click on a location pin and follow the link to find out more information about the cache such as difficulty, logged finds and clues. For a traditional cache, the exact coordinates will be given and this can be copied into Google Maps or similar to help you locate the cache. As a start, try traditional caches that are low in difficulty. There are also geocaching apps available for phones that you can investigate.

Activity 2: Design a Geocache Hunt Create Your Own Hunt Let’s create our own multi-stage geocache hunt around your school. We will not use the geocaching website to log these hunts as we do not want to create public caches on school grounds. We will design the caches to be located by other students in your class or in other classes at your school for a limited time, so we don’t have to maintain the cache. Let’s create a 3-stage puzzle hunt. The cache at the ﬁrst location should give a puzzle with an answer to unlock the coordinates for the second location. The cache at the second location should give a puzzle with an answer to unlock the coordinates to locate the ﬁnal cache. The code can be given to the hunt creator as proof you found the cache and they can then give the next puzzle or final cache treasure box.

Break up into groups of 3 − 5 people and discuss what you will use as cache containers and what some great things to put in the ﬁnal ‘cache’ as a reward would be. Discuss where some great places would be to hide geocaches around the school. Remember we need 3 places. How could you disguise the cache so that it is trickier to ﬁnd? Investigation: Geocaching mathspace.co

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What puzzles will you use at locations 1 and 2? These should give a one-number or word answer that can be used to unlock a file with the coordinates for the next location. Perhaps a mathematical puzzle? Hide your geocaches and record the exact coordinates. For locations on a small scale such as this, you will need to record many decimal places for the latitude and longitude. These can be found using a GPS device or Google Maps. For example, dropping a pin in Google Maps (Right-click and select “What’s here?”) will show the location with 6 decimal places. Be sure to zoom in to select the location with a high degree of accuracy if using Google Maps. Put the exact coordinates in three separate files and lock file 2 and 3 with a password (the answer to your puzzles). Most documents such as Word, PDF or OneNote have the ability to password-protect. These should be placed on a shared location or emailed to the students involved in the hunt. Once set up, exchange the coordinates of the first location with another group and HAPPY HUNTING.

Investigate After designing your hunt, answer these questions: 1.

Present the coordinates for your three cache locations in decimal degrees, written to at least six decimal places.

Describe the puzzle you designed for location 1 and explain how its solution provides access to the coordinates for location 2.

Reﬂect on the challenges your group faced in accurately determining the coordinates for your locations. What steps did you take to ensure precision?

Extension Incorporate Distance Use the coordinates of your cache locations to determine the distance to travel between locations. This can be done using a GPS-enabled device, an online calculator (www.movable-type.co.uk/scripts/latlong.html) or Google Maps.

Investigate

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Can you develop clues that involve the distance? For example, “From this point, travel 55 metres due North to ﬁnd your next clue.”

Using the formulas from the previous lesson, calculate the distance between your ﬁrst and second cache. How does this calculated value compare to the distance provided by a digital tool like Google Maps? Discuss any potential reasons for discrepancies.

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5.03 International time zones and time differences After this lesson, you will be able to… • understand the meaning of Greenwich Mean Time (GMT), International Date Line, and Coordinated Universal Time (UTC) • calculate time differences between two places on Earth

International time zones and time differences The world is divided into standard time zones, most of which are offset from a central reference time by whole hours ranging from UTC−12 to UTC+12. However, there are also non-standard time zones such as UTC+13 and UTC+14, as well as several time zones offset by 30 or 45 minutes. A time zone is a region of the Earth that uses the same standard time, usually determined by its position relative to lines of longitude. All time zones are deﬁned in relation to a global reference time. Two terms are commonly used for this purpose: Greenwich Mean Time (GMT) and Coordinated Universal Time (UTC). Greenwich Mean Time (GMT) The mean solar time at the prime meridian, located in Greenwich, London. It was once the world’s standard for timekeeping and is still used in some time zones, particularly in the UK and parts of Africa. Coordinated Universal Time (UTC) The modern international standard for timekeeping. It is based on highly accurate atomic clocks and is occasionally adjusted using leap seconds to stay in sync with the Earth’s slightly irregular rotation.

Although GMT and UTC are often used interchangeably, especially in everyday settings, UTC is more precise and is the oﬃcial time standard used worldwide for scientiﬁc and international purposes.

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In this map, the colour of the region is used to indicate the time zone.

F H

M J

E G

B A

+13 +14 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 UTC +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12

A. Canberra B. Brisbane C. Tokyo D. Perth E. Singapore I. Ouagadougou J. Dakar F. Moscow G. Johannesberg H. Rome K. London L. Dublin M. Casablanca The coloured bar along the bottom of the map shows how time zones are meant to be divided, based on lines of longitude. These divisions are usually made every 15 degrees. However, comparing this to the actual time zones shown on the map reveals that countries do not always follow these lines exactly. There are several reasons for this. Some countries or regions choose time zones that better suit their government, economy, or way of life, even if this means not following the ideal lines based on longitude. Time zones are based on how far they are behind or ahead of GMT/UTC. The zone of Greenwich Mean Time is considered to be 0. Countries to the west of the prime meridian are behind GMT and use a negative (−) GMT/UTC. Countries to the east of the prime meridian are ahead of GMT and use a positive (+) GMT/UTC. For example, Argentina and Paraguay are −3 GMT, which means these countries are 3 hours behind Greenwich Mean Time. So if it was midday in England, it would be 9:00 a.m. in Paraguay. As we move west, the time decreases. Time zones to the west of GMT/UTC use a negative sign (−). As we move east, the time increases. Time zones to the east of GMT/UTC use a positive sign (+). The time difference between two locations can be calculated by subtracting their UTC offsets. A positive result means the ﬁrst location is ahead of the second, and a negative result means it is behind. • Example 1 (opposite sides of UTC 0): The time difference between Paris (UTC+1) and New York (UTC−5) is calculated as (+1) − (−5) = 1 + 5 = 6 hours. • Example 2 (same side of UTC 0): The time difference between Perth (UTC+8) and Tokyo (UTC+9) is calculated as (+8) − (+9) = −1 hour. This means Perth is 1 hour behind Tokyo.

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On the opposite side of the world to the prime meridian is the International Date Line. The International Date Line marks the beginning and end of a calendar day (24 hours). It passes through the middle of the Paciﬁc Ocean at about 180° line of longitude, but it has been drawn around some territories and island groups, so that the date line would not pass through the middle of a country or town. Hawaii

Eastern hemisphere International Date Line

10°N 0

Tokelalu Samoa Fiji

Kiribati

Equator

American Samoa

10°S

Tahiti Australia

Tonga Pacific Ocean

New Zealand

160°W 160°E

180°

30°S

140°W

Western hemisphere

Countries to the left of the International Date Line, that is, countries in +12 GMT/UTC, spend most of the time (23 hours) a whole day ahead of countries to the right of the International Date Line, that is, countries in −11 GMT/UTC. Countries with +14 GMT/UTC are more than a day ahead of some of their neighbours. When travelling east across the International Date Line, the date goes backward by one day (a day is gained). When travelling west across the International Date Line, the date advances by one day (a day is lost).

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Exploration Use the time zone map to explore how time zones work around the world:

F H

M J

E G

B A

+13 +14 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 UTC +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12

A. Canberra B. Brisbane E. Singapore C. Tokyo D. Perth I. Ouagadougou J. Dakar F. Moscow G. Johannesberg H. Rome K. London L. Dublin M. Casablanca 1. Identify a country with a positive UTC offset (e.g., +8). What does this positive offset indicate about its time compared to the time in London (UTC 0)? 2. Identify a country with a negative UTC offset (e.g., −5). What does this negative offset indicate about its time compared to the time in London (UTC 0)? 3. Using the map, determine the UTC offsets for Moscow and New York. Calculate the difference between these two offsets. What does this calculated value represent? 4. Examine the path of the International Date Line on the second map. Why is it not a perfectly straight vertical line?

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Example 1 Answer these questions: a Calculate the time difference between Paris, France (UTC+1) and New York, USA (UTC−5).

Create a strategy The total time difference between two locations on opposite sides of the prime meridian is determined by adding the absolute values of their UTC offsets.

Apply the idea The time difference can be calculated by subtracting the UTC offsets. Time difference = (+1) − (−5)

Subtract the UTC offsets

=1+5

Simplify

= 6 hours

Evaluate

The time difference between Paris and New York is 6 hours.

b If it is 4:00 p.m. on Friday in Paris, determine the local time and day in New York.

Create a strategy New York has a negative UTC offset, so its time is behind Paris. Subtract the 6-hour time difference from the time in Paris. It may be easier to ﬁrst convert the time to 24-hour format.

Apply the idea First, convert 4:00 p.m. to 24-hour time, which is 16:00. Now, subtract the 6-hour time difference. New York time = 16 : 00 − 6 hours = 10 : 00 Since 10:00 is in the morning and we did not cross midnight, the day is still Friday. The time in New York is 10:00 a.m. on Friday.

c If it is 2:30 a.m. on Wednesday in Paris, determine the local time and day in New York.

Create a strategy Subtract the 6-hour time difference from the time in Paris. Since the number of hours being subtracted is larger than the current time, 24 hours will need to be borrowed from the previous day.

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Apply the idea The time in Paris is 02:30 on Wednesday. A subtraction of 6 hours is required. Subtracting 2 hours and 30 minutes brings the time to midnight. This leaves another 3 hours and 30 minutes to subtract. This means the time will be on the previous day, Tuesday. Time = 24 : 00 − 3 hours 30 minutes = 20 : 30 The time in New York is 20:30, or 8:30 p.m., on Tuesday.

Example 2 This table shows the time of several locations relative to UTC: UTC offset

Location

−5

New York, Cuba, Peru

−4

Chile, Barbados, Brazil (west), Bolivia

−3

Argentina, Brazil (east), Uruguay, Greenland

−2

South Sandwich Islands

−1

Azores

UK, Ireland, Iceland, Portugal, Ghana, Liberia, Mali

Algeria, Angola, Chad, Bosnia, Croatia

Finland, Greece, Lebanon, Egypt, South Africa

Russia (west), Saudi Arabia, Kenya, Madagascar, Iraq

+3.5

Iran

Mauritius, United Arab Emirates, Armenia

+4.5

Afghanistan

Pakistan, Maldives, Kazakhstan

+5.5

India, Sri Lanka

a Determine the time difference between Kenya and Mauritius.

Create a strategy Use the table to determine the UTC offset of each location and determine the difference.

Apply the idea Mauritius is 4 hours ahead of UTC, while Kenya is 3 hours ahead of UTC. Time difference = (+4) − (+3) = 1 hour

Subtract the UTC offsets Evaluate

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b If it is 17: 25 on the 15th day of the month in Kenya, determine what time and day it is in Mauritius, giving your answer in 24-hour time.

Create a strategy Identify which location is ahead in time, then add or subtract the time difference.

Apply the idea Mauritius is ahead of Kenya as the UTC offset is larger for Mauritius than it is for Kenya. Mauritius is 4 hours ahead of UTC, compared to Kenya which is 3 hours ahead of UTC. Therefore, the time difference calculated from part (a) is added to the time in Kenya to calculate the time in Mauritius. Using the algorithm method:

The time in Mauritius is 18:25 on the same day, the 15th of the month.

Example 3 A traveller takes a ﬂight from Sydney, Australia (UTC+10) to Los Angeles, USA (UTC−8), crossing the International Date Line. Explain whether the traveller gains or loses a calendar day upon arrival.

Create a strategy To travel from Sydney to Los Angeles, the traveller must ﬂy east across the Paciﬁc Ocean, crossing the International Date Line. Use the rule for travelling east across the IDL to determine if a day is gained or lost.

Apply the idea The traveller is ﬂying east from Sydney to Los Angeles, crossing the International Date Line. When travelling east across the International Date Line, the date goes backward by one day. So, the traveller gains a calendar day.

Reﬂect and check Gaining a day means the calendar shifts back by one date. For example, if the traveller departed Sydney on Saturday, they would arrive in Los Angeles on Friday of the same week.

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Ex 2

This table shows the time of several locations relative to UTC: UTC Offset

370

Location

−5

New York, Cuba, Peru

−4

Chile, Barbados, Brazil (West), Bolivia

−3

Argentina, Brazil (East), Uruguay, Greenland

−2

South Sandwich Islands

−1

Azores

UK, Ireland, Iceland, Portugal, Ghana, Liberia, Mali

Algeria, Angola, Chad, Bosnia, Croatia

Finland, Greece, Lebanon, Egypt, South Africa

Russia (West), Saudi Arabia, Kenya, Madagascar, Iraq

+3.5

Iran

Mauritius, United Arab Emirates, Armenia

+4.5

Afghanistan

Pakistan, Maldives, Kazakhstan

+5.5

India, Sri Lanka

Using the table, determine which location is ahead in time: South Sandwich Islands or Brazil (West).

Calculate the time difference between South Sandwich Islands and Brazil (West).

When it is 6:00 p.m. on the 17th day of the month in South Sandwich Islands, determine the time and day in Brazil (West).

Using the table, determine which location is ahead in time: Uruguay or Algeria.

Calculate the time difference between Uruguay and Algeria.

If it is 21:25 on the 13th day of the month in Uruguay, determine the time and day in Algeria.

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In this map, the colour of the region is used to indicate the time zone:

G V

I F M

P O

+13 +14 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 UTC +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12

A. Canberra E. Beijing J. Johannesberg P. Santiago V. San Francisco

B. Brisbane F. Singapore L. Rome Q. Washington DC W. Anchorage

C. Tokyo G. Moscow M. Rio de Janeiro R. Toronto

D. Perth I. Cairo O. Buenos Aires T. Denver

Determine the time difference between Toronto and Rio de Janeiro.

Determine the time difference between Beijing and Johannesburg.

Determine the time difference between Moscow and Canberra.

Determine the time difference between Toronto and Moscow.

Determine the time difference between Cairo and Tokyo.

Consider the map shown: Nuuk

Moscow

San Francisco New York

Bamako Caracas

Cape Town

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

UTC +1

How many hours is San Francisco behind Cape Town?

Determine the time in San Francisco when it is 17:00 in Cape Town.

What time is it in Bamako when it is 10:00 a.m. in Nuuk?

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Consider the map shown:

Moscow

Abakan

Paris Tokyo Beijing

Cairo Accra

Dobo

Hobart −1

UTC

Identify the two cities that are in the same time zone.

How many hours is Hobart ahead of Paris?

Determine the time in Cairo when it is 11:00 a.m. in Beijing.

Determine the time in Hobart when it is 2:00 p.m. in Moscow.

+10

+11

+12

Katie wants to know the time in Quebec (UTC−5) when it is 11:00 p.m. on Saturday in Honolulu (UTC−10): a

Calculate the time difference between the two cities.

Determine the time in Quebec when it is 11:00 p.m. on Saturday in Honolulu.

What day is it in Quebec when it is 11:00 p.m. on Saturday in Honolulu?

Which of these locations will be the ﬁrst to celebrate New Year’s Eve? A

Canberra (UTC+10)

Perth (UTC+8)

Darwin (UTC+9.5)

Which of these cities will have a time of 11:00 p.m. when it is 4:00 p.m. UTC? A

Cairo (UTC+2)

Rio de Janeiro (UTC−3)

Singapore (UTC+7)

Which of these cities will have a time after 3:00 a.m. when it is 2:00 a.m. in Cairo (UTC+2)? A

Singapore (UTC+7)

Washington (UTC−5)

Rio de Janeiro (UTC−3)

An astronomer in Beirut observes a solar ﬂare at 11:00 p.m. local time (UTC+3). Determine the local time their colleague in Pretoria would have seen the same ﬂare if their time zone is UTC+2.

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Ex 3

Valerie in Hobart has a phone conversation with Oprah in Manila. Manila is 2 hours behind Hobart time: a

If Valerie calls at 10:30 a.m., determine what time it is for Oprah.

If Oprah calls at 7:30 p.m., determine what time it is for Valerie.

Isabelle in Brisbane is expecting a phone call from Gwen in Manila. Manila is 2 hours behind Brisbane time: a

If Isabelle calls at 12:30 a.m., determine what time it is for Gwen.

If Gwen calls at 11:45 p.m., determine what time it is for Isabelle.

Ellie in Darwin is expecting a phone call from Avril in Shanghai. Shanghai is 1.5 hours behind Darwin time: a

If Ellie calls at 12:15 a.m., determine what time it is for Avril.

If Avril calls at 11:45 p.m., determine what time it is for Ellie.

Han, who is in Melbourne (UTC+10) at the moment, has an interview over Skype with the human resources manager of an American trading ﬁrm scheduled for 4:00 p.m. on Sunday, Memphis (UTC−6) time: a

Calculate the time difference between the two cities.

Determine the time in Melbourne when it is 4:00 p.m. on Sunday in Memphis.

What day will the interview fall on in Melbourne?

The rugby World Cup ﬁnal is broadcast live across the world. One particular year, it was held in Wales and the game started at 4:00 p.m. local time. Where Beth lives, it is 17 hours behind Wales. Determine the time and day the game started in her local time.

In your own words, explain the difference between Greenwich Mean Time (GMT) and Coordinated Universal Time (UTC).

What is the purpose of the International Date Line? Explain what happens to the date when a traveller crosses it from west to east.

A location is in the time zone UTC−6. Explain what this means in relation to the time at the prime meridian (UTC 0).

Why is the International Date Line not a perfectly straight line?

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Complex familiar 22

The table shows the current time in major capital cities beginning with the letter M:

City

Time

When it is 15:10 Minsk time, determine the time in Madrid. Give your answer in 24-hour time.

Madrid

Thu 1:58 a.m.

Managua

Wed 5:58 p.m.

When it is 11:20 a.m. Montevideo time, determine the time in Minsk.

Manila

Thu 7:58 a.m.

Melbourne

Thu 9:58 a.m.

Mexico City

Wed 6:58 p.m.

Miami

Wed 7:58 p.m.

Minneapolis

Wed 6:58 p.m.

Minsk

Thu 2:58 a.m.

Montevideo

Wed 8:58 p.m.

Montreal

Wed 7:58 p.m.

Moscow

Thu 3:58 a.m.

Mumbai

Thu 5:58 a.m.

A company wants to set up a business meeting between four of its employees. The employees are in these cities: Cairo (UTC+2), Amsterdam (UTC+1), London (UTC 0), Perth (UTC+8). If it is 1:00 p.m. in Perth, determine the time in these cities: a

Cairo

Amsterdam

London

Three scientists collaborating on the same project all need to get together for weekly meetings. Each scientist is in a different location. Ryan is located in Rome (UTC+1). Henry is located in Cairo (UTC+2). Annabelle is located in San Francisco (UTC−8): a

Which scientist will have the earliest local time?

Which of the scientists will have the latest local time?

Annabelle suggests 8:00 p.m. in San Francisco. Determine the time it would be in the other two cities: i

Rome

Cairo

Ryan suggests 8:00 a.m. in Rome. Determine the time it would be in the other two cities: i

374

Cairo

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San Francisco

5.04 Australian time zones and daylight savings After this lesson, you will be able to… • identify Australia’s standard time zones • calculate the time difference between Australian locations • solve practical time conversion problems

Australian time zones Australia is divided into 3 time zones. They are: • Australian Eastern Standard Time (AEST): covering Queensland, New South Wales, Australian Capital Territory, Victoria and Tasmania. • Australian Western Standard Time (AWST): covering Western Australia, which is 2 hours behind AEST. • Australian Central Standard Time (ACST): covering South Australia and the Northern Territory. (half an hour behind AEST) Western Standard Time (AWST) 2 hours behind AEST

Central Standard Time (ACST) 0.5 hours behind AEST

WA (AWST) UTC+8

NT (ACST) UTC+9.5 SA (ACST)

Eastern Standard Time (AEST)

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

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Daylight Saving Time Daylight Saving Time (DST) is used by many countries, including Australia, New Zealand, the USA and the UK during their summer months to make better use of daylight in the evenings. In Australia, only some states observe daylight saving time. In these states, clocks are moved forward by one hour on the first Sunday in October, as the days become longer during spring. Clocks are then set back to standard time on the first Sunday in April, as the days begin to shorten. This means that during DST, observing regions are an extra hour ahead of the standard time zones. Currently, Queensland, the Northern Territory and Western Australia do not use Daylight Saving Time. When converting between standard time and daylight saving time, it is important to understand the following abbreviations and adjustments: • AEST and ACST stand for Australian Eastern Standard Time and Australian Central Standard Time respectively, which is used in states such as Queensland during the non-daylight saving period. • AEDT and ACDT stand for Australian Eastern Daylight Time and Australian Central Daylight Time, which are used in states like New South Wales, Victoria, Tasmania, the Australian Capital Territory and South Australia during Daylight Saving Time. • To convert AEDT to AEST or ACDT to ACST (daylight saving time to standard time), subtract one hour. • To convert AEST to AEDT or ACST to ACDT (standard time to daylight saving time), add one hour. This applies only to regions that observe Daylight Saving Time. States and territories such as Queensland, Western Australia, and the Northern Territory remain on standard time year-round. Time zones in Australia become more complex during daylight saving time, as not all states observe the change. Queensland, the Northern Territory, and Western Australia remain on standard time. As shown on the map, during daylight saving time, Australia is divided into five distinct time zones: Western Standard Time (AWST) 2 hours behind AEST

Central Standard Time (ACST) 0.5 hours behind AEST

Central Daylight Time (ACDT) 0.5 hours ahead AEST

Eastern Daylight Time (AEDT) 1 hour ahead AEST

WA (AWST) UTC+8

NT (ACST) UTC +9.5 SA (ACST)

No DST

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

WA (AWST) UTC+8

Eastern Standard Time (AEST)

NT (ACST) UTC +9.5 SA (ACST)

DST (Daylight Saving Time)

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

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Example 2 The time in Perth is 14:15. Determine the time in these cities: AWST (UTC+8)

ACST (UTC+9.5)

2 h behind

0.5 h behind

AEST (UTC+10) Compared to AEST

ACDT (UTC+10.5)

AEDT (UTC+11)

0.5 h ahead

1 h ahead

QLD

QLD WA

WA NSW

VIC NO DST

TAS

NSW VIC

DST

TAS

a Sydney without DST

Create a strategy

Apply the idea

Use the map showing the time zones with no DST to determine the time difference.

Without DST, Sydney is 2 hours ahead of Perth. When the time in Perth is 14:15 the time in Sydney will be 2 hours later, that is 16:15.

b Sydney during DST

Create a strategy

Apply the idea

Use the map showing the time zones during DST to determine the time difference.

During DST, Sydney is 3 hours ahead of Perth. When the time in Perth is 14:15 the time in Sydney will be 3 hours later, that is 17:15.

Example 3 If it is 10:13 in Sydney (AEST), determine the time in Australian Eastern Daylight Time (AEDT).

Create a strategy During Daylight Saving Time, the time moves one hour ahead of the standard time.

Apply the idea

Reﬂect and check

Add one hour to 10:13.

AEDT is UTC+11, and AEST is UTC+10, so the difference is +1 hour.

Therefore, 10:13 AEST would be 11:13 AEDT.

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Australia is divided into 3 time zones: • AEST (Eastern Standard Time) encompasses QLD, NSW, the ACT, VIC and TAS. • ACST (Central Standard Time) encompasses the NT and SA, and is half an hour behind AEST. • AWST (Western Standard Time) encompasses WA and is two hours behind AEST. a

If it is 8:16 p.m. AEST, determine: i

Ex 2

If it is 12:10 a.m. AEST, determine, giving your answers in 24-hour time: The time in ACST

The time in AWST

If it is 1:16 p.m. in ACST, determine the time in AWST.

During which months does daylight saving time operate in Australia?

During which months is daylight saving time not in effect in Australia?

During which season do the clocks go 1 hour forward?

During which season do the clocks return to standard time?

Determine whether these Australian states or territories observe Daylight Saving Time (DST): a

Victoria

Western Australia

New South Wales

Northern Territory

Queensland

South Australia

When the time is 14:15 in Perth determine the time in: a

Sydney during No DST

Sydney during DST

Western Standard Time (AWST) 2 hours behind AEST

Central Standard Time (ACST) 0.5 hours behind AEST

Central Daylight Time (ACDT) 0.5 hours ahead AEST

Eastern Daylight Time (AEDT) 1 hour ahead AEST

WA (AWST) UTC+8

NT (ACST) UTC +9.5 SA (ACST)

No DST

382

The time in AWST

If it is 7:27 p.m. in ACST, determine the time in AWST.

The time in ACST

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

Mathspace QLD 12 General Mathematics 2025 mathspace.co

WA (AWST) UTC+8

Eastern Standard Time (AEST)

NT (ACST) UTC +9.5 SA (ACST)

DST (Daylight Saving Time)

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

Complex familiar 13

Indonesia is divided into three time zones: • Indonesia Western Time Zone (IWTZ) • Indonesia Eastern Time Zone (IETZ) • Indonesia Central Time Zone (ICTZ) Its neighbouring country Papua New Guinea (PNG) has another time zone as shown in this map:

Brunei Malaysia Singapore

Palau Papua New Guinea

Indonesia 1 2 Australia 9

19:00 +7 IWTZ a

PNG

IWTZ

IETZ

iii

PNG

IWTZ

ICTZ

iii

PNG

IWTZ

IETZ

iii

PNG

IWTZ

ICTZ

iii

PNG

IWTZ

ICTZ

iii

IETZ

Consider that only IWTZ is on Daylight Saving Time (which means the time has been brought forward 1 hour). If it is 5:11 p.m. in IWTZ, determine the time in: i

384

iii

Consider that Papua New Guinea is on Daylight Saving Time. If it is 7:24 p.m. in Papua New Guinea, determine what time it is in: i

IETZ

Consider that only IETZ is operating on Daylight Saving Time. If it is 8:24 a.m. in IETZ, determine what time it is in: i

Consider that only ICTZ is operating on Daylight Saving Time. If it is 10:13 a.m. in ICTZ, determine what time it is in: i

ICTZ

If it is 9:21 p.m. in IETZ, determine what time it is in: i

22:00 +10

If it is 2:48 a.m. in ICTZ, determine what time it is in: i

21:00 +9 IETZ

If it is 6:16 a.m. in IWTZ, determine what time it is in: i

20:00 +8 ICTZ

ICTZ

IETZ

Mathspace QLD 12 General Mathematics 2025 mathspace.co

iii

PNG

Australia and its neighbouring countries are located across several time zones as shown in this map:

Vietnam Papua New Guinea

1 2 Australia 9

19:00 7

20:00 8

21:00 9

22:00 10

If it is 5:42 a.m. in Papua New Guinea, determine the time in: a 15

IETZ

Vietnam

AEST

AWST

Four friends plan to have a video conference at 8:00 p.m. Sydney time:

Brunei Malaysia Singapore

Palau Papua New Guinea

Indonesia 1 2 Australia 9

19:00 +7 IWTZ

20:00 +8 ICTZ

21:00 +9 IETZ

22:00 +10

Use the above map of time zones to determine the time each person should join the meeting: a

Roald who lives in Denpasar, (Central) Indonesia

Yuri who lives in Jakarta, (West) Indonesia

Nadia who lives in Alice Springs, NT

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Complex unfamiliar This map shows the time zones across the Australian States and can be used for Question 16 and 17: Western Standard Time (AWST) 2 hours behind AEST

Central Standard Time (ACST) 0.5 hours behind AEST

WA (AWST) UTC+8

NT (ACST) UTC+9.5 SA (ACST)

Eastern Standard Time (AEST)

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

Using the map provided, calculate: a

386

Luke and Maria board a train at 7:58 p.m. on Saturday from Sydney, New South Wales towards Adelaide, South Australia. i

If the trip to Adelaide took 20 hours and 22 minutes, determine the time they arrived in Adelaide local time.

Luke boards a train from Adelaide, South Australia at 5:45 p.m. on Sunday to Darwin, Northern Territory. If his trip took 28 hours and 36 minutes, determine the time he arrived in Darwin local time.

iii

Maria boards a train from Adelaide, South Australia at 4:55 p.m. Sunday towards Perth, Western Australia. If the trip took her 19 hours and 20 minutes, determine the time she arrived in Perth local time.

Han and Amelia board a train at 8:52 p.m. on Thursday from Sydney, New South Wales towards Adelaide, South Australia. i

They arrived in Adelaide, South Australia at 6:48 p.m. Friday. Determine the duration of their train trip.

Han boards a train from Adelaide, South Australia at 8:43 p.m. on Friday to Darwin, Northern Territory. He arrives at 8:11 p.m. Saturday. Determine the duration of his trip.

iii

Amelia boards a train from Adelaide, South Australia at 7:53 p.m. Friday towards Perth, Western Australia. She arrives at 10:54 a.m. on Saturday. Determine the duration of her train trip.

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Answer these questions using the map provided: a

A ﬂight leaves Hooker Creek Airport, NT at 6:10 a.m. It takes 2 hours and 15 minutes to reach Sydney airport, NSW. Determine the time the ﬂight arrives in NSW local time.

A ﬂight leaves Sydney Airport in NSW at 10:55. It takes 5 hours and 25 mins to reach Perth airport, WA. Determine the time the ﬂight arrives in WA local time.

An aeroplane leaves Perth Airport in Western Australia at 9:16 p.m. and takes 4 hours and 22 minutes to reach Brisbane airport, Queensland. Determine the time the aeroplane arrives in Brisbane local time.

An aeroplane leaves Brisbane Airport in Queensland at 3:24 a.m. and takes 6 hours and 34 minutes to arrive at Darwin Airport in Northern Territory. Determine the time the aeroplane arrives in Darwin local time.

If an aeroplane leaves Darwin Airport in Northern Territory at 10:48 p.m. and takes 6 hours and 28 minutes to arrive at Perth Airport in Western Australia. Determine the time the aeroplane arrives in Perth local time.

Harry lives in Melbourne and wants to catch the live broadcast of a match between the Hawks and Eagles at Optus Stadium in Perth. If the match is at 4:00 p.m. local time in Perth, what time will Harry need to tune into the broadcast?

Tobias lives in Adelaide and wishes to call in for a radio competition for a national broadcaster based in Sydney. If the competition opens at 7:00 a.m. Sydney local time, what is the earliest local time Tobias can call from Adelaide?

Many Aboriginal and Torres Strait Islander peoples observe seasonal calendars that are based on local ecological knowledge rather than the four seasons of the European model. Choose one of these Aboriginal seasonal calendars to research: • The Gariwerd seasonal calendar (Victoria) • The Yawuru seasonal calendar (Western Australia) • The D’harawal seasonal calendar (New South Wales) Based on your research, answer these questions: a

Identify how many seasons are in the calendar you chose.

Describe one of the seasons from your chosen calendar. What natural events, such as weather patterns or plant and animal activity, signal the start of this season?

Explain one way this seasonal system is different from the four-season (Summer, Autumn, Winter, Spring) model.

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The attached map shows the time zones across the Australian states: Western Standard Time (AWST) 2 hours behind AEST

Central Standard Time (ACST) 0.5 hours behind AEST

Central Daylight Time (ACDT) 0.5 hours ahead AEST

Eastern Daylight Time (AEDT) 1 hour ahead AEST

WA (AWST) UTC+8

NT (ACST) UTC +9.5 SA (ACST)

No DST

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

NT (ACST) UTC +9.5 SA (ACST)

DST (Daylight Saving Time)

QLD (AEST) UTC+10 NSW (AEST) VIC (AEST) TAS (AEST)

A ﬂight leaves Hobart Airport, TAS at 10:55 a.m. during DST. It takes 4 hours and 30 minutes to reach Perth airport, WA. Determine the time the ﬂight arrives in WA local time.

A ﬂight leaves Alice Springs Airport, NT at 5:15 a.m. during DST. It takes 3 hours and 10 minutes to reach Hobart airport, TAS. Determine the time the ﬂight arrives in TAS local time.

If an aeroplane leaves Broome Airport in Western Australia at 7:30 p.m. and arrives at Albury (operating on Daylight Saving Time) Airport in Victoria at 3:52 a.m., determine its ﬂight time.

If an aeroplane leaves Albury Airport in Victoria at 7:14 p.m. and arrives at Adelaide Airport in South Australia (operating on Daylight Saving Time) at 9:10 p.m., determine its ﬂight time.

If an aeroplane leaves Adelaide Airport in South Australia at 9:53 a.m. and arrives at Broome Airport in Western Australia at 3:41 p.m., determine its ﬂight time.

Avril lives in Perth and wants to catch the live broadcast of a cricket match at Melbourne Cricket Ground. If the match is at 5:00 p.m. local time in Melbourne, and it is Daylight Saving Time, what time will Avril need to tune into the broadcast?

Tickets go on sale for the Australian tour of a band in the summer (Daylight Saving Time) at 7:00 a.m. in Sydney.

Complete this table with the times for when each person should log in to buy online tickets in their local time.

388

WA (AWST) UTC+8

Eastern Standard Time (AEST)

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Location

Time

Michael

Townsville, QLD

Katrina

Whyalla, SA

⬚

Sally

Coral Bay, WA

⬚ ⬚

5.05 Longitude and time difference After this lesson, you will be able to… • understand the link between longitude and time • determine the number of degrees of longitude for a given time difference • comprehend how changes in longitude affect time

Longitude and time difference The Earth rotates 360° every 24 hours, which means it turns 15° each hour (360 ÷ 24 = 15). This steady rotation links longitude directly to time. Since the Earth rotates from west to east, locations to the east experience sunrise and sunset earlier than those to the west. To help standardise time across the globe, the Earth has been divided into 24 main time zones, each roughly 15° of longitude wide. Each time zone represents one hour of time difference from Coordinated Universal Time (UTC) or Greenwich Mean Time (GMT), which is based on the prime meridian at 0° longitude in Greenwich, England. The general rule is: Every 15° of longitude corresponds to a 1-hour difference in time. Conversely, a 1-hour time difference between two locations corresponds to a 15° difference in longitude. Time difference (hours) = Longitude difference (°) ÷ 15° Longitude difference (°) = Time difference (hours) × 15° Locations east of the prime meridian (0° longitude) have later local times than UTC. They are ahead in time. Locations west of the prime meridian have earlier local times than UTC and are behind in time. This means that if a person travels eastward across the globe, hours are added to their starting time, and if they travel westward, hours are subtracted from their starting time to find the new local time. However, this method gives only an approximate time difference. Recall that actual time zones may differ slightly because countries may shift their oﬃcial time zone to suit their population, some regions adopt daylight saving time, temporarily moving clocks forward by one hour, time zones often follow political or geographic boundaries, not strictly longitude lines.

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Exploration Use an online map or globe to determine the longitudes of the cities listed in the table. Then, calculate the time difference between them, assuming no daylight saving time is in effect. Two cities 1 and 2

Longitude 1 - Longitude 2

Time difference

⬚

Brisbane - Perth

⬚

Sydney - Tokyo New York - London

⬚

Dubai - Cape Town

⬚

Example 1 Estimate the time difference between Sydney, Australia (34°S, 151°E), and Manila, Philippines (15°N, 121°E).

Create a strategy Use the longitude of each location to determine the difference in degrees. Then divide by 15 to estimate the number of hours between the time zones. The location further east will be ahead in time.

Apply the idea Only longitude is used to calculate time differences. The longitudes of Sydney and Manila are 151°E and 121°E, respectively. First, determine the difference in longitudes by subtracting the longitudes: 151°E − 121°E = 30° Secondly, divide by 15° to estimate the time difference: 30° ÷ 15 = 2 hours Finally, determine which location is ahead in time: Sydney is 2 hours ahead of Manila.

Reflect and check In this case, the time difference was a whole number. If a calculation resulted in a decimal (e.g., 2.7 hours), rounding to the nearest whole hour (3 hours) provides a good estimate of the official time zone difference. This is because official time zones are generally set in full-hour increments.

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Example 2 It is 3:15 p.m. on Monday in Los Angeles, USA (34°N, 118°W). What is the approximate local time and day in New Delhi, India (29°N, 77°E)?

Create a strategy Use the longitude of each location to determine the difference in degrees, divide by 15 to estimate the number of hours between the time zones, calculate the time and day in New Delhi by adding the time difference, as New Delhi is east of Los Angeles.

Apply the idea The longitudes of Los Angeles and New Delhi are 118°W and 77°E, respectively. Since the locations are on opposite sides of the prime meridian (one west and one east), the total difference in longitude is found by adding their values. 118°W + 77°E = 195° 195° ÷ 15 = 13 hours 3:15 p.m. + 13 hours = 4:15 a.m. the next day

Write the equation Divide by 15° to estimate the time difference Evaluate

Therefore, if it is 3:15 p.m. on Monday in Los Angeles it is 4:15 a.m. on Tuesday in New Delhi.

Example 3 The time difference between City A and City B is 5 hours. Determine the difference in their longitude.

Create a strategy Use the rule that each hour of time difference corresponds to 15° of longitude. Multiply the time difference in hours by 15 to calculate the longitude difference in degrees.

Apply the idea Longitude difference (°) = Time difference (hours) × 15°

Write the equation

= 5 × 15°

Substitute time difference

= 75°

Evaluate

The difference in longitude is 75°.

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Ex 2

Estimate the time difference between New York City, USA (41°N, 74°W) and London, UK (51°N, 0°).

Estimate the time difference between Perth, Australia (32°S, 116°E) and Auckland, New Zealand (37°S, 174°E).

It is 6:00 a.m. on Tuesday in Buenos Aires, Argentina (35°S, 58°W). What is the approximate local time and day in Tokyo, Japan (36°N, 140°E)?

Determine the time in a city located at (35°S, 31°W) when it is 3:00 p.m. in another city located at (7°N, 104°E).

The local time in London, UK (51°N, 0°) is 10:45 p.m. on Friday. Estimate the time and day in Perth, Australia (32°S, 116°E).

Eileen lives in London, but is away on holiday at a location with a longitude of 135°E:

Calculate the time difference between her location and London (GMT).

She will be able to call home when it is 1:10 p.m. London time (GMT). Determine the local time at her holiday location.

Her ﬂight leaves at 10:10 a.m. local time at her holiday location. Determine the London time her ﬂight will depart.

Maria is away at a conference. Her location is (24°S, 77°E), while her hometown is (27°S, 122°E). Calculate the local time at home when the conference ﬁnishes at 5:00 a.m.

Two satellites are to be launched simultaneously, one at location A (3°N, 54°W) and the other at location B (35°S, 111°E). If the launch time at location B is 17:40, determine the launch time at location A.

Ex 3

Two locations are 6 hours apart in local time. Determine the approximate difference in longitude between the two locations.

Two locations are 3.5 hours apart in local time. Determine the approximate difference in longitude between the two locations.

A live global event is broadcast simultaneously. Viewers in City X see it at 1:00 p.m. and viewers in City Y see it at 8:00 p.m. Determine the approximate difference in longitude between the two cities.

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5.06 Distance and time zone applications After this lesson, you will be able to… • determine the angular distance between two points on Earth given their latitude and longitude • calculate the surface distance between two locations along a great circle or a small circle • use Coordinated Universal Time (UTC) offsets to determine the time difference between any two international locations • solve complex, multi-step problems involving international travel itineraries, accounting for flight duration and DST

Calculate distance using latitude and longitude The Earth is mapped using a grid system of latitude and longitude lines. Latitude lines, or parallels, measure the distance north or south of the equator. Longitude lines, or meridians, measure the distance east or west of the prime meridian. Lines of longitude are part of great circles, which are circles on the Earth’s surface that have the same centre as the Earth itself. The equator is the only line of latitude that is a great circle. All other parallels of latitude are small circles. The shortest distance between two points on the Earth’s surface is along a great circle.

Interactive exploration Discover this concept in action online

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The angular distance is the angle between two points, as measured from the centre of the Earth. To calculate it: • For two points on the same meridian of longitude: add their latitudes if they are in different hemispheres (N and S) and subtract their latitudes if they are in the same hemisphere. • For two points on the same parallel of latitude: add their longitudes if they are in different hemispheres (E and W) and subtract their longitudes if they are in the same hemisphere. Once the angular distance is known, the surface distance can be calculated. The value 111.2 is used as an approximation for the number of kilometres in one degree of latitude, based on the Earth’s average circumference. To calculate the surface distance along a great circle (a meridian), use this formula:

D = 111.2 × α D

is the surface distance in kilometres (km)

is the angular distance in degrees

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To calculate the surface distance along a small circle (a parallel of latitude), the formula is adjusted to account for the shorter distance between meridians:

D = 111.2 × cos(θ ) × α D

is the surface distance in kilometres (km)

is the parallel of latitude in degrees

is the angular distance in degrees

Example 1 Calculate the surface distance between the two locations, rounded to the nearest kilometre: • Honolulu, Hawaii is located at (21°N, 158°W). • Dakar, Senegal is located at (21°N, 17°W).

Create a strategy Determine the angular distance by subtracting the longitudes then use the formula: D = 111.2 × cos (θ ) × α

Apply the idea Subtract the longitudes since they are both in the Western Hemisphere to calculate the angular distance: 158° − 17° = 141° Calculate the surface distance: D = 111.2 × cos (θ ) × α

Write the formula for small circle distance

= 111.2 × cos (21°) × 141°

Substitute the latitude and the angular distance

= 14 638

Evaluate and round

Reﬂect and check Notice the use of cos 21° in the formula. This is because the distance between each degree of longitude decreases as the latitude increases. If these cities were on the Equator (latitude 0°), the distance would be 111.2 × 141 ≈ 15 679 km, which is signiﬁcantly larger.

Example 2 Calculate the surface distance between Cairo, Egypt (30°N, 31°E) and Harare, Zimbabwe (18°S, 31°E), rounded to the nearest kilometre.

Create a strategy The locations are on the same meridian, so the great circle distance formula is used. First, calculate the angular distance by adding the latitudes since they are in different hemispheres. Then, use the formula D = 111.2 × α to calculate the surface distance.

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Some regions observe daylight saving time (DST), advancing clocks by one hour during summer months to make better use of natural daylight. When solving problems, it is critical to account for DST by adjusting the UTC offset (e.g., a standard UTC+10 zone becomes UTC+11 during DST). Many Aboriginal and Torres Strait Islander peoples use seasonal calendars based on environmental signs, such as plant flowering or animal migration, rather than fixed hours or months. For example, the Noongar people of south-western Western Australia recognise six distinct seasons, each corresponding to specific weather patterns and natural events. Time is observed through natural cues like sunrise and sunset, and activities are guided by the environment. These systems provide an important cultural perspective on how time can be understood and measured in relation to the local environment and its cycles.

Exploration Use this table of cities and their time zone information to answer the questions that follow: City

Standard UTC offset

Daylight Saving Time (DST)

London, UK

UTC+0

Yes (UTC+1)

New York, USA

UTC-5

Yes (UTC-4)

Tokyo, Japan

UTC+9

Perth, Australia

UTC+8

1. What is the time difference between London and Tokyo during January (when London is not on DST)? 2. What is the time difference between London and Tokyo during July (when London is on DST)? 3. A live sporting event begins at 8:00 p.m. in New York during its DST period. At what local time should a viewer in Perth tune in to watch it live?

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Example 3 Ava is planning a short trip during the summer holidays. She lives in Sydney (UTC+10), which observes daylight saving time (DST), and she will visit Auckland, New Zealand (UTC+12), which also observes DST, and Brisbane (UTC+10), which does not observe DST. Her itinerary is as follows: • She departs Sydney at 7:00 a.m. AEDT (Australian Eastern Daylight Time) on 5th October and arrives in Auckland after a 3-hour flight. • Two days later, she flies from Auckland to Brisbane, departing at 2:00 p.m. NZDT (New Zealand Daylight Time) with a flight duration of 4 hours. a Determine the local arrival time in Auckland.

Create a strategy Convert the departure day and time from Sydney (AEDT) to the equivalent time in Auckland (NZDT), then add the ﬂight duration to this converted time to determine the local arrival day and time.

Apply the idea The location furthest east is Auckland, so it is ahead of Sydney. Sydney is on AEDT (UTC+11) and Auckland is on NZDT (UTC+13) in early October due to daylight savings. Calculate the time difference between Auckland and Sydney: 13 − 11 = 2 hours Subtract UTC offsets Auckland is 2 hours ahead of Sydney. Convert the departure time from Sydney local time (AEDT) to Auckland local time (NZDT): 7:00 a.m. AEDT + 2 hours = 9:00 a.m. NZDT Add time difference Add the ﬂight duration of 3 hours to the Auckland time: 9:00 a.m. NZDT + 3 hours = 12:00 p.m. NZDT Add ﬂight time Ava arrives in Auckland at 12:00 p.m. NZDT on the 5th October.

Reﬂect and check The ﬂight duration can be added to the departure time in Sydney, and then converted to Auckland local time, yielding the same result.

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5.06 Practice questions Simple familiar Ex 1

Ex 2

•

Nairobi, Kenya is located at (1°S, 37°E) (UTC+3).

•

Quito, Ecuador is located at (1°S, 79°W) (UTC−5).

Calculate the surface distance between the two locations, rounded to the nearest kilometre.

If it is 7:00 a.m. in Quito, determine the local time in Nairobi.

•

Los Angeles, USA is located at (34°N, 118°W).

•

Athens, Greece is located at (34°N, 24°E).

Calculate the surface distance between the two locations, rounded to the nearest kilometre.

If it is 8:45 a.m. in Athens, determine the local time in Los Angeles.

•

Cape Town, South Africa is located at (34°S, 18°E).

•

Buenos Aires, Argentina is located at (34°S, 58°W).

Calculate the surface distance between the two locations, rounded to the nearest kilometre.

A direct ﬂight travels from Cape Town to Buenos Aires at an average speed of 900 km/h. Estimate the ﬂight time, in hours and minutes.

If it is 6:15 p.m. in Cape Town, determine the local time in Buenos Aires.

Two research teams are recording summer solstice sunrise data from different locations. One team is stationed at Uluru (25°S, 131°E), and the other is based at Darwin (13°S, 131°E): a

Determine the shortest distance between the two locations, using their latitude and longitude coordinates.

If a light aircraft travels directly from Uluru to Darwin at an average speed of 300 km/h, how long will the journey take (in hours and minutes)?

Calculate the surface distance, to the nearest kilometre, between Point A (45°N, 120°W) and Point B (20°N, 120°W).

City A is located at 50°N, 80°W and City B is located at 20°N, 80°W. Calculate the surface distance between the two cities, rounded to the nearest kilometre.

A marketing executive in Perth (UTC+8) needs to make a phone call to a supplier in Tokyo (UTC+9). If it is 2:00 p.m. in Perth, what is the local time in Tokyo? Neither city observes daylight saving time.

An AFL match is being broadcast live from Melbourne (UTC+11 during DST). If the match starts at 7:30 p.m. AEDT, what time should viewers in Darwin (UTC+9:30) tune in?

A live tennis match begins at 2:00 p.m. in London. During this time of year, London is observing daylight saving time (UTC+1). What time should a viewer in Dubai (UTC+4) watch the match live?

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Ex 3

Maya is attending an international youth conference. She lives in Darwin (UTC+9:30), which does not observe daylight saving time (DST). She will travel to Singapore (UTC+8), which also does not observe DST, and then to Melbourne (UTC+10), which does observe daylight saving time. This is her itinerary: • She departs Darwin at 6:30 a.m. ACST (Australian Central Standard Time) on 15th October and arrives in Singapore after a 4.5-hour ﬂight. • Two days later, she departs Singapore at 1:00 p.m. SGT (Singapore Time) for Melbourne. The ﬂight takes 7 hours and 30 minutes.

402

Determine the local arrival time in Singapore.

Determine the local arrival time in Melbourne.

Consider this flight schedule for Becky, who is planning a holiday. She has organised her flights into this table which shows her flying out of Perth (GMT+8), heading to New York (GMT−5) via Hong Kong (GMT+8) and then to Amsterdam (GMT+2). All times given are local times: Flight

From

Depart

Arrive

Departure date

QA 735

Perth (PER)

Hong Kong (HKG)

1:10 a.m

9:15 a.m.

13th Nov 2018

AA 724

Hong Kong (HKG)

New York (JFK)

10:10 a.m.

1:50 p.m.

13th Nov 2018

CAS 1226

New York (JFK)

Amsterdam (AMS)

5:35 p.m.

07:20 a.m.

29th Nov 2018

How long was her ﬂight from New York to Amsterdam?

When she was in New York, Becky wanted to call her parents back home. Determine the date and time it would be in Perth if she called her parents on the 16th November at 8:00 p.m. in New York?

Which of these best describes how some Aboriginal peoples organise their understanding of time? A

They follow the same four seasons as the Western calendar: summer, autumn, winter, and spring.

They use clocks and time zones to decide when to begin daily activities.

They observe changes in the environment, such as winds, animal behaviour, and plant cycles, to mark different seasons.

They adjust their clocks during daylight saving time like the rest of Australia.

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A global media company is planning a live New Year’s Eve broadcast, showing the moment the clock strikes midnight in 3 major cities. The broadcast will be based in their New York studio (UTC−5). The cities featured are: • Sydney (UTC+11, observing DST) • London (UTC+0) • New York (UTC−5) Determine the local time in the New York studio for the midnight celebrations in Sydney and London.

Complex unfamiliar 18

In Arnhem Land (Northern Territory), the Yolngu people traditionally recognise 6 seasons based on local environmental changes, rather than the 4 seasons of the Western calendar. During Rarrandharr (the build-up season, October-December), Yolngu communities often begin work early in the morning, when it is cooler, and rest during the hottest part of the day. A weather scientist in Sydney wants to schedule a daily video call at 9:00 a.m. Sydney time with a Yolngu ranger based near Nhulunbuy, NT, during Rarrandharr season. • Sydney is on Australian Eastern Daylight Time (AEDT) in October-December, which is UTC+11. • Nhulunbuy (NT) remains on Australian Central Standard Time (ACST) all year, which is UTC+9:30.

If Yolngu rangers typically begin work around sunrise (about 5:30 a.m. local time) and prefer to avoid working in the hottest part of the day, would a 9:00 a.m. call in Sydney be suitable? Explain your reasoning.

Suggest a more appropriate meeting time for both parties, respecting the Yolngu seasonal routine and accounting for the time difference.

A sea captain set sail from Greenwich (0° longitude) with a marine chronometer that was accurate at the time of departure but lost 4 minutes per day. After 30 days at sea, the captain observed that local noon (the time when the Sun was highest in the sky) occurred at 8:32 a.m. according to the chronometer. Based on this observation, determine the longitude the captain assigned to the island where they landed.

404

A remote research team on the Tropic of Cancer (23.5°N) and longitude 95°E is studying solar movement. They want to schedule an experiment to start exactly at solar noon: a

Determine the standard time zone they are likely in and the standard time at which solar noon occurs at their location.

Explain how the team could use a shadow stick and trigonometry to verify local noon experimentally.

Discuss how and why the timing of solar noon would differ if they moved to 95°W while remaining on the same latitude.

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5 Chapter review SF

Perth is located at approximately (32°S, 116°E) and Geraldton is located at (29°S, 116°E): a

Determine the angular distance between Perth and Geraldton.

Assuming the Earth’s radius is 6371 km, calculate the shortest distance between the two cities along the surface, to the nearest kilometre.

A business meeting is scheduled for 11:00 a.m. in Sydney (AEST, UTC+10) on a day when daylight saving is not in effect. Determine the local time for a colleague joining from Perth (AWST, UTC+8).

The time difference between City A and City B is 4.5 hours, with City B being ahead of City A. Estimate the difference in longitude between the two cities.

Two cities, A and B, are located on the same parallel of latitude at 40°S. City A has a longitude of 15°E and City B has a longitude of 55°E: a

Calculate the distance between City A and City B travelling along the 40°S parallel. Use 6371 km for the Earth’s radius and round to the nearest kilometre.

If a plane travels between the cities at an average speed of 750 km/h, how long would the ﬂight take? Give your answer in hours and minutes, to the nearest minute.

A national project team needs to schedule a video conference during summer when daylight saving time is active. The team members are in: • Brisbane (QLD, UTC+10, no DST) • Adelaide (SA, UTC+9:30, observes DST) • Perth (WA, UTC+8, no DST) The meeting is scheduled for 2:00 p.m. Adelaide local time (ACDT). Determine the local time for the team members in Brisbane and Perth.

A flight departs from City X (25°N, 80°W) and travels to City Y (25°N, 10°E). The total flight time is 8 hours. If the flight arrives at City Y at 9:00 p.m. local time, calculate the local time of departure from City X.

The Yawuru people of the Broome area in Western Australia recognise six seasons. One of these seasons is ‘Man-gala’, the wet season, which typically occurs from December to March. During this time, cultural activities might be adjusted due to heavy rain and high humidity. A researcher in Canberra (UTC+10) wants to schedule a regular video call with a Yawuru elder in Broome (UTC+8) during January, when Canberra is observing Daylight Saving Time (UTC+11). The researcher proposes a call time of 4:00 p.m. Canberra time: a

Determine the local time in Broome when the call would take place.

Given that ‘Man-gala’ is often the hottest part of the day, explain whether this is a suitable time and suggest an alternative if needed.

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6 Loans, investments and annuities 1 Chapter outline 6.01 6.02 6.03 6.04 6.05 6.06 6.07 6.08

Simple interest - Review Compound interest using sequences Compound interest using technology Effective annual interest rate Recurrence relations for present value annuities Formula for present value annuities Spreadsheet & TVM solver Applications of present value annuities Chapter 6 review

410 417 430 440 449 458 465 473 481

6.01 Simple interest - Review After this lesson, you will be able to… • recall the formula for simple interest • calculate simple interest for a given principal, rate, and time • solve practical problems involving simple interest • understand the components of simple interest (principal, interest rate, and time)

Simple interest When money is invested or borrowed, the lender earns an extra amount, and the borrower pays an extra amount. This is called interest. Simple interest is calculated only on the initial amount, known as the principal. The interest earned per period remains constant and does not get added to the principal for future interest calculations (i.e., it does not ‘compound’).

Exploration An investment of $1000 earns simple interest at a rate of 10% per annum. The interest earned each year and the total value of the investment are shown in the table: Year

Interest earned

Total value

$100

$1100

$100

$1200

$100

$1300

$100

$1400

$100

$1500

1. What is noticeable about the interest earned each year? 2. How would the growth in the total value of the investment be described? 3. Can a rule be developed to calculate the total interest earned after n years?

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The amount of simple interest, I, earned or paid is given by the formula:

I = Pin I

is the interest earned or paid

is the initial amount, or principal

i is the interest rate per period (expressed as a decimal or fraction) n

is the number of periods

The total amount, An, after n periods is:

An = P + I This expression shows the total value of the investment or loan at the end of n time periods.

Example 1 An investment of $6350 earns simple interest at a rate of 6% per annum: a Calculate the simple interest earned after 13 months.

Create a strategy Use the simple interest formula, I = Pin. Ensure the time period n is in years to match the annual interest rate.

Apply the idea The values are P = 6350, i = 0.06, and n =

years. Write the formula

Substitute the values

Evaluate

The interest earned is $412.75.

b Calculate the total value of the investment after 13 months.

Create a strategy Add the interest earned to the original principal using the formula A = P + I.

Apply the idea A=P+I

Write the formula

= 6350 + 412.75

Substitute the values for principal and interest

= 6762.75

Evaluate

The total value of the investment is $6762.75.

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Simple interest and sequences Simple interest involves adding a constant amount of interest each period, which means the total value of the investment or loan grows linearly. This pattern can be modelled using an arithmetic sequence. Let An be the total amount after n periods. The sequence starts with the principal, A0 = P. The constant amount added each period is the simple interest for one period, which is the common difference, d: d=P×i The value of the investment or loan can be modelled using a recurrence relation. The recursive formula is: An + 1 = An + d, where A0 = P This formula shows that the value for the next period, An + 1, is found by adding the interest, d, to the current value, An. To determine the value after n periods directly, an explicit formula can be used: An = A0 + nd Substituting A0 = P and d = Pi gives: An = P + n(Pi) = P (1 + in)

Example 3 Manpreet invests 56 000 INR (Indian rupees) into an account that pays 6.9% p.a. simple interest: a By what amount will the investment increase each year?

Create a strategy Calculate the simple interest for one year using I = Pin.

Apply the idea The values are P = 56 000, i = 0.069, and n = 1. I = Pin

Write the formula

= 56 000 × 0.069 × 1

Substitute the values

= 3864

Evaluate

The investment will increase by 3864 INR each year.

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b Write a recurrence relation to model the investment’s value, where Vn is the value after n years.

Create a strategy Use the recursive form An + 1 = An + d, with A0 = P, where d is the annual interest.

Apply the idea The initial investment is V0 = 56 000. The annual interest (common difference) is d = 3864. The recurrence relation is: Vn + 1 = Vn + 3864, where V0 = 56 000

c Write an explicit rule to determine the value of the investment after n years.

Create a strategy

Apply the idea

Use the explicit formula An = P (1 + in) or An = A0 + nd.

Using Vn = V0 + nd, with V0 = 56 000 and d = 3864, the explicit rule is: Vn = 56 000 + 3864n

d Determine the value of the investment after 7 years.

Create a strategy Substitute n = 7 into the explicit rule from the previous part.

Apply the idea V7 = 56 000 + 3864 × 7

Substitute n = 7 into the rule

= 56 000 + 27 048

Evaluate the multiplication

= 83 048

Evaluate

The balance after 7 years is 83 048 INR.

e Determine how many whole years it takes for the value to exceed 108 164 INR.

Create a strategy Set up the inequality Vn > 108 164 using the explicit rule and solve for n.

Apply the idea Set up the inequality

Subtract 56 000 from both sides

Divide both sides by 3864

Evaluate

Since n must be a whole number of years, the value will exceed 108 164 INR after 14 years.

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Ex 3

Sophie invests $10 000 into an investment account that pays 5% p.a. simple interest: a

By what amount will the account increase each year?

Complete the recurrence relation for Sophie’s situation, where Tn is the balance at the end of the nth year and T0 is the initial investment:

Tn + 1 = Tn + ⬚, where T0 = ⬚

Complete the explicit rule that can be used to determine the balance at the end of n years: Tn = ⬚ + n × ⬚ which simplifies to Tn = ⬚ + ⬚n

Determine the balance after 5 years.

Determine how many whole years it takes for the balance to exceed $14 900.

Layla invests $22 000 into a ﬁxed-term savings account that earns 10% p.a. simple interest: a

By what amount will the investment grow each year?

Determine the total value of the investment after 5 years.

Determine how many whole years it will take for the investment to exceed $40 000.

Complex familiar 11

The recurrence relation Vn + 1 = Vn + 11.5, V0 = 1150 models the value of an investment ($) after n + 1 months: a

Is the investment type simple interest or compound interest?

Determine the annual interest rate.

Determine the balance of the account after 8 years.

The recurrence relation Sn + 1 = Sn + 350, S0 = 5000 models the total amount saved ($) in a savings account after n + 1 years: a

Is the savings type simple interest or compound interest?

Determine the annual interest rate.

Determine the balance of the account after 5 years.

This graph shows the value of an investment that earns simple interest over a period of 5 years: a

Identify the initial principal of the investment.

Determine the amount of interest earned each year.

Calculate the annual simple interest rate.

Write a recurrence relation to model the value of the investment, where An is the value after n years.

Value ($) 2500 2000 1500 1000 500

Time (years) 0

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Complex unfamiliar 14

A principal of P dollars earns simple interest at an annual rate of 5% p.a. over 3 years. The total amount after interest is $6900: a

Write an equation to represent the total amount in terms of P.

Determine the value of P.

Two investments have the same principal and simple interest rate of 3.5% p.a. Investment A runs for 2 years, and Investment B runs for 3 years. The difference in their total amounts is $210: a

Express the total amount of each investment in terms of the principal, P.

Calculate the principal amount.

6.02 Compound interest using sequences After this lesson, you will be able to… • use a recurrence relation to model a compound interest loan or investment • understand the components of the recurrence relation for compound interest (An+1, An, r, i) • use the compound interest formula A = P(1 + i)n to model a compound interest loan or investment • identify the variables (A, P, i, n) in the compound interest formula

Compound interest formula When money is invested or borrowed, the person who invests earns extra money, and the person who borrows pays extra money for using that money. This extra amount is called interest. With compound interest, the interest is calculated not only on the initial amount, the principal, but also on the interest that has already been added in previous periods. This causes the total amount to grow faster over time, because the interest itself earns more interest. Unlike simple interest, where the amount increases by the same value each period, compound interest increases by a growing amount. The total gets larger more quickly as each period builds on a bigger balance.

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The formula used to calculate the total amount after n periods is:

A = P (1 + i)n A

is the total amount

is the principal (initial amount)

i is the interest rate per compounding period (expressed as a decimal or fraction) n

is the number of compounding periods

This formula is used to determine how much an investment or loan will grow when interest is added again and again over time. To calculate the interest, I, earned on an investment or charged on a loan, subtract the principal from the ﬁnal amount:

I=A−P Example 1 A $1110 investment earns interest at 2.2% p.a., compounded annually, over 18 years: a Calculate the value of this investment rounded to the nearest cent.

Create a strategy Use the compound interest formula: A = P (1 + i)n

Apply the idea P = 1110, i = 0.022, and n = 18. Substitute these into the formula: A = P (1 + i)n

Write the formula 18

= 1110 × (1 + 0.022) = 1110 × (1.022)

= $1642.25

Substitute the values Evaluate the addition Evaluate and round

Therefore, the value of the investment is $1642.25, rounded to the nearest cent. b How much interest is earned over the 18 years? Give the answer to the nearest dollar.

Create a strategy Use the formula I = A − P, that represents the interest as the difference between the total amount and the original principal.

Apply the idea I=A−P

Write the formula

= 1642.25 − 1110

Substitute the values of A and P

= $532.25

Evaluate

Therefore, the interest earned is $532, rounded to the nearest dollar.

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Use of a spreadsheet to calculate compound interest Spreadsheets offer a practical way to explore how compound interest builds over time. By setting up a table, it is possible to track how the balance changes each period, calculate the interest added, and see how the total grows. This method makes calculations easier, shows how changing interest rates or compounding periods affect the results, and allows for testing different scenarios.

Interactive exploration Discover this concept in action online

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Example 3 This spreadsheet shows the balance, in dollars, in a savings account in 2024, where interest is compounded monthly: A

Month

Balance at beginning of month

Interest

Balance at end of month

July

8000

160

August

8160

163.20

8323.20

September

8323.20

8489.66

October

169.79

8659.45

November

8659.45

173.19

8832.64

a Calculate the value of X, Y, and Z.

Create a strategy Use the spreadsheet to solve the unknown value.

Apply the idea Calculate the balance at the end of July, X, using the formula, B2 + C2. X = B2 + C2

Write the formula

= 8000 + 160

Substitute the given values

= $8160

Evaluate

Calculate the interest, Y, using the formula, D4 − B4. Y = D4 − B4

Write the formula

= 8489.66 − 8323.20

Substitute the given values

= $166.46

Evaluate

Calculate the balance at the beginning of the month, Z, using the formula, D5 − C5. Z = D5 − C5

422

Write the formula

= 8659.45 − 169.79

Substitute the given values

= $8489.66

Evaluate

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b Use the numbers for July to calculate the monthly interest rate.

Create a strategy Use the interest rate formula:

Apply the idea Write the formula

Substitute the given values

Evaluate

Therefore, the monthly interest rate is 2%.

c Write a recursive rule for the balance Bn that gives the balance at the end of the nth month, including the initial condition B0, where July is the ﬁrst month.

Create a strategy Use the recursive formula: An + 1 = rAn, where A0 = P

Apply the idea The investment is increased by 2% each month. Therefore, the balance at the end of a month is 102% or 1.02 of the balance at the end of the previous month. The initial value of investment is $8000. So, the complete recurrence relation is: Bn + 1 = 1.02Bn, B0 = 8000

d Write an explicit rule, An, that gives the balance at the end of the nth month, with July being the ﬁrst month.

Create a strategy Use the compound interest formula: An = P (1 + i)n

Apply the idea For this scenario, the values are P = 8000, and i = 0.02. An = P (1 + i)n

Write the formula n

= 8000(1 + 0.02) = 8000(1.02)

Substitute the given values Evaluate

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Ex 2

The balance of an investment, in dollars, at the beginning of each quarter where interest is compounded quarterly is given by An + 1 = 1.02An, A0 = 5000: a

Determine the quarterly interest rate.

Determine the annual interest rate.

Calculate the balance at the beginning of the second year.

Calculate the balance at the beginning of the third year.

The balance of an investment, in dollars, at the end of each month where interest is compounded monthly is given by An + 1 = 1.015An, A0 = 2000. The investment began at the beginning of January 2020: a

Determine the monthly interest rate.

Use appropriate technology to calculate the balance at the end of the ﬁrst year.

Use the compound interest formula to calculate the balance at the end of the ﬁrst year and conﬁrm the answer from the previous part.

Use appropriate technology to determine in which month and year the investment is worth double the initial amount invested.

The value of land which is currently priced at $520 per square metre, is expected to grow by 7.9% each year for the next 4 years: a

Determine the recurrence relation for this situation.

Calculate the expected value of the land at the end of 4 years.

Callum invests $5700 into an investment account that pays 3.2% per annum, compounded annually: a

Determine the recurrence relation for this situation.

Write an explicit rule that can be used to calculate the balance at the end of n years.

Calculate the balance after 9 years.

Determine how many whole years it takes for the balance to exceed $10 208.

$3900 is invested for 3 years at a rate of 10% per annum, compounding annually. The balance and interest for the ﬁrst 2 years are shown in the table: a

Calculate the value of: i

iii

Balance

Interest earned

First year

$3900

$390

Second year

$4290

$429

Third year

Fourth year

−

Calculate the total interest earned over the 3 years.

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Ex 3

$8000 is invested at 6% p.a., compounded annually: a

Determine the recurrence relation for this situation.

Complete this table: Time period (years)

Value at beginning of time period

Value at end of time period

Interest earned in time period

$8000

⬚

3 c 11

$8988.80

$8988.80 $9528.13

Calculate the total interest earned over the 3 years.

⬚ ⬚

Erica invests $50 000 into an investment account that pays 2.8% per annum, compounded annually: a

Determine the recurrence relation for this situation.

Write an explicit rule that can be used to calculate the balance at the end of n years.

Determine the balance after 21 years.

Determine how many whole years it takes for the balance to exceed $109 843.

$5000 is invested at the beginning of the year in an account that earns interest at 3% per annum, compounded annually: a

How much money is in the account at the end of the ﬁrst year?

Write a recursive rule, Vn + 1, that gives the balance in the account at the end of year n, including the initial investment V0.

Mae invested $1800 at 6% p.a., compounded annually, over 3 years. Use appropriate technology to calculate these values:

426

The recurrence relation for this situation

The interest for the ﬁrst year

The total amount of the investment after the ﬁrst year

The interest for the second year

The total amount invested after the second year

The interest for the third year

The total amount invested after the third year

The total amount of interest earned over the 3 years

The interest as a percentage of the initial investment, rounded to two decimal places

The simple interest earned after 3 years

How much more compound interest than simple interest would have accumulated over these 3 years?

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Sharon has $25 000 that she wishes to invest for a period of time without touching it. She chooses to invest this money in an account offering 4.25% p.a. compound interest: a

Calculate the value of the investment in dollars after 7 years if interest is compounded monthly.

Calculate the number of years required to double her investment if interest is compounded daily. Round your answer to two decimal places. Assume there are 365 days in a year.

Rani invests 27 500 INR into an investment account that pays 6.3% compound interest per annum, compounded monthly: a

Determine the recurrence relation for this situation.

Write an explicit rule that can be used to calculate the balance at the end of n months.

Calculate the balance after 4 years.

Determine how many whole months it takes for the balance to exceed 35 543 INR.

Convert your answer from part (d) to years, rounded to two decimal places.

$2000 is invested at the beginning of the year in an account that earns interest at 4% per annum, compounded quarterly: a

How much money is in the account at the end of the ﬁrst year?

Write a recursive rule, Vn + 1, that gives the balance in the account at the end of the nth quarter, including the initial investment V0.

Han opened a savings account at the beginning of February 2025, where the interest is compounded monthly. His statements for March, April and May show his account balance at the beginning of each month: a

How much did Han deposit into this savings account when he opened it?

Month

Balance

Calculate the monthly interest rate of his investment.

March

$2550

Calculate the annual interest rate of his investment.

April

$2601

Write a recursive rule, Vn + 1, that gives the balance in the account at the beginning of the nth month, where n = 0 is the initial deposit.

May

$2653.02

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Neil and Carl both inherit $12 000 and put their money in compound interest-bearing accounts for a period of 5 years: Neil places his money in an account with an interest rate of 2.75% p.a. compounded monthly:

Number of months

...

Value of investment ($)

$12 027.50

...

$13 766.65

Calculate the values of these: i

Write a recursive rule for Neil’s investment Vn + 1 in terms of Vn, where Vn describes the value of the investment after the nth month (in exact form) and the initial investment V0.

Carl places his money in an account which earns interest compounded daily. At the end of the 5 years, Carl’s balance is the same as Neil’s balance. Calculate the interest rate per annum for Carl’s investment account as a percentage. Round your answer to three decimal places. Assume there are 365 days in a year.

Does the difference in compounding periods mean that Carl’s interest rate per annum is higher or lower than Neil’s?

This table shows the balance, in dollars, in a savings account in 2024, where interest is compounded quarterly: A

Quarter

Balance at beginning of quarter

Interest

Balance at end of quarter

$6000

120

$6120

122.40

$6242.40

124.85

$6367.25

⬚

428

Calculate the quarterly interest rate.

Complete the row for Quarter 4.

Write a recursive rule, Bn + 1, that gives the balance at the beginning of the nth quarter after the beginning of 2024, including the initial investment B0.

Calculate the balance at the beginning of the third year.

How many whole quarters after the beginning of 2024 will the balance be at least double the initial investment of $6000?

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This table gives the account balance of an investment after each week: a

Calculate the weekly interest rate as a percentage.

Determine the recursive rule for the investment Bn + 1 in terms of Bn, where Bn describes the value of the investment after the nth week. Include the initial investment B0.

Calculate the values of these: i

If interest was to be calculated weekly and added to the account monthly, would the balance be higher, lower or the same after 4 years? If the interest rate was doubled, what would the effect be on the amount of interest earned?

Week

Balance

$15 000

$15 675

$16 380.38

$17 117.49

$17 887.78

$18 692.73

...

Investor A puts $10 000 into a savings account offering 5% p.a., compounded annually. Investor B puts $10 000 into a different account offering 4.8% p.a., compounded monthly. a

Determine the balance in Investor A’s account after 10 years.

Determine the balance in Investor B’s account after 10 years.

Determine who has the greater balance after 10 years and by how much.

Complex unfamiliar 22

This table shows the balance, in dollars, in a savings account in 2024 where interest is compounded quarterly. Some values are missing: A

Quarter

Balance at beginning of quarter

Interest

$3060

⬚

$3060

$3121.20

⬚

62.42

$3183.62

63.67

$3247.29

D Balance at end of quarter

⬚

Write a recursive rule, Bn + 1, that gives the balance at the beginning of the nth quarter, including the initial investment B0. 23

Maria opened a savings account at the beginning of April 2024, where the interest is compounded quarterly. Her statements for June, September and December show her account balance at the beginning of each quarter. Write a recursive rule, Vn + 1, that gives the balance in the account at the end of the nth quarter, including the initial investment V0.

Month

Balance

June

$7800

September

$8112

December

$8436.48

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6.03 Compound interest using technology After this lesson, you will be able to… • solve practical problems involving compound interest loans or investments using technology • determine the total amount of a compound interest loan or investment using technology • calculate the total interest, principal, and interest rate per year or per compounding period for compound interest loans or investments using technology • analyse the effect of the interest rate and number of compounding periods on the total amount of a compound interest loan or investment using technology

Compound interest using technology In addition to spreadsheets, compound interest scenarios can be explored using a range of digital tools, including Time Value of Money (TVM) solvers. These are available on many calculators and online platforms. By entering known values such as the principal, interest rate, number of compounding periods per year, and duration, these tools apply or rearrange the compound interest formula to calculate the unknown value. There are also a variety of online calculators designed for specific financial purposes, such as comparing investment options or estimating loan repayments. These tools support deeper exploration of compound interest in practical, real-world contexts. TVM solvers and calculator tools typically use standard notation, making them a convenient and efficient way to model different financial scenarios: N

Total number of payments

Interest rate per annum (as a percentage, not a decimal)

Present value (the principal or starting amount)

PMT

Payment made every period (set to 0 if there are no regular payments)

Future value (the amount at the end or final amount)

P/Y

Number of payments per year

C/Y

Number of compounding periods per year

TVM solvers such as ﬁnancial calculators and apps follow speciﬁc conventions when working with money values. One key convention relates to the signs of the present value (PV ) and future value (FV ).

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In TVM solvers, the sign of the amount entered depends on the direction of the money ﬂow: • Borrowing: The bank gives you money now, so the PV is entered as positive. Since you’ll owe money in the future, the FV will be negative. • Investing: You give your money to the bank, so the PV is negative. In return, you receive money in the future, so the FV is positive. This sign convention helps the solver distinguish between money going out and money coming in. TVM solvers can also handle regular payments, such as monthly contributions to a savings account or loan repayments. If there are no regular payments, PMT is set to 0. Users must also enter the number of payments per year (P /Y ) and the number of compounding periods per year (C/Y ). When payments and compounding occur at the same frequency, the total number of periods (N ) is calculated by multiplying the number of years by C/Y.

Example 1 $13 000 is borrowed at an interest rate of 2.5% p.a. compounded semi-annually. Calculate the amount owed after 3.5 years. Round your answer to the nearest cent.

Create a strategy Use a ﬁnancial solver or technology to calculate the future value.

Apply the idea The inputs for the ﬁnancial solver are: N=7

Total periods: 3.5 × 2 = 7

I% = 2.5

The annual interest rate as a percentage

PV = 13 000

Positive because the borrowed amount is received

PMT = 0

There are no regular payments

FV = ?

This is the value to be calculated

P /Y = 2

Set to match C/Y for simplicity

C/Y = 2

Compounded semi-annually

Solving for FV returns approximately −14 181.056. The value is negative as it represents money owed. The amount owed after 3.5 years is $14 181.06, rounded to the nearest cent.

Reﬂect and check The compound interest formula can be used to verify the amount owed after 3.5 years: A = P (1 + i)n, where P = 13 000, i = A = 13 000

, and n = 7

≈ $14 181.06

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Example 2 Neil invests $900 in a term deposit with a rate of 2.3% p.a. compounded daily. How many years will it take for the investment to at least double in value? Assume there are 365 days in a year.

Create a strategy Use a ﬁnancial solver to calculate N and convert it to years.

Apply the idea The inputs for the ﬁnancial solver are: N=?

This is the value to be calculated

I% = 2.3

The annual interest rate as a percentage

PV = −900

Negative as the investment amount is given away

FV = 1800

The investment doubles in value

PMT = 0

There are no regular payments

P /Y = 365

Set to match C/Y

C/Y = 365

Compounded daily

Solving for N returns approximately 11 015 days. To convert this to years, divide by 365: Divide total days by 365

Evaluate

Since the investment must at least double, it will take 31 years.

Example 3 How much money must be invested today at an interest rate of 5% p.a. compounded quarterly to have $25 000 in 10 years? Round your answer to the nearest cent.

Create a strategy Use a ﬁnancial solver to calculate the present value (PV ), which is the initial investment amount.

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Apply the idea The inputs for the ﬁnancial solver are: N = 40

Total periods: 10 × 4 = 40

I% = 5

The annual interest rate as a percentage

PV = ?

This is the value to be calculated

PMT = 0

There are no regular payments

FV = 25 000

Positive as the future value is received back

P /Y = 4

Set to match C/Y

C/Y = 4

Compounded quarterly

Solving for PV returns approximately −15 252.684. The value is negative as it represents the initial investment (money given away). The amount that must be invested today is $15 252.68.

Example 4 An investment of $8000 grew to $10 500 over 5 years with interest compounded monthly. Determine the annual interest rate, rounded to two decimal places.

Create a strategy Use a ﬁnancial solver to calculate the annual interest rate (I%).

Apply the idea The inputs for the ﬁnancial solver are: N = 60

Total periods: 5 × 12 = 60

I% = ?

This is the value to be calculated

PV = −8000

Negative as the investment amount is given away

PMT = 0

There are no regular payments

FV = 10 500

Positive as the final value is received back

P/Y = 12

Set to match C/Y

C/Y = 12

Compounded monthly

Solving for I% returns approximately 5.4856%. The annual interest rate is 5.49% p.a., rounded to two decimal places.

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Example 5 A person wants to invest $15 000 for 7 years: • Option A offers an interest rate of 4.2% p.a. compounded monthly. • Option B offers 4.5% p.a. compounded monthly. a Calculate the future value for both options.

Create a strategy Use a ﬁnancial solver twice, once for each option, to calculate the future value (FV ).

Apply the idea For Option A, the inputs for the ﬁnancial solver are:

For Option B, the inputs for the ﬁnancial solver are:

N = 84

I% = 4.2

I% = 4.5

PV = −15 000

PMT = 0

FV = ?

P /Y = 12

C/Y = 12

Solving for FV gives $20 111.41.

Solving for FV gives $20 517.84.

Reﬂect and check With the same principal and compounding period, a higher interest rate results in a higher future value.

b Determine how much more interest is earned with the better option.

Create a strategy Subtract the future value of Option A from the future value of Option B. The difference in future values is equal to the difference in interest earned, since the principal is the same.

Apply the idea Difference = 20 517.84 − 20 111.41 = $406.43

Subtract the future values Evaluate

Option B earns $406.43 more in interest.

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Example 6 A business is taking out a $50 000 loan for 4 years at an interest rate of 5.6% p.a.: • Option 1 compounds interest quarterly. • Option 2 compounds interest fortnightly. Assume there are 26 fortnights in a year. a Calculate the total amount to be repaid for both options.

Create a strategy Use a ﬁnancial solver to calculate the future value for each option. This represents the total amount to be repaid.

Apply the idea For Option 1 (Quarterly), the inputs for the ﬁnancial solver are:

For Option 2 (Fortnightly), the inputs for the ﬁnancial solver are:

N = 16

N = 104

I% = 5.6

PV = 50 000

PMT = 0

FV = ?

P /Y = 4

P /Y = 26

C/Y = 4

C/Y = 26

Solving for FV gives −$62 442.23. The total amount to repay is $62 442.23.

Solving for FV gives −$62 522.61. The total amount to repay is $62 522.61.

b Recommend which option the business should choose, and explain why.

Create a strategy For a loan, the better option is the one that results in less interest paid, which means a lower future value.

Apply the idea Option 1 (quarterly compounding) results in a total repayment of $62 442.23, while Option 2 (fortnightly compounding) results in a total repayment of $62 522.61. The business should choose Option 1 because it results in paying less interest over the life of the loan.

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A savings account has a balance of $25 500 after 10 years of interest at 1.6% p.a. compounded quarterly. Calculate the initial deposit.

Ex 2

$1000 is placed in a term deposit with a rate of 5% p.a. compounded annually. How many whole years will it take for the balance to increase in value to at least $1500?

Ex 3

Maximilian wants to have $1500 at the end of 5 years. The bank offers 7.8% p.a. compounded annually. Calculate how much money he should invest now.

A collection of new shoes is valued at $1460. Determine the expected cost after 6 years if prices increase at an annual inﬂation rate of 6%.

A stamp collection is currently valued at $4770. If it appreciates at a rate of 10% per year, what will its value be after 6 years?

The price of oil is currently $76 per barrel. If oil prices are expected to increase by 7.1% per year due to dwindling supplies, what will the price of a barrel be in 29 years?

If prices rise at an inﬂation rate of 6.1% per year, what will a $250 basket of groceries cost in 9 years?

In 2018, a bottle of orange juice cost $2.60. If prices increased at an inﬂation rate of 11% per year, what would the price be in 2024?

Harry is expecting to receive a $7000 Christmas bonus in 6 months. His credit card charges interest at 14% p.a. compounded weekly with no interest free period. What is the maximum amount he can spend on his credit card now if he wants to fully pay it off with his bonus?

A $2000 investment grows to $4000 over 2 years with interest compounded monthly. What annual interest rate was applied? Give your answer as a percentage rounded to two decimal places.

Ex 4

Complex familiar 16

$218 is invested at 2% p.a. compounded annually for 5 years. After 5 years, the total amount (principal plus interest) is reinvested at 3% p.a. compounded annually for another 2 years. Calculate the ﬁnal value of the investment.

What interest rate per annum would be required to double an investment of $1000 in 2 years if the interest is compounded daily? Assume there are 365 days in a year. Give your answer as a percentage, rounded to two decimal places.

What interest rate per annum would be required to double an investment of $P in 4 years time, if the interest is compounded quarterly? Give your answer as a percentage, rounded to two decimal places.

Determine how many whole weeks it will take an investment of $P to at least double in value at an interest rate of 4.2% p.a. compounded quarterly?

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Ex 5

Ex 6

A block of land is currently worth $610 for every square metre. The price is expected to go up by 6% each year for the next 4 years. If a family wants to buy a 350 m2 block in 4 years, how much will it cost them in total? Round your answer to the nearest dollar.

Laura is deciding between two investment options: a high-risk investment which attracts interest at 10% p.a. compounded annually, and a low-risk investment at 5% p.a. compounded annually. She has $50 000 to invest and her goal is to grow it to $60 000 within 3 years. Which investment, if any, will help her reach her goal?

A loan of $15 000 is taken out at 6% p.a. compounded monthly for 5 years. After 2 years, the interest rate changes to 6.5% p.a. compounded monthly. Calculate the total interest paid over the entire 5-year term.

An investment of $10 000 is made for 8 years at an annual interest rate of 4%. Calculate the difference in the ﬁnal value of the investment if interest is compounded annually versus if it is compounded monthly.

Jack is aiming to save $35 000 after 10 years of investing at an interest rate of 0.1% per week compounded weekly. Calculate the amount he needs to deposit as a principal. Assume there are 52 weeks in a year.

An investment of $20 000 grows to $30 000 with interest compounded quarterly: a

If this growth took 10 years, calculate the annual interest rate.

If the interest rate was instead 5% p.a., determine how many years it would take to reach the $30 000 goal.

Complex unfamiliar 26

Neil invests $x in a term deposit with a rate of 2.3% p.a. compounded daily. How many years will it take for the investment to at least double in value? Assume there are 365 days in a year.

Vanessa plans to borrow $25 000 on the 2nd of January 2026 at an interest rate of 4.3% per annum, compounded annually, to pay for her undergraduate studies. Two years later, she plans to borrow another $10 000 under the same terms to start postgraduate study. She will not make any repayments until after her studies are complete. On what date will the total amount she owes reach or exceed $40 000? Give your answer, rounded to the nearest day.

If an investment doubles in value over time, does the original amount affect how long it takes for the investment to double in value? Use a TVM solver to explain your ﬁndings.

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6.04 Effective annual interest rate After this lesson, you will be able to… • calculate the effective annual rate of interest, ieffective • use the effective annual rate of interest to compare interest paid on loans or investments when interest is paid or charged for different compounding periods • solve practical problems involving compound interest loans or investments, including determining the total amount of the loan or investment • determine the total interest, principal, interest rate per year, and interest rate per compounding period for compound interest loans or investments

Effective annual interest rate When a bank advertises a loan with an interest rate of 6% per year, compounded monthly, this does not mean that only 6% interest is paid over the year. Because the interest is applied monthly, and each month’s interest is calculated on an increasing balance, the total interest paid over the year is actually higher than 6%. The effective annual rate of interest is the actual annual rate of interest paid on a loan or earned on an investment once the effect of compounding is taken into account. It allows for a fair comparison of ﬁnancial products that have different compounding periods (e.g., monthly, quarterly, daily). The advertised rate is known as the nominal interest rate.

Exploration An investment of $1000 is made for one year at a nominal interest rate of 5% per annum. Use the compound interest formula to calculate the ﬁnal amount and total interest earned if the interest is compounded: • Annually • Quarterly • Daily Based on your calculations: 1. In which scenario is the most interest earned? 2. For each scenario, what is the interest earned as a percentage of the principal? How does this ‘actual’ percentage compare to the advertised 5% nominal rate?

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The table shows that the amount of interest earned increases as the number of compounding periods increases: Compounding periods per year, k

1 (annually)

4 (quarterly)

365 (daily)

= 1000(1.05)

≈ 1000(1.05 095)

≈ 1000(1.05 127)

Final amount

$1050.00

$1050.95

$1051.27

Interest

$50.00

$50.95

$51.27

Calculation using A = P (1 + i)n

Effective rate × 100% = 5%

× 100% ≈ 5.10%

× 100% ≈ 5.13%

When interest is compounded annually, the effective rate is the same as the nominal rate. For all other compounding periods, the effective rate is higher than the nominal rate. To calculate the effective annual interest rate directly, use the formula:

ieffective = (1 + i)k – 1 ieffective is the effective annual rate of interest, expressed as a decimal i is the interest rate per compounding period, where i = k

and r is the nominal annual rate

is the number of compounding periods per year

The effective rate is essential for making informed ﬁnancial decisions. • When investing, a higher effective rate is better as it means more interest is earned. • When borrowing, a lower effective rate is better as it means less interest is paid.

Example 1 James invested $3000 at 4.6% p.a., compounded daily: a Calculate the amount of interest earned in the first year, assuming there are 365 days in a year. Round the answer to the nearest cent.

Create a strategy First, calculate the final amount using the compound interest formula. Then, subtract the principal from the final amount to find the interest earned.

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Apply the idea Write the formula

Substitute P = 3000, r = 0.046, k = 365 and t = 1

Evaluate and round to two decimal places

Now, calculate the interest: Interest = A − P

Write the formula

= 3141.21 – 3000

Substitute the values

= 141.21

Evaluate

The interest earned in the ﬁrst year is $141.21.

b Calculate the effective annual interest rate. Express it as a percentage rounded to two decimal places.

Create a strategy Divide the interest earned in the ﬁrst year by the principal and express the result as a percentage.

Apply the idea Write the formula

Substitute the values

Evaluate and round to two decimal places

Reﬂect and check Alternatively, use the effective annual interest rate formula:

As a percentage, this gives an effective annual interest rate of 4.71%.

Example 2 An investment earns interest at a nominal rate of 7.2% p.a., compounding semiannually. Calculate the effective annual interest rate, rounded to two decimal places.

Create a strategy Use the formula ieffective = (1 + i)k − 1, where i is the rate per compounding period and k is the number of periods per year.

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Apply the idea The nominal rate is r = 0.072. Interest is compounded semiannually, so k = 2. Write the formula

Substitute r = 0.072 and k = 2

Evaluate

As a percentage rounded to two decimal places, the effective annual interest rate is 7.33%.

Example 3 An investment account has an effective annual interest rate of 6.4%, with interest compounded monthly. Calculate the nominal annual interest rate, rounded to two decimal places.

Create a strategy Use the effective interest rate formula, substitute the known values, and rearrange to solve for the nominal rate, r.

Apply the idea The known values are ieffective = 0.064 and k = 12 (compounded monthly). Write the formula

Substitute the known values

Add 1 to both sides

Take the 12th root of both sides

Subtract 1 from both sides

Multiply both sides by 12 and swap sides

Evaluate and round

The nominal annual interest rate is approximately 6.21%.

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Example 4 An investment ﬁrm offers two options: • Option A: 6.2% p.a., compounded quarterly. • Option B: 6.15% p.a., compounded monthly. By calculating the effective annual rate for each, determine which option is the better investment.

Create a strategy Calculate the effective annual interest rate for each option. The option with the higher effective rate is the better investment.

Apply the idea Option A: Nominal rate r = 0.062, compounded quarterly so k = 4. Substitute values into the formula

Evaluate

The effective rate for Option A is approximately 6.35%. Option B: Nominal rate r = 0.0615, compounded monthly so k = 12. Substitute values into the formula

Evaluate

The effective rate for Option B is approximately 6.33%. Comparing the two rates, 6.35% > 6.33%. Therefore, Option A is the better investment as it has a higher effective annual interest rate.

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Ex 2

Ex 3

Calculate the effective annual interest rate, rounded to two decimal places: a

A $1720 investment earns interest at 4.6% p.a., compounded quarterly, over 3 years.

A $5380 investment earns interest at 4% p.a., compounded monthly, over 17 years.

A $2460 investment earns interest at 4.3% p.a., compounded semiannually, over 11 years.

A $6490 investment earns interest at 3.2% p.a., compounded weekly, over 3 years.

A $5000 investment earns interest at 2.8% p.a., compounded daily, over 19 years. Assume there are 365 days in a year.

Buzz’s investment of $1590 earns interest at 4.7% p.a., compounded daily, over 19 years. Assume there are 365 days in a year.

Sam’s investment of $3090 earns interest at 2.5% p.a., compounded weekly, over 4 years. Assume there are 52 weeks in a year.

Mo invested $3000 at 4.6% p.a., compounded daily. Assume there are 365 days in a year.

Calculate the effective annual rate of interest, rounded to one decimal place: a

A loan that attracts interest at a nominal rate of 6.7% p.a., compounded monthly.

A loan that attracts interest at a nominal rate of 1.8% p.a., compounded annually.

An investment that earns interest at a nominal rate of 9.1% p.a., compounded quarterly.

An investment that earns interest at a nominal rate of 7.2% p.a., compounded semiannually.

An investment that earns interest at a nominal rate of 4.3% p.a., compounded daily.

A loan of $3200 has earned interest of $768 after the ﬁrst year. What is the effective annual interest rate on this loan?

Calculate the nominal annual interest rate of these effective rates. Express your answer as a percentage, rounded to two decimal places:

446

3.2% p.a., compounded quarterly

2.7% p.a., compounded monthly

13.9% p.a., compounded quarterly

13.1% p.a, compounded monthly

An investment has a nominal interest rate of 12% p.a.. Calculate: a

The quarterly interest rate

The monthly interest rate

The effective annual interest rate if interest is compounded quarterly, rounded to two decimal places.

An investment has a nominal interest rate of 10.4% p.a.. Calculate: a

The weekly interest rate

The effective annual interest rate, rounded to two decimal places

An investment has a nominal interest rate of 7.3% p.a.. Calculate: a

The daily interest rate

The effective annual interest rate, rounded to two decimal places

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 4

Mae made an investment at a rate of 5% p.a.. Calculate the effective annual interest rate for Mae’s investment, rounded to four decimal places, if the interest was compounded: a

Annually

Semiannually

Quarterly

Monthly

Weekly, assume there are 52 weeks in a year

Daily, assume there are 365 days in a year

Valentina is deciding between two investment options, both over a 2-year period: • Option A offers 3.6% p.a., compounded quarterly • Option B offers 3.55% p.a., compounded monthly Calculate the effective annual rate for both options to determine which option is the better investment.

Sami is comparing two savings accounts offered by different banks: • Bank A offers 4.1% p.a., compounded monthly • Bank B offers 4.05% p.a., compounded weekly Calculate the effective annual rate for both options to determine which account would give Sami a better return on his savings over time.

Explain why an investment earning interest at 6% p.a., compounded monthly, provides a better return than one earning 6% p.a., compounded annually.

Complex familiar 14

Dave’s investment of $17 380 earned 6.2% p.a. simple interest for the ﬁrst 5 years and 0.29% p.a. compound interest for the next 8 years, compounded daily: a

Calculate the value of the investment after 5 years.

Calculate the ﬁnal value of the investment.

Calculate the total amount of interest earned.

Calculate the average amount of interest earned per year.

Express the average interest per year as a percentage of the original investment. Round your answer to two decimal places.

An investment offers an effective annual interest rate of 11% p.a., with interest compounded semiannually. A loan has the same effective rate, but interest is compounded quarterly. Calculate the nominal annual interest rate for the loan, rounded to two decimal places.

Two different lending institutions are offering different rates on their loans: • Betta Bank is offering 4.4% p.a., compounded monthly. • Lucky Lending is offering 4.2% p.a., compounded quarterly. Which institution has the better offer? Justify your answer with calculations.

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6.05 Recurrence relations for present value annuities After this lesson, you will be able to… • describe how the balance of a reducing balance loan or pension changes over time • model a present value annuity using the recurrence relation An+1 = rAn - d • use a recurrence relation to calculate the balance of an annuity after a small number of compounding periods

Reduce balance loans or pensions using recurrence relations The present value of an annuity is the amount of money needed now to fund a series of future payments, taking into account the interest that will be added over time. For example, imagine someone takes out a car loan for $20 000. They agree to repay this amount in equal monthly payments over several years. At the start of each month, interest is added to the remaining balance, and then a ﬁxed payment is made. This type of setup is called a reducing balance loan, because the balance gets smaller over time as repayments are made. This kind of situation is an example of an ordinary annuity, where periodic payments (like loan repayments or pension withdrawals) are made at the end of each period — usually weekly, monthly, or yearly — after interest has been applied. This table shows a $5000 loan with a 0.5% monthly interest rate and $200 payments. Each period, interest is added before the payment reduces the balance: Loan balance over time Period

Starting balance

$5000.00

Interest added

Payment

Ending balance

$0.00

$5000.00

$25.00 (0.5% of $5000.00)

$200.00

$4825.00

$24.13 (0.5% of $4825.00)

$200.00

$4649.13

$23.25 (0.5% of $4649.13)

$200.00

$4472.38

Here are some important terms when discussing a $10 000 car loan repaid monthly over 3 years with a monthly interest rate of 1%: Present value The current worth of a future sum of money or stream of cash ﬂows, discounted at a speciﬁc interest rate. Example: The $10 000 borrowed today to buy the car is the present value of all future repayments.

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Ordinary annuity A series of equal payments made at the end of each compounding period, with interest calculated before the payment. Example: The monthly car loan repayments are made after interest is added for the month. Recurrence relation A mathematical equation that deﬁnes a sequence where each term is expressed as a function of previous terms. Example: A relation that shows that each month, interest is added to the car loan balance and then the monthly payment reduces it. Reducing balance loan A loan where periodic payments reduce the outstanding balance, with interest calculated on the remaining balance before each payment. Example: After each car repayment, the balance gets smaller, so the interest next month is calculated on a lower amount. Periodic payment A ﬁxed amount paid or received at regular intervals (e.g., monthly or yearly). Example: The car loan requires monthly payments of the same amount over 3 years. Compounding period The time interval at which interest is calculated and added to the principal. Example: Interest on the car loan is calculated monthly — the compounding period is 1 month. Interest rate per compounding period The rate of interest applied to the principal for each compounding period, denoted as i. Example: The loan charges 1% interest per month — this is the monthly interest rate.

Exploration Consider a loan of $1000 for a laptop with a 1% monthly interest rate. Repayments of $100 are made at the end of each month, after interest is added. Explore how the loan balance changes: 1. Start with $1000. Add 1% interest, then subtract $100. What is the balance after 1 month? 2. Using the balance calculated in the previous step, add 1% interest again, then subtract $100. What is the balance after 2 months? 3. Determine a rule to calculate the next month’s balance from the current month’s balance.

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To model how the balance changes from 1 period to the next, a recurrence relation can be used:

An + 1 = rAn − d An + 1 is the total amount at the beginning of the (n + 1)th period is the total amount at the beginning of the nth period An r is the compounding factor, where r = 1 + i, and i is the interest rate per compounding period d

is the periodic payment

Here, rAn calculates the balance after adding interest, and subtracting d accounts for the payment. Balance($)

Balance($)

3000

2500

2000

1500

1000

500 0

500

Period (months) 1

Loan with 1% interest and $150 payments

Period (months) 1

Loan with 1% interest and $200 payments

The recurrence relation is used to calculate balances iteratively and provides a foundation for exploring more direct methods for calculating present values.

Example 1 A pension pays $1000 annually, with a 2% annual interest rate. The initial balance is $10 000. Describe how the balance changes over the ﬁrst year.

Create a strategy Explain the process of a reducing balance pension by identifying the interest added and payment withdrawn in the ﬁrst year.

Apply the idea

Reﬂect and check

The pension starts with a balance of $10 000. At the end of the year, 2% interest is calculated on this balance, which is 0.02 × 10 000 = $200. This increases the balance to $10 200.

The balance decreases because the payment exceeds the interest earned, which is typical for a reducing balance pension.

Then, the annual payment of $1000 is withdrawn, reducing the balance to 10 200 − 1000 = 9200. The balance after one year is $9200.

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Example 2 A $5000 loan has a 0.5% monthly interest rate and $200 monthly payments: a Write the recurrence relation that models the balance of the loan.

Create a strategy Calculate r = 1 + i, where i is the monthly interest rate. Use the formula An + 1 = rAn − d, with the given payment d.

Apply the idea The interest rate is i = 0.005 (since 0.5% = 0.005), so r = 1 + 0.005 = 1.005. The payment is d = 200. The recurrence relation is: An + 1 = 1.005An − 200, A0 = 5000

Reﬂect and check The formula adds interest (1.005An) before subtracting the payment, matching the ordinary annuity structure.

b Calculate the balance after 1 month.

Create a strategy Use the recurrence relation with the initial balance A0 = 5000 to determine A1.

Apply the idea An + 1 = rAn – d A1 = 1.005 × 5000 − 200

Write the formula Substitute A0 = 5000

= 5025 − 200

Evaluate the multiplication

= 4825

Evaluate

The balance after 1 month is $4825.

Reﬂect and check Interest of $25 was added before the payment, showing the reducing balance process.

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c Calculate the balance after 2 months.

Create a strategy Apply the recurrence relation with A1 = 4825 to determine A2.

Apply the idea An + 1 = rAn − d

Write the formula

A2 = 1.005 × 4825 − 200

Substitute A1 = 4825

= 4849.125 − 200

Evaluate the multiplication

= 4649.125

Evaluate

The balance after 2 months is $4649.13.

Reﬂect and check The balance decreases as the payment exceeds the interest added, which is typical for a reducing balance loan.

Example 3 A loan of $3000 accrues interest at a rate of 1% per month. The loan can be repaid using two different payment plans: Plan A, with monthly repayments of $150, or Plan B, with monthly repayments of $200. a For each repayment plan, write a recurrence relation that models the loan balance, where interest is applied before each payment.

Create a strategy Calculate r = 1 + i with the given interest rate, and identify An + 1 = rAn − d for each payment d.

Apply the idea The interest rate is i = 0.01, so r = 1 + 0.01 = 1.01. For Plan A, d = 150: An + 1 = 1.01An − 150, A0 = 3000 For Plan B, d = 200: An + 1 = 1.01An − 200, A0 = 3000

Reﬂect and check Both relations use the same interest factor, but the larger repayment reduces the balance more quickly each month.

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b Calculate the loan balance after 3 months for both repayment plans. Compare the results and determine which plan results in a lower outstanding balance after the third month.

Create a strategy Apply each recurrence relation An + 1 = rAn − d for 3 months, starting with A0 = 3000, and compare the results.

Apply the idea For Plan A: A1 = 1.01 × 3000 – 150

Calculate balance for Month 1

= 2880 A2 = 1.01 × 2880 − 150

Calculate balance for Month 2

= 2758.80 A3 = 1.01 × 2758.80 – 150

Calculate balance for Month 3

≈ 2636.39 For Plan B: A1 = 1.01 × 3000 – 200

Calculate balance for Month 1

= 2830 A2 = 1.01 × 2830 – 200

Calculate balance for Month 2

= 2658.30 A3 = 1.01 × 2658.30 – 200

Calculate balance for Month 3

≈ 2484.88 Plan

Payment

Balance after 3 months

$150

$2636.39

$200

$2484.88

Plan B, with the $200 repayment, results in a lower outstanding balance after the third month.

Reﬂect and check The $200 payment reduces the balance by about $151.51 more, showing that larger payments clear loans faster if the interest rate is the same.

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Ex 3

For the recurrence relation, An + 1 = 1.01An − 100 with A0 = 10 000, determine what happens to the balance after one period. A

The balance increases.

The balance stays the same.

The balance decreases.

Determine whether each statement about ordinary annuities is true or false: a

In an ordinary annuity, interest is added after the periodic payment.

A reducing balance loan’s balance always decreases each period.

A loan of $5000 accrues interest at a rate of 2% per month. The loan can be repaid using two different payment plans: monthly repayments of either $200 or $250: a

For each repayment plan, write a recurrence relation that models the loan balance, where interest is applied before each payment.

Calculate the loan balance after four months for both repayment plans.

Determine which plan results in a lower outstanding balance after the fourth month.

A loan of $2000 accrues interest at a rate of 3% per month. The loan can be repaid using two different payment plans: monthly payments of either $100 or $125: a

For each repayment plan, write a recurrence relation that models the loan balance, where interest is applied before each payment.

Determine which plan results in a lower outstanding balance after the fourth month.

A loan of $6000 has a 0.6% monthly interest rate and $250 monthly payments. Calculate the balance after 2 months.

A pension starts at $12 000 with a 2% annual interest rate and $1500 annual payments:

Write the recurrence relation to model the balance.

Calculate the balance after 1 year.

A loan of $5000 has a 0.8% monthly interest rate. Calculate the balance after 1 month for each periodic payment: a

$300

$350

A car loan of $15 000 has a 0.5% monthly interest rate and $600 monthly payments: a

Write the recurrence relation to model the balance.

Calculate the balance after 3 months.

Using a loan of $1000 with a 1% monthly interest rate and $50 monthly payment, explain why interest is calculated before the payment in an ordinary annuity.

A student calculates the balance of a $4000 loan with a 0.5% monthly interest rate and $200 payment by subtracting the payment before adding interest:

456

Identify the error in the student’s method.

Calculate the correct balance after 1 month.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Match each loan scenario to its recurrence relation: a

$10 000 loan, 1% monthly interest, $500 payments

$5000 loan, 0% monthly interest, $200 payments

$8000 loan, 0.8% monthly interest, $400 payments

An + 1 = An − 200, A0 = 5000

An + 1 = 1.008An − 400, A0 = 8000

iii

An + 1 = 1.01An − 500, A0 = 10 000

Complex familiar 18

A loan of $7000 has a 0.9% monthly interest rate. Determine by how much more a $400 monthly payment reduces the balance after 2 months compared to a $350 payment. Round the answer to the nearest cent.

For a recurrence relation where the periodic payment equals the interest added each period (d = iAn), explain what happens to the balance over time. Verify with an example using a $5000 loan with a 1% monthly interest rate.

Complex unfamiliar 20

A loan of $10 000 has a 1.1% monthly interest rate. Determine the minimum monthly payment, d, required to reduce the balance by at least $600 after 1 month.

A pension starts at $25 000 with a 3% annual interest rate and $4000 annual payments. Determine how many years it takes until the balance ﬁrst falls below $18 000.

Justify why the recurrence relation An + 1 = rAn − d is suitable for modelling a reducing balance loan but would need to be modiﬁed to model an interest-only loan.

This table shows the balance of a loan over 2 periods: Period (months)

Balance

$8000.00

$7796.00

$7589.55

Identify the recurrence relation that models the loan.

Determine the balance after 4 months.

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6.06 Formula for present value annuities After this lesson, you will be able to… • identify the components of the present value annuity formula • use the present value annuity formula to calculate the initial value of a loan or pension • convert annual interest rates and time periods to match the compounding frequency

Present value annuities using a formula A present value annuity is the current value of a series of ﬁxed, regular payments, where interest is calculated before each payment. These periodic payments form an ordinary annuity. The present value annuity formula calculates this amount directly, which is more eﬃcient than using a recurrence relation. This is useful for scenarios like loans or retirement pensions where equal payments are made over time.

Exploration Imagine planning to receive $1000 at the end of each year for 3 years, with an annual interest rate of 5%: 1. How many steps would it take to calculate the present value using the recurrence relation An + 1 = (1 + i)An − d? 2. Why might a single formula be more eﬃcient than iterating the recurrence relation for a 30-year loan?

The present value annuity formula calculates the total present value in one step, accounting for the periodic payment, interest rate, and number of periods.

APV

is the present value of the annuity

is the periodic payment

i is the interest rate per compounding period, as a decimal n 458

is the number of compounding periods

Mathspace QLD 12 General Mathematics 2025 mathspace.co

This timeline shows payments of d at the end of each compounding period. Interest at rate i is applied before each payment. The formula discounts these future payments to their present value, APV. Ordinary annuity timeline APV = $500.00 i = 0.05

i = 0.05 i = 0.05

i = 0.05

d = $500.00 d = $500.00 d = $500.00 d = $500.00 d = $500.00 d = $500.00 d = $500.00 d = $500.00

4 Time (Periods)

The formula is eﬃcient because it combines all iterative steps from the recurrence relation into one calculation. Ensure i and n match the compounding period, such as monthly or yearly.

Example 1 A retirement pension pays $5000 at the end of each year for 10 years, with an annual interest rate of 4%: a Calculate the present value of the pension.

Create a strategy Identify the variables d, i, and n, then substitute them into the formula:

Apply the idea Here, d = 5000, i = 0.04, and n = 10. Substitute into the formula: Substitute the values

Evaluate the power

Evaluate the fraction

Evaluate and round

The present value is $40 554.50.

Reﬂect and check This means $40 554.50 today is equivalent to receiving $5000 annually for 10 years at 4% interest.

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b Interpret the present value in the context of the pension.

Create a strategy Explain what the calculated APV represents in terms of the pension’s current worth.

Apply the idea

Reflect and check

To be able to withdraw $5000 each year for 10 years at 4% interest, an amount of $40 554.50 needs to be invested today.

This amount is less than the total of all future payments ($50 000), because money received in the future is not worth as much as money received today.

Example 2 A loan requires monthly payments of $200 for 2 years, with an annual interest rate of 6%: a Convert the annual interest rate to a monthly rate.

Create a strategy Divide the annual interest rate by 12 to determine the monthly rate.

Apply the idea

Reﬂect and check

The annual interest rate is 6% = 0.06. The monthly rate is:

Ensure the rate matches the compounding period to avoid errors in the formula.

Divide by 12

Simplify

The monthly interest rate is 0.5% or 0.005.

b Calculate the present value of the loan.

Create a strategy Identify d, i, and n, then use the formula monthly payments.

. Note that n is the total number of

Apply the idea Here, d = 200, i = 0.005, and n = 2 × 12 = 24. Substitute into the formula: Substitute the values

Evaluate the fraction

Evaluate and round

The present value is $4512.57.

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Reﬂect and check This value represents the loan amount that could be borrowed today if it were to be repaid with $200 monthly for 2 years.

Example 3 A pension fund offers $3000 annually for 5 years, with an interest rate of 3% per year: a Calculate the present value with the given values.

Create a strategy Substitute d = 3000, i = 0.03, and n = 5 into the formula:

Apply the idea Substitute the values

Evaluate the fraction

Evaluate and round

The present value is $13 739.12.

Reﬂect and check This calculation is faster than using the recurrence relation, which would require 5 iterative steps.

b Recalculate the present value if the pension lasts 7 years.

Create a strategy Use the same formula with n = 7, keeping d = 3000 and i = 0.03.

Apply the idea Substitute the values

Evaluate the fraction

Evaluate and round

The present value is $18 690.85.

Reﬂect and check The present value increased because more payments over a longer period add to the annuity’s worth.

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Ex 1

Calculate the present value, APV, of an ordinary annuity for each scenario, where payments are made at the end of each year: a

d = $1000, i = 5%, n = 3

d = $1500, i = 6%, n = 4

Ex 3

d = $2000, i = 4%, n = 5

A retirement pension pays $3000 at the end of each year for 8 years, with an annual interest rate of 5%: a

Calculate the present value of the pension.

Interpret the result from part (a) in the context of the pension.

A scholarship fund pays $2500 at the end of each year for 6 years, with an annual interest rate of 4.5%: a

Calculate the present value of the scholarship payments. Round your answer to the nearest dollar.

Interpret the result in the context of the scholarship.

Convert an annual interest rate of 12% to: a

Ex 2

A monthly interest rate

A quarterly interest rate

A loan requires monthly payments of $250 for 3 years, with an annual interest rate of 5%: a

Convert the annual interest rate to a monthly rate.

Calculate the present value of the loan.

A customer buys a laptop for $2160 through a payment plan requiring monthly payments of $180 for 12 months. A ﬁnance company offers this plan at an annual interest rate of 6%: a

Convert the annual interest rate to a monthly rate.

Calculate the present value of the purchase.

A family chooses to pay $750 monthly for 2 years to cover private school tuition. A thirdparty ﬁnance provider is involved and charges 5% annual interest: a

Convert the annual interest rate to a monthly rate.

Calculate the present value of the tuition payments.

Determine whether each statement about the present value annuity formula is true or false: a

The present value is always less than the sum of periodic payments.

i is the annual interest rate.

n represents the number of years.

The formula applies only to payments made at the start of each period.

Calculate the present value of an ordinary annuity with a single payment of $5000 at the end of one year, with an interest rate of 4%.

A pension fund offers annual payments of $2500 for a certain number of years, with interest compounded annually at a rate of 4%: a

Calculate the present value of the pension if the payments last 4 years.

Recalculate the present value if the payments last 5 years instead.

Explain how increasing the number of years affects the present value.

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A loan requires monthly payments of $500 for 2 years, with an annual interest rate of 6%: a

Convert the annual interest rate to a monthly rate.

Calculate the present value, APV, of the loan.

Verify if the total payments exceed the present value.

A pension pays $2000 at the end of each quarter for 5 years, with an annual interest rate of 8%. Calculate the present value of the pension.

An ordinary annuity pays $1000 annually at an interest rate of 5%. Calculate the present value for: a

n = 5 years

Which has a higher present value, and why?

n = 10 years

Calculate the present value of an ordinary annuity with annual payments of $3000 for 6 years, at an interest rate of 2.5% per year.

A car loan requires monthly payments of $300 for 3 years, with an annual interest rate of 4.8%. Calculate the present value of the loan.

Complex familiar 17

An ordinary annuity pays $100 monthly for 10 years, with a monthly interest rate of 0.1%: a

Calculate the present value, APV.

Explain why the present value is close to the sum of the payments.

A scholarship fund offers payments of $4000 annually for 8 years, with an annual interest rate of 5%: a

Calculate the present value if the payments are made annually.

Calculate the present value if the fund instead pays the equivalent amount monthly (i.e., $4000 ÷ 12 per month) for 8 years at the same annual interest rate.

Which option has a higher present value, and why?

A business must choose between two ordinary annuity investments to accept: • Annuity A: $1500 annually for 6 years at 4% annual interest. • Annuity B: $1000 annually for 10 years at 5% annual interest. a

Calculate the present value of each annuity.

Recommend which annuity to choose, and explain your reasoning.

A person wants to take out a car loan of $20 000. They can get an interest rate of 6% per annum, compounded monthly, and wish to pay it off over 5 years. Calculate the monthly repayment they would need to make.

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Complex unfamiliar Explain why the present value annuity formula,

, fails when i = 0, and

propose an alternative method to calculate the present value in this case. 22

A lottery winner is offered a choice between two prizes: • Option A: A lump sum payment of $500 000 today. • Option B: An annuity that pays $35 000 at the end of each year for 20 years. The winner knows they can invest money safely at an interest rate of 4% p.a., compounded annually. By calculating the present value of the annual payments, determine which option is ﬁnancially better for the winner. Justify your answer.

6.07 Spreadsheet & TVM solver After this lesson, you will be able to… • use technology to efficiently find an unknown variable in an annuity • interpret a spreadsheet amortisation table to track the balance of a loan • calculate the total amount paid and the total interest paid over the life of a loan

Spreadsheet and TVM solver An ordinary annuity involves a series of equal payments made at regular intervals, where interest is calculated before each payment. While the present value can be found using the formula, technology such as spreadsheets and TVM solvers make solving practical problems easier by automating these calculations.

APV is the present value of the annuity (the initial loan amount) d

is the periodic payment

i is the interest rate per compounding period, as a decimal n

is the number of compounding periods

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Exploration A car loan of $10 000 is taken out with 5% annual interest, paid monthly over 3 years. This scenario can be modelled in a spreadsheet with columns for period, payment, interest, and balance, starting with the initial balance of $10 000. A

Period

Payment

Interest

Balance

−

10 000.00

294.92

41.67

9746.75

294.92

40.61

9492.44

294.92

39.55

9237.07

294.92

38.49

8980.64

294.92

37.42

8723.14

...

294.92

1.99

Total

10 617.12

617.12

185.59

1. How does increasing the periodic payment, d, affect the total interest paid? 2. What is the resulting balance after 12 months? , to verify the present value, APV, for this

3. Use the formula, loan.

A TVM solver, found on ﬁnancial calculators, simpliﬁes annuity calculations. The required inputs include values such as:

466

Total number of periods

Annual interest rate (as a percentage)

Present value

PMT

Periodic payment

Future value (usually 0 for loans)

P /Y

Number of payments per year

C/Y

Number of compounding periods per year

Mathspace QLD 12 General Mathematics 2025 mathspace.co

TVM solver N

3.50

25 000

PMT

− 400.23

P/Y

C/Y

This TVM solver shows inputs for a $25 000 loan at 3.5% interest over 6 years. The output PMT = $400.23 is the monthly payment.

Spreadsheets can model how a loan is paid off over time by showing the balance, interest, and principal repaid each period. This process is called amortisation. To ﬁnd the full cost of a loan, the total amount paid is calculated by multiplying the periodic payment (PMT ) by the total number of payments (N ). The total interest is then the total amount paid minus the initial loan amount (PV ).

Example 1 For a loan of $20 000 with 4% annual interest, paid monthly over 5 years: a Calculate the periodic payment, d, using the annuity formula.

Create a strategy Use the annuity formula to determine d.

Apply the idea The monthly interest rate is i =

, and n = 5 × 12 = 60 months.

Use the present value formula:

Substitute the values into the formula

Evaluate the bracket

Rearrange to solve for d

Evaluate and round

The monthly payment is $368.33.

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b Verify the result using a spreadsheet.

Create a strategy Set up a spreadsheet to model the loan and conﬁrm the payment reduces the balance to 0.

Apply the idea A spreadsheet model is used to verify the payment amount. The initial balance is entered as $20 000. For each of the 60 payment periods, interest is calculated using the formula Interest =

× Balance, and added to the balance before the payment of $368.33 is subtracted. Period

Payment

Interest

Balance

$20 000

$368.33

$66.67

$19 698.34

⋮

$368.33

$1.22

$0.05

After the ﬁnal payment, the balance is approximately $0, which conﬁrms the calculated payment is correct for paying off the loan.

Reﬂect and check The spreadsheet shows the loan is paid off, matching the formula’s result. Rounding d may cause a tiny ﬁnal balance, which can be adjusted in the last payment.

Example 2 A student loan of $30 000 has 4.5% annual interest, paid monthly over 7 years: a Use a TVM solver to determine the periodic payment, d.

Create a strategy Input the TVM solver values: N, I%, PV, FV, P /Y, C/Y, and solve for PMT.

Apply the idea N

7 × 12 = 84

4.5

30 000

PMT

P /Y

C/Y

Solving for PMT on the TVM solver gives − 447.66. Therefore, the monthly payment is $447.66.

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Reﬂect and check The negative value for PMT indicates a cash outﬂow (money being paid out), which is correct for a loan payment.

b Calculate the total payments and total interest.

Create a strategy Multiply PMT × N for total payments, then subtract PV for total interest.

Apply the idea Total payments = 447.66 × 84

Use PMT × N

= 37 603.44

Evaluate

Total interest = 37 603.44 − 30 000

Subtract PV

= 7603.44

Evaluate

The total amount paid is $37 603.44, and the total interest is $7 603.44.

c Verify the payment using a spreadsheet.

Create a strategy Model the loan in a spreadsheet with a $30 000 initial balance, monthly payment of $447.66, and interest rate

= 0.00 375. Check if the balance reaches 0 after 84 periods.

Apply the idea An amortisation schedule in a spreadsheet can verify this payment. The interest for each period is calculated on the previous balance (Interest = 0.00 375 × Previous balance) and added, then the payment of $447.66 is subtracted to find the new balance. After repeating this for 84 periods, the final balance is nearly $0, confirming the payment is correct. A small residual amount, like $0.02, is due to rounding. Period

Payment

Interest

Balance

$30 000

$447.66

$112.50

$29 664.84

⋮

$447.66

$1.67

$0.02

The payment is correct, as the loan is nearly paid off.

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Ex 1

For a loan of $15 000 with 5% annual interest, paid monthly over 5 years: a

Calculate the periodic payment, d, using the annuity formula.

Verify the result using a spreadsheet.

For a loan of $10 000 with 4% annual interest, paid monthly over 4 years: a

Calculate the periodic payment, d, using a spreadsheet.

Verify the result using the annuity formula.

For a $10 000 loan with monthly repayments, determine whether the statement is true or false about TVM solver inputs: a

The present value (PV ) is entered as positive.

The payment (PMT ) is entered as positive.

The future value (FV ) is entered as negative.

The sign of each value depends on whether the amount is received or paid.

A loan requires monthly payments of $300 over 36 months. Calculate the total amount paid.

For a $12 000 loan with 5% annual interest, monthly payments over 3 years: a

Calculate the monthly interest rate, i.

Write the spreadsheet formula for interest in period 1, with balance b = $12 000.

Write the formula for the new balance after period 1, given payment, d.

A $25 000 car loan at 4% annual interest is paid monthly over 5 years. Use a TVM solver to calculate the monthly payment, d.

A loan with monthly payments of $400 over 48 months has a present value of $18 000. Calculate the total interest.

A $20 000 loan at 6% annual interest has monthly payments of $608.44. Complete the spreadsheet table for periods 1 − 3.

Period

Payment

Interest

Balance

$20 000

$608.44

⬚

$608.44

⬚

A $15 000 loan at 3.5% annual interest is paid monthly: a

Using a TVM solver, calculate the monthly payment for a 4-year term.

Using a TVM solver, calculate the monthly payment for a 5-year term.

Identify which payment is lower.

A $10 000 loan at 5% annual interest is being repaid with monthly payments of $299.71. Calculate the balance after the ﬁrst payment has been made.

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Ex 2

A $22 500 loan at 5% annual interest is paid monthly over 6 years: a

Use a TVM solver to determine the periodic payment, PMT, rounded to two decimal places.

Calculate the total interest.

Verify the balance after period 1 using a spreadsheet.

Complex familiar 15

A student inputs I% = 0.06 for a 6% annual interest loan with monthly payments in a TVM solver. Determine the error.

A $30 000 loan at 4.8% annual interest has a monthly payment of $600. Using a TVM solver: a

Determine the number of months, N, needed to pay off the loan, rounding up to the next whole month.

Calculate the total interest.

Complex unfamiliar 17

For a $18 000 loan at 4% annual interest over 4 years, with monthly payments, explain why a TVM solver might give a slightly different ﬁnal balance than a spreadsheet. Suggest a way to minimise this difference.

A $40 000 loan with 5.5% annual interest is repaid over 8 years with monthly payments:

Calculate the monthly payment using a TVM solver.

If the payment increases by $100, estimate the new loan term, N, in months, rounding up to the next whole month.

Explain how increasing the payment affects total interest.

A spreadsheet for a $15 000 loan, repaid monthly, at 4% per annum interest shows incorrect balances. Identify the error and correct the balance for period 1. Period

Payment

Interest

Balance

$15 000

$355.22

$600

$15 244.78

Explain why the total interest of a loan increases with a longer term, even if monthly payments decrease. Illustrate with an example.

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6.08 Applications of present value annuities After this lesson, you will be able to… • use the present value annuity formula to model the present value of an ordinary annuity in practical situations • solve for different variables in annuity problems, including the present value, periodic payment, or loan balance • compare different financial scenarios by calculating and interpreting the total interest paid

Applications of present value annuities An ordinary annuity involves a series of equal payments made at regular intervals, where interest is applied before each payment. This structure is common for ﬁnancial products like loans and pensions.

Exploration Explore how a loan balance changes over time. Imagine a small loan of $500 with a monthly interest rate of 10% and a regular monthly payment of $150. Month

Opening balance

Interest added (10%)

Balance before payment

Payment

Closing balance

$500.00

$50.00

$550.00

$150.00

$400.00

$40.00

$440.00

$150.00

$290.00

⬚

Use the process from the ﬁrst two months to complete the table for Month 3.

1. What two calculations are performed each month to determine the new closing balance? 2. How does this process show the loan balance reducing over time, even though interest is added?

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Apply the idea Start with A0 = 20 000, i = 0.004 (so r = 1.004), and d = 400. For month 1: A1 = 1.004 × 20 000 – 400

Substitute values into An + 1 = rAn − d

= 20 080 − 400

Evaluate the multiplication

= 19 680

Evaluate

For month 2: A2 = 1.004 × 19 680 − 400

Use the previous balance A1

= 19 758.72 − 400

Evaluate the multiplication

= 19 358.72

Evaluate

For month 3: A3 = 1.004 × 19 358.72 – 400

Use the previous balance A2

≈ 19 436.15 − 400

Evaluate the multiplication

= 19 036.15

Evaluate

The balance after 3 months is $19 036.15.

Example 2 A retiree wants a pension paying $1000 monthly for 20 years, with an annual interest rate of 3%: a Calculate the present value needed to fund the pension using the annuity formula.

Create a strategy Use the formula

with n = 240 and i =

= 0.0025.

Apply the idea Substitute d = 1000, i = 0.0025, n = 240

Evaluate the fraction

Evaluate and round

The present value is approximately $180 475.60.

Reﬂect and check This means $180 475.60 must be invested today to fund $1000 monthly payments for 20 years at 3% annual interest.

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b Calculate total payments and total interest.

Create a strategy Determine total payments as 1000 × 240. Calculate the total interest by subtracting present value from total payments.

Apply the idea Total payments = 1000 × 240

Multiply payment by periods

= 240 000

Evaluate

Total interest = 240 000 − 180 475.60 = 59 524.40

Subtract present value Evaluate

The total payments are $240 000, and the total interest is $59 524.40.

Reﬂect and check The interest shows the extra amount paid due to the time value of money. Comparing interest across different rates could help retirees choose better pension plans.

Example 3 A family takes out a home loan of $600 000 to be repaid over 30 years. The interest rate is 5.4% per annum, compounded monthly. Calculate their required monthly payment, to the nearest cent.

Create a strategy Rearrange the present value annuity formula to calculate the periodic payment, d. The formula is .

Apply the idea Identify the known values: • Present value: APV = $600 000 • Interest rate per period: i =

= 0.0045

• Number of periods: n = 30 × 12 = 360 Substitute these values into the rearranged formula: Substitute the values

Simplify the numerator and evaluate the power

Simplify the denominator

Evaluate and round

The required monthly payment is $3369.98.

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Ex 2

Complete the table for a $8000 loan with 0.4% monthly interest, i = 0.004, and $150.00 payments, showing the balance, interest, and payment for each month for months 0 − 2 using the recurrence relation. Balance

Interest

Payment

$8000.00

1 2

⬚

$150.00

⬚

$150.00

A retiree needs $800 monthly for 15 years at a 3.6% annual interest rate. Calculate: a

Month

The present value

The total payments

A loan is repaid with 48 monthly payments of $350. Determine the total repayments.

Complex familiar 9

Ex 3

A $25 000 car loan has a 0.6% monthly interest rate and $500 monthly payments: a

Calculate the balance after 3 months using the recurrence relation.

Calculate the interest paid in month 3.

A loan of $10 000 is to be repaid over 2 years with monthly payments. The interest rate is 6% per annum, compounded monthly. Using the rearranged present value formula, , calculate the required monthly payment.

A $30 000 loan has a 4.8% annual interest rate over 5 years. Calculate the monthly payment using the present value formula.

An annuity consists of monthly payments of $600 for 6 years. Interest is calculated at a rate of 6% per annum, compounded monthly. Calculate:

478

Total repayments

Total interest

Present value

Match each scenario to the correct formula or value: i

Calculating loan amount for ﬁxed payments

Tracking loan balance monthly

An + 1 = rAn − d

iii

Total cost of payments

Total repayments minus present value

Extra cost of borrowing

d×n

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A graph shows present value versus number of periods for a $200 monthly payment at 0.5% monthly interest:

9000

Present value ($)

8000 7000

Estimate the present value for 12 months, rounded 6000 to two decimal places.

Estimate the present value for 24 months, rounded to two decimal places.

Explain why the present value needed increases as the number of periods increases.

5000 4000 3000 2000 1000 0

Months 12

A $10 000 loan has a 0% monthly interest rate and $200 monthly payments: a

Calculate the balance after 3 months using the recurrence relation.

Explain why the present value formula simpliﬁes when i = 0.

A pension requires $1200 monthly for 10.5 years at 2.4% annual interest: a

Calculate the present value.

Explain how to handle the half-year in practice.

A $50 000 loan can be repaid over: a

Five years at 4% annual interest with monthly payments Calculate the monthly payment and total interest.

Seven years at 3.6% annual interest with monthly payments Calculate the monthly payment and total interest.

c 18

Recommend an option, justifying your choice.

A student is asked to calculate the present value of a $400 monthly payment for 4 years at 6% annual interest. They provide an answer of $19 200, using the incorrect method APV = 400 × 48: a

Identify the conceptual error in the student’s method.

Calculate the correct present value.

A retiree wants to withdraw $2500 monthly for 20 years from a pension fund. Their ﬁnancial plan requires their investment to earn 4.2% per annum, compounded monthly: a

Calculate the lump sum (present value) they need to have saved to meet this goal.

The retiree successfully saves the amount calculated in part (a). However, the best interest rate they can secure is only 3.6% per annum, compounded monthly. Calculate the new, smaller monthly withdrawal this fund can actually sustain for the full 20 years.

Calculate the monthly shortfall between their desired withdrawal and the actual withdrawal possible with the lower interest rate.

A person has $200 000 in a retirement account earning 4.2% p.a. compounded monthly. They withdraw $1200 per month. Use the recurrence relation to determine the account balance after the ﬁrst 3 withdrawals.

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6 Chapter review SF

The value of an investment, Vn, after n years can be modelled by the recurrence relation Vn + 1 = 1.035Vn, where V0 = $12 000: a

Identify the principal amount of the investment.

Determine the annual compound interest rate as a percentage.

Calculate the value of the investment after 2 years.

An investment of $15 000 is made for 5 years. Compare the ﬁnal value of the investment if it is calculated using: a

Simple interest at 4.5% per annum

Compound interest at 4.5% per annum, compounded annually

Determine the difference in the interest earned between these two investment types.

An investor deposits $25 000 into an account that earns interest at 3.6% p.a., compounded monthly. After 3 years, the interest rate increases to 4.8% p.a., compounded monthly. Calculate the total value of the investment at the end of 5 years.

A student is choosing a savings account to invest their money. They are presented with two options: • Bank A: Offers a nominal rate of 4.2% p.a., compounded quarterly. • Bank B: Offers a nominal rate of 4.15% p.a., compounded monthly. By calculating the effective annual interest rate for each option, determine which bank offers the better return on investment. Show your calculations.

A person takes out a personal loan of $20 000. The loan accrues interest at a rate of 6% per annum, compounded monthly. The person makes monthly repayments of $450. Write a recurrence relation, in the form An + 1 = rAn − d, that models the balance of the loan, An, after n months.

Upon winning a small lottery, a person is offered a choice between two prizes: • Option 1: A lump sum of $150 000 paid immediately. • Option 2: An annuity that pays $10 000 at the end of each year for 20 years. The winner can invest their money at a guaranteed rate of 4.5% p.a., compounded annually. By calculating the present value of the annuity, determine which option is ﬁnancially superior and justify the answer.

A family takes out a home loan of $480 000 at an interest rate of 5.4% per annum, compounded monthly. They plan to pay the loan off over 30 years: a

Use a TVM solver or the present value formula to determine the required monthly repayment.

Calculate the total amount the family will repay over the life of the loan.

Calculate the total interest charged on the loan.

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Example 1 An investor starts with an initial balance of $5000 and makes monthly payments of $200 into an annuity with an annual interest rate of 6%, compounded monthly. Use a recurrence relation to calculate the balance after 3 months, rounded to two decimal places.

Create a strategy Deﬁne the recurrence relation using the formula, An + 1 = rAn + d, calculate the monthly interest rate, and iterate the relation for 3 periods starting from A0 = 5000.

Apply the idea Calculate the growth rate: Calculate the monthly interest rate from the annual rate of 6%

Evaluate

Calculate the compounding factor

Evaluate

The recurrence relation is An + 1 = 1.005An + 200, with A0 = 5000. Calculate the iterations: A1 = 1.005 × 5000 + 200 = 5225 A2 = 1.005 × 5225 + 200 = 5451.125 A3 = 1.005 × 5451.125 + 200 = 5678.38

Calculate the accumulated after 1 month Evaluate Calculate the accumulated after 2 months Evaluate Calculate the accumulated after 3 months Evaluate and round to two decimal places

The balance after 3 months is $5678.38.

Reﬂect and check To verify, note that each month, interest is calculated on the previous accumulated amount before adding the payment. For A1, interest is 5000 × 0.005 = 25, so the balance becomes 5000 + 25 + 200 = 5225, matching the calculated value of A1. Note that the rounding to two decimal places is only performed in the last stage of the calculation.

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7.01 Practice questions Simple familiar 1

Ex 1

Consider the recurrence relation for an ordinary annuity, An + 1 = rAn + d: a

Identify what An + 1 represents.

Identify what d represents.

An annuity has an annual interest rate of 6%, compounded monthly. Calculate the monthly interest rate, i.

For i = 0.005, calculate the compounding factor, r.

A savings account starts with $2000 and receives monthly payments of $100. The annual interest rate is 4%, compounded monthly. Calculate the balance after 4 months, rounded to two decimal places.

An investor starts with $3000 and makes monthly payments into an annuity with an annual interest rate of 5%, compounded monthly. Calculate the balance, rounded to two decimal places, for these payment amounts and periods:

Monthly payment of $150, after 2 months

Monthly payment of $150, after 3 months

Monthly payment of $200, after 2 months

Monthly payment of $200, after 3 months

A retirement fund starts with $5000 and receives quarterly payments of $400. The annual interest rate is 4%, compounded quarterly. Calculate the balance, rounded to two decimal places, for these periods: a

Ex 2

After 2 quarters

After 3 quarters

A savings plan starts with $1000 and receives monthly payments of $120. The annual interest rate is 6%, compounded monthly. Calculate the balance after 4 months, rounded to two decimal places.

An annuity starts with $6000 and receives monthly payments of $250. The annual interest rate is 4.8%, compounded monthly. Calculate the balance after 6 months, rounded to two decimal places.

A fund starts with $4000 and receives quarterly payments of $300. The annual interest rate is 3.6%, compounded quarterly. Calculate the balance after 3 quarters, rounded to two decimal places.

An investor starts with $8000 and makes monthly payments of $200 into an annuity with an annual interest rate of 5%, compounded monthly. Calculate the balance after 5 months, rounded to two decimal places.

A savings account has no initial balance and receives monthly payments of $150. The annual interest rate is 4%, compounded monthly. Calculate the balance after 3 months, rounded to two decimal places.

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An annuity starts with $10 000 and receives monthly payments of $500. The annual interest rate is 6%, compounded monthly. Calculate the balance after 8 months, rounded to two decimal places.

A fund starts with $2500 and receives quarterly payments of $200. The annual interest rate is 4.5%, compounded quarterly. Calculate the balance after 4 quarters, rounded to two decimal places.

An investment of $10 000 receives payments of $800 every 6 months. If the annual interest rate is 3.5% compounded semi-annually, calculate the value of the investment after 2 years.

Determine whether these statements about recurrence relation, An + 1 = rAn + d, are true or false: a

The compounding factor, r, is always equal to the interest rate, i.

Interest is calculated on the previous balance before adding the payment.

If d = 0, the recurrence relation models compound interest.

Complex familiar 14

A savings plan involves monthly payments of $250 for 3 months with an annual interest rate of 5%, compounded monthly: a

Calculate the future value, rounded to two decimal places.

If the annual interest rate increases to 6%, recalculate the future value, rounded to two decimal places, and determine the additional interest earned compared to the original plan.

A savings plan has an initial balance of $2000 and receives monthly payments of $100. The annual interest rate is 5%, compounded monthly: a

Calculate the balance after 6 months, rounded to two decimal places.

If the plan started with an initial balance of $0, what would the balance be after 6 months, rounded to two decimal places?

Liam starts a savings account with $500 and plans to deposit $150 each month. The account earns 4.2% annual interest, compounded monthly: a

Write a recurrence relation to model the value of Liam’s savings.

Use the relation to calculate the balance after 4 months.

Chloe receives a bonus of $1500 and invests it in an account. She decides to add $200 of her own money at the end of each month. The account offers an annual interest rate of 5.4%, compounded monthly. Calculate how much Chloe will have in the account after her third deposit.

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Complex unfamiliar 18

An annuity starts with $2000. For the ﬁrst 2 months, the owner deposits $100 per month and the interest rate is 4% per annum, compounded monthly. After the second month, the owner increases the deposit to $150 and the interest rate changes to 4.8% per annum, compounded monthly. What is the value of the annuity after a total of 4 months, rounded to two decimal places?

An investment starts with an initial balance of $10 000 and receives monthly payments of $500. The annual interest rate is 6% compounded monthly. Calculate the total interest earned over the ﬁrst 4 months, rounded to two decimal places.

7.02 Future value annuity formula After this lesson, you will be able to… • identify the components of the future value annuity formula • use the future value annuity formula to calculate the total amount accumulated in an investment • convert annual interest rates and time periods to match the compounding frequency .

Future value annuity formula The future value of an ordinary annuity is the total amount accumulated from regular payments made at the end of each period, with interest compounded on the balance before each new payment is made.

AFV is the total amount after n compounding periods d is the periodic payment made at the end of each compounding period i is the interest rate per compounding period, expressed as a decimal n

is the number of compounding periods

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Example 1 A person invests $500 at the end of each year into an account with an annual interest rate of 4%, compounded annually. Calculate the future value after 5 years.

Create a strategy , substituting d = 500, i = 0.04, and n = 5

Use the future value annuity formula, to calculate AFV.

Apply the idea Write the formula

Substitute the values

Evaluate the addition

Evaluate the power without intermediate rounding

Evaluate the subtraction

Evaluate the division

Evaluate and round to two decimal places

The future value after 5 years is $2708.16.

Reflect and check The result is reasonable since 5 × 500 = 2500 without interest, and the additional $208.16 reflects compound interest over 5 years.

Example 2 A savings plan requires monthly payments of $200 into an account with an annual interest rate of 3%, compounded monthly. Calculate the future value after 2 years.

Create a strategy Calculate the monthly interest rate and number of compounding periods, then use the future value annuity formula.

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7.02 Practice questions Simple familiar 1

Calculate the monthly interest rate for an annual interest rate of: a

9.2%

8.3%

11.5%

Calculate the number of compounding periods for each investment length: 5 years, compounded monthly

3 years, compounded quarterly

2 years, compounded weekly

10 years, compounded annually

4.5 years, compounded fortnightly

Ex 1

Calculate the quarterly interest rate for an annual interest rate of: a

3.6%

A Queensland student saves $300 at the end of each year in a savings account with an annual interest rate of 5%, compounded annually. Calculate the future value of the savings after 4 years using the future value annuity formula.

A Brisbane resident invests $2000 at the end of each year into a superannuation fund with an annual interest rate of 7%, compounded annually. Calculate the future value after 3 years.

Ex 2

A person deposits $100 at the end of each month into an account with an annual interest rate of 6%, compounded monthly. Calculate the future value after 2 years.

A Cairns family deposits $500 annually into an investment account with an annual interest rate of 3.5%, compounded annually. Calculate the future value after 6 years.

A Gold Coast worker saves $150 at the end of each quarter in an account with an annual interest rate of 4%, compounded quarterly. Calculate the future value after 5 years.

A Townsville resident saves $250 every six months in an account with an annual interest rate of 5.2%, compounded semi-annually. Calculate the future value after 4 years.

A person saves $50 every fortnight into an account with an annual interest rate of 3.9%, compounded fortnightly. Calculate the future value after 3 years.

A person makes monthly payments of $100 into an account earning 6% p.a. compounded monthly for 2 years:

494

Calculate the future value of the annuity after 2 years.

Determine the interest earned over the two-year period.

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Roald invests $1700 each year for 25 years into an account with an interest rate of 4.5% p.a. compounded annually. Determine the future value of this annuity at the end of the investment period.

To start saving money for retirement, Christa invests $375 each quarter at an interest rate of 9% p.a. compounded quarterly. Determine the future value of this annuity after 25 years.

Fred invests $300 each quarter for 30 years into an account with an interest rate of 4% p.a. compounded quarterly. Determine the future value of this annuity at the end of the investment period.

Complex familiar 15

A Queensland couple plans to save for a house deposit by depositing $400 at the end of each month into an account with an annual interest rate of 4.8%, compounded monthly: a

Calculate the future value after 3 years.

If they want to save at least $16 000 in 3 years, will their current plan be suﬃcient? Justify your answer.

A Sunshine Coast business owner invests $1000 at the end of each quarter into a retirement fund with an annual interest rate of 6.5%, compounded quarterly: a

Calculate the future value after 5 years with the current payment.

Calculate the additional future value that would be gained if the quarterly payment was $1200 for the 5 years.

Two annuities have monthly payments of $200 for 2 years. Plan A has an annual interest rate of 4%, and Plan B has 5%, both compounded monthly. Calculate the difference in their future values after 2 years, rounded to two decimal places.

Amira invests $200 at the end of each month into an account that earns 4.8% interest per year, compounded monthly for the ﬁrst 3 years. After 3 years, she increases her monthly deposit to $300, and the account earns 6% interest per year, compounded monthly for the next 2 years. Calculate the total future value of Amira’s investment at the end of 5 years.

Two friends invest for the same total time: • Eli deposits $200 per month for 6 years at 4.5% interest per year, compounded monthly. • Lena deposits $150 per month for 3 years at 4.5% interest per year, compounded monthly, then increases to $250 per month for the next 3 years at the same interest rate. Who ends up with more money at the end of 6 years? Justify your answer.

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Let An represent the balance at the end of month n. The value of the investment can be modelled by the recurrence relation: An + 1 = 1.005An + 100, where the initial investment A0 is $5000. This is an application of the general form An + 1 = rAn + d, where the compounding factor r = 1.005 and the deposit d = $100.

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The original investment amount, the annual interest rate, how often the interest is compounded and the regular deposit amount are all variables that can be changed. Note that the extra deposits always match the number of compounding periods each year. For the example, the amount invested is $5000, the annual interest rate is 6%, the number of compounding periods is 12 per year and the deposit amount is $100. Therefore, the values in cells B1 to B4 are correct for the given example. However, the data only goes to the 12th month, and it is required to go up to the 24th month to determine the value of the annuity after 2 years. Activity 1: • Double-click on the cells in the above spreadsheet to see how the values are being calculated. (To exit the formula view for a cell, press the Esc key.) • Extend column A so that the time period goes up to 24. • Extend column D so that the deposit is 100 for all 24 time periods. • Row by row, drag the formulas in columns B, C, and E to generate the values for each time period. (To do this, click on the cell, then click and drag the small square in the bottom-right corner of the cell downwards.) Verify that the value in cell E30, representing the balance after 24 months, is $8178.99. Activity 2: • Use the applet to double the interest rate. Describe the effect this has on the final balance after 24 months. • Now halve the monthly contribution, making it $50 per month. Does this cancel out the effect of doubling the interest rate? Explain your answer. • With this new interest rate and contribution, use trial and error to determine a new original investment amount that generates a final amount similar to $8178.99. • $6000 is invested at an interest rate of 8% compounded quarterly, where contributions of $250 are made every quarter. Use the applet to determine the number of quarters required for the annuity’s value to first exceed $12 000.

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7.03 Practical applications of annuities After this lesson, you will be able to… • solve practical financial problems by selecting and applying the appropriate future value annuity tool • calculate the required periodic payment needed to reach a specific savings goal • analyse the performance of an annuity by calculating the total payments made and the total interest earned

Periodic payments In certain practical applications of annuities, it is necessary to determine the periodic payment required to accumulate a speciﬁed amount by a given time. The periodic payment can be calculated by rearranging the future value annuity formula:

d is the periodic payment made at the end of each compounding period AFV is the future value, the desired amount to accumulate after n compounding periods i is the interest rate per compounding period, expressed as a decimal n

is the number of compounding periods

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7.03 Practice questions Simple familiar 1

Maya wants to save up for a holiday to Europe in 4 years. She plans to deposit the same amount into a high-interest savings account at the end of every month. The account offers an interest rate of 6% per annum, compounded monthly. Maya’s goal is to have $12 000 saved by the time she leaves. Identify the value of each variable required for the future value annuity formula to calculate Maya’s monthly deposit.

Ex 1

Determine whether these statements about the future value of an ordinary annuity are true or false: a

The future value includes only the sum of the periodic payments.

Total interest is the difference between the future value and total payments.

The compounding factor r is always greater than 1.

A student wants to save $8000 for university fees in 3 years by making monthly payments into an annuity with an annual interest rate of 5%, compounded monthly. Calculate the monthly payment required, rounded to two decimal places.

A retiree aims to save $12 000 in 5 years by making monthly payments into an annuity with an annual interest rate of 3.5%, compounded monthly. Calculate the monthly payment required, rounded to two decimal places.

A couple wants to save $18 000 in 4 years by making quarterly payments into an annuity with an annual interest rate of 5.2%, compounded quarterly. Calculate the quarterly payment required, rounded to two decimal places.

Ex 2

504

A Gold Coast resident wants to save $25 000 for a boat purchase in 5 years by making monthly payments into an annuity with an annual interest rate of 4.5%, compounded monthly: a

Calculate the monthly payment required, rounded to two decimal places.

Calculate the total amount paid into the annuity over 5 years.

Ben wants to save $8000 for a holiday. Ben deposits $600 quarterly for 3 years in an annuity with an annual interest rate of 4%, compounded quarterly: a

Calculate the future value of the annuity.

Determine if Ben has enough money after 3 years.

Calculate the total payments made.

Calculate the total interest earned.

Lisa wants to save $10 000 in 4 years by making monthly payments into an annuity with an annual interest rate of 3.6%, compounded monthly. Calculate, rounded to two decimal places: a

The monthly payment

The total interest earned

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The total payments made

Carlos saves $1000 semi-annually for 5 years in an annuity with an annual interest rate of 5%, compounded semi-annually. Calculate, rounded to two decimal places: a

The future value of the annuity

The total interest earned

The total payments made

Priya wants to save $20 000 in 6 years by making quarterly payments into an annuity with an annual interest rate of 4.8%, compounded quarterly. Calculate, rounded to two decimal places: a

The quarterly payment

The total interest earned

The total payments made

A savings plan requires annual payments of $1500 for 8 years at an annual interest rate of 6%, compounded annually. Calculate, rounded to two decimal places: a

The future value of the annuity

The total interest earned

The total payments made

Raj wants to save $30 000 in 5 years by making semi-annual payments into an annuity with an annual interest rate of 5.2%, compounded semi-annually. Calculate, rounded to two decimal places: a

The semi-annual payment

The total interest earned

The total payments made

Regular semi-annual payments with an interest rate of 1.5% p.a., compounded semi-annually for 10 years. Determine the minimum regular payment needed to accumulate a total of $15 000.

Regular monthly payments with an interest rate of 3% p.a., compounded monthly for 30 years. Determine the minimum regular payment needed to accumulate a total of $280 000.

Mae is saving for a new car that costs $22 000. To accumulate enough money to buy the car in 7 years, how much money should she invest each month in an annuity with a 4.5% p.a. interest rate, compounded monthly?

Complex familiar 16

An investor wants to save for a $15 000 goal in 4 years with monthly payments into an annuity at an annual interest rate of 5%, compounded monthly. The investor can afford up to $300 monthly: a

Calculate the monthly payment required, rounded to two decimal places.

Determine whether the investor’s budget is suﬃcient and explain the ﬁnancial implications if the payment is set at $250 monthly.

Sarah and Tom each invest in an annuity to save for a future goal: • Sarah invests $300 at the end of each month for 10 years in an account that earns 4.5% p.a. compounded monthly. • Tom invests $250 at the end of each month for 12 years in an account that earns 5% p.a. compounded monthly. Compare the two annuities and explain who earns more interest. 7.03 Practical applications of annuities mathspace.co

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A fund receives quarterly payments of $200 for 5 years, with an annual interest rate of 3.6% compounded quarterly. Calculate the interest earned as a percentage of the total payments made, rounded to two decimal places.

Elena has two savings plans for a $25 000 goal in 5 years: • Plan A: Monthly payments at an annual interest rate of 4.8%, compounded monthly • Plan B: Quarterly payments at an annual interest rate of 5%, compounded quarterly Which plan should Elena choose? Explain your answer.

A community in Toowoomba invests $1200 quarterly into an annuity to fund a park renovation, with an annual interest rate of 4.8%, compounded quarterly. After 3 years, they need $20 000: a

Calculate the future value of the annuity after 3 years, rounded to two decimal places.

Calculate the new quarterly payment required for the remaining 1 year to reach $20 000, assuming the same interest rate, rounded to two decimal places.

Complex unfamiliar 21

A student claims that doubling the number of periods will halve the required payment to reach the same future value. Is this true? Justify your answer.

To fund a business, Emma needs to save $120 000 in 5 years. Her plan is to make regular end-of-month deposits into an account earning 5.4% p.a. compounded monthly. First, calculate the monthly deposit she would need to make to achieve this goal. Then, consider this scenario: Emma makes these calculated deposits for the ﬁrst 3 years. After 3 years, she moves her investment to a new account earning 6.6% p.a. compounded monthly, continuing to contribute the same monthly deposit for the ﬁnal 2 years. Calculate the total amount she actually contributed and the total interest she actually earned over the 5 years.

Liam starts saving for a house deposit by investing $500 per month into an annuity earning 4.2% p.a. compounded monthly. After exactly 4 years, he inherits $10 000 and adds it directly to his annuity account. He continues to deposit $500 per month for another 3 years. Calculate the total value of Liam’s house deposit after the full 7 years.

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7.04 Perpetuity formula After this lesson, you will be able to… • use the perpetuity formula,

, where A is total amount, d is periodic

payment and i is interest rate per compounding period • solve practical problems involving perpetuities • determine the total amount of a perpetuity • determine the periodic payment of a perpetuity • determine the interest rate per compounding period of a perpetuity

Perpetuity formula A perpetuity is a type of annuity that provides ﬁxed payments at regular intervals indeﬁnitely. The payment amount is equal to the interest earned during each interval, meaning the principal remains constant over time. The total amount, or present value, of a perpetuity is the lump sum required today to generate these payments without ever depleting the principal.

A is the total amount (present value) of the perpetuity d is the periodic payment made or received each period i is the interest rate per compounding period expressed as a decimal The formula for a perpetuity assumes that payments begin one period after the initial investment and that the interest rate remains constant over time. It is commonly used in scenarios such as perpetual scholarships, endowments, or trust funds, where ﬁxed payments are required indeﬁnitely. To apply the formula, ensure d and i match the compounding period (e.g., monthly payment with monthly interest rate). The formula can be rearranged to solve for d (d = A × i) or needed.

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Example 1 A Queensland community foundation wants to establish a perpetuity to pay $1500 monthly to support local arts programs. The fund earns an annual interest rate of 6%, compounded monthly: a Calculate the total amount required for the perpetuity.

Create a strategy Use the perpetuity formula, A =

, to determine the total amount.

Apply the idea Calculate the monthly interest rate i from the annual rate: Divide the annual interest rate of 6% by 12 months Evaluate Calculate the total amount required: Write the formula

Substitute the values

Evaluate

The total amount required is $300 000.

b Calculate the monthly payment possible if the foundation has $350 000 available.

Create a strategy Rearrange the perpetuity formula to solve for d. d=A×i

Apply the idea Calculate the monthly payment for A = 350 000 and i = 0.005: d=A×i

Write the formula

= 350 000 × 0.005

Substitute the values

= 1750

Evaluate

The monthly payment possible is $1750.

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7.04 Practice questions Simple familiar 1

Ex 1

Consider the perpetuity formula, A =

What does A represent?

What is the role of i in the formula?

A perpetuity has an annual interest rate of 4.8%, compounded monthly. Calculate the monthly interest rate i.

If d = 1000 and i = 0.005, calculate A.

Determine whether these statements about perpetuity are true or false: a

The total amount A is depleted over time by the payments.

The formula assumes payments continue indeﬁnitely.

The periodic payment d can be calculated as d = A × i.

A community fund pays monthly amounts to support a local program, with an annual interest rate of 6%, compounded monthly. Calculate the following values, rounded to two decimal places: a

The total amount required for a monthly payment of $1200

The monthly payment possible from a fund of $300 000

The total amount required for a monthly payment of $800

The monthly payment possible from a fund of $200 000

A university endowment pays quarterly amounts to fund research, with an annual interest rate of 4%, compounded quarterly. Calculate the following values, rounded to two decimal places: a

The total amount required for a quarterly payment of $2500

The quarterly payment possible from a fund of $400 000

A perpetuity has a total amount of $500 000 and an annual interest rate of 4.8%, compounded monthly. Calculate the monthly payment possible, rounded to two decimal places.

A community endowment has $250 000 and an annual interest rate of 6%, compounded annually. Calculate the annual payment possible, rounded to two decimal places.

Asghar invests his superannuation payout of $220 000 in a perpetuity that pays 6% per annum, compounding annually. Calculate the annual payment possible, rounded to two decimal places.

Samih has just won a $5 000 000 jackpot and decides to invest it in a perpetuity that pays 3% per annum, compounding monthly. Calculate the monthly payment possible, rounded to two decimal places.

A trust fund pays $6000 annually to a charity, with an annual interest rate of 3%, compounded annually. Calculate the total amount required, rounded to two decimal places.

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A scholarship fund pays $5000 annually forever, with an annual interest rate of 5%, compounded annually. Calculate the total amount required, rounded to two decimal places.

A scholarship fund pays $1000 quarterly, with an annual interest rate of 5%, compounded quarterly. Calculate the total amount required, rounded to two decimal places.

A perpetuity pays $2000 monthly, with an annual interest rate of 4.2%, compounded monthly. Calculate the total amount required, rounded to two decimal places.

A perpetuity pays $750 monthly, with an annual interest rate of 3.6%, compounded monthly. Calculate the total amount required, rounded to two decimal places.

Glen won the lotto and has invested the money in a perpetuity paying 8% per annum compounded quarterly. He is able to pay himself $25 000 every quarter from his account without using any of the principal. How much money did he win?

Ex 2

A trust fund pays $2400 monthly, with a total amount of $480 000. Calculate the annual interest rate, expressed as a percentage.

A trust fund pays $3000 quarterly, with a total amount of $360 000. Calculate the quarterly interest rate. Express your answer as a percentage, rounded to two decimal places.

Calculate the interest rate per compounding period for these perpetuities. Express the rates as a percentage, rounded to two decimal places: a

A perpetuity with $250 000 pays $10 000 annually. Determine the annual interest rate.

A fund worth $120 000 pays $3600 semi-annually. Determine the semi-annual interest rate.

A perpetuity valued at $400 000 pays $8000 quarterly. Determine the quarterly interest rate.

Complex familiar 18

Jenny receives $600 000 from an inheritance and wishes to invest the money so that her interest payments cover her monthly living expenses of $1500 per month. The investment will be compounded monthly. Determine the annual interest rate that is needed for this investment.

A scholarship fund pays $8000 annually, with an annual interest rate of 4%, compounded annually:

Calculate the total amount required, rounded to two decimal places.

If the interest rate decreases to 3%, recalculate the total amount required and determine the additional amount needed compared to the original fund, rounded to two decimal places.

Two perpetuities pay $1000 monthly. Fund A has an annual interest rate of 5%, and Fund B has 4%, both compounded monthly. Calculate the total amount required for each fund and determine the difference between these amounts, rounded to two decimal places. 7.04 Perpetuity formula mathspace.co

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A community fund wants to pay $2000 monthly forever, with an annual interest rate of 6%, compounded monthly. The fund has $350 000 available. Determine whether the available amount is suﬃcient, and if not, calculate the shortfall and suggest one ﬁnancial strategy to address it.

An investor has $500 000 to establish a perpetuity and is considering two options: • Option A: Provides a monthly payment of $2000 • Option B: Provides an annual payment of $25 000

Calculate the required annual interest rate, compounded monthly, for Option A. Express your answer as a percentage, rounded to two decimal places.

Calculate the required annual interest rate for Option B. Express your answer as a percentage, rounded to two decimal places.

Justify which option requires a higher interest rate to be ﬁnancially viable.

$16 000 is invested in a perpetuity at 3% per annum, compounded annually. A constant amount is withdrawn from the account at the end of each year. This perpetuity can be deﬁned recursively by An + 1 = aAn − b, A0 = c, where An + 1 is the amount remaining in the account after n + 1 years. Determine the values of a, b and c.

Complex unfamiliar 24

Alex wants to establish a perpetual scholarship fund that will provide $10 000 every year, starting one year after the ﬁnal contribution is made. The goal is for the scholarship to continue indeﬁnitely, with the annual payments funded entirely by the interest earned. To build up the required fund, Alex plans to make equal annual deposits for 15 years, starting one year from now. The account earns 5% interest per annum, compounded annually. What should Alex’s annual deposit be to accumulate this amount in 15 years?

A university has two separate endowment funds: • Fund A contains $300 000, earning 4.5% p.a. compounded annually. • Fund B is intended to produce the same annual perpetual payment as Fund A, but it earns only 3.75% p.a. compounded annually. If Fund B currently has $280 000, how much additional money must be added to it to generate the same annual payment as Fund A?

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7 Chapter review SF

A student starts a savings account for their university textbooks with an initial deposit of $1000. They contribute $120 at the end of each month. The account earns 3.6% p.a. interest, compounded monthly: a

Write a recurrence relation in the form An + 1 = rAn + d to model the value of this investment.

Use the recurrence relation to calculate the balance of the account after 4 months.

A small business owner is saving for new equipment by depositing $450 every quarter into an annuity. The investment earns interest at a rate of 5.4% per annum, compounded quarterly, for 5 years: a

Calculate the future value of the annuity after 5 years.

Calculate the total interest earned over the period.

A family wants to save $30 000 for a home deposit over 5 years. They ﬁnd an account that offers an interest rate of 4.8% per annum, compounded monthly. Calculate the monthly payment they must make to reach their goal.

Michael is saving for a new car and has a goal of $19 000. His current plan is to deposit $350 per month for 4 years into an account earning 3.9% p.a., compounded monthly: a

Calculate the future value of Michael’s investment after 4 years.

Will Michael reach his goal? Determine the surplus or shortfall.

Two friends, Olivia and Noah, each start a superannuation fund and plan to invest for 8 years: • Olivia deposits $250 per month into an account earning 4.2% p.a. compounded monthly. • Noah deposits $300 per month into an account earning 3.6% p.a. compounded monthly. By comparing their future values and total interest earned, determine who has the better investment strategy after 8 years. Justify your conclusion with calculations.

A local sports club is considering two perpetuity options: a

A benefactor has provided a lump sum of $350 000. If this is invested in a perpetuity earning 4% p.a. compounded quarterly, what is the maximum quarterly payment the club can receive?

Alternatively, to fund ongoing maintenance costs of $2000 per month, how much money would need to be invested in a perpetuity that earns 6% p.a. compounded monthly?

An educational trust fund has a principal of $600 000 and is set up as a perpetuity to provide monthly scholarships. The fund is required to pay out $2200 each month. Determine the minimum annual interest rate, compounded monthly, that the fund must achieve to be sustainable. Express your answer as a percentage.

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8.01 Introduction to graphs and networks After this lesson, you will be able to… • understand the meaning of a graph in the context of networks • recognise the different components of a graph, including vertices (nodes) and edges (arcs) • identify specific types of graph elements such as loops and the degree of a vertex • comprehend the meaning of various graph classifications, including simple, complete, bipartite, directed (digraph), and weighted graphs • describe a network as a representation of practical situations using graphs

Introduce graphs and their components What do roads, human ancestry, currency exchange, planning a wedding, and the internet have in common? They can all be represented with a mathematical structure called a graph, or more speciﬁcally a network. A network is used to illustrate connections between objects, allowing for the exploration and analysis of complex relationships.

Pink

Belconnen

y Po

4 Arda

Cambray

Fernam

Mordan Range 3

Dolby

Tiller's

l La

l She

Green Eastfarthing

Gilgandra

A vertex (or node) is the fundamental building block of a network and is drawn as a dot or circle. In the network graph provided, the vertices are used to represent towns.

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An edge (or arc) is a line segment that begins and ends at a vertex. An edge represents a relationship between the objects or ideas that it links together. Edge Vertex

Vertex

A graph is a collection of vertices with edges drawn between them. Some special features can be seen in graphs. An edge that starts and ends at the same vertex is called a loop. When there are two or more edges that start and ﬁnish at the same two vertices, these are referred to as multiple edges. Multiple edges Two or more edges that start and ﬁnish at the same two vertices.

This graph has a loop at vertex A and multiple edges between vertices D and F. Vertex G is an isolated vertex as it does not attach to any other vertices.

B The degree of a vertex is the number of edges that connect to it. Loops are counted twice, as they connect to the vertex at both ends. The degree of a vertex is often written as deg(V ), where V is the vertex label. A vertex is referred to as odd or even depending on its degree. If a vertex has an even degree, it is an even vertex. If it has an odd degree, it is an odd vertex. An isolated vertex has a degree of 0.

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Networks can be drawn in different ways but still represent the same information. If the vertices and their connections are identical, the graphs are considered the same.

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Example 1 For the graph shown: A

a Identify the number of vertices and edges in the graph.

Create a strategy

Apply the idea

Count the number of vertices (dots) and edges (lines) in the graph.

Number of vertices = 5

Number of edges = 7

b Determine the degree of each vertex.

Create a strategy For each vertex, count the number of edges connected to it. Remember to count the loop at vertex A twice.

Apply the idea deg(A) = 1 + 1 + 1 + 2 = 5 (one edge each to A, C and D, plus the loop counted twice) deg(B) = 0 (isolated vertex) deg(C ) = 1 + 1 + 1 = 3 (one edge to A, one edge to D and one edge to E) deg(D) = 1 + 1 + 1 = 3 (one edge to A, one edge to C and one edge to E) deg(E) = 1 + 1 + 1 = 3 (one edge to A, one edge to C and one edge to D)

c Identify any odd, even, or isolated vertices.

Create a strategy Use the degrees calculated in the previous part to classify each vertex.

Apply the idea Odd vertices (degree is odd): A and C Even vertices (degree is even and not zero): D and E Isolated vertices (degree is zero): B

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A complete graph is a simple graph in which every vertex is connected to every other vertex by a single edge. A complete graph with n vertices is denoted Kn.

A bipartite graph is a graph whose vertices can be divided into two separate sets (for instance, Set 1 and Set 2), such that every edge in the graph connects a vertex from Set 1 to a vertex in Set 2. There are no edges connecting two vertices within the same set.

Exploration A common way to test if a graph is bipartite is to try and colour its vertices with only two colours (e.g., green and blue). A graph is bipartite if it is possible to colour the vertices so that no two connected vertices have the same colour.

Graph 1

Graph 2

1. For Graph 1, can the vertices be coloured with just two colours so that no connected vertices share the same colour? 2. Now, try to do the same for Graph 2. Is it possible? Why or why not? 3. A graph that can be two-coloured is a bipartite graph. Which of the graphs shown is bipartite?

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Example 2 Consider the graphs shown:

Graph A

Graph B

a Identify which graph is a simple graph. Justify the answer.

Create a strategy A simple graph cannot have any loops or multiple edges. Examine each graph for these features.

Apply the idea Graph A is not a simple graph because it has a loop and multiple edges. Graph B has no loops and no multiple edges, so it is a simple graph.

b Identify which graph is a complete graph. Justify the answer.

Create a strategy A complete graph is a simple graph where every vertex is connected to every other vertex.

Apply the idea Graph A is not a simple graph, so it cannot be a complete graph. Graph B is a simple graph. It has 4 vertices, and each vertex is connected to the other 3 vertices. Therefore, Graph B is a complete graph, speciﬁcally K4.

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8.01 Practice questions Simple familiar 1

For each network, identify the number of vertices and the number of edges: a

Ex 1

For the network shown, determine the degree of each vertex: a c e

deg(A) = ⬚

deg(C ) = ⬚

deg(E) = ⬚

b d

deg(B) = ⬚

deg(D) = ⬚

Identify whether each network is a simple graph or not: a

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Ex 2

Identify which of these graphs is complete: a

Ex 3

Is Graph 2 a subgraph of Graph 1? Graph 1

Identify which of these graphs are bipartite: a

Graph 2

Consider the weighted directed graph shown: a

Determine the weight of the edge from vertex S to vertex Q.

Calculate the total weight of the entire network.

Q 14

S T 10

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Deﬁne these terms: a

Loop

Multiple edges

How many edges are in a complete graph with 5 vertices?

Complex familiar 10

Consider the graph shown:

Is the graph simple? Explain your answer.

Is the graph connected? Explain your answer.

Which vertex has the highest degree? Identify its degree.

Determine whether a directed or undirected network would be best to model each situation: a

An animal food chain (e.g., hawk eats snake)

A map of two-way roads connecting towns

The results of a round-robin tennis tournament where each player plays every other player once

Consider the weighted graph shown. A subgraph is formed by the vertices P, Q, S, T and all edges connecting them. Calculate the total weight of this subgraph.

6 R 7 Q

9 4

S 5

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Complex unfamiliar The Handshaking Lemma specifies that for any graph, the sum of the degrees of all vertices is equal to twice the number of edges: A a Verify the Handshaking Lemma for the graph shown. B Show your working.

Is it possible for a graph to have 7 vertices, each with a degree of 3? Justify your answer using the Handshaking Lemma.

C An undirected, connected graph has 5 vertices labelled A, B, C, D, E. The graph has these properties:

• The degrees of the vertices are 2, 3, 3, 4, 4. • The graph is not a simple graph. • The two vertices with degree 4 are not connected to each other by an edge. Draw one possible network graph that satisﬁes all these conditions.

8.02 Network representations After this lesson, you will be able to… • construct a network diagram to represent practical situations • construct an adjacency matrix from a given graph • construct an adjacency matrix from a given digraph

Represent networks with adjacency matrices While drawing a graph is a useful way to visualise a network, it can be ineﬃcient for complex systems. An adjacency matrix is a powerful tool that represents the connections in a graph using a square grid of numbers. The matrix has one row and one column for each vertex in the network. The numbers, or entries, in the matrix show the number of edges connecting each pair of vertices. In an undirected graph, an edge between two vertices (e.g., A and B) works in both directions. The matrix is therefore symmetrical across its main diagonal ( from top-left to bottom-right). The entry for (row A, column B) is the same as the entry for (row B, column A).

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This image shows an undirected graph and its symmetric adjacency matrix. The number ‘1’ in row A, column B shows the edge between A and B. The ‘1’ in the main diagonal at (row C, column C ) represents the loop at vertex C. In a directed graph (digraph), the matrix entries represent one-way connections. The convention is that an entry in row i and column j shows the number of edges going from vertex i to vertex j. Because the connections are one-way, the matrix for a directed graph is usually not symmetrical.

To Y

From

Y Constructing a graph from a matrix: A network graph can also be drawn if its adjacency matrix is given.

1. Count the number of rows to determine the number of vertices needed. Draw and label them. 2. Read across each row to draw the edges. An entry of ‘1’ in row A, column B means an edge is drawn from A to B (with an arrow if the graph is directed). 3. A number on the main diagonal (e.g., at row C, column C ) indicates a loop at that vertex. 4. An entry greater than ‘1’ (e.g., ‘2’) indicates multiple edges.

Example 1 Construct the adjacency matrix for the undirected network shown. P Q S R

Create a strategy Construct a 4 × 4 matrix with rows and columns labelled P, Q, R, S. For each cell, record the number of edges connecting the row vertex to the column vertex. Remember that loops are recorded on the main diagonal.

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Apply the idea

Reﬂect and check Check the matrix for accuracy. The graph is undirected, so the matrix should be symmetric around the main diagonal (e.g., the entry for P → R is 1, and the entry for R → P is also 1). There are loops at P and S, so there is a ‘1’ at positions (P → P ) and (S → S) on the diagonal.

Example 2 Draw the directed graph represented by this adjacency matrix: To

From

Create a strategy First, draw the three vertices X, Y, and Z. Then, read each row of the matrix to add the directed edges. Remember that the entry in row i, column j represents an edge from i to j.

Apply the idea • Row X has all zeros, so no edges start from vertex X. • Row Y has a ‘2’ in the X column. This means there are two edges from Y to X. • Row Z has a ‘1’ in the X column and a ‘1’ in the Y column. This means there is one edge from Z to X, and one edge from Z to Y. The resulting directed graph is: Z X

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Ex 2

Draw the undirected graph represented by this adjacency matrix:

Draw the directed graph represented by this adjacency matrix:

An adjacency matrix is shown:

Does the matrix represent a directed or undirected graph? Explain why.

Does the graph contain any loops? Explain why.

Does the graph contain any multiple edges? Explain why.

Complex familiar 7

532

The map shows the major roads connecting various towns. Some roads are one-way, as indicated by arrows: A

Draw a directed graph to represent the roads connecting the towns.

Construct the adjacency matrix for your graph.

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For each adjacency matrix, describe the associated graph by selecting all appropriate terms from the list: i

Simple

Directed

Complete

Bipartite

An incomplete adjacency matrix for a simple, undirected graph is given. The sum of the degrees of all vertices in the graph is 10:

Determine the values of x, y, and z.

Complex unfamiliar 10

This matrix represents a social network where an entry in row i, column j shows the number of messages sent from person i to person j :

Who sent the most messages in total?

Who received the most messages in total?

If Ben sends two more messages to Ali, show the updated adjacency matrix.

A student was asked to draw a simple, connected graph with 4 vertices. They created the graph and corresponding adjacency matrix shown, but made one error in each: A

Identify the error in the student’s graph drawing that violates the conditions.

Identify the error in the adjacency matrix that means it does not represent the graph they drew.

Provide the correct adjacency matrix for a simple, connected graph containing the edges: A, B, B, C, and C, D. 8.02 Network representations mathspace.co

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8.03 Planar graph and Euler’s formula After this lesson, you will be able to… • understand the meaning of a planar graph • understand the meaning of a face in the context of planar graphs • apply Euler’s formula to solve problems relating to planar graphs

Planar graph and Euler’s formula A graph is called a planar graph if it can be drawn on a ﬂat surface without any of its edges crossing over each other. The actual drawing of a planar graph with no crossing edges is called a planar representation.

Graph with crossing edges

Planar representation (no crossing edges)

The graph on the left has crossing edges, but it is still a planar graph because it can be redrawn without crossings, as shown on the right. F4 When a graph is drawn in its planar representation, it divides the plane into regions called faces (or regions). The area completely outside the graph is also counted as one face.

F3 F2

For any connected planar graph, there is a relationship between the number of vertices (v), edges (e), and faces ( f ). This relationship is known as Euler’s formula:

v+f−e=2 v f e

is the number of vertices is the number of faces is the number of edges

This formula can be used to determine one property if the other two are known, or to test if a graph meets a necessary condition for being a connected planar graph.

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Example 1 Consider the graph shown, which represents a cube:

a Determine the number of vertices (v), edges (e), and faces ( f ) in the graph.

Create a strategy Count the number of dots (vertices), lines (edges), and enclosed regions ( faces), including the outer region.

Apply the idea • By counting the dots, the number of vertices is v = 8. • By counting the lines, the number of edges is e = 12. • By counting the enclosed regions (including the outside), the number of faces is f = 6. There are 5 inner faces and 1 outer face. Therefore, the graph has 8 vertices, 12 edges, and 6 faces.

b Verify that Euler’s formula holds for this graph.

Create a strategy Substitute the values for v, f, and e into Euler’s formula, v + f − e = 2, and check if the equation is true.

Apply the idea v + f − e = 8 + 6 − 12 =2

Substitute the values of v, f, and e Evaluate

The result is 2, so Euler’s formula holds for this graph.

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Ex 2

This graph is planar. Draw an equivalent planar representation of it (a version with no crossing edges).

A connected planar graph has 7 vertices and 10 edges. How many faces does it have?

A connected planar graph has 6 vertices and 6 faces. How many edges does it have?

A connected planar graph has 12 edges and 8 faces. How many vertices does it have?

Could a connected planar graph have 5 vertices, 7 edges, and 3 faces? Justify your answer using Euler’s formula.

Complex familiar 7

For this graph: a

Draw an equivalent planar representation of the graph.

Count the number of vertices, edges, and faces in your planar representation.

Verify that Euler’s formula, v + f − e = 2, holds for this graph.

A connected planar graph has 5 vertices with degrees 2, 3, 3, 4, and 4: a

Calculate the total number of edges, e.

Using Euler’s formula, calculate the number of faces, f .

This graph represents the vertices and edges of a tetrahedron (a pyramid with a triangular base). Show that Euler’s formula is satisﬁed for this graph.

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Complex unfamiliar A simple, connected planar graph is made up of v vertices and e edges. Every face in the graph is bounded by exactly 3 edges (i.e., every face is a triangle):

Explain why 3f = 2e.

By substituting this relationship into Euler’s formula, show that e = 3v − 6.

The “Utilities Problem” involves connecting three houses (H1, H2, H3) to three utilities (Water, Gas, Electricity) without any lines crossing. The graph representing this is known as K3, 3.

This is a famous non-planar graph. By counting its vertices and edges, use Euler’s formula to explain why it cannot be drawn without crossing edges if every face were bounded by at least 4 edges (a property of bipartite graphs like this one).

8.04 Walks, trails and paths After this lesson, you will be able to… • understand the meaning of walk, trail, and path • understand the meaning of open and closed walks, trails, and paths • understand the meaning of a connected graph and a bridge • recognise circuits and cycles

Walks, trails, and paths Moving around a network can be described in several speciﬁc ways. The distinctions between these terms depend on whether vertices or edges are repeated. A walk is the most general way of moving around a graph. It is a sequence of vertices where each adjacent pair in the sequence is connected by an edge. Walks can repeat both vertices and edges. • An open walk starts and ends at different vertices. • A closed walk starts and ends at the same vertex.

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A trail is a walk that does not repeat any edges. Vertices can still be repeated. • An open trail is a trail that starts and ends at different vertices. • A circuit is a closed trail. It starts and ends at the same vertex without repeating edges. A path is a walk that does not repeat any vertices (except possibly the start and end vertex). A path cannot repeat edges either, which means all paths are also trails. • An open path starts and ends at different vertices. • A cycle is a closed path. It starts and ends at the same vertex without repeating other vertices.

Trails eated ed p re Paths (No repeated vertices)

s) ge

(No

Walks

This diagram shows the relationship between these terms. All paths are trails, and all trails are walks, but the reverse is not always true.

Example 1 Consider the network shown: C

B Start A

D Finish E

G a Classify the sequence A − B − C − F − D − A as a walk, trail, or path. Is it open or closed?

Create a strategy Check if any vertices or edges are repeated to determine whether the sequence is a walk, trail, or path. Check if the sequence starts and ends at the same vertex to determine if it is open or closed.

Apply the idea The sequence starts and ends at vertex A, so it is closed. The vertices are A, B, C, F, D, A. No vertices are repeated (except the start/end), so it is a path. Since it is a closed path, it is a cycle. All paths are also trails and walks.

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8.04 Practice questions Simple familiar 1

Match each deﬁnition to the correct term: a

Ex 1

Walk

A trail where no vertex is repeated (except start/end).

Trail

A walk that starts and ends at the same vertex.

Path

iii

A general sequence of vertices connected by edges.

Circuit

A walk where no edge is repeated.

Cycle

A closed path.

Consider the graph shown:

For each sequence, determine if it is a walk, trail, or path: a

A−B−D−C−A

E−C−D−B−C−A

A−C−E−D−B

D Ex 2

Determine the length of the shortest path from vertex A to vertex D in the weighted graph.

Determine the length of the shortest path from vertex X to vertex Z in the weighted undirected graph.

4 X

Y 5

2 W

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From the graph, provide one example of: a

A cycle that starts at vertex P

An open trail that is not a path

What is the difference between a trail and a path?

Complex familiar 7

The table shows the time (in minutes) it takes to travel between different points in a theme park: Entrance

Ferris wheel

Rollercoaster

Food court

Entrance

−

Ferris wheel

−

Rollercoaster

−

Food court

−

Draw a weighted graph to represent this information.

Identify the shortest time to walk from the Entrance to the Rollercoaster.

This weighted graph shows the cost, in dollars, of bus tickets between several towns:

Identify the cheapest route from town A to town C and determine its cost.

2 D

5 6 E

3 F

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For the graph shown, identify:

A cycle of length 3

An open path that visits 5 vertices

The bridge in the graph

Complex unfamiliar 10

The network shows the possible paths and travel times in minutes for a delivery driver starting at the Depot (D): D

2 5

C 4

F 3

The driver must deliver packages to location A and location E before returning to the Depot. They can visit the locations in any order. Identify the route with the minimum possible travel time and determine the total time. 11

The shortest path from A to D in the graph is A − B − D with length 12: A 4

B 2

3 E

7 D

One new edge can be added with a weight of 2 to the network. Where should this edge be placed to create a new, shorter path from A to D? Determine the new shortest path and its length.

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Modelling the problem The map can be simpliﬁed into a network graph. Each town becomes a vertex (or node), and each road becomes an edge. The number on each edge is its weight, which represents the travel time in minutes. The corresponding network is shown: F

12 A

8 16

C 14

The objective now is to determine the shortest path from vertex E (Eastfarthing) to vertex A (Arda).

Step 1: Identify possible paths Using the network graph, identify at least three different paths that the ambulance could take to get from vertex E to vertex A. Write down the sequence of vertices for each path. For example, one possible path is E → G → F → B → A. Step 2: Calculate the weight of each path For each path you identiﬁed in Step 1, calculate its total weight (total travel time) by adding the weights of all the edges along the path. For the example path E → G → F → B → A, the total weight is 6 + 12 + 22 + 4 = 44 minutes. Step 3: Compare paths and determine the shortest path Compare the total weights of the paths you have calculated. Which one is the shortest so far? Continue to look for other possible paths. Calculate their weights and check if any are shorter than the best one you have determined. Repeat this process until you are conﬁdent you have determined the shortest path. Step 4: Provide your conclusion Write a concluding sentence that determines the shortest path from Eastfarthing to Arda and its total travel time in minutes.

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8.05 Eulerian and Hamiltonian graphs After this lesson, you will be able to… • understand the meaning of Eulerian trail, semi-Eulerian graph, Eulerian circuit, and Eulerian graph • recognise the conditions for the existence of Eulerian trails, semi-Eulerian graphs, Eulerian circuits, and Eulerian graphs • solve practical problems involving semi-Eulerian graphs and Eulerian graphs

Eulerian graphs (traversing edges) Some graphs are considered traversable, which means it is possible to trace every edge of a graph exactly once without lifting the drawing instrument. This concept is named after Leonhard Euler. The key is to focus on visiting every edge. There are two types of traversable graphs: • An Eulerian trail is a trail that travels along every edge in a graph exactly once. It is an open trail, meaning it starts and ends at different vertices. A graph that has an Eulerian trail is called a semi-Eulerian graph. • An Eulerian circuit is a circuit that travels along every edge in a graph exactly once. It is a closed trail, meaning it starts and ends at the same vertex. A graph that has an Eulerian circuit is called an Eulerian graph. Whether a connected graph is Eulerian or semi-Eulerian can be determined by looking at the degree of its vertices. Type of graph

Condition

Path characteristics

Eulerian

All vertices have an even degree

A circuit can be started from any vertex

Semi-Eulerian

Exactly two vertices have an odd degree

The trail must start at one odd vertex and end at the other

Neither

More than two vertices have an odd degree

The graph is not traversable

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Example 1 Consider the network shown:

a Determine if the network is Eulerian, semi-Eulerian, or neither. Explain your reasoning.

Create a strategy First, determine the degree of each vertex.

Apply the idea The degree of each vertex is: • deg(A) = 2 • deg(B) = 3 • deg(C ) = 3 • deg(D) = 4 • deg(E ) = 2 • deg(F ) = 2 The graph has exactly two vertices with an odd degree (B and C ). Therefore, the graph is semi-Eulerian.

b Write down a possible Eulerian trail for the network.

Create a strategy An Eulerian trail must start at one of the odd-degree vertices (B or C ) and end at the other. Trace a path that uses every edge exactly once.

Apply the idea Since there are multiple valid routes, here are two possible Eulerian trails starting at B: • B−A−C−D−F−E−D−B−C • B−D−E−F−D−C−A−B−C

Reflect and check Notice that both trails start and end at the two vertices with odd degrees (B and C ) and use all 8 edges exactly once. Your own valid trail might be different from these examples.

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Ex 2

The graph shown has a Hamiltonian cycle. Write down one possible Hamiltonian cycle starting from vertex A.

D 4

Does this graph contain a Hamiltonian path? If so, write one down.

What is the key difference between an Eulerian trail and a Hamiltonian path?

The graph shown is semi-Eulerian: a

Which two vertices must be the start and end points of the Eulerian trail?

Write down one possible Eulerian trail.

Complex familiar 7

The map shows the bridges connecting landmasses in a historic city. A security guard must walk along every bridge exactly once as part of their nightly patrol: a

552

Model this problem as a graph where landmasses are vertices and bridges are edges. Determine the degree of each vertex.

Can the guard start and ﬁnish at the same point without repeating a bridge? Explain why or why not.

Describe a possible route for the guard. Where must they start and ﬁnish?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A salesperson needs to visit every city shown on the map, starting from city A and returning to A at the end. The numbers on the edges represent the travel time in hours. Determine the Hamiltonian cycle that minimises the total travel time.

A 5

8 10 B

E 7 2

6 D

This graph is not Eulerian because it has 4 vertices of odd degree (B, H, E, G ):

12 3

What is the minimum number of edges that need to be duplicated (re-traced) to make the graph traversable (Eulerian)? Which pairs of vertices should be connected by duplicating the shortest paths between them?

D F

Complex unfamiliar 10

Draw a simple, connected graph for each of these conditions: a

The graph is Eulerian but not Hamiltonian.

The graph is Hamiltonian but not Eulerian.

A chess knight is on a 3 × 3 chessboard. A knight’s move is ‘L-shaped’ - 2 squares in one direction (horizontal or vertical) and then 1 square in a perpendicular direction.

Model the chessboard as a graph where each square is a vertex and a possible knight’s move is an edge. Can a knight start at a corner square, visit every other square exactly once, and end at the centre square (a Hamiltonian path)? Justify your answer by drawing the graph and determining a path if one exists.

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8 Chapter review SF

Consider the directed network graph shown: a

Determine the total degree (the sum of the in-degree and out-degree) of each vertex.

Identify any odd, even, or isolated vertices.

Construct an adjacency matrix for the directed graph shown.

From the two graphs shown, identify which is a complete graph and which is a bipartite graph: Graph 1

Graph 2

R P

S SF

Draw the undirected graph represented by this adjacency matrix:

A connected planar graph has 8 vertices and 5 faces. Use Euler’s formula to determine the number of edges in the graph.

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Consider the graph shown, which has crossing edges: a b

Redraw the graph so that it is planar (has no crossing edges).

Verify that Euler’s formula (v + f − e = 2) holds for your planar representation.

A D E

For this weighted network, identify the shortest path from vertex A to vertex D and identify its length.

For the graph shown, determine if it is Eulerian, semi-Eulerian, or neither. Justify your answer by stating the degree of each vertex.

The map shows several locations in a small town and the available walking paths between them. Arrows indicate one-way paths: a

A tourist wants to start at the Library (L) and visit every other location exactly once. Identify a possible Hamiltonian path.

Is it possible for the tourist to start at the Library, visit every other location exactly once, and then return to the Library? Justify your answer. C

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Graphs 1 and 2 are shown: Graph 1

Graph 2

Show that Euler’s formula applies to Graph 1.

What feature in the drawing should be changed to represent Graph 2 as planar?

Redraw Graph 2 to represent it as a planar graph.

(QCAA 2023 Paper 1 Q20) 14

At 9:00 a.m., a security guard begins their patrol of the eight work sites represented in the network diagram, starting and ending at site A. They drive at 40 km/h on the roads between sites and check every site once for 15 minutes. The length (km) of each road corresponds to the terms of the arithmetic sequence tn = t1 + 2(n − 1), where t1 = 1. Determine the earliest possible time the security guard can ﬁnish their patrol, and identify the route they must follow. B t3

t1 + t2

2t2

2t1 G

t2 F

t1 + t2

2t2

t3 t2

(QCAA 2023 Paper 2 Q5)

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Here are all the spanning trees, with their weights. Notice that all the spanning trees have the same number of edges. The spanning tree for a network with n vertices will have (n − 1) edges. 6 10

5 10

3 3

2 3

6 2 3

3 4

10 4 2 3

5 2

Example 3 Can this network have a spanning tree?

Create a strategy Check whether all the vertices are connected.

Apply the idea To have a spanning tree, the network must be connected. In a connected graph any vertex can be reached from any other vertex either directly or by travelling via other vertices. Vertices D and E are disconnected from vertices A, B, and C. Therefore it cannot have a spanning tree.

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9.01 Practice questions Simple familiar Ex 1

Determine whether each of these graphs is a tree. If it is not a tree, explain why: a

C B d

F e

B C

D g

Ex 3

Determine whether these statements are true about trees: a

Adding one edge to any tree creates a new tree.

A tree has no Eulerian or Hamiltonian circuits.

If you remove any edge from a tree it creates a disconnected graph.

A tree has at least one Hamiltonian circuit, but not necessarily an Eulerian circuit.

Determine whether these networks can have a spanning tree: a

D G

F C

E E

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Ex 4

Draw a spanning tree for this network. H

Draw two spanning trees for each of the weighted networks: a

5 A B

9 C 6

F 8

5 E

d 3

A 7

9 B

B 4 3

E A e

5 10

B 6 A

3 1

C 9 E

6 5

A spanning tree has 12 vertices, how many edges will it have?

A spanning tree has 16 edges, how many vertices will it have?

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4 D

6 F

E 9

A network graph contains 10 vertices: a

How many vertices could be in a tree that is a subgraph of the original network?

How many vertices would be in a spanning tree that is a subgraph of the original network?

Which edges could be removed to create a tree in these graphs? a

a h 3 10

d 5 e

a 3

3 c

f e

Determine a spanning tree for each network from vertex A: a

D E G D c

B A

D C

How many edges need to be removed from these graphs to create a spanning tree? a

A D

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Complex familiar Ex 2

Mitchell is tracing his family lineage back three generations: • His mother’s name is Rebecca, and his father’s name is Connor. • Rebecca’s parents are Mabel and Patrick, and Connor’s parents are Juliet and Mark. • Mitchell does not know Mabel’s parents’ names, but he knows that Patrick’s parents are April and Todd. • On the other side, Mitchell knows that Juliet’s parents are Daisy and Leonard, and Mark’s dad is called Richard. • Mitchell does not know the name of Mark’s mother, however. If the initial vertex represents Mitchell, create a family tree of the last three generations from the information above. Make sure to include any ancestors that he doesn’t know the name of using the word “Unknown”.

Bianca is writing a report and needs to create a table of contents. Her report consists of five chapters, each broken down into sections: • The first chapter is her introduction, so it has just one section. • Her second and third chapters both have three sections, while the fourth chapter has six sections. • The fifth and final chapter is the conclusion which again has only one section. Create a tree that represents the structure of Bianca’s table of contents.

Complex unfamiliar 14

The network graph represents the electrical cables between different amusement sites at a local town show. The vertices represent each tent and the edges represent the distance (metres) in cables between the sites. If all sites only have to be connected to at least one other site, what is the minimum amount of cabling needed? A

D 6 G

6 F

6 4

9 3

I 5

1 J

If the sum of a weighted spanning tree is the same as the sum of all the edges in the entire graph, what does this indicate about the original graph?

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Create a strategy Apply Prim’s algorithm step-by-step to determine the minimum spanning tree.

Apply the idea Step 1: Start at Vertex F • Tree vertices: {F} • Total weight: 0 A

14 26

14 12

12 10 10 14 33

Step 3: Add the second edge • The available edges from {F, E} are: FD(12), FB(13), FA(21), ED(10), EG(14), and EC(33). • The lowest weighted edge is ED(10). • Tree vertices: {F, E, D} • Total weight: 10 + 10 = 20 A

7 G

14 26

14 12

Step 2: Add the ﬁrst edge • The edges leaving F are FE(10), FD(12), FB(13), and FA(21). • The lowest weighted edge is FE(10). • Tree vertices: {F, E} • Total weight: 10

F 12 10 10 14

Step 4: Add the third edge • The available edges from {F, E, D} are: FB(13), FA(21), EG(14), DC(12), and DB(14). • The lowest weighted edge is DC(12). • Tree vertices: {F, E, D, C} • Total weight: 20 + 12 = 32

15 C

14 12 D

12 10 10 14 33

14 26

14 26 13

A B

7 G C

13 14 12 D

21 F

12 10 10 14 33

7 G

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Step 5: Add the fourth edge • The available edges are: FB(13), FA(21), EG(14), DB(14), CG(33), and CB(15). • The lowest weighted edge is FB(13). • Tree vertices: {F, E, D, C, B} • Total weight: 32 + 13 = 45

Step 7: Add the sixth edge • The available edges are: EG(14), BH(26), CG(33), and AG(8). • The lowest weighted edge is AG(8). • Tree vertices: {F, E, D, C, B, A, G} • Total weight: 59 + 8 = 67 A

14 26

14 12

F 12 10 10 14

G C

13 12 D

572

F 12 10 10 14

Mathspace QLD 12 General Mathematics 2025 mathspace.co

7 G

Step 8: Add the ﬁnal edge • The available edges are: BH(26) and GH(7). • The lowest weighted edge is GH(7). • All 8 vertices are now in the tree. • Total weight: 67 + 7 = 74 A

F 12 10 10 14

12 D

14 26

Step 6: Add the ﬁfth edge • The available edges are: BA(14), BH(26), FA(21), EG(14), and CG(33). • The lowest weighted edges are BA(14) or EG(14). Either of these can be selected, for this example BA has been selected. • Tree vertices: {F, E, D, C, B, A} • Total weight: 45 + 14 = 59

14 26

10 D

7 G

The minimum spanning tree consists of the edges {FE, ED, DC, FB, BA, AG, GH} and the minimum total weight is 74.

The weighted network shows possible connections (and associated costs, in millions of dollars) for connecting rural towns to a common cabled internet service:

To minimise the cost of connecting each town to the network, the minimum spanning tree must be determined: a

5 9

C 2

How many edges will be a part of a spanning tree for this network?

Construct the minimum spanning tree.

What is the minimum cost of connecting all of the towns together?

7 6

The direct ﬂight routes between 5 major cities of the same country are shown in the given graph. Each edge is marked with the distance (in km) travelled on each route: B 740 A 720

500

720

C 640

520

D E

560

To optimise their investment, the operating airline wants to determine the minimum spanning tree for the graph:

How many edges will be a part of the spanning tree for this network?

Construct the minimum spanning tree.

Using these ﬂight routes, what is the minimum distance connecting these major cities?

The rangers of a national park want to clear walking tracks to six landmarks in the area. The potential tracks between landmarks are shown in the graph, along with the number of days required to clear each track for use:

8 E

6 A

How many edges will be a part of the spanning tree for this network?

Construct the minimum spanning tree.

What is the minimum number of days required to clear the track?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

To make a network of tracks in the shortest time, the rangers want to determine a minimum spanning tree for the graph:

574

The cost of train tickets between different towns varies. The network represents the possible routes between towns and the costs of the train tickets: a b c

D 3

How many edges will be a part of the spanning tree for this network? Construct the minimum spanning tree for the given network.

Determine the total cost of train tickets when travelling along the minimum spanning tree?

E A

If the minimum spanning tree is equal to 9, determine the value of m.

5 m

B 3

4 2

E 8

If the total value of the minimum spanning tree is 10, determine the value of n.

2 C

n 5

5 2

Complex familiar Ex 2

At the university, there is a new plan to connect all the campus buildings using underground ﬁber optic cables. The given network was suggested by the engineer: The project accountant wanted to minimise the cost of the project. He realised that there are unnecessary cables that could be removed to minimise cable length.

G 12

7 F

4 5

Using Prim’s algorithm, construct the network that represents the least cost.

3 B

10 A

2 H

K 2

4 D

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In a certain area, new underground cables need to be laid to connect electric substations. The graph displays the current connection layout: J

15 F

11 E 16

17 24

16 8

H 19

14 A

Since the project has a limited budget, the engineers have to design a connection layout that uses the minimum cable length. Using Prim’s algorithm, construct the graph that represents the connection layout with the least cost. 11

An aeroplane company wants to redesign their flight routes to make it cost-efficient. The network displays the current flight routes:

160

175

Construct a graph that will minimise the cable length.

3 6

C E

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10 4

576

151 A

To minimise the cost of cables, the council wants to connect the houses in such a way that they minimise the cable length.

148

134

In a certain area, the council wanted to lay down new electric cables. The given graph displays the map of the area:

100

117 126

Using Prim’s algorithm, construct the graph that will achieve this.

137

In order for the ﬂight routes to be eﬃcient, the sum of the route distances needs to be minimised.

153

111

K 1

4 I

Construct the minimum spanning tree for each weighted network: a

10 B C 6 6 7 3 9 A D 1

b A

B 7

2 8 A B 5

2 D

6 G

9 3

17 B

24 22

A D

The bus company decided to redesign its travel routes after new roads were built. The network shows the existing routes connecting the towns and the respective times taken to cover the route:

Since the project has a limited budget, the engineers have to minimise the length of cabling used. Construct the graph that represents the least costly connections.

G 45

10 21

32 B

13 E

21 17

D 34

A large amusement park is trying to implement a new security system such that all security posts are connected by electric cabling. The network shows all possible connections and their respective lengths:

The bus company wants to be time eﬃcient with its routes. Construct the graph that will achieve this.

15 H

5 4

27 31

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Urban planners are trying to connect houses in a certain area.

A 22

The network represents all possible routes between houses in the area, and the distances between them.

The urban planners wish to create roads that are most distance eﬃcient, and allow them to connect each house.

17 H

21 26

13 13

H 12

To minimise the cost of the operation, the water authority wants to minimise the length of the pipes. Construct the graph that will achieve this.

The water authority wants to lay water pipes along the roads in order to put a ﬁre hydrant at every node on the network shown on the network. The numbers represent the length of the pipe that connects any two nodes:

Construct the graph that is the most distance eﬃcient.

A 22

Complex unfamiliar 18

578

For a part of the National Broadband Network, optical ﬁbre communications cables must be run between several towns. The distances (in km) between the towns are listed in the table. A “−” indicates that there is no direct route between those towns: Wickepin

Lake Grace

Kulin

Pingaring

Hyden

Dumbleyung

Wickepin

−

134

−

Lake Grace

134

−

Kulin

−

103

Pingaring

−

Hyden

−

Dumbleyung

103

146

−

Determine the minimum length of cabling required.

If a change in property access regulations makes it possible to shorten the cabling required between Lake Grace and Pingaring to 50 km, by how much will the length of cabling decrease?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

A 3D printer has a movable print head that produces hot, melted plastic to build up layers of the 3D object. The layers are very thin, and it takes a long time for a printed model to be completed, so the designers of the 3D printers try to make the print head move the shortest possible time to move between separate parts. A particular design has 5 legs, with the time (in seconds) to traverse directly between pairs of legs given by the table:

Leg 1

Leg 2

Leg 3

Leg 4

Leg 5

Leg 1

−

1.0

1.4

0.5

1.9

Leg 2

1.0

−

0.7

0.6

1.2

Leg 3

1.4

0.7

−

1.0

Leg 4

0.5

0.6

−

1.5

Leg 5

1.9

1.2

1.0

1.5

−

Determine the minimum time to traverse all 5 legs.

There was no time speciﬁed to traverse between Leg 3 and Leg 4 because this path was obstructed by another part of the design. How much faster could all 5 legs be traversed if it was possible to traverse directly between Leg 3 and Leg 4 in just 0.2 seconds?

The electrical wiring in an industrial plant needs to connect several areas of the plant. Distances (in metres) between the areas are shown: Generator

Crushing

Leaching

Filter

Kiln

Grading

Generator

−

Crushing

−

Leaching

−

Filter

−

Kiln

−

Grading

−

Determine the minimum length of wiring required.

If an access way is built, such that it is no longer possible to use the direct connection between the Generator and Filter, calculate the new minimum length of wiring.

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9.03 Applications of minimum spanning trees After this lesson, you will be able to… • solve practical problems involving minimum spanning trees • minimise the length of cable needed to provide power from a single power station to substations

Applications of minimum spanning trees Minimum spanning trees are a powerful tool for solving real-world optimisation problems. A minimum spanning tree is used to connect all vertices in a weighted graph with the smallest possible total weight for the included edges. In practical scenarios, the “weight” of an edge can represent various quantities like cost, distance, or time. Calculating the minimum spanning tree translates to determining the most efficient solution, such as: • The lowest cost to build a network. • The shortest distance to connect locations. • The fastest time to establish connections. This principle applies to many different fields and industries. 1. Infrastructure: Laying electrical cables, water pipes, or internet fibre optics to connect multiple locations. The goal is to use the least amount of material, which minimises cost. 2. Transportation: Designing a network of roads or railway lines to connect several towns or points of interest with the minimum total length. 3. Logistics: Planning routes for a delivery service that must have the potential to reach all warehouses in a region using a minimum road network. 4. Communications: Designing efficient computer networks that connect all devices with the minimum amount of cabling or identifying the most efficient data pathways.

Example 1 A company is tasked with connecting several small villages on an island to a new solar power station. The weighted network graph shows the possible connections, where vertices represent the power station (P ) and villages (A, B, C, D, E ), and edge weights represent the cost in thousands of dollars of laying the cables.

B 12

C 4 11

Use Prim’s algorithm to determine the minimum cost to connect all the villages to the power station.

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E 8

Create a strategy Apply Prim’s algorithm to the graph to determine the minimum spanning tree and the minimum cost of laying the cables.

Apply the idea Apply Prim’s algorithm step-by-step, starting from vertex P, to build the minimum spanning tree. 1. Start at vertex P. The lowest weighted edge from P is PA with a weight of 10. Highlight edge PA to add it to the tree. 2. The tree now includes vertices P and A. The lowest weighted edge connecting to either A or P connects to B with a weight of 5. Add edge AB. 3. The tree now includes vertices P, A, and B. The lowest weighted edge connecting to any of the included vertices is BC with a weight of 9. Add edge BC. 4. The tree includes vertices P, A, B, and C. The lowest weighted edge connecting to the included vertices is CD with a weight of 4. Add edge CD. 5. The tree includes vertices P, A, B, C, and D. The only remaining vertex is E. The lowest weighted edge connecting to E is CE with a weight of 7. Add edge CE. All vertices are now connected. The minimum spanning tree consists of edges PA, AB, BC, CD, and CE. 15

B 12

C 4 11

E 8

Calculate the total weight (cost): Minimum cost = 10 + 5 + 9 + 4 + 7 = 35

Add the weights of the edges Evaluate the sum

The weight represents thousands of dollars, so the minimum cost is $35 000.

Reflect and check The final tree connects all 6 vertices using 5 edges, which follows the rule that a spanning tree has (n − 1) edges for n vertices. This confirms a valid spanning tree. Using Prim’s algorithm allows the smallest cost of cabling to be determined.

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Example 2 A city council is installing new water fountains in a large park. The table shows the distances between the water main, the fountains and junctions which connect the fountains to the water main: From

Distance (m)

Water main (W)

Junction 1 (J1)

Water main (W)

Fountain A (A)

Junction 1 (J1)

Fountain A (A)

Junction 1 (J1)

Junction 2 (J2)

Junction 1 (J1)

Fountain B (B)

Junction 2 (J2)

Fountain B (B)

Junction 2 (J2)

Fountain C (C)

Fountain A (A)

Fountain B (B)

Fountain C (C)

a Draw a weighted network graph to model the possible pipe connections.

Create a strategy To create the graph, represent each location (W, J1, J2, A, B, C) as a vertex. Then, for each entry in the table, draw an edge between the corresponding vertices and label it with the given distance.

Apply the idea The locations W, J1, J2, A, B, and C become the vertices of our graph. The distances become the weights of the edges connecting them. W 35

20 10

A 30

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12 B

C 15

b Determine the minimum length of pipe needed to supply water to all the fountains.

Create a strategy Use the visual model and Prim’s algorithm to determine the minimum spanning tree, which represents the shortest total length of pipe required to connect all locations.

Apply the idea Apply Prim’s algorithm, starting from the Water main (W). 1. Start at vertex W. The lowest weighted edge is WJ1 (20). Add WJ1. 2. From W or J1, the lowest weighted edge is J1 A(10). Add J1 A. 3. From W, J1, A, the lowest weighted edge is J1 J2 (25). Add J1 J2. 4. From W, J1, A, J2, the lowest weighted edge is J2 B(12). Add J2 B. 5. From the current tree, the lowest weighted edge to the last vertex C is BC(15). Add BC. The minimum spanning tree uses edges WJ1, J1 A, J1 J2, J2 B, and BC. W

20 10

C 15

The total minimum length is the sum of these weights: Minimum length = 20 + 10 + 25 + 12 + 15 = 82 m

Add the weights of the edges Evaluate the sum

The minimum length of pipe needed is 82 m.

Reflect and check Creating a visual model of the network graph is often a crucial first step as it turns a list of data into a clear diagram where the algorithm can be applied. Without the graph, it would be very difficult to see the connections and determine the optimal path.

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Example 3 An IT technician is setting up a computer network in a new oﬃce. The main server (S) must be connected to all workstations (A, B, C, D). The cost of cabling in dollars is shown on the weighted graph. The technician has proposed two different layouts, shown: Layout 1:

45 A

D 60

S 50

Layout 2: D

A S

a Determine if both layouts are minimum spanning trees.

Create a strategy Calculate the total cost for each proposed layout. Then, determine the true minimum spanning tree using Prim’s algorithm and compare its cost to the layouts.

Apply the idea First, calculate the cost of each layout. For Layout 1, the edges are SA(20), AB(25), BC(35), and CD(35). Cost = 20 + 25 + 35 + 35 = $115

Add the weights of the edges in Layout 1 Evaluate the sum

The total cost for Layout 1 is $115. For Layout 2, the edges are SA(20), AB(25), AD(45), and CD(35). Cost = 20 + 25 + 45 + 35 = $125

Add the weights of the edges in Layout 2 Evaluate the sum

The total cost for Layout 2 is $125. The layouts have different costs, so they cannot both be minimum spanning trees. However, it is possible that neither is the actual minimum spanning tree for this network. Determine the true minimum spanning tree using Prim’s algorithm starting at S. 1. Start at S. The lowest weighted edge is SA with a weight of 20. Add edge SA. 2. From the tree (S, A), the lowest weighted edge is AB with a weight of 25. Add edge AB. 3. From the tree (S, A, B), the lowest weighted edge is BC with a weight of 35. Add edge BC. 4. From the tree (S, A, B, C), the lowest weighted edge to the ﬁnal vertex D is CD with a weight of 35. Add edge CD. The total cost is 20 + 25 + 35 + 35 = 115. This matches Layout 1. Therefore, only Layout 1 is a minimum spanning tree.

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9.03 Practice questions Simple familiar 1

Does a minimum spanning tree for a network always need to include the edge with the lowest weight in the entire graph?

The diagram shows a minimum spanning tree for a network of water pipes connecting four locations. The weights represent the length of pipe in metres.

The weighted graph represents the distances in kilometres between ﬁve towns. Identify the layout that represents the minimum spanning tree for the network:

D A

4 A

D 7

Determine the total weight of this minimum spanning tree.

D 9

D A

C E

The diagram shows a network of paths in a zoo with distances in metres. A minimum spanning tree is highlighted: 50

Bear

Aquarium

Reptile 30 Lion

Bird 20 Monkey

Determine the minimum total length of paths required to connect all the attractions.

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The ﬁrst three edges of a minimum spanning tree, determined using Prim’s algorithm starting at vertex A, are highlighted in the graph:

B 2

Complete the algorithm to determine the full minimum spanning tree and identify its total weight.

4 Ex 1

A technician is using Prim’s algorithm to determine the minimum length, in metres, of cable needed to connect the computers in an oﬃce, starting from the Server (S).

8 C

Determine the minimum spanning tree for this network.

6 7

B 4

A minimum spanning tree would be most useful for solving which of these problems? A

Calculating the shortest driving route from a house to a beach.

Installing a water pipe system to connect ﬁve houses to the water main using the least amount of pipe.

Planning a delivery route that visits every customer once and returns to the depot.

Determining the maximum amount of data that can be sent through a computer network.

A network is needed to connect 10 remote monitoring stations to a central hub. How many edges will be included in a minimum spanning tree?

The graph shows the distances, in kilometres, of possible tracks between campsites in a national park. The park rangers want to connect all campsites using the shortest possible total length of tracks:

5 Entrance 7

2 E 9

1 F

Determine the minimum total length of tracks required.

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A graph shows the cost, in millions of dollars, to build high-speed rail links between cities: Brisbane 20 Sydney

Perth

Canberra 5 40

Adelaide

Melbourne

Determine the minimum cost to connect all cities.

Can the project be completed within a budget of $80 million? Justify your answer.

Complex familiar Ex 2

588

An internet provider plans to connect ﬁve towns with ﬁbre optic cable. The costs of laying cable between the towns, in tens of thousands of dollars, are given in the table: From

Cost ($10 000s)

Draw a weighted network graph to represent this information.

Use Prim’s Algorithm to determine the minimum spanning tree for this network. Draw the resulting tree.

What is the minimum cost to connect all ﬁve towns?

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Ex 3

The network shows the time in hours to build emergency access roads between several remote communities after a ﬂood:

5 A

7 8

4 6

6 E

Determine the minimum time required to build a road network that provides access to all communities.

If the road between communities B and F can not be built, how does this impact the total number of hours required to create the access roads?

A farm needs irrigation pipes run from a pump (P ) to four fields (F1, F2, F3, F4). The possible connection distances are: • P to F1 is 30 m • P to F2 is 40 m • F1 to F2 is 15 m • F1 to F3 is 50 m • F2 to F3 is 25 m • F2 to F1 is 20 m • F3 to F4 is 10 m

Draw a weighted graph to represent this scenario.

Determine the minimum length of pipe required.

Consider the weighted network graph: a b

Determine the total weight of a minimum spanning tree for the network.

A 2

Is there another minimum spanning tree with the same total weight? If so, draw it.

4 D

A student has highlighted what they think is the minimum spanning tree for this network: a

Explain the fundamental error the student made.

Determine the correct minimum spanning tree and its total weight.

A 4

D C

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A regional council wants to connect a group of rural properties to a communications tower. They are considering two technologies: a standard cable network and a new wireless relay system. The costs (in thousands of dollars) for each option are shown in the two graphs: Graph 1: Cable network

Graph 2: Wireless relay

20 P1 10 P3 T 12

15 T

P3 8 P4

Determine the minimum cost to connect all properties using the Cable network.

Determine the minimum cost to connect all properties using the Wireless relay system.

Based purely on minimising the initial connection cost, which technology should the council choose? Justify your answer.

A tour company is setting up a system of zip lines between platforms in a forest. The cost to build each zip line is shown in the graph: 10 Start

P1 4

8 P3

15 3 P4

Finish 9

Determine the minimum cost to build a network connecting all platforms.

After planning, the engineers discover that the zip line between P1 and P2 is impossible to build due to unstable trees. Determine the new minimum cost to connect all platforms without this speciﬁc zip line.

Complex unfamiliar 18

A new, very cheap road (edge) is built between two previously unconnected towns in a network. Will this new road always be part of the new minimum spanning tree for the network? Explain why or why not.

An existing network of roads, represented by a minimum spanning tree, connects several towns. A new town is built, along with potential new roads connecting it to some of the existing towns. Does the original minimum spanning tree remain part of the new, larger minimum spanning tree? Explain your reasoning.

If all the edge weights in a connected graph are distinct (i.e., no two edges have the same weight), will the minimum spanning tree always be unique? Justify your answer with a logical explanation.

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9.04 Project networks After this lesson, you will be able to… • construct a project network diagram • represent the durations of activities in a project network diagram • represent the interdependencies of activities in a project network diagram

Project networks and activity tables Many projects, from making a cup of tea to building a skyscraper, involve a series of activities that must be completed in a speciﬁc order. The tasks that must be completed before a new task can begin are called prerequisite activities, predecessors or dependency. This information can be organised in an activity table, which lists each activity, its duration, and its prerequisites. Note that this can also be called a precedence table. Activity

Description

Prerequisite

Time (days)

Get manuscript from author

−

Figure out tour dates

−

Format manuscript for printing

Print books

Book venues for tour

To visualise the flow and prerequisites of a project, a project network can be constructed. This is a weighted, directed graph with some specific conventions: • The edges (arcs) represent the activities and are labelled with the activity letter and its duration. • The vertices (nodes) represent the start and end points of activities. They show how activities are connected. • The network has a single starting point and a single finishing point.

Example 1 Given this network, complete the activity table: D, 6

A, 6 Start

E, 6

B, 1

C, 5

F, 1

H, 3

Finish

G, 6

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Activity

Duration

Predecessor

⬚

B C D E F G H

⬚ ⬚

⬚

⬚ ⬚

Create a strategy For each activity, the number written next to the letter indicates its duration. A, 6 means that activity A has a duration of 6 units of time. To determine the prerequisite for an activity, observe which other activities lead into that specific activity.

Apply the idea A, B and C have no predecessors, so mark these with a “−”. These have durations of 6, 1 and 5 respectively. Activity D has a predecessor of A, and duration 6. Continue working through the network to complete the activity table. Activity

Duration

Predecessor

−

D, E

F, G

Reﬂect and check Note that Activities A, B and C have no prerequisite activities. These three activities are at the very start of the project, therefore do not need to wait for any other activity to be ﬁnished before commencing.

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Consider the network shown: A, 10

Start

D, 5

B, 3

E, 7

Finish

C, 1

Which activities are prerequisite for activity D?

What are the initial activities?

What is the ﬁnal activity?

Which activities have no prerequisite?

For the network shown: A

G Finish

Start B

Which activities are prerequisite for activity E?

What are the initial activities?

What are the ﬁnal activities?

Which activities have no prerequisite?

Consider the network shown: B, 4 Start

A, 5

C, 7

E, 9

F, 10

Finish

D, 2 a

Which activities are prerequisite for activity A?

What is the initial activity?

What is the ﬁnal activity?

Which activity has no prerequisite?

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Consider the network shown: E

G Finish

Start B

Which activity is a dependency for activity E?

What are the initial activities?

What are the ﬁnal activities?

Which activities have no prerequisite?

Consider the network shown: D

A Start

Finish

C F

Ex 1

Which activities are prerequisite for activity G?

What are the initial activities?

What is the ﬁnal activity?

Which activities have no prerequisite?

Use the network diagram to complete the precedence table: D F

B Start

H C

G E Activity

Prerequisite

⬚

B C D E F G H

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⬚ ⬚ ⬚ ⬚ ⬚ ⬚ ⬚

Finish

Use the network diagram to complete the precedence table: B, 7

Start

A, 3

C, 10

E, 1

F, 2

G, 5

Finish

D, 9 Activity

Duration

Prerequisite

⬚

D E

⬚

G 9

⬚

The process of preparing a meal consists of tasks A to K as shown in the network diagram: D A Start

H E

Finish G

Complete the activity chart showing prerequisites. Task

Time (minutes)

Prerequisite

⬚

⬚ ⬚

⬚ ⬚ ⬚

⬚ ⬚ ⬚ ⬚ ⬚

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For each network shown, construct a precedence table: a

A, 6

C, 6

G, 8

E, 3

Finish

Start D, 5

B, 1

F, 9

H, 1

b D, 1

A, 3

F, 8 H, 6

E, 2

Start B, 4 c

C, 9

G, 1

C, 5

A, 6

Finish

G, 8

E, 4

Start

Finish B, 3

F, 3

D, 4

E, 5

C, 3

A, 5

H, 8

F, 6

Start

D, 4

G, 7

H, 3

B, 2 11

For each network shown, construct a precedence table: a

D, 5

A, 5 Start

F, 3

H, 1

E, 6

B, 8

Finish

G, 6 C, 6 b

A, 7

F, 7

D, 3 Start

C, 2

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G, 5

Finish

D, 9

A, 2

F, 6

E, 9 Start

H, 2

B, 4

Finish

G, 8 C, 4

Complex familiar 12

Draw a network representation for each activity table shown: a

Ex 2

Task Prerequisite

Task

Prerequisite

−

A, B

B, C

A, D

F, E

Construct a project network for each activity chart shown: a

Activity Duration Prerequisite

Activity

Duration

Prerequisite

−

B, C

Activity

Duration

Prerequisite

Activity Duration Predecessor

−

A, B, C

B, C

C, D

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Construct a project network for each activity chart shown: a

Activity Duration Prerequisite

Activity

Duration

Prerequisite

−

A, B, C

E, F

Activity

Duration

Prerequisite

Activity Duration Prerequisite

−

C, D

F, E

To assemble a ﬂat-pack desk from a kit requires various activities. These are shown in the table, together with the number of minutes each activity takes and the order in which each activity must be done: Activity

Description

Duration (minutes)

Prerequisite

Assemble the drawer

−

Fix the legs to the desktop

−

Attach the drawer

Attach the sides

Attach the back

Attach the network sockets

D, E

Connect the network wiring

C, F

Using the provided information, construct a project network.

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The table contains information for a project to convert an old bus into a restaurant: Activity

Description

Immediate Predecessor

Time (weeks)

Purchase old bus and strip interior

−

Purchase equipment

−

Hire and train staff

I, M

Select and purchase location

Prepare location

Obtain food and beverage licenses

Move the bus to the selected location

Connect electricity and water

G, F

Decorate the interior

H, B

Stock bar and kitchen

I, M

Advertise promotional events

G, F

Install kitchen equipment and dining tables/chairs

G, F

Promotional events

J, K, C

Using the provided information, construct a project network.

Complex unfamiliar 17

A company is creating a prototype of a new product they wish to manufacture. The table speciﬁes the amount of time, in hours, certain tasks take in the production of the prototype: a

Using the provided information, construct a project network.

Which activity can only start after tasks D, F, and H are all complete?

Task

Duration

Prerequisite

−

D, F, H

N, P, Q

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9.05 Forward and backward scanning After this lesson, you will be able to… • use forward scanning to determine the earliest starting time (EST) for each activity in a project • use backward scanning to determine the latest starting time (LST) for each activity in a project • locate the critical path/s for a project using ESTs and LSTs

Forward scanning and earliest start times (EST) After constructing a project network, the next step is to analyse it to determine the project’s timeline. This is done by calculating the earliest and latest start times for each stage of the project.

EST

LST

To prepare the network for scanning, two boxes are added to each A, 3 vertex. The left box will hold the earliest start time (EST), and the right box will hold the latest start time (LST).

D, 5

Forward scanning

Earliest start time (EST)

The process of working from the Start vertex to the Finish vertex to calculate the earliest possible start time for each activity.

The earliest time an activity can begin, assuming all its predecessor activities have been completed.

To perform a forward scan: 1. The EST for the Start vertex is always 0. 2. Move forwards through the network. The EST for the vertex at the end of an activity is the EST of the vertex at its start plus the activity’s duration. 3. When multiple paths lead to a vertex, its EST is determined by the maximum of the calculated values from each path. For this network: • The EST for the Start vertex is 0. • The earliest start time (EST) for activity C, which begins after A, is the EST of the Start vertex plus the duration of A: 0 + 3 = 3. • The earliest start time (EST) for activity D, which begins after B, is the EST of the Start vertex plus the duration of B: 0 + 2 = 2.

3 A, 3 0

C, 5

Start

Finish B, 2

D, 8 2

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• To determine the EST for the Finish vertex, the completion time for each incoming path must be calculated: • Path via C: The time is the EST for C plus its duration: 3 + 5 = 8. • Path via D: The time is the EST for D plus its duration: 2 + 8 = 10. • The EST for the Finish vertex is the maximum of these values: max(8, 10) = 10.

Example 1 Consider the project network shown. All activity durations are in days:

C, 5

A, 6 Start

E, 8 B, 4

Finish

D, 3

Perform a forward scan to determine the earliest start time (EST) for each activity.

Create a strategy Start at the ‘Start’ vertex with an EST of 0. Work forwards, adding the activity duration to the previous EST. At any vertex with multiple incoming paths, choose the maximum calculated value as the EST.

Apply the idea The EST for the start of activities A and B is 0. The EST for the start of activity C is 0 + 6 = 6. The EST for the start of activity D is 0 + 4 = 4. The start of activity E has two paths leading to it. • Path from activity C: EST = 6 + 5 = 11 • Path from activity D: EST = 4 + 3 = 7 The maximum value is 11, so the EST for the start of activity E is 11. The EST for the Finish vertex is 11 + 8 = 19. 6 C, 5

A, 6 0

Start

E, 8 B, 4

D, 3 4

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Finish

Example 2 The forward scanning was already done on this network: 6 C, 5

A, 6 0

Start

E, 8 B, 4

Finish

D, 3 4

Determine the latest start time (LST) for every vertex in the network.

Create a strategy Perform a backward scan, starting with LST equal to the EST at the Finish vertex. Subtract durations along each path, and at vertices with multiple outgoing arrows, take the minimum value.

Apply the idea Finish: LST = 19 LST for start of activity E = 19 − 8 = 11 LST for start of activity C = 11 − 5 = 6 LST for start of activity D = 11 − 3 = 8 Start vertex: LST = min(6 − 6, 8 − 4) = min(0, 4) = 0. 6 6 C, 5

A, 6 0 0

Start

E, 8 B, 4

D, 3 4 8

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Finish

19 19

Ex 1

For each project network shown, perform a forward scan to determine the earliest start time (EST) for each vertex: a C, 4

A, 7 Start

E, 5 B, 3

D, 6

b A, 2

C, 9

Start

Finish B, 4

D, 6

Using the network diagram shown, calculate the LST for vertex 2. 1

E, 8 Start F, 3

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Finish

Ex 2

For each project network shown, the forward scan has been completed. Perform a backward scan to ﬁnd the latest start time (LST) for each vertex: a

5 C, 7

A, 5 0

Finish

Start

D, 8

B, 4 4 b

Start

D, 5

B, 6

Finish

A, 3 C, 4

E, 7 7

What is the earliest start time (EST) of the Start vertex in any project network? A

It depends on the ﬁrst activity’s duration.

It is always a negative number.

In a network, a forward scan has been completed and the EST for the Finish vertex is 25. What is the LST for the Finish vertex?

Complex familiar 10

For the project network shown, perform a full forward and backward scan.

A, 2

C, 4 F, 1

Start

D, 6 B, 3 E, 5

G, 4

Finish

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A marketing campaign is represented by the network diagram, with durations in days:

A, 3

C, 5

Start

F, 2

D, 6 B, 1

Finish E, 4

G, 9

Calculate the EST and LST for each vertex. 12

For the project network shown, calculate the EST and LST for each vertex. C, 3 A, 4

E, 6 Finish

Start F, 1

B, 2 D, 8

Analyse this complex project network. Calculate the EST and LST for each vertex of this network.

B, 5 Start

D, 2 G, 2

A, 1 C, 3

E, 6 F, 4

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Finish H, 5

Calculate the EST and LST for each vertex of this project network.

R, 3

P, 6 Start

T, 7 Q, 4

Finish

S, 5

Complex unfamiliar 15

This network diagram has been fully scanned, but there is an error in one of the LST calculations. Identify the vertex with the incorrect LST, and provide the correct value.

A, 3 0 0

C, 5

Start

D, 6 B, 2

8 8 3

F, 4 Finish

2 2 4

E, 5

12 12

G, 3

4 7 9

In the network, the duration of activity C is unknown and is represented by the variable x:

A, 5

C, x Finish

Start B, 7

D, 4

Determine the EST for the Finish vertex in terms of x.

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The earliest starting time (EST) for the Finish vertex is determined by the path with the longest duration. In the network shown, the duration of activity E is represented by the variable y. Determine the value of y for which the two paths (A-C and B-E) have equal total durations.

A, 4

C, 8

Start

Finish B, 3

E, y

9.06 Critical path After this lesson, you will be able to… • use the critical path to determine the minimum time for a project to be completed • understand how the critical path impacts project completion time • solve practical problems involving the minimum time for project completion

Critical path The critical path is the sequence of activities in a project that connects vertices where the earliest start time and the latest start time are equal. These are the activities that cannot be delayed without affecting the overall completion time of the project. The critical path represents the longest path through the network in terms of total duration, and its length determines the minimum completion time for the entire project. Any delay along this path will directly delay project completion, making it essential to identify and closely monitor these activities during planning and scheduling. There are two simple methods to determine the minimum completion time once a network has been fully scanned: • Method 1 (From the forward scan): The minimum completion time is the earliest start time (EST) of the Finish vertex. This is the most direct method. • Method 2 (By summing durations): The minimum completion time can also be found by adding up the durations of all the individual activities that lie on the critical path. This is a useful way to check whether the answer is correct. 4 6

C, 13

17 19 E, 6

A, 4 0 0

Start

Finish 25 25 6 6

14 14

B, 6

F, 11 D, 8

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Using Method 1, the EST and LST of the Finish vertex is 25. Therefore, the minimum completion time is 25 days. Using Method 2, the critical path is B − D − F. The durations of these activities are 6, 8, and 11. The sum is 6 + 8 + 11 = 25. This conﬁrms the minimum completion time is 25 days. Some projects may have more than one critical path. If this occurs, all critical paths will have the same total duration. This duration is still the minimum completion time for the project. 10 10 D, 6

B, 8 2 2 0 0

Start

16 16

A, 2

F, 4

7 7 C, 5

Finish 20 20

E, 9

Example 1 The network diagram for a project has been fully scanned, and the critical path is highlighted. All durations are in hours: 4 4 C, 7

A, 4

11 11 0 0

Start B, 3

E, 6

3 6

Finish

17 17

D, 5

a Determine the minimum completion time for the project by inspecting the Finish vertex.

Apply the idea The EST of the Finish vertex is 17. Therefore, the minimum completion time for the project is 17 hours.

b Verify the minimum completion time by summing the durations of the activities on the critical path.

Apply the idea The critical path consists of activities A, C, and E. The durations are: • Activity A: 4 hours

• Activity C: 7 hours

• Activity E: 6 hours

Total duration = 4 + 7 + 6 = 17. The sum of the critical activities is 17 hours, which matches the EST of the Finish vertex.

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Example 2 Calculate the minimum completion time for the project network shown.

C, 6

A, 5 Start

E, 3 B, 2

Finish

D, 8

Create a strategy Perform a forward scan to identify the minimum completion time at the ﬁnish vertex. Then, perform a backward scan to identify the critical path. Verify the result by adding the durations of the activities along the critical path and ensuring it matches the project’s minimum completion time.

Apply the idea Step 1: Perform forward and backward scans Forward scan ESTs: • EST(start of A and B) = 0 • EST(start of C) = 5 • EST(start of D) = 2

• EST(start of E) = max(5 + 6, 2 + 8) = max(11, 10) = 11 • EST(Finish) = 11 + 3 = 14

Therefore, minimum completion time is 14 weeks. Backward scan (LSTs): • LST(Finish) = 14 • LST(start of E) = 14 − 3 = 11 • LST(start of D) = 11 − 8 = 3

• LST(start of C) = 11 − 6 = 5 • LST (Start) = min(5 − 5, 3 − 1) = min(0, 2) = 0

5 5 C, 6

A, 5 0 0

11 11

Start

E, 3

2 3 B, 2

Finish 14 14

D, 8

Step 2: Identify the critical path The vertices where EST = LST are the Start, activity A, activity C, activity E and Finish. The critical path is A − C − E. Step 3: Verify the minimum completion time The EST of the Finish vertex is 14. Sum the critical activities: 5 + 6 + 3 = 14. The minimum completion time for the project is 14 weeks.

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9.06 Practice questions Simple familiar Ex 1

The fully scanned network diagram for a project is shown: 5 5 C, 12

A, 5

17 17 0 0

Start

E, 11

Finish

28 28

6 7 B, 6

D, 10

Identify the minimum completion time for the project.

List the activities on the critical path and determine the sum of their durations to verify the minimum completion time.

The length of the critical path in a project is found to be 32 days. What is the minimum completion time for the project?

The network diagram shows the completed forward scan for a project. What is the minimum completion time? 23 D, 10

B, 14 9 0

Start

Finish

A, 9 C, 12

E, 6

The critical path for the project shown is highlighted:

A, 8

C, 8

Start

E, 5 B, 7

Finish

D, 10

Calculate the length of the critical path to determine the minimum completion time.

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From the fully scanned network, what is the minimum time required to complete the project? 3

7 C, 5

A, 3

8 12 0 0

Start

F, 2 1

B, 1

D, 6

E, 4

Finish 5 5

14 14

G, 9

A simple, fully scanned network is shown: 8 8 0 0

Start

B, 10

A, 8

Finish

18 18

Identify the minimum completion time from the Finish vertex.

Identify the critical path activities and determine the sum of their durations.

A project has three possible paths from Start to Finish. The total duration for each path is listed: • Path 1 (A − C − F ): 15 hours • Path 2 (B − D − F ): 18 hours • Path 3 (B − E − G): 17 hours What is the minimum completion time for the project?

Complex familiar 8

For the project network shown, determine the minimum time (in days) required for completion:

A, 3 B, 5

Finish

Start C, 6

D, 4

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Ex 2

Consider the project network shown:

C, 2

A, 6 Start B, 5

D, 4

Perform a full forward and backward scan on the network.

Identify the critical path.

Identify the minimum completion time for the project.

The network diagram represents the tasks required to organise a school concert, with durations in days. What is the shortest possible time from starting the organisation to being ready for the concert?

Rehearsals, 14

Audition, 5

Dress rehearsals, 2

Start Book venue, 10

Ex 3

Finish

E, 6

Finish

Promote, 5

The network has two critical paths:

B, 6 Start

D, 7 Finish

A, 3 C, 4

E, 9

Determine the minimum completion time for the project and verify that both critical paths have the same length.

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Perform a full analysis on the project network to determine the minimum completion time.

E, 8

B, 4 Start

C, 5

Finish

D, 3

A, 2

F, 6

An activity table for a project is shown. Draw the network diagram and determine the minimum completion time for the project: Activity

Duration (days)

Predecessors

−

C, D

The network shows a project plan where the duration of activity D is unknown, represented by x. If the minimum completion time for the project is 25 days and A − D − E is the only critical path, what is the value of x?

A, 10

D, x

Start

E, 5 B, 8

Finish

C, 9

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Complex unfamiliar 15

In the network shown, the duration of the non-critical activity C is unknown, represented by y. The minimum completion time for the project is 22 weeks. What is the maximum possible integer value of y? 12 A, 12 0 0

D, 10 Finish

Start

22 22

C, y

B, 8 8 16

A project network has two critical paths. The duration of Path 1 is 2x + 5 days. The duration of Path 2 is x + 12 days. Determine an expression for the minimum completion time of the project in terms of x.

A classmate claims, “The critical path is always the shortest path through a network because it represents the most eﬃcient way to do things.” Explain why this statement is incorrect.

If the minimum completion time for a project is 40 hours, is it possible for one of the critical activities to have a duration of 45 hours? Explain your reasoning.

Draw a simple project network diagram with at least 5 activities that has a minimum completion time of 22 weeks and contains at least one non-critical path. Label all activities and their durations.

Draw a simple project network diagram with 4 activities (A, B, C, D) that meets these criteria: • The minimum completion time is 15 days. • The critical path is A − D.

In the network, the durations of activities C and F are unknown, represented by p and q respectively. The minimum completion time for the project is 20 days, and the critical path is B − D − F.

A, 6

C, p

Start

F, q B, 5

Determine the values of p and q. 620

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D, 7

Finish

9.07 Float times After this lesson, you will be able to… • calculate float times for non-critical activities

Critical and non-critical activities In project networks, the critical path shows the minimum time a project will take to complete. Any delay to an activity on this path will delay the entire project. Activities that are not on the critical path do not have the same earliest and latest start times. This means these non-critical activities have some ﬂexibility about when they commence. Critical activity

Non-critical activity

An activity that has no room for delay. If a critical activity is delayed, the entire project will be delayed.

An activity that has some ﬂexibility. It can be delayed by a certain amount of time without affecting the project’s overall completion date.

To formally identify which activities are critical, forward and backward scanning is used to determine the earliest start times (EST) and latest start times (LST). 7 9

A, 7 0 0

C, 5 14 14

Start

E, 3

B, 6

Finish

17 17

D, 8

6 6 After completing the forward and backward scanning, the critical path can be established. Any activity on the critical path is considered a critical activity, so any activity not on the critical path is considered a non-critical activity.

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The information from the project network can be organised into a table to get a clear overview of the project’s ﬂexibility. Activity

Duration

EST

LST

Float time

5 6 A, 5 0 0

C, 4

Start

Finish B, 7

10 10

D, 3 7 7

Interactive exploration Discover this concept in action online

mathspace.co

Example 2 This network diagram shows the activities required to complete a project. The earliest start time (EST) and latest start time (LST) for each activity are: 3 5 A, 3 0 0

B, 7 Finish

Start C, 8

12 12

D, 4 8 8

Calculate the ﬂoat time for Activity B and Activity D.

Create a strategy Use the formula: Float time = LST − EST for each activity.

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Apply the idea For Activity B: Float(B) = LST − EST

Write the formula

=5−3

Substitute LST = 5 and EST = 3

Evaluate

The ﬂoat time for Activity B is 2. For Activity D: Float(D) = LST − EST

Write the formula

=8−8

Substitute LST = 8 and EST = 8

Evaluate

The ﬂoat time for Activity D is 0. This means Activity D is critical.

Example 3 Analyse this project network to determine the ﬂoat times for all activities:

A, 4

C, 4

Start B, 3

E, 5 D, 5

a Determine the EST and LST for each activity.

Create a strategy Perform a forward and backward scan on the project network.

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Finish

Apply the idea Forward scan: EST(Start) = 0 EST(Start of activity A) = 0 + 4 = 4 EST(Start of activity B) = 0 + 3 = 3 The start of activity E has two paths leading to it. • Path from activity C: EST = 4 + 5 = 9

• Path from activity D: EST = 3 + 5 = 8

The maximum value is 9, so the EST for the start of activity E is 9. EST(Finish) = 9 + 5 = 14 Backward scan: LST(Finish) = 14 LST (Start of activity E) = 14 − 5 = 9 LST (Start of activity D) = 9 − 5 = 4 LST(Start of activity C) = 9 − 5 = 4 LST(Start) = min(4 − 3, 4 − 4) = min(1, 0) = 0 4 4 C, 5

A, 4 0 0

9 9

Start

E, 5

B, 3

Finish

14 14

D, 5 3 4

The minimum completion time is 14 days. b Create a table and calculate the ﬂoat time for every activity.

Create a strategy Draw a table with columns for the activity, EST, LST and ﬂoat time.

Apply the idea Activity

EST

LST

Float time

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9.07 Practice questions Simple familiar Ex 1

For each network with the critical path highlighted, list the non-critical activities: a

3 7

C, 1

4 8 E, 3

A, 3 0 0

Start B, 2 2 2

Finish

15 15

E, 2

H, 4 G, 6

Start 9 9 D, 7

I, 9

F, 3

23 23

12 8

Identify the correct formula for calculating the ﬂoat time of an activity. A

Float time = EST – LST

Float time = LST − EST

Float time = LST + EST

Float time = Duration − EST

Calculate the ﬂoat time of this activity:

C, 23

17 36 4

Finish

17 17

Ex 2

16 16

11 13

2 2

H, 1

7 7

C, 4

B, 2

11 11

F, 4

A, 1 0 0

D, 5

G, 4

2 40 59

This network diagram has the earliest start times (EST) and latest start times (LST) labelled for each activity. Calculate the ﬂoat time for these activities: a

Activity A

Activity B

Activity C

3 5

0 0

A, 3

C, 7

Start

Finish

B, 4

D, 8

12 12

4 4

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Ex 3

This table shows the EST and LST for each activity in a project. Calculate the ﬂoat time for each activity and identify the critical path. Activity

Duration

EST

LST

Float time

⬚

⬚ ⬚ ⬚ ⬚

For this project network diagram, with durations in weeks: C, 4 E, 6

A, 3 Start

Finish

B, 5

F, 5 D, 2

Conduct a forward scan to determine the EST for each node and the minimum project completion time.

Conduct a backward scan to determine the LST for each node.

Create a table to calculate the ﬂoat time for every activity.

Identify the critical path.

Analyse this project network, with durations in hours: I, 7 G, 8 K, 5

Start H, 6

628

Finish

J, 10

Determine the minimum completion time for the project.

Calculate the ﬂoat time for all activities.

Identify the critical path.

In a project to build a garden shed, the activity ‘Paint the walls’ has a ﬂoat time of 4 days. What does this mean for the project manager? Mathspace QLD 12 General Mathematics 2025 mathspace.co

Determine whether these activities would lead to a delay of the deadline of the whole project: a

Delaying a critical activity

Delaying a non-critical activity by a duration less than its ﬂoat time

Delaying a non-critical activity by a duration more than its ﬂoat time

Complex familiar 10

Ex 4

A project has the activities, durations (in days), and immediate predecessors shown in this table: Activity

Duration

Predecessors

−

C, D

Draw the network diagram for this project.

Calculate the ﬂoat time for Activity C and Activity D.

To prepare for a school presentation, a student must complete several tasks: Research (A, 3 hours), Write Script (B, 4 hours, after A), Create Slides (C, 5 hours, after A), and Practice (D, 2 hours, after B and C): a

Draw the project network diagram.

Determine which activities have a ﬂoat time.

In a software development project, the activity ‘Test Module B’ has a ﬂoat of 5 hours. If the testing is delayed by 3 hours because the server is down, will the overall project be completed on time?

For this project network, calculate the ﬂoat time for all non-critical activities. N, 8 L, 10 P, 6

Start M, 12

Finish

O, 5

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A student analysed this network and calculated the ﬂoat time for Activity C as 3 days. This is incorrect. Identify the student’s mistake, explain it, and provide the correct ﬂoat time.

A, 4

C, 5

Start

Finish

B, 8

D, 2

Explain why any activity with a ﬂoat time of zero must be on the critical path. Use the deﬁnitions of EST and LST in your answer.

Complex unfamiliar 16

In this network, the ﬂoat time for Activity G is known to be 2 days. The critical path is D-F. Determine the missing duration of Activity G, represented by x.

D, 7

F, 9

Start

Finish

E, 4

G, x

A construction project has two parallel, non-critical paths for installing plumbing and electrical systems. The plumbing path has a total ﬂoat of 6 days, and the electrical path has a total ﬂoat of 4 days. If a materials delivery delay causes a 5-day setback that affects both paths equally, will the project’s overall completion time be delayed? If so, by how much?

In this network, the project manager wants to increase the ﬂoat time for Activity E to at least 3 days by reducing the duration of another single activity. Which activity’s duration should be shortened, and by how much, to achieve this without changing the project’s minimum completion time of 15 days?

B, 4 Start

Finish

A, 3 C, 7

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E, 7

F, 5

Consider two activities, X and Y, that start at the same time and run in parallel. If Activity X has a duration of d1 and Activity Y has a duration of d2, where d1 > d2, write an expression for the ﬂoat time of Activity Y in terms of d1 and d2.

Planning a music festival involves the tasks in the table. Durations are in weeks. Task

Duration

Predecessors

A : Book venue

−

B : Hire bands

C : Arrange sound/lighting

D : Marketing

E : Ticket sales

F : Final setup

D, E

The sound company (Task C) says they might be delayed by 2 weeks. The lead band (Task B) is threatening to pull out unless they are paid earlier, which would delay the start of marketing (Task D) by 1 week. Which is a bigger problem for the festival planner? Justify your answer.

9.08 Applications of critical path analysis After this lesson, you will be able to… • solve small-scale practical problems involving critical path analysis • identify critical paths in practical problems • determine the minimum time for a project to be completed in practical problems

Project time reduction Project networks can be used to determine the minimum time required to complete a project. However, this time may not always be fast enough. For example, a software company might need to launch an app by a speciﬁc date to coincide with a major event, or a construction company might face penalties for ﬁnishing a project late. This is achieved by reducing the length of time of individual activities, known as crashing. Crashing activities, however, almost always comes at a cost. This could involve paying workers for overtime, hiring more staff, or paying for express shipping on materials. This creates a time-cost trade-off. The goal of many practical problems is to achieve a desired reduction in project time for the minimum possible additional cost.

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The strategy for minimum-cost time reduction is as follows: 1. Perform a full critical path analysis using scanning to determine the earliest start times (EST) and latest start times (LST) in order to establish the initial completion time and critical path(s). 2. To reduce the completion time reduce the time of activities on the critical path. 3. Select the activity (or combination of activities, if multiple critical paths exist) with the lowest crash cost for each day that is reduced. 4. After each reduction re-evaluate the network. Check if the critical path has changed. 5. Repeat steps 2 – 4 until the target completion time is reached.

Example 1 The network diagram for a project is shown, with durations in days. The table gives the cost to crash each activity: A, 5

C, 8

Start

Finish B, 4

D, 6

Activity

Cost to crash per day

$200

$300

$180

$250

a Determine the initial critical path and minimum completion time.

Create a strategy Perform a forward and backward scan to establish the critical path.

Apply the idea Length of path A-C: 5 + 8 = 13 days Length of path B-D: 4 + 6 = 10 days The critical path is A-C and the minimum completion time is 13 days.

b The manager needs to reduce the project time by 2 days. Determine the minimum cost to achieve this, showing the step-by-step crashing decisions.

Create a strategy To reduce the project time, crash activities on the critical path (A-C). Crash the cheaper activity ﬁrst. Check if the critical path changes after each reduction.

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Apply the idea The critical path is A-C. The crash costs are $200 per day for A and $180 per day for C. Activity C is cheaper. To reduce by one day: • Crash activity C by 1 day • Cost: $180 • New duration of A-C: 5 + 7 = 12 days • Path B-D is still 10 days. A-C is still critical. To reduce a second day: • The critical path is still A-C. Crash the cheapest activity, C, again. • Crash activity C by another day • Cost: $180 • New duration of A-C: 5 + 6 = 11 days • Path B-D is still 10 days. A-C is still critical. The total cost to reduce the project time by 2 days is $180 + $180 = $360.

Example 2 Consider a simple project with the network diagram and crash costs shown in the table. The initial project cost is $10 000: A, 4

B, 6

Start

Finish C, 3

Activity

Crash cost per day

$400

$500

D, 5

Determine the minimum possible project completion time by crashing and calculate the total cost for that completion time.

Create a strategy Determine the critical path and ﬂoat time to calculate the cost of reducing the project time.

Apply the idea The critical path is A-B with a duration of 10 days. The path C - D has a duration of 8 days, so it has a float of 2 days. The project time can only be reduced by 2 days before path C - D becomes critical. The maximum possible reduction is 2 days. Crash activities on path A - B by 2 days. Activity A is cheaper to crash. Crash activity A by 2 days Additional cost: 2 × $400 = $800 New minimum completion time: 10 − 2 = 8 days New total project cost: $10 000 + $800 = $10 800

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Example 3 A project network has been analysed, and the ﬂoat times for its non-critical activities are shown in the table, along with potential savings if the activities are extended: B, 5 Start

A, 5

E, 7

C, 8

Finish

D, 3 Activity

Float time (days)

Saving per day of extension

$100

$80

What is the maximum amount of money that can be saved without delaying the project? Which activity or activities should be extended?

Create a strategy To maximise savings, extend the non-critical activity with the highest saving per day. It can be extended by an amount up to its ﬂoat time.

Apply the idea Activity B offers a saving of $100 per day. Activity D offers a saving of $80 per day. Activity B provides the greater saving. Activity B has a float time of 3 days. It can be extended by the full 3 days. Maximum saving from Activity B: 3 days × $100 per day = $300 Check the savings from Activity D. It can be extended by its full float time of 5 days. Saving from activity D: 5 days × $80 per day = $400 Reevaluate the project. Even though activity B offers more savings per day, activity D offers a larger total saving because it has more float time. As these activities are on separate parallel paths, both can be extended. Total maximum saving = $300 + $400 = $700 To maximise savings, both activity B and activity D should be extended by their full float times.

Reflect and check It is important to consider both the saving per day and the available float time. A lower daily saving might result in a larger total saving if there is significantly more float time available.

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Example 4 Consider this project with an initial cost of $12 000: A, 6

B, 4 Finish

Start C, 4

D, 3

Extension savings

Crashing costs Activity

Cost per day

Activity

Saving per day

$400

$150

$300

$200

a The project is currently on schedule but $500 over budget. Can the budget be met by extending non-critical activities? Show your calculations.

Create a strategy First, perform a critical path analysis to determine the critical path and the ﬂoat times for the noncritical activities C and D. Then, calculate the maximum possible savings from extending C and D.

Apply the idea A full analysis shows the critical path is A-B with a duration of 10 days. The path C-D has a duration of 7 days. The ﬂoat time for the path C-D is 10 − 7 = 3 days. This ﬂoat is shared by activities C and D. To maximise savings, extend the activity with the higher saving per day, which is D ($200/day). Extend the path by a total of 3 days. Maximum saving: 3 days × $200 per day = $600 Since the potential saving ($600) is greater than the budget overrun ($500), the budget can be met.

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A non-critical activity has a ﬂoat time of 5 days. Extending its duration saves $50 per day. What is the maximum cost saving possible from this activity without delaying the project?

A project has two critical paths. To reduce the project time by one day, what must a project manager do?

A table shows the crash costs for two activities on a project’s single critical path: Activity

Crash cost per day

$250

$200

Which activity should be crashed to save one day at the minimum cost? Ex 1

A project manager needs to save one day on a project. The crash costs for the three activities on the critical path are shown in the table: Activity

Cost per day to crash

$100

$150

$120

Which activity should be crashed to save one day at the minimum cost? Ex 2

This network has a critical path of A-D-F. The cost to crash each critical activity by one day is given in the table: A, 3

D, 8 F, 6

Start B, 5

Finish

E, 4

Activity

Crash cost per day

$500

$350

$400

Determine the cheapest way to reduce the project time by 1 day. 10

A project network has two critical paths: A-C and B-D. The cost to crash each activity by one week is shown in the table: Activity

Crash cost per week

$2000

$2500

$3000

$1500

Determine the most cost-effective way to reduce the project completion time by 1 week.

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For this project network, durations are expressed in hours: A, 8

C, 10

Start

Finish D, 13

B, 6

Ex 3

Identify the critical path and the minimum completion time.

The cost to crash activity D is $100 per hour. Determine the cost to reduce the project time by 2 hours.

A network analysis shows these ﬂoat times and potential savings for non-critical activities: Activity

Float time (days)

Saving per day

$120

$200

What is the maximum possible cost reduction for the project without delaying it? 13

A project has two critical paths: A-D and B-E. The cost to crash each activity is shown: Activity

Crash cost per day

$600

$500

$700

$400

Determine the most cost-effective way to reduce the project completion time by 1 day.

Complex familiar 14

To set up a new oﬃce, these tasks must be done: • Lease agreement (A, 10 days) • Order furniture (B, 3 days, after A) • IT setup (C, 5 days, after A) • Final inspection & handover (D, 1 day, after B and C) The cost to crash activity A is $500/day and for C is $400/day: a

Draw the network diagram and identify the critical path.

Calculate the cost to reduce the completion time by 2 days.

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A project manager has a budget of $1000 for crashing. The network and crash costs are provided. The critical path is A-C-E, another path B-D takes 14 days. What is the greatest reduction in project time that can be achieved within this budget? Show your step-by-step logic. C, 6 E, 5

A, 6

Finish

Start B, 8

D, 6

Activity

Crash cost per day

$450

$250

$300

A project is over budget by $400. The network analysis provides these information about non-critical activities. Can the project be brought back on budget without being delayed? If so, which activities should be extended and by how much? Activity

Float time (days)

Saving per day

$150

$50

Organising a charity run involves the tasks in this table: Task

Normal time (days)

Crashed time (days)

Crash cost per day

A: Get council permit

$200

B: Design route

$150

C: Promote event

$100

D: Organise volunteers

$80

These information applies: • B and C can start after A. • D can start after B. • The event is held after C and D are complete.

640

Determine the minimum completion time and the critical path.

The event must be brought forward by 4 days, to a new completion time of 20 days. Determine the minimum cost to achieve this new deadline.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 4

A company is launching a new product. The table shows the tasks, times, costs, and dependencies. The initial project budget is $50 000:

Task

Normal time (weeks)

Crashed time (weeks)

Crash cost/ week

Extension saving/ week

Dependencies

A: Research and devt

$2000

−

B: Market research

−

$500

−

C: Production setup

$3000

D: Marketing plan

$1500

$400

Final launch occurs after C and D are complete. a

Determine the minimum completion time and the critical path.

The launch date needs to be moved up by 3 weeks (to 11 weeks). Determine the minimum cost to meet this new deadline.

Starting from the original plan, a marketing consultant offers a $1000 discount if their task D can be extended by 3 weeks. Should the project manager accept this offer? Justify your answer.

Complex unfamiliar 19

A network’s critical path is B-D-F, which takes 18 days. Another path, A-C-E, takes 16 days. The project manager decides to crash activity D, which reduces its duration by 3 days: a

Calculate the actual reduction in the project’s completion time.

Explain the discrepancy between the time saved on activity D and the overall project time reduction.

A project manager claims, “I can always reduce the project time most eﬃciently by crashing the activity that can be shortened by the most days.” Explain why this is not always the best strategy for reducing project time at minimum cost.

A complex project has three different paths with these durations: • Path 1 (Critical): 22 hours • Path 2: 21 hours • Path 3: 19 hours The project manager needs to reduce the completion time by 3 days. A table of crash costs is provided. Devise the cheapest crashing strategy. Activity

Path

Crash cost per hour

$100

$120

$80

$90

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9 Chapter review SF

For each graph, determine if it is a tree. If it is not a tree, provide a brief explanation: a

B C

H I

G E

A connected network graph has 15 vertices. How many edges must its spanning tree contain?

A local music festival is connecting its main stages with temporary power lines. The costs, in hundreds of dollars, to run a direct cable between stages are given in the table. A dash (−) indicates a direct connection is not possible:

Main

Rock

Indie

Dance

Acoustic

Main

−

Rock

−

Indie

−

Dance

−

Acoustic

−

Use Prim’s algorithm, starting from the ‘Main’ stage, to determine the minimum spanning tree that connects all stages. Draw the resulting network.

Calculate the minimum cost to connect all the stages.

Consider this project network diagram with activity durations in hours: A, 3

C, 4

Start B, 5

642

E, 6

Finish

D, 2

What are the immediate prerequisites for activity E?

Which activities can begin at the start of the project?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Draw a project network diagram for the project described in the precedence table. Include activity letters and durations. Activity

Duration (days)

Prerequisites

—

B, C

In a project network, two activities, P and Q, lead into the same vertex. The EST for the start of P is 5, and its duration is 7. The EST for the start of Q is 8, and its duration is 3. What is the earliest start time (EST) for the vertex where P and Q ﬁnish?

An activity G, with a duration of 6 weeks, is the only activity leading from vertex 3 to vertex 4. If the latest start time (LST) at vertex 4 is 22 weeks, what is the LST for vertex 3?

A classmate claims, “The critical path is the one with the most activities, because more steps always takes longer.” Use a simple example network to explain why this statement is incorrect.

In the network shown, the duration of activity C is unknown, represented by x. Durations are in hours: A, 7

B, 9

Start

Finish C, x

D, 10

Determine the range of values for x for which the path C-D is the sole critical path. SF

What is the ﬂoat time for any activity that lies on the critical path of a project?

For the fully scanned network, calculate the ﬂoat time for activity C. 4 4

0 0

A, 4

D, 6

Start

Finish

B, 2

C, 5

10 10

2 5

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This network diagram represents the tasks needed to set up a school celebration. Durations are in hours: C, 5 E, 6

A, 2 Start

Finish F, 3

B, 3 D, 4

Perform a full forward and backward scan to determine the EST and LST for each vertex.

Identify the critical path and the minimum time to set up the celebration.

Calculate the ﬂoat time for activity D.

A project has a minimum completion time of 25 days. An activity on the critical path, which has a duration of 8 days, is crashed by 3 days. Assuming no other path becomes critical, what is the new minimum completion time?

A project has a critical path of A-D-F with a completion time of 28 days. A parallel path B-E has a completion time of 25 days. The manager needs to reduce the project time by 5 days. The costs to crash the critical activities are shown: Activity

Crash cost per day

$400

$250

$300

Determine the most cost-effective way to achieve this reduction and calculate the total cost. Show your step-by-step logic. CU

A project has two critical paths: A-C-E and B-D-E. Activity E is common to both. To reduce the project completion time by 1 day, the manager considers two options: • Option 1: Crash the common activity E for $500. • Option 2: Crash activity A (cost $200) and activity B (cost $250) simultaneously. Which is the more cost-effective option to reduce the project time by 1 day? Justify the choice.

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10.01 Introduction to flow networks After this lesson, you will be able to… • define the key components of a flow network: source, sink, edge, and capacity • identify the source and sink nodes in a given flow network diagram

Source and sink nodes A flow network is a directed graph that shows how something moves through a system, like water through pipes, electricity through wires, or data through the internet. Each arrow (called an edge) has a capacity, which is the maximum flow it can carry. In any flow network, there are two important special nodes: Source node (S)

Sink node (T )

The node where the flow starts. It has only outgoing edges (arrows going out), and no incoming edges. It only sends ﬂow but does not receive any.

The node where the flow ends. It has only incoming edges (arrows coming in), and no outgoing edges. It only receives ﬂow without sending any.

An example is a warehouse in a transport network where goods start their journey.

An example is a delivery location that receives goods but does not send them elsewhere.

Source node S 8

B 4

10 T Sink node

Nodes having both incoming and outgoing edges are called intermediate nodes, and help carry the ﬂow through the network. Delivery hubs are places where the goods are received sorted, and sent out again. Note that the intermediate nodes are not moving objects and hence vehicles delivering the product are not intermediate nodes. Vehicles travel along the edges that connect the two nodes.

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Example 1 Consider the ﬂow network shown:

S 12

8 B

A 10

6 T

a Identify the source and sink nodes.

Create a strategy Check for nodes with only outgoing edges (source) or only incoming edges (sink).

Apply the idea

Reﬂect and check

Node S has edges to A and B but no incoming edges, so it is the source.

Verify that S has no incoming edges and T has no outgoing edges in the diagram.

Node T has edges from A and B but no outgoing edges, so it is the sink.

b Which nodes in this ﬂow network are intermediate nodes (neither source nor sink)?

Create a strategy

Apply the idea

Identify the nodes that have both incoming and outgoing edges.

Nodes A and B are intermediate because they have both incoming and outgoing edges.

Example 2 Consider the ﬂow network:

a Which node is the source for this ﬂow network?

Create a strategy Determine the node which only has edges pointing away from it.

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Given this diagram, determine whether the statement is true or false: a

The source node is M.

The sink node is R.

There are 4 intermediate nodes.

R Ex 1

For the water distribution network shown, identify: a

The source node

The sink node

The intermediate nodes

The capacity of the edge from S to B

S 8

10 A

B 6

7 T

Using the network diagram shown, determine the capacity of these edges: a

Edge P → Q

Edge Q → S

Edge R → S

Edge S → T

P 10

12 Q

R 8

7 S 15 T

A ﬂow network has four nodes: a source (S), a sink (T ), and two intermediate nodes (A and B). Draw a possible diagram for this network, including directed edges and example capacities.

Consider the ﬂow network with these directed edges and capacities: S → A = 15, S → B = 6, A → T = 9, and B → T = 5 a

Create a diagram that represents the network.

What is the maximum possible ﬂow directly out of the source node S?

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Complex familiar Ex 2

A ﬂow network is described by the information in the table:

From

Capacity

Draw a diagram of this network.

Depot

Identify the source and sink of the network.

Depot

Factory

A small logistics network is modelled as a ﬂow network. A depot (D) sends goods to two distribution centres (C1 and C2). From the distribution centres, goods are sent to a regional hub (H). The depot is the start of the network and the hub is the end: a

Draw a diagram to represent this ﬂow network.

Identify the source, sink, and intermediate nodes.

For the network shown, what is the total capacity of all edges leading directly into the sink node T ?

7 12 B

Complex unfamiliar 12

The diagram shows a network of one-way roads connecting several intersections. The council has stated that the roadworks depot is at intersection A and all traffic eventually flows towards the main highway access at intersection F. No road flows back to the depot. Based on this information, identify the source and the sink of the traffic network. Justify your answer. B

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10.02 Capacity of a cut After this lesson, you will be able to… • define a cut as a partition of a network that separates the source and sink • identify the forward-flowing edges that cross a given cut • calculate the capacity of a specific cut

Cuts in flow networks Cut in a ﬂow network A partition of the vertices into two sets: one containing the source node (denoted S) and the other containing the sink node (denoted T ). The edges crossing from the source set to the sink set form the cut.

S 10

A 12

The dashed line shows a cut dividing vertices into two sets: {S, A} and {B, T }. Edges crossing the cut are S → B, A → B and A → T.

Interactive exploration Discover this concept in action online

mathspace.co

The cut is used to separate the network into two parts, one with the source and one with the sink, to help measure how much ﬂow can pass from one side to the other.

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Example 1 In a ﬂow network, vertices are partitioned into sets {S, A} and {B, T }. Edges from S → B and A → T cross between sets. Is this a valid cut?

Create a strategy Check if the source S and sink T are in different sets and if the edges cross from the source set to the sink set.

Apply the idea

Reﬂect and check

The source S is in {S, A}, and the sink T is in {B, T }, so they are in different sets. Edges S → B and A → T go from the source set to the sink set. This is a valid cut.

If there was an edge flowing backward, such as B → A, it would cross the partition but would not be included in the cut’s definition, which only counts forward-flowing edges.

Example 2 A ﬂow network has vertices S, A, B, C, T and directed edges: S → A, S → B, A → B, A → C, B → C, and C → T.

Identify the edges in the cut with source set {S, A, B} and sink set {C, T }.

14 11

6 B

15 T

Create a strategy List all edges starting in the source set {S, A, B} and ending in the sink set {C, T }.

Apply the idea

Reﬂect and check

Check each edge: • S → A: Both in source set, not in cut. • S → B: Both in source set, not in cut. • A → B: Both in source set, not in cut. • A → C: Crosses from source set to sink set, include. • B → C: Crosses from source set to sink set, include. • C → T : Both in sink set, not in cut.

Verify that S is in the source set and T is in the sink set, and only edges crossing the partition from the source side to the sink side are listed.

The cut includes edges A → C and B → C.

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A ﬂow network has vertices S, A, B, T and directed edges with capacities: S → A = 5, S → B = 7, A → T = 6, and B → T = 9. The vertices are partitioned into sets {S, B} and {A, T }: a

Identify the edges in the cut.

Calculate the capacity of the cut.

Consider this ﬂow network diagram: A

10 S

8 T

The dashed line shows a partition with S in one set and {A, T } in the other. a 4

Ex 2

Calculate the capacity of the cut.

A ﬂow network has vertices S, A, B, T and directed edges: S → A = 10, S → B = 12, A → B = 5, and B → T = 15. The partition is {S, A, B} and {T }: a

Identify the edges in the cut.

Calculate the capacity of the cut.

A ﬂow network has vertices S, A, B, T and directed edges: S → A = 9, A → B = 4, B → T = 11, and A → T = 8. The partition is {S} and {A, B, T }: a

Is this a valid cut? Explain why or why not.

Identify the edges in the cut and calculate its capacity.

Consider this ﬂow network diagram: 15

S 10

A T

The dashed line shows a partition into sets {S, A} and {B, T }. a 7

Identify the edges in the cut.

Calculate the capacity of the cut.

For the network shown, calculate the capacity of the cut that separates the vertices into sets {S, A, C} and {B, D, T }. S

12 B

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Complex familiar 8

A data network has source S, sink T, and nodes A, B, C. Edge capacities (in Mbps) are: S → A = 25, S → B = 18, A → C = 10, B → C = 12, A → T = 15, and C → T = 20 Calculate the capacity of the cut separating {S, A, B} and {C, T }.

In a logistics network, source S (warehouse) supplies depots A and B, which deliver to destination T. Edge capacities (in tons per day) are: S → A = 30, S → B = 20, A → B = 10, A → T = 25, and B → T = 15. Calculate the capacity of the cut separating {S, B} and {A, T }.

A ﬂow network models water supply with vertices S (reservoir), A, B, C (pumping stations), and T (city). The partition is {S, A, B} and {C, T }: S 60 50 B

25 C 30

a Ex 3

Identify the edges in the cut.

T b

Calculate the capacity of the cut.

A road network is described by these edges and capacities (in vehicles per hour): S → A = 50, S → B = 60, A → C = 40, B → C = 30, and C → T = 80 Draw this network and calculate the capacity of the cut that separates {S, A, B} from {C, T }.

A ﬂow network has a cut partitioning the vertices into {S, A} and {B, T }. The edges crossing from the source set to the sink set are S → B and A → T. The capacity of S → B is 15. The total capacity of the cut is 40. What is the capacity of the edge A → T ?

Complex unfamiliar 13

A communication network has source S, sink T, and nodes A, B, C. Edge capacities (in Gbps) are: S → A = 50, S → B = 40, A → B = 15, A → C = 20, B → C = 25, B → T = 30, and C → T = 35. Two cuts are proposed: Cut 1: {S, A} and {B, C, T }, Cut 2: {S, A, B} and {C, T }. Calculate the capacity of each cut: a

658

Capacity of Cut 1

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Capacity of Cut 2

In a power distribution network, source S (power plant) supplies substations A, B, and C, which deliver to a city T. Edge capacities (in MW) are: S → A = 100, S → B = 80, S → C = 60, A → B = 30, A → C = 40, B → C = 20, B → T = 50, and C → T = 70. A cut is made such that {S, A, B, C} and {T }. Calculate the capacity of this cut and explain its relevance to power distribution planning.

A network’s edges are deﬁned in the table. For a cut that partitions the vertices into {S, B} and {A, C, T }, determine the capacity.

From

Capacity

For the network shown, a cut partitions the vertices into sets {S, A} and {B, T }. Calculate the capacity of this cut and justify your method. 15

B 10

20 A

10.03 Maximum flow and minimum cut After this lesson, you will be able to… • explain the max-flow min-cut theorem • determine the minimum cut and maximum flow

The max-flow min-cut theorem In any flow network, the primary goal is often to determine the absolute maximum amount of flow that can travel from the source to the sink. This is known as the maximum flow.

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While a network may have many paths and large capacities on individual edges, the overall flow is always limited by one or more bottlenecks. These bottlenecks can be identified by using cuts. As a reminder, a cut is a partition of the network’s vertices into two sets, one containing the source and one containing the sink. The capacity of a cut is the sum of the capacities of all edges flowing from the source’s set to the sink’s set. A network can have many possible cuts, each with a different capacity. The cut with the smallest possible capacity is called the minimum cut. This cut represents the tightest bottleneck in the entire network. Cut 1

Minimum cut A

T 8

(Cap. 18)

(Cap. 16)

The relationship between the maximum flow and the minimum cut is defined by a fundamental principle called the max-flow min-cut theorem. Maximum flow = Capacity of minimum cut This theorem states that the maximum flow possible through a network is exactly equal to the capacity of the minimum cut. The maximum flow is therefore found by identifying and calculating the capacity of the minimum cut.

Example 1 The network shows two cuts, Cut A and Cut B: Cut B

Cut A

T 10

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Problems with the minimum cut To solve practical problems involving maximum flow, the primary task is to determine the minimum cut of the network. While there can be many possible cuts, a systematic approach helps to identify the minimum cut efficiently. To determine the maximum flow of a network: 1. Identify all possible cuts that separate the source from the sink. Start by looking for simple “vertical” cuts near the source and sink. 2. Calculate the capacity of each of these cuts. Remember to only sum the capacities of edges that flow from the source side to the sink side. 3. Identify the smallest capacity calculated. This is the capacity of the minimum cut. 4. The maximum flow of the network is equal to this minimum cut capacity. This process allows for identifying bottlenecks. The edges that form the minimum cut are the ones restricting the overall flow. To improve the network’s maximum flow, the capacity of one or more of these specific edges must be increased.

Example 2 This graph represents a road map, showing the maximum number of vehicles that can travel on each road per hour: 145

Balgownie

Fairy Meadow 133

117

161 Wollongong 140 Unanderra

133 129

129 Port Kembla

160

Kanahooka

110

a What is the maximum number of vehicles that can travel from Balgownie to Kanahooka per hour?

Create a strategy To determine the maximum ﬂow, identify all possible cuts and calculate their capacities. The minimum of these capacities will be the maximum ﬂow.

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10.03 Practice questions Simple familiar 1

For the ﬂow network: A

Identify the vertex that is the source.

Identify the vertex that is the sink.

For the network shown, which of the cuts are valid? 2

C1 1

C4 C2

For the network shown, calculate the capacity of each cut: 4 40

2 35

664

20 10

45 C1

40 1

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5 c

For the ﬂow network shown: C1

23 C2

4 3

17 5 C3

Identify all the cuts shown on the network that are valid.

Calculate the capacity through each of the valid cuts.

What is the maximum ﬂow for this network?

For the network shown, the source is vertex 1 and the sink is vertex 5: 5 10

20 10

20 4 30

50 30

20 1

Calculate the capacity of the cut leaving the source.

Calculate the capacity of the cut entering the sink.

Determine the maximum ﬂow for the network.

This network shows the ﬂow of oil (in litres/hour) from a storage tank (vertex 1) to its sink (vertex 7): 132 1 2 101

105

110

105 3

4 125

107

119 143

7 Determine the maximum ﬂow of oil through the pipes. 10.03 Maximum flow and minimum cut mathspace.co

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Complex familiar 7

The ventilation system of an apartment is represented by this graph, showing maximum ﬂow in litres/second: D

24 F

49 18

G 11

B 13

Determine the maximum ﬂow of air through the system from source A to sink G. 8

This graph represents rainwater drains and their maximum capacity in litres/hour: 32

2 23

13 40

3 16

29 6 11

Determine the maximum ﬂow of rainwater from the source (vertex 1) to the sink (vertex 6). 9

The traﬃc ﬂow (in hundreds of cars per hour) through a city centre is shown in the network: 12

A 10 B

12 E 19

D 8

H 30

Calculate the maximum ﬂow through this network from source F to sink H.

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For the network shown, determine which of the four cuts shown has the smallest capacity: C1

2 90

30 C2

4 60

A network of water pipes is shown, with the maximum ﬂow capacity of each pipe given in litres per minute: Q

20 S

20 C

20 50

B 20

P 50

Ex 2

Ex 1

F 40

Determine the ﬂow through each of the cuts P, Q, R, and S.

What is the maximum ﬂow for this network?

The graph represents the delivery network between Balgownie and Kanahooka, showing the maximum number of delivery trucks that can pass along each road segment at one time: 145

Balgownie

Fairy Meadow 133 161

117

Wollongong 140

Unanderra

113 129

112 Port Kembla

160

Kanahooka

110

Calculate the maximum number of delivery trucks that can travel from Balgownie to Kanahooka at one time.

The road from Wollongong to Port Kembla is closed. What is the new maximum ﬂow of the network? 10.03 Maximum flow and minimum cut mathspace.co

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Complex unfamiliar 13

A circuit is represented by this network, with the currents given in mA: 1.58

1.58 R3

1.23

Wire Crossing

1.53 R1

Capacitor

2.72

1.49

1.28 R2

Input 2

Power in

1.28

1.53

Power out

1.28

Input 1 1.53

Calculate the maximum current that can ﬂow between power in and power out.

Explain any signiﬁcant feature of the capacities of different cuts in this network.

Calculate the maximum ﬂow of this network from source to sink: 12 11

R 11

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1 12

10.04 Assignment and allocation After this lesson, you will be able to… • understand the meaning of assignment and allocation problems • recognise the characteristics of a bipartite graph • construct a bipartite graph to represent an allocation problem • represent possible assignments for an allocation problem using a tabular or matrix form • use a bipartite graph to represent possible assignments for an allocation problem

Bipartite graphs Bipartite graph A type of graph in which the vertices are divided into two distinct sets, and every edge connects a vertex from one set to a vertex in the other. No edges exist between vertices within the same set.

In allocation problems, one set typically represents resources (such as workers), and the other set represents tasks. An edge between a resource and a task indicates that the resource is capable of completing that task. This type of graph provides a visual way to represent possible assignments, making it easier to explore and compare allocation options.

In each bipartite graph, the vertices are divided into two groups: blue and orange. The orange vertices are connected only to blue vertices, and the blue vertices are connected only to orange ones. There are no edges between vertices of the same colour.

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Tabular and matrix representation While bipartite graphs are useful visually, the same information can be shown in a table or a matrix. These formats are often more practical for analysis. A tabular representation lists one set of items (e.g., workers) as rows and the other set (e.g., tasks) as columns. Each cell shows whether an assignment is possible. While this can be shown with a  for a possible assignment and a  for an impossible one, it is more common to use numbers. A 1 means the assignment is possible (an edge exists in the graph). A 0 means the assignment is not possible (no edge exists). This table of 1s and 0s is also known as a matrix representation or an adjacency matrix:

where Mij is the entry in row i and column j of the matrix. The matrix format is useful for computational analysis of allocation problems.

Example 2 Three volunteers, Ali, Ben, and Chloe, are available for two event shifts, a morning shift and an afternoon shift: • Ali is available for the morning shift only. • Ben is available for both shifts. • Chloe is available for the afternoon shift only. Create a tabular and matrix representation of the volunteers’ availability.

Create a strategy Construct a table (or matrix) with rows for the volunteers and columns for the shifts. Place a 1 in a cell if the volunteer is available for that shift, and a 0 otherwise.

Apply the idea The table has 3 rows for the volunteers and 2 columns for the shifts. • For Ali: Available for the morning shift gives 1, unavailable for the afternoon shift gives 0. • For Ben: Available for both shifts gives 1 for both columns. • For Chloe: Unavailable for the morning shift gives 0, available for the afternoon shift gives 1. The tabular representation is: Morning shift

Afternoon shift

Ali

Ben

Chloe

The matrix representation is:

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Ex 1

Three tutors Ali, Beth, and Carlos are available to supervise two study sessions, Math and Science: • Ali can supervise Math only. • Beth can supervise both Math and Science. • Carlos can supervise Science only. Draw a bipartite graph to represent the possible assignments between tutors and study sessions.

Consider the bipartite graph showing possible assignments between delivery drivers and delivery zones. How many drivers can work in Zone Z2?

Bipartile graph: Driver assignments to delivery zones D1

Three students, Mr. Diaz, Ms. Evans, and Mr. Patel can be assigned to two projects, Project X and Project Y : • Mr. Diaz can do Project X. • Ms. Evans can do both projects. • Mr. Patel can do Project Y.

Draw a bipartite graph to represent the possible assignments, labelling students as S1, S2, S3 and projects as P1, P2.

List the possible assignments indicated by the edges.

Refer to this tabular representation: Which worker can only do Task A?

None

Task A

Task B

Task A

Task B

⬚

Complete the table by adding  or  based on this information: W1 can do Task A, W2 can do both tasks, and W3 can do Task B only.

W1 W2 W3

⬚

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Ex 2

Three students — Samantha (S1), Leo (S2), and Maya (S3) — are being considered for two research projects, P1 and P2 : • Samantha can work on Project P1. • Leo can work on both projects. • Maya can only work on Project P2.

Construct the table with rows for students and columns for projects, using 1 for a possible assignment and 0 otherwise.

What does a 0 in the cell for S1 and P2 mean?

In the table, which task can be assigned to the most workers?

Task A

Task B

Task C

Task A

Task B







Task C

All tasks are equal







A logistics company assigns three drivers (W1, W2, W3) to three zones (T1, T2, T3). The matrix representation is given by:

Which entry shows that W2 can work in Zone T3? A 12

M11

M23

M32

M12

What does a 0 in a matrix cell mean in the context of a bipartite graph? A

The assignment is completed.

The worker must be assigned.

The assignment is not possible.

The task is unassigned.

A matrix representation for assigning students to projects is:

How many projects can Student S3 work on?

Which students can work on Project 1?

Complex familiar 14

Four students — Lina, James, Noah, and Emily — are being assigned to three group projects, P1, P2, and P3 : • Lina can work on Projects P1 and P2. • James can work on Projects P2 and P3. • Noah can work on Projects P1 and P3. • Emily can work on Projects P1, P2, and P3.

674

Construct the matrix with rows for students and columns for projects.

What does the entry M23 = 1 indicate?

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A bipartite graph assigns four students to three projects, with this matrix representation:

How many students can work on Project 2?

Which students can work on Project 3?

Two students, Ms. Lee and Mr. Khan, can be assigned to three projects, Project M, Project N, and Project O: • Ms. Lee can do Projects M and N. • Mr. Khan can do Projects N and O.

Draw a bipartite graph, labeling students as S1, S2 and projects as P1, P2, P3.

Create a tabular representation.

A tabular representation for assigning students to projects is: a

Draw the corresponding bipartite graph.

Create the matrix representation.

Three students can be assigned to two projects. The bipartite graph has edges S1 → P1, S2 → P2, S3 → P1, and S3 → P2: a

Create a tabular representation.

Create a matrix representation.

Complex unfamiliar 19

Four students, Ms. Garcia, Mr. Singh, Ms. Wong, and Mr. Tran, can be assigned to three projects, Project A, Project B, and Project C: • Ms. Garcia can do Project A. • Mr. Singh can do Projects A and B. • Ms. Wong can do Projects B and C. • Mr. Tran can do Project C. a

Create a matrix representation for the possible assignments, labelling students as S1, S2, S3, S4 and projects as P1, P2, P3.

Explain how the matrix helps in analysing the allocation problem.

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A bipartite graph assigns students to projects, with the tabular representation: P1

Draw the bipartite graph.

If each project must be assigned at least one student and each student can only be assigned to one project, is it possible to assign students to cover all three projects? Explain with an example.

A matrix representation for assigning students to projects is:

Create the corresponding tabular representation.

Draw the bipartite graph.

Suggest one change to the assignments to ensure Project 1 has at least two students, and update the matrix.

10.05 Optimisation by inspection After this lesson, you will be able to… • determine the optimum (minimum) assignment for small-scale practical problems by inspection • determine the optimum (maximum) assignment for small-scale practical problems by inspection • solve small-scale practical problems involving optimisation by inspection

The minimum assignment An assignment problem involves allocating resources (such as workers) to tasks in a way that optimises an objective, such as minimising the total cost or time taken to complete the tasks. For small-scale problems (e.g., 3 × 3 matrices), the minimum assignment can be determined by inspection, which means checking all possible assignments manually to identify the one with the lowest total cost.

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The maximum assignment In some assignment problems, the goal is not to minimise cost but to maximise a certain value, such as profit, output, or efficiency. In these cases, the objective is to determine the maximum assignment, which refers to the combination of resource-to-task pairings that results in the highest possible total value. Just like in cost-based problems, small-scale maximum assignment problems (e.g., 3 × 3 matrices) can be solved by inspection, which involves checking all valid assignments and selecting the one that gives the greatest overall return. This method is effective for small-scale problems, where every possible assignment can be compared directly. It is important to focus on the overall total — not just the highest individual values — to ensure the most efficient use of available resources.

Example 2 Three salespeople (P, Q, R) are to be assigned to three regions (1, 2, 3). The proﬁt matrix shows the expected proﬁt, in thousands of dollars, for each salesperson in each region. Determine the maximum assignment by inspection. 1

Create a strategy

Apply the idea

List all possible assignments for salespeople P, Q, and R to regions 1, 2, and 3. Calculate the total proﬁt for each assignment and identify the one with the maximum proﬁt.

There are 3! = 6 possible assignments: 1. P → 1, Q → 2, R → 3 : 20 + 22 + 23 = 65 2. P → 1, Q → 3, R → 2 : 20 + 19 + 21 = 60 3. P → 2, Q → 1, R → 3 : 15 + 17 + 23 = 55 4. P → 2, Q → 3, R → 1 : 15 + 19 + 16

= 50

5. P → 3, Q → 1, R → 2 : 18 + 17 + 21

= 56

6. P → 3, Q → 2, R → 1 : 18 + 22 + 16 = 56 The maximum total proﬁt is 65 thousand dollars, achieved with the assignment P → 1, Q → 2, R → 3.

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Ex 1

Ex 2

680

The time matrix (in hours) for assigning workers A, B, and C to tasks 1, 2, and 3 is shown: Task 1

Task 2

Task 3

List all possible assignments for the three workers to the three tasks.

Calculate the total cost for the assignment A → Task 1, B → Task 2, and C → Task 3.

By inspecting all possible assignments, determine the assignment that minimises the total cost.

A small bakery assigns three staff members (X, Y, Z) to three jobs (Cake, Bread, Pastry). The time taken (in minutes) is shown in this matrix: Cake

Bread

Pastry

Calculate the total time for the assignment X → Cake, Y → Bread, and Z → Pastry.

Determine the assignment that minimises the total time by inspection.

Three machines (M1, M2, M3) are assigned to produce three products (P1, P2, P3). The cost (in dollars) for each assignment is shown in this matrix: a

Calculate the total cost for the assignment M1 → P1, M2 → P2, M3 → P3.

Determine the assignment that minimises the total cost by inspection.

Three volunteers (P, Q, R) are assigned to three charity events (E1, E2, E3). The eﬃciency scores are shown in this matrix: E1

Calculate the total eﬃciency for the assignment P → E1, Q → E2, R → E3.

Determine the assignment that maximises the total eﬃciency by inspection.

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Four students (S1, S2, S3, S4) are assigned to four group projects (G1, G2, G3, G4). The time (in hours) for each task is shown in this matrix: G1

How many possible assignments are there for this 4 × 4 matrix?

Calculate the total time for the assignment S1 → G1, S2 → G2, S3 → G3, S4 → G4.

Determine the assignment that minimises the total time by inspection.

Four tutors (T1, T2, T3, T4) are assigned to four classes (C1, C2, C3, C4). The effectiveness scores are: C1

Calculate the total effectiveness for the assignment T1 → C1, T2 → C2, T3 → C3, T 4 → C 4.

Determine the assignment that maximises the total effectiveness by inspection.

Complex familiar 10

A small business assigns three workers (A, B, C) to three tasks (X, Y, Z) with the cost matrix (in dollars): X

Determine both the minimum and maximum cost assignments by inspection.

Explain why the business might want to know both the minimum and maximum cost assignments.

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Three delivery drivers (D1, D2, D3) are assigned to three routes (R1, R2, R3). The time matrix (in minutes) is: R1

Suppose the company wants to minimise the total delivery time, but Driver D1 cannot be assigned to Route R3 due to vehicle restrictions.

List all feasible assignments considering the restriction.

Determine the assignment that minimises the total time by inspection, respecting the restriction.

Consider the cost matrix for three workers (W1, W2, W3) and three tasks (T1, T2, T3): T1

A student claims that the assignment W1 → T3, W2 → T1, and W3 → T2 always gives the minimum cost because it uses the smallest cost in each row. a

Calculate the total cost for the student’s proposed assignment.

Is the student’s claim correct for this speciﬁc problem? Justify your answer by ﬁnding the true minimum assignment.

Complex unfamiliar 13

Four workers (A, B, C, D) are assigned to four tasks (1, 2, 3, 4). The cost matrix (in dollars) is: Task 1

Task 2

Task 3

Task 4

The manager wants to minimise the total cost but notices that two assignments yield the same minimum cost.

682

Determine the two assignments that minimise the total cost by inspection.

Discuss the implications of having two optimal assignments for the manager’s decisionmaking.

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10.06 Introduction to the Hungarian algorithm After this lesson, you will be able to… • set up a cost matrix to represent an assignment problem • perform row and column reduction to create zeros in the matrix • identify an optimal assignment if one exists after the initial reduction • calculate the total value of an assignment using the original matrix

Set up the assignment problem Hungarian algorithm A method used to solve assignment problems - situations where a number of resources (e.g. workers, machines, or people) must each be assigned to a set of tasks in a way that optimises a total outcome (such as minimising cost or maximising proﬁt).

These problems are represented using a square matrix, where: • Each row corresponds to a resource (e.g. a worker) • Each column corresponds to a task (e.g. a job) • Each entry vij represents the cost or value of assigning resource i to task j The aim is to assign each resource to exactly one task, and each task to exactly one resource, such that the total value is optimised.

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Example 2 Apply row and column reduction to prepare the matrix for the Hungarian algorithm:

Create a strategy First, subtract the smallest value in each row from every entry in that row (row reduction). Then, using the new matrix, subtract the smallest value in each column from every entry in that column (column reduction).

Apply the idea Step 1: Row reduction The minimum value in Row A is 5. Subtract 5 from each element in Row A: (5 − 5, 8 − 5, 7 − 5) = (0, 3, 2). The minimum value in Row B is 4. Subtract 4 from each element in Row B: (4 − 4, 6 − 4, 5 − 4) = (0, 2, 1). The minimum value in Row C is 8. Subtract 8 from each element in Row C: (9 − 8, 10 − 8, 8 − 8) = (1, 2, 0). The matrix after row reduction:

Step 2: Column reduction The minimum value in Column 1 is 0. The column remains unchanged. The minimum value in Column 2 is 2. Subtract 2 from each element in Column 2: (3 − 2, 2 − 2, 2 − 2) = (1, 0, 0). The minimum value in Column 3 is 0. The column remains unchanged. Therefore, the ﬁnal reduced matrix is:

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10.06 Practice questions Simple familiar 1

Ex 1

Ex 2

Determine whether the statement is true or false: a

The Hungarian algorithm is used to determine the worst possible assignment in an allocation problem.

The matrix used in the Hungarian algorithm must be square, meaning it has the same number of rows and columns.

In the assignment matrix, the entry vij represents the value (such as cost or beneﬁt) of assigning resource i to task j.

The ﬁrst step of the Hungarian algorithm is to randomly choose an assignment and check if it’s optimal.

Three workers (A, B, C) are to be assigned to three tasks (1, 2, 3). The cost matrix (in dollars) is:

Which dimension represents resources, and which represents tasks?

What does each matrix entry represent?

What is the goal of this assignment problem?

For each matrix, perform row reduction, then column reduction: a

688

Three park rangers (R1, R2, R3) are assigned to three trails (T1, T2, T3). The values represent the time (in hours) to complete a patrol:

Perform row reduction on the matrix.

Perform column reduction on the row-reduced matrix.

Determine the optimal assignment and the minimum total hours.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 3

Three ferries (F1, F2, F3) are assigned to three routes (R1, R2, R3). The cost matrix shows the fuel cost (in dollars) for each assignment:

Use the Hungarian algorithm to determine the optimal assignment that minimises the total fuel cost.

Calculate the total cost of the assignment F1 → R1, F2 → R2, F3 → R3 to verify it is not optimal.

Three volunteers (V1, V2, V3) are assigned to three activities (A1, A2, A3). The matrix shows the time (in minutes) each volunteer takes for each activity:

Perform row and column reductions on the matrix.

Determine the optimal assignment and the minimum total time.

Three staff members (S1, S2, S3) are assigned to three duties (D1, D2, D3). The cost matrix (in hours) is given:

Determine the optimal assignment using the Hungarian algorithm to minimise total hours.

Complex familiar 8

A logistics company needs to assign four delivery drivers (D1, D2, D3, D4) to four different routes (R1, R2, R3, R4). The matrix shows the estimated time (in minutes) for each assignment:

Use the Hungarian algorithm to determine the optimal assignment that minimises the total delivery time.

Calculate the minimum total time for the deliveries.

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A software company is assigning ﬁve programmers (P1, …, P5) to ﬁve coding modules (M1, …, M5). The matrix shows the estimated time in hours for each programmer to complete each module:

Using the Hungarian algorithm, determine one optimal assignment to minimise the total programming time.

Calculate the minimum total hours required.

Three boats (B1, B2, B3) are assigned to three docks (D1, D2, D3). The cost matrix is provided:

Determine the optimal assignment and the minimum total cost.

Explain what the optimal assignment means in this context.

Complex unfamiliar 11

Explain why the Hungarian algorithm guarantees an optimal solution for assignment problems. Refer to the role of zeros in the reduced matrix and the line-covering step.

The committee for a local fete needs to assign three food stalls (Pizza, Tacos, Burgers) to three locations (A, B, C). The daily operational cost for each assignment is given: • Pizza stall: $50 for Location A, $40 for Location B, $65 for Location C • Taco stall: $45 for Location A, $55 for Location B, $50 for Location C • Burger stall: $60 for Location A, $45 for Location B, $55 for Location C

690

Construct the 3 × 3 cost matrix for this problem.

Determine the optimal assignment of stalls to locations to minimise the total cost.

Calculate the minimum total daily cost.

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10.07 Maximisation with the Hungarian algorithm After this lesson, you will be able to… • convert a maximisation problem into a minimisation problem by creating an opportunity cost matrix • apply the complete Hungarian algorithm to the transformed matrix to determine the optimal assignment • solve practical problems that require maximising a total value, such as profit or productivity

The Hungarian algorithm for maximisation The Hungarian algorithm is a method to determine the optimal assignment in allocation problems. While it is often used to minimise a total value (like cost), it can also be adapted to maximise a total value, such as productivity or proﬁt. In a maximisation problem, the matrix entries represent values where higher numbers are better. The goal is to assign each row (e.g., worker) to a unique column (e.g., task) to achieve the highest possible total score. The key to solving a maximisation problem is to ﬁrst convert it into an equivalent minimisation problem, which can then be solved using the standard steps of the algorithm.

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Example 1 Three workers, Ms. Lee, Mr. Patel, and Ms. Wong, can be assigned to three tasks, Task A, Task B, and Task C. The productivity scores are given in the matrix:

Identify the matrix structure for this maximisation problem.

Create a strategy Interpret the matrix rows as workers, columns as tasks, and entries as productivity scores. Ensure the matrix is square for the Hungarian algorithm.

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Example 3 An airline wants to assign three pilots ( X, Y, Z ) to three flight routes (1, 2, 3) to maximise the total efficiency rating. The original efficiency ratings are shown in the matrix on the left. After transforming the problem for minimisation, the opportunity cost matrix on the right was created: Original Efficiency Matrix:

Cost Matrix:

Apply the Hungarian algorithm to the cost matrix to determine the optimal assignment, and calculate the maximum total efficiency rating.

Create a strategy Follow the Hungarian algorithm steps on the cost matrix: row reduction, column reduction, cover zeros, and assign zeros. Use the original efficiency matrix to compute the total score.

Apply the idea Step 1: Row reduction Subtract the minimum value of each row from all elements in that row.

Step 2: Column reduction Subtract the minimum value of each column from all elements in that column. Columns 1 and 2 have a minimum of 0. Column 3 has a minimum of 1.

Step 3: Cover zeros and assign The zeros can be covered by 3 lines. Since n = 3, an optimal assignment can be made. • Row 1 has one zero: Assign Pilot X to Route 1. • Row 2 has one zero: Assign Pilot Y to Route 2. • With Routes 1 and 2 taken, the only remaining assignment for Pilot Z (Row 3) is Route 3. The optimal assignment is: Pilot X → Route 1, Pilot Y → Route 2, Pilot Z → Route 3 Step 4: Calculate maximum value Use the original efficiency matrix to determine the total rating: Total rating = Rating ( X → 1) + Rating (Y → 2) + Rating (Z → 3)

694

Write the equation

= 10 + 12 + 10

Substitute the values

= 32

Evaluate

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Complex familiar Ex 1

Ex 2

Ex 3

696

Three workers, Ms. Carter, Mr. Nguyen, and Ms. Kim can be assigned to three tasks, Task X, Task Y, and Task Z. The productivity scores are:

Identify the matrix structure (rows, columns, and entry meaning).

What is the largest entry in the matrix?

A factory manager needs to assign three machines (M1, M2, M3) to three different production runs (R1, R2, R3). This matrix shows the expected output (in units per hour) for each assignment:

Transform the maximisation problem into a minimisation problem by creating a cost matrix.

Perform row reduction on the cost matrix.

A 3 × 3 cost matrix after transformation is:

Perform row reduction.

Perform column reduction on the resulting matrix.

A manager wants to assign three employees (E1, E2, E3) to three projects (P1, P2, P3). The matrix shows the proﬁt (in hundreds of dollars) generated by each assignment. The goal is to maximise the total proﬁt:

Convert the proﬁt matrix into an opportunity cost matrix by subtracting each element from the largest value in the matrix.

Apply the Hungarian algorithm to the cost matrix to determine the assignment that maximises proﬁt.

Calculate the maximum possible proﬁt.

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A cost matrix after row and column reduction is:

Cover all zeros with the minimum number of lines. How many lines are needed?

Is an optimal assignment possible? Explain.

A cost matrix after row reduction is:

Perform column reduction.

Cover zeros with the minimum number of lines.

A reduced cost matrix is:

Assign zeros to determine the optimal assignment for workers W1, W2, W3 to tasks T1, T2, T3.

If the original matrix scores for these assignments are 7, 9, 7, what is the total productivity score?

Complex unfamiliar 12

A 3 × 3 productivity matrix is:

Transform the maximisation problem into a minimisation problem.

Perform row and column reduction on the cost matrix.

A 5 × 5 productivity matrix is:

Transform the maximisation problem into a minimisation problem.

Perform row reduction and explain how this step helps in determining the optimal assignment.

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10.08 Apply the Hungarian algorithm: Minimisation After this lesson, you will be able to… • determine if a matrix is optimal by covering all zeros with the minimum number of lines • perform the matrix adjustment step if the number of lines is less than the matrix size • apply the complete Hungarian algorithm, including the adjustment step, to solve any minimisation problem

The Hungarian algorithm for minimisation The Hungarian algorithm is a systematic method to solve minimisation assignment problems. While row and column reduction create zeros in the cost matrix, an optimal assignment is only possible if these zeros are in “independent” positions. The complete algorithm includes a check for optimality and a procedure to improve the matrix if an optimal assignment cannot yet be made. The full Hungarian algorithm follows these steps: Step 1: Row reduction - For each row, determine the smallest entry and subtract it from every entry in that row. Step 2: Column reduction - Using the new matrix from Step 1, determine the smallest entry in each column and subtract it from every entry in that column. Step 3: Check for optimality - Cover all zeros in the reduced matrix using the minimum possible number of horizontal or vertical lines. If the number of lines is equal to the size of the matrix (n), an optimal solution exists. Proceed to Step 5. Step 4: Matrix adjustment - If the number of lines is less than n, the matrix must be adjusted: • Determine the smallest element that is not covered by any line. • Subtract this element from all uncovered elements. • Add this element to all elements that are at the intersection of two lines. • Return to Step 3. Step 5: Determine the optimal assignment - Select a set of n zeros such that each row and each column has exactly one zero selected. Calculate the total cost of this assignment using the values from the original matrix.

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Example 1 Use the Hungarian algorithm to determine the minimum cost assignment for this 4 × 4 cost matrix:

Create a strategy Apply the ﬁve steps of the Hungarian algorithm: row reduction, column reduction, check for optimality, matrix adjustment (if necessary), and determine the optimal assignment.

Apply the idea Steps 1 and 2: Row and column reduction After row reduction, then column reduction, the resulting matrix is:

Step 3: Check for optimality Cover all zeros with the minimum number of lines. This can be done with 3 lines (e.g., covering Column 2, Column 3, and Row 2). Since the number of lines (3) is less than the matrix size (n = 4), an optimal assignment cannot be made yet. Step 4: Matrix adjustment The smallest uncovered element is 1. Subtract 1 from all uncovered elements and add 1 to the elements at the intersections of lines (at positions (2, 2) and (2, 3)). The adjusted matrix is:

Step 3 (Revisited): Check for optimality Covering the zeros in the new matrix now requires 4 lines. Since the number of lines (4) equals the matrix size, an optimal assignment is now possible. Step 5: Determine the optimal assignment Select a unique zero from each row and column. The optimal assignment is: • W1 → T2 • W 2 → T4 • W 3 → T3 • W 4 → T1 Calculate the total cost using the original matrix: Total cost = 7 + 7 + 6 + 6 = 26 Therefore, the minimum cost is 26.

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Complex familiar 4

Four workers (W1, W2, W3, W4) are assigned to four tasks (T1, T2, T3, T4) at a Brisbane event. The costs (in minutes) are:

Perform row and column reductions.

Determine the optimal assignment and total cost.

Four machines (M1, M2, M3, M4) are assigned to four jobs (J1, J2, J3, J4) in a Queensland factory. The costs (in dollars) are:

Determine the optimal assignment and minimum cost using the Hungarian algorithm. 6

A company assigns four projects (P1, P2, P3, P4) to four teams (T1, T2, T3, T4). The cost matrix is shown. After row and column reduction, the minimum number of lines to cover all zeros is less than 4:

Perform row and column reduction to get the initial reduced matrix.

Perform the matrix adjustment step to determine a new matrix where an optimal assignment can be made.

From the adjusted matrix, determine the optimal assignment and the minimum cost.

10.08 Apply the Hungarian algorithm: Minimisation mathspace.co

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Ex 1

A non-square matrix represents costs for three workers and four tasks. Add a dummy row to make it square, then determine the optimal assignment and minimum cost.

Determine the minimum cost assignment for this 4 × 4 matrix, which requires the matrix adjustment step.

Complex unfamiliar 9

Five volunteers (V1, V2, V3, V4, V5) are assigned to ﬁve tasks (T1, T2, T3, T4, T5). The costs (in minutes) are:

Determine the optimal assignment and total cost. 10

Explain why the Hungarian algorithm guarantees an optimal solution for minimisation problems. Use the concept of zeros in the reduced matrix.

In an assignment problem, what does it signify if there is more than one way to select a set of n independent zeros for the optimal assignment?

A cost matrix is reduced, and the minimum number of lines to cover all zeros is found to be less than n. When adjusting the matrix, why is the smallest uncovered value added to the elements at the intersection of two lines?

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Mathspace QLD 12 General Mathematics 2025 mathspace.co

10.09 Solve complex assignment problems After this lesson, you will be able to… • adapt a non-square assignment problem by adding a dummy row or column • model a problem with restricted assignments by using a large value (M ) in the cost matrix • solve complex, multi-step assignment problems that may involve non-square matrices or restricted assignments

Complex assignment problems Complex assignment problems build on the standard Hungarian algorithm by introducing additional challenges. These can include larger matrices, an unequal number of resources and tasks, or speciﬁc restrictions on assignments. The core ﬁve steps of the algorithm remain the same: 1. Row reduction 2. Column reduction 3. Check for optimality (covering zeros) 4. Matrix adjustment (if needed) 5. Determine the optimal assignment Complexities often arise in the setup or during the matrix adjustment step. Non-square matrices

Restricted assignements

The Hungarian algorithm requires a square matrix. If the number of resources and tasks differs, a ‘dummy’ row or column must be added to make the matrix square. The costs or values in this dummy row/column are always 0.

Sometimes a speciﬁc assignment is impossible (e.g., a worker is not qualiﬁed for a task). To handle this in a minimisation problem, replace the cost of that assignment with a very large number, represented by M. This ensures the algorithm will never select that assignment as part of the optimal solution.

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Example 1 Four machines (M1, M2, M3, M4) are to be assigned to four jobs (J1, J2, J3, J4). The cost matrix is shown:

Determine the minimum cost assignment.

Apply the idea Steps 1 and 2: Row and column reduction After performing row and then column reduction, the matrix becomes:

Step 3: Check for optimality The minimum number of lines to cover all zeros is 3. Since 3 < 4, the matrix needs adjustment. Step 4: Matrix adjustment The smallest uncovered element is 6. Subtract 6 from all uncovered elements and add it to elements at intersections.

Now, 4 lines are required to cover all zeros. An optimal assignment can be made. Step 5: Optimal assignment The optimal assignment is M1 → J3, M2 → J2, M3 → J1, M4 → J4. Therefore, the total cost is 69 + 37 + 11 + 23 = 140.

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Example 2 Three workers (A, B, C) are available for four jobs (1, 2, 3, 4). The cost matrix is shown:

Determine the minimum cost assignment.

Apply the idea Step 1: Create a square matrix Add a dummy worker (Row D) with costs of 0 to make the matrix 4 × 4.

Steps 2 and 3: Row and column reduction Row reduction is performed, followed by column reduction. The ﬁnal reduced matrix is:

Steps 4 and 5: Cover zeros and assign The minimum number of lines to cover all zeros is 2. After matrix adjustment, an optimal assignment can be found. The optimal assignment is A → 3, B → 1, C → 4. Worker D is assigned to the remaining job (Job 2), which means Job 2 is the one that does not get done. The minimum cost is calculated from the original assignments: 9 (A → 3) + 8 (B → 1) + 12 (C → 4) = 29

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10.09 Practice questions Simple familiar 1

Match each term to its deﬁnition for complex assignment problems using the Hungarian algorithm: a

Complex assignment problem

Hungarian algorithm

Cost matrix

Optimal assignment

A method to determine the optimal assignment for maximisation or minimisation

Selecting one 0 per row and column to optimise the total value or cost

iii

A matrix representing costs, used directly in minimisation or transformed for maximisation

Optimally assigning resources to tasks using a square matrix

Determine whether these statements are true or false for the Hungarian algorithm in complex assignment problems: a

For maximisation, the matrix is transformed by subtracting each entry from the largest entry.

The algorithm can be applied to non-square matrices without modiﬁcation.

Covering zeros with lines equal to the matrix size indicates an optimal assignment is possible.

In a minimisation problem using the Hungarian algorithm, what does a 0 in the ﬁnal matrix represent?

A 3 × 3 eﬃciency matrix for assigning machines to jobs is:

What is the largest entry in the matrix?

Complex familiar 5

Three machines, labelled M1, M2, M3, can be assigned to three jobs, labelled J1, J2, J3. The eﬃciency scores are:

Identify the matrix structure (rows, columns, and entry meaning).

What is the largest entry in the matrix? 10.09 Solve complex assignment problems mathspace.co

707

708

A manager needs to assign three consultants ( A, B, C ) to three projects (1, 2, 3). This matrix shows the suitability score for each assignment, where a higher score is better:

Transform the maximisation problem into a minimisation problem by creating a cost matrix.

Perform row reduction on the cost matrix.

A 3 × 3 cost matrix for a minimisation problem is:

Perform row reduction.

Perform column reduction on the resulting matrix.

A cost matrix after row and column reduction is:

Cover all zeros with the minimum number of lines. How many lines are needed?

Adjust the matrix to increase the number of zeros.

A cost matrix for a 4 × 4 minimisation problem is:

Perform row reduction.

Perform column reduction on the resulting matrix.

A cost matrix after adjustment is:

Cover all zeros with the minimum number of lines. How many lines are needed?

Assign zeros to determine the optimal assignment for machines M1, M2, M3 to jobs J1, J2, J3.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Ex 1

Ex 2

A 4 × 4 eﬃciency matrix is:

Transform the maximisation problem into a minimisation problem.

Perform row reduction on the cost matrix.

A 3 × 3 cost matrix after row and column reduction is:

Cover all zeros with the minimum number of lines.

Adjust the matrix to increase the number of zeros.

A 4 × 4 cost matrix for a minimisation problem is:

Apply the Hungarian algorithm to determine the optimal assignment.

Calculate the total cost using the original matrix. Is this guaranteed to be the minimum? Explain.

A delivery company has three available vans (V1, V2, V3) but needs to make deliveries to four suburbs (S1, S2, S3, S4). The cost matrix shows the fuel cost (in dollars) for each van to travel to each suburb. Since one suburb cannot be served, the company wants to determine the assignment that minimises the cost for the three suburbs that are served:

Set up the initial square cost matrix required to solve this problem using the Hungarian algorithm.

Determine the optimal assignment and the minimum possible cost.

Which suburb will not be served in this optimal assignment?

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Ex 3

Three auditors (A1, A2, A3) are to be assigned to three departments (D1, D2, D3). The matrix shows the time (in hours) required for each audit. However, Auditor 2 is not qualiﬁed to audit Department 2:

Set up the initial cost matrix, accounting for the restriction.

Use the Hungarian algorithm to determine the optimal assignment and the minimum total time.

Complex unfamiliar 16

710

Five machines, labelled M1, M2, M3, M4, M5, can be assigned to five jobs, labelled J1, J2, J3, J4, J5. The efficiency scores are:

Using technology, transform the maximisation problem into a minimisation problem and perform row reduction.

Explain why a 5 × 5 matrix requires ﬁve lines to cover all zeros for an optimal assignment.

A 5 × 5 cost matrix for a minimisation problem is:

Perform row reduction and column reduction.

Explain how row and column reduction help in solving complex assignment problems.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

10 Chapter review SF

For the ﬂow network diagram shown, identify: a

The source node

The sink node

All intermediate nodes

P 3

2 S

T 3

3 U

Consider the ﬂow network shown. A cut, C1, partitions the vertices into sets {S, A} and {B, T }. Calculate the capacity of this cut.

T 18

20 18 A

Explain the max-ﬂow min-cut theorem.

For the ﬂow network shown, three cuts (C1, C2, and C3) are indicated by dashed lines: a

Calculate the capacity of each of the three cuts.

What is the maximum ﬂow for this network? Justify your answer.

C2 A 7

10 S

C 8

5 B

6 D

For the water supply network shown, determine the maximum ﬂow from the reservoir to the town in megalitres per day (ML/day). Show the minimum cut on a diagram and determine its capacity. P1 30

15 P3

Reservoir

Town

20 10

40 P2

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The maximum ﬂow of the network shown is 22 units. A proposal is made to upgrade the capacity of edge B → D from 6 to 10 units: A

12 T

5 12

D B

Determine the new maximum ﬂow of the network after the upgrade. Justify your answer by referencing the minimum cut. SF

Three friends, Amy, Ben, and Chloe, are available to help with three tasks: decorations, food, and music: • Amy can help with decorations and music. • Ben can help with food and music. • Chloe can only help with decorations. Draw a bipartite graph to represent the possible assignments of friends to tasks.

This bipartite graph shows which of four machines (M1, M2, M3, M4) can be used for three different jobs (J1, J2, J3): M4

Construct a matrix representation for this assignment problem. Use a 1 for a possible assignment and a 0 otherwise. The rows should represent the machines and the columns should represent the jobs. SF

Two baristas, Leo and Mia, are assigned to make two types of coffee, a latte and a cappuccino. The table shows the proﬁt (in dollars) for each assignment: Latte

Cappuccino

Leo

2.50

2.00

Mia

2.20

2.40

By inspecting all possible assignments, determine the assignment that maximises the total proﬁt. 712

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Perform the row reduction step of the Hungarian algorithm on this cost matrix:

This matrix shows the eﬃciency scores for assigning three workers to three tasks. Convert this maximisation problem into a minimisation (opportunity cost) matrix, ready for the Hungarian algorithm:

A council needs to assign three contractors (A, B, C) to three road repair jobs (J1, J2, J3). This cost matrix shows the quote (in thousands of dollars) from each contractor for each job.

Use the Hungarian algorithm to determine the assignment that minimises the total cost, and state this minimum cost. CF

A marketing manager wants to assign three salespeople (Ali, Beth, Carlos) to three regions (North, South, West). This table shows the predicted sales (in tens of thousands of dollars) for each assignment: North

South

West

Ali

Beth

Carlos

Use the Hungarian algorithm to determine the assignment that maximises total sales, and determine this maximum sales ﬁgure. CF

Four technicians (T1, T2, T3, T4) are to be assigned to repair four different machines (M1, M2, M3, M4). This matrix shows the time in hours required for each assignment. However, technician T2 is not qualiﬁed to work on machine M3:

Determine the optimal assignment that minimises the total repair time and identify this minimum time.

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Answers 1.01 Introduction to bivariate data Simple familiar 1 a Univariate data involves the analysis of a single variable, focusing on its distribution, central tendency, and spread. Example: The heights of students in a class. b Bivariate data involves two variables and explores the relationship or association between them. Example: The relationship between hours studied and test scores. 2 a Univariate

b Bivariate

c Bivariate 3 a Univariate data

b Bivariate data

c Univariate data

d Bivariate data

4 a Univariate data

b Bivariate data

c Bivariate data 5 a Variables: Hours trained and performance scores. Data type: Bivariate data. b Variable: Fuel economy (litres per 100 km). Data type: Univariate data. c Variables: Daily caloric intake and weight change. Data type: Bivariate data. 6 a Univariate data. It contains information about a single variable (height). b Bivariate data. It contains information about two variables (hours studied and test score) and shows a relationship between them. 7 Bivariate data. It shows data for two variables (Temperature and Ice Cream Sales) and is presented to explore a potential relationship between them. 8 Univariate data. It presents data for a single variable (Age) and does not explore a relationship with another variable. Complex Familiar 9 a Bivariate data. The variables are daily water intake and overall energy levels. The study aims to investigate a relationship between these two variables.

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Mathspace QLD 12 General Mathematics 2025 mathspace.co

b Univariate data. The variable is book genre. The study focuses on the distribution of a single variable within a population. 10 a Univariate data. The variable is the number of steps. The data is about a single characteristic of individuals. b Bivariate data. The variables are amount of sleep and reaction time. The study is explicitly looking for a relationship between these two variables. 11 a Possible question: “Is there a relationship between the amount of carbon emissions produced by a country and its average annual temperature?” b Possible question: “Does the price of a product inﬂuence the quantity consumers purchase?” 12 a Possible question: “Is there a relationship between the amount of time students spend on online learning platforms and their academic performance scores?” b Possible question: “Does the number of hours of practice per week correlate with an athlete’s success rate in competitions?” 13 The researcher’s claim is incorrect. This is a univariate study. The single variable being analysed is ‘test scores’. Although there are two groups, the study compares this one variable between the groups, rather than investigating a relationship between two different variables for each student. Complex unfamiliar 14 The researcher’s claim is incorrect. This study collects bivariate data. Although the data is collected from a single group of students, the study is investigating the relationship between two distinct variables: ‘total volunteer hours’ and ‘current GPA’. The aim is to see if a correlation exists between these two variables, which is a key characteristic of bivariate data analysis. 15 This statement is not always true. Counterexample: Student ID

101

102

103

Height (cm)

165

170

162

Explanation: In this table, ‘Student ID’ and ‘Height (cm)’ are two columns of numbers. However, Student ID is merely a label or identiﬁer and not a variable that is being related to height. The data primarily focuses on describing the heights of students, which is a single variable. Therefore, this is an example of univariate data, even with two columns of numbers. Bivariate data requires two variables where a relationship or association is being investigated. 16 a Univariate data. In Phase 1, the primary variable being collected is the ‘number of cars passing a speciﬁc intersection per hour’. While this data might be collected over time, the focus is on the distribution and patterns of this single variable (car count) without explicitly relating it to another variable within each observation. b Bivariate data. In Phase 2, the data collected includes both the ‘number of cars passing the intersection’ AND the ‘average travel time for vehicles’ during peak hours. The study implicitly seeks to understand the relationship between these two variables (e.g., how car count affects travel time), making it bivariate.

association between offering delivery and an increase in sales.

1.02 Describe data Simple familiar 1

Pop

Rock

Morning

Afternoon

Evening

Total

Under 30

30-50

Over 50

Total

Male

Female

Total

120

Vaccinated

Not vaccinated

Total

4 a

Not Vegetarian Total vegetarian Children

Adults

Total

2. Variables: • Variable 1 (Explanatory): Availability of delivery service (Categorical: “Delivery Offered” or “No Delivery Offered”). • Variable 2 (Response): Daily sales revenue (Numerical: in dollars).

b 5

• For two weeks, do not offer delivery and record the daily sales revenue. • For the next two weeks, offer delivery and record the daily sales revenue.

c Cats

Dogs None

Total

Urban

120

Rural

120

Total

120

240

3. Data Collection Method: The owner could run an experiment over a set period (e.g., four weeks), controlling for other factors like the day of the week.

Not Total microchipped

Microchipped

17 Study design for bivariate data: 1. Investigative Question: “Is there a relationship between the availability of a delivery service and the daily sales revenue of the café?”

Classical Total

6 a 16

b 22

c 26

7 a 25%

b 30%

c 70%

d 20%

By comparing the sales data from both periods, the owner can analyse if there is an

Answers mathspace.co

717

Sports club

Art club

Debate club

Total

Grade 9

60.0%

30.0%

10.0%

100.0%

Grade 10

41.7%

33.3%

25.0%

100.0%

Grade 11

30.8%

38.5%

30.8%

100.0%

Grade 12

33.3%

44.4%

22.2%

100.0%

Total

40.9%

36.4%

22.7%

100.0%

d There is a signiﬁcant drop in the percentage of students from Year 7 to Year 12. 15 a Gender b B. The explanatory variable (Gender) should be on the x-axis to compare the music preferences for each gender. c Yes, there appears to be an association. For example, males show a higher percentage for Rock and Roll and Rap music compared to females, while females show a higher percentage for Classical and Country and Western music.

Satisﬁed Neutral Dissatisﬁed

Total

Car

55.6%

30.8%

22.2%

40.0%

Public Transport

33.3%

61.5%

44.4%

45.0%

Bike

11.1%

7.7%

33.3%

15.0%

Total

100.0%

b School Pathway

Total

c B: There appears to be no association between the variables. The numbers in each category of phone are similar.

10 a

Support Do not support Males

Females

Total

103

150

b 56

c 51.4%

11 a 25.6%

12 a 46.09%

c 74.4% =

c 31.02%

Complex familiar 13 a Height b Play netball

Do not play netball

Total

Height ≥ 170 cm

31%

69%

100%

Height < 170 cm

30%

70%

100%

14 a Year group b i A = 2 iii C = 55 c i 60%

718

ii B = 3 iv D = 150 ii 4%

Mathspace QLD 12 General Mathematics 2025 mathspace.co

16 a

iPhone Samsung Other Total ATAR

100

non-ATAR

100

Complex unfamiliar 17 a Column percentages would be more appropriate. The explanatory variable is ‘Age Group’ (listed as columns), and to see how this inﬂuences the response variable ‘Preferred Promotion Method’ (rows), it is best to compare percentages within each age group column. b

Under 25

25-45

Over 45

Total

41.7%

66.7%

81.3%

65.0%

SMS

58.3%

33.3%

18.8%

35.0%

Total

100.0%

c There appears to be a strong association. Younger age groups (Under 25) show a preference for SMS promotions (58.3%), while older age groups (25–45 and Over 45) show an increasing preference for Email promotions (66.7% and 81.3% respectively). This indicates that as age increases, the preference shifts from SMS to Email.

18 a

Total

Arts

150

Sports

150

Total

130

100

300

b 26.7%

c 42.9%

However, the column percentage table reveals that among all Sport B injuries, 55.6% were female and 44.4% were male. The percentages clarify the relationship by showing the proportion relative to a speciﬁc total, which raw counts alone do not provide.

d 43.3%

e Yes, there appears to be an association. While both programs have the same total number of students (150 each), the distribution across academic results differs. For example, 26.7% of Arts students achieved HD, compared to 20.0% of Sports students. Conversely, Sports students had a higher percentage of Distinction results (46.7%) than Arts students (40.0%). These differences suggest that academic performance might vary depending on program participation. Complex unfamiliar 19 a

20 a Variables and Categories: • Variable 1: Social Media Usage (e.g., Categories: “< 3 hours per day”, “≥ 3 hours per day”) • Variable 2: Academic Grades (e.g., Categories: “High (A/B)”, “Medium (C/D)”, “Low (E/F)”) Data Collection: A survey administered to a random sample of students. Questions would include estimated daily social media use and self-reported (or school-veriﬁed, with consent) academic grades. b If constructed percentages (e.g., row percentages where social media usage is explanatory) show a noticeably higher percentage of students using “≥ 3 hours/day” falling into the “Low” academic grade category compared to students using “< 3 hours per day”, this would support an association.

Sport A injuries

Sport B injuries

Total injuries

Male injuries

42.9%

57.1%

100.0%

Female injuries

28.6%

71.4%

100.0%

Total

35.7%

64.3%

100.0%

Sport A injuries

Sport B injuries

Total injuries

Male injuries

60.0%

44.4%

50.0%

Female injuries

40.0%

55.6%

50.0%

• Socio-economic background (e.g., access to resources, need for part-time work)

Total

100.0%

These factors could inﬂuence both social media use and academic grades independently.

b The percentage tables provide a clearer picture by accounting for the total number of athletes in each group. While the raw count of Sport B female injuries (25) is higher than male (20), the row percentage table shows that 71.4% of female athletes’ injuries occurred in Sport B, compared to 57.1% for males. This suggests a higher propensity for Sport B injuries among female athletes.

However, association does not imply causation. A causal link would mean that increased social media use directly causes lower grades. This claim might be misleading. Confounding variables could include: • Time spent on homework/study • Participation in extracurricular activities • Amount of sleep

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719

21 a

Under 30

30-50

Over 50

Total

Morning Brew

37.5%

50.0%

12.5%

100.0%

Evening Roast

28.6%

42.9%

100.0%

Total

33.3%

40.0%

26.7%

100.0%

Under 30

30-50

Over 50

Total

Morning Brew

60.0%

66.7%

25.0%

53.3%

Evening Roast

40.0%

33.3%

75.0%

46.7%

Total

100.0%

b Interpreting different percentage tables (row vs. column) is crucial because the base for the percentage changes, affecting the conclusion. 1. For “Which age group is most likely to prefer ‘Evening Roast’?”, row percentages are more appropriate. This allows comparison of the makeup of customers who drink ‘Evening Roast’. From the row percentage table, the ‘Over 50’ group makes up 42.9% of all ‘Evening Roast’ drinkers, which is the largest proportion. 2. For “Which coffee blend is preferred by the ‘30–50’ age group?”, column percentages are more appropriate. Within the ‘30–50’ age group, 66.7% prefer ‘Morning Brew’ compared to 33.3% who prefer ‘Evening Roast’. This indicates ‘Morning Brew’ is preferred by this age group.

1.03 Associations between categorical variables Simple familiar 1 a Explanatory: Amount of sleep, Response: Test scores b Explanatory: Age, Response: Music genre preference c Explanatory: Type of fertiliser, Response: Crop yield

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2 a 40% b Yes, there appears to be an association. A much higher percentage of Night Shift workers prefer Coffee (85%) compared to Day Shift workers (60%). 3

Studies chemistry

Not chemistry

Total

Studies physics

Not physics

Total

Beach

City Break

Under 40

40 and Over

4 a

b 5 a

Support

Do not support

Total

Under 50

100

50 and over

100

Total

150

200

b 100

c 70%

6 a 66.7%

c 62.5% 7 a 63.33%

c 46.67% 8 a Year level b i 15 c i 30%

ii 30

iii 30

iv 100

ii 60%

d There appears to be an association. A much higher percentage of Year 12 students (60%) work in hospitality compared to Year 11 students(30%). 9 a Height b Row percentage frequency table

Complex familiar

Plays basketball

Does not play

Total

Height ≥ 180 cm

45%

55%

100%

Height < 180 cm

20%

80%

100%

d Yes, there appears to be a strong association. A much higher percentage of taller students (45%) play basketball compared to shorter students (20%). 10 a Has a backyard b Row percentage frequency table. The explanatory variable (backyard presence) is in the rows. c

Does Owns a not own pet a pet

Total

Has a backyard

75%

25%

100%

No backyard

20%

80%

100%

d There appears to be an association. Residents with a backyard are much more likely to own a pet (75%) compared to those without a backyard (20%). Conversely, residents without a backyard are more likely not to own a pet (80%). 11 a Age Group b B c Yes. The data suggests there is a clear association, as the corresponding segments are of different sizes. For example, a much larger percentage of the ‘Under 25’ group prefers laptops, while a larger percentage of the ‘25 and over’ group prefers TV. 12 a

Summer

Winter

Total

Seated

100

General Admission

100

Total

110

200

b Ticket Type

c C

13 a Gender Cardio

Weight

Total

Male

30%

70%

100%

Female

73%

27%

100%

c Yes, there appears to be an association as the numbers are quite different. Men seem to prefer weights, while women seem to prefer cardio. d No, association is not causation. There appears to be an association but we cannot say whether one variable causes the other. 14 a Type of part-time job b Yes, as the corresponding segments are of different sizes. For example, the ‘High Stress’ segment is much larger for Hospitality jobs compared to Retail or Other jobs. c No. While there is an association, we cannot say that the job type causes the stress. Other factors, such as the number of hours worked or the specific workplace environment, could also be influencing stress levels. 15 a November b June c Book loans d Yes, there appears to be an association for certain book types. The segment for Business book increases over time while the segment for Children’s book decreases over time. 16 a Fitness level b Yes c The corresponding segments are different in size, suggesting that fitness level increases as the number of cigarettes smoked per day decreases. 17 a There appears to be an association. Generally as year group increases the percentage of students watching cartoons decreases. b There appears to be an association. Generally as year group increases the percentage of students watching documentaries increases.

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721

c There appears to be an association between the year group and genre as the percentage of students in each year group who like the same genre is different. Complex unfamiliar

Satisﬁed Neutral Dissatisﬁed Total Car

32%

24%

44%

100%

Bus

33%

42%

25%

100%

Train

70%

20%

10%

100%

Cycle

50%

17%

33%

100%

Interpretation: While trains do show the highest satisfaction (70%), the overall impact on city-wide satisfaction depends on the proportion of residents using each transport mode. The statement implies causation (encouraging train use will increase satisfaction) based on association, which is not necessarily true. Other factors (e.g., journey length, cost, comfort) could also inﬂuence satisfaction. The conclusion is therefore limited by only considering one aspect of association and implying causation where only association is evident. b To make a more robust recommendation, the council would need: • The total number of commuters for each transport mode in the city to understand the scale of each group. • Data on why people choose a certain transport mode (e.g., availability, cost, speed, convenience). • Consideration of lurking variables that might inﬂuence satisfaction (e.g., commute distance, time of day). • A controlled experiment or longitudinal study to attempt to establish causation, rather than just association.

722

b The explanatory variable is ‘Age Group’. A row percentage table should be used.

18 a The council’s conclusion is partially valid but oversimpliﬁed. To properly evaluate, a row percentage frequency table should be used as ‘Mode of Transport’ is the explanatory variable.

19 a Is there an association between a donor’s age group and their preferred donation method?

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Online Mail Phone

Total

18-30 years

71%

18%

11%

100%

31-50 years

45%

30%

24%

99%

51-70 years

15%

53%

32%

100%

71+ years

48%

44%

100%

c Based on the data, the ‘Online Only’ campaign might be effective for the 18–30 age group as 71% already prefer online. However, for the 31–50 age group, while online is still the most preferred (45%), a significant portion (30%) prefer mail. An ‘Online Only’ campaign might exclude a notable segment of the 31–50 age group, potentially reducing donations from this demographic. The decision should be evaluated further to balance efficiency with potential donor loss from non-preferred methods. 20 a The explanatory variable is ‘App Usage’. This is because whether an employee uses the app (or not) is hypothesised to influence their productivity level, while productivity level is not expected to influence app usage. b Yes, the graph supports the company’s claim of an association. There appears to be an association between app usage and productivity. Employees who used Focus Mode show a higher percentage of High Productivity (60%) compared to those who did not use the app (30%). This suggests a positive association between using the app and higher productivity. c No, the company cannot definitively state that Focus Mode causes increased productivity based solely on this graph. This graph only shows an association, not causation. Potential limitations or other factors not accounted for (lurking variables) could include:

• Self-selection bias: Employees who are already highly motivated or disciplined might be more likely to use the app, and also more likely to be productive, regardless of the app’s effect. • Nature of work: Different roles might have different baseline productivity levels or app applicability. • Other interventions: Employees might be using other productivity tools or strategies that are not accounted for. • External factors: Personal factors, workload ﬂuctuations, or team dynamics could inﬂuence productivity. Therefore, while there’s an association, a causal link cannot be established without further controlled studies. 21 a Explanatory Variable: Proximity to Green Spaces (rows).

2. Refute (Lurking Variable): People with higher socio-economic status might be able to afford to live in areas with more green spaces, and also have better access to healthcare, healthier lifestyles, and less ﬁnancial stress, which are the true drivers of their improved mental well-being, not just the green space itself. 22 a The statement is misleading because it uses raw counts instead of percentages, which makes it diﬃcult to compare across different news sources that have varying total numbers of users. To understand the association, it’s more appropriate to use row percentages (since ‘News Source’ is likely the explanatory variable) to see the proportion of engagement levels within each news source category. For TV, 45% of users

Low High Moderate wellwellwell-being being being

Total

High proximity

70%

100%

Medium proximity

40%

20%

100%

Low proximity

which is the most frequent category for TV users.

15%

30%

55%

100%

b This conclusion is ﬂawed and misleading. It compares raw counts of ‘High Engagement’ without considering the total number of residents for each news source. While 5 is the lowest count, Newspaper readers might be a much smaller group overall. To accurately assess engagement by news source, percentage comparisons within each news source category are needed (row

20%

10%

b Reported stress levels by part-time job type 100%

High well-being

80% 60%

Moderate well-being

40%

Low well-being

20%

High

have medium engagement

Interpretation: This table would suggest a strong association because as proximity to green spaces decreases, the percentage of residents with high well-being decreases signiﬁcantly, while the percentage with low well-being increases. For instance, 70% of residents with high proximity report high well-being, compared to only 15% for those with low proximity.

c 1. Support (Causation): Exposure to natural environments has known physiological beneﬁts (e.g., reduced stress hormones), and provides opportunities for physical activity and social interaction, which directly contribute to well-being.

Medium

percentages). For example,

× 100% = 17%

of Newspaper readers have high engagement. This percentage should be compared to the high engagement percentages for TV (30%) and Online (29%), which shows Newspaper readers are indeed less likely to be highly engaged, but the conclusion based on raw counts is poorly justiﬁed. c A 100% stacked column graph would be most effective. The explanatory variable (Primary News Source) should be on the

Low

Proximity to green spaces

Answers mathspace.co

723

horizontal axis, and the vertical axis should range from 0–100% for civic engagement levels. This graph clearly shows the proportion of each engagement level within each news source category. The primary message it would convey is that there is an association between news source and civic engagement, with TV and Online users showing higher proportions of ‘High Engagement’ and lower proportions of ‘Low Engagement’ compared to Newspaper readers, who have a higher proportion of ‘Low Engagement’. This visual makes it easy to compare the distributions of engagement across different news sources.

1.04 Explanatory and response variables Simple familiar

7 The correct order is A, C, and B. • A shows a weak positive correlation because the points are widely scattered and do not closely follow a straight line. • C shows a moderate positive correlation, with points clustering more tightly around a line but still showing some noticeable spread. • B shows a strong positive correlation, where the points are tightly grouped around an upward-sloping line, indicating a very consistent positive relationship between the variables. 8 a No

b Yes

e Yes

f Yes

9 a Yes 10

1 a Explanatory variable: Gender, Response variable: Favourite dog breed

c Explanatory variable: Country of birth, Response variable: Language spoken at home

Possible answers include: Type of drinks consumed (alcoholic or non-alcoholic) b Answers may vary. Possible answers include: Education level

b No

b Explanatory variable: Parent’s eye colour, Response variable: Child’s eye colour

2 a Answers may vary.

8 4 2

Week 0

2 3 4 5 6 7

11 a i No

ii Not applicable

b i Yes

ii Positive

c i Yes

ii Positive

d i No

ii Not applicable

12 a False

c Answers may vary.

b True

Possible answers include: Type of job

13 Strong, positive, linear

3 a i Incorrect

14 a Test scores

ii Correct

b The explanatory variable is the number of ticket sales and the response variable is the revenue from the show. 4 a Height of the ball b Amount of time passed 5 Time spent exercising 6 a Linear c No relationship

724

b Non-linear d Non-linear

Mathspace QLD 12 General Mathematics 2025 mathspace.co

d Yes

Height

14 10

d Explanatory variable: Amount of sleep, Response variable: Alertness level

c No

c False

b Daily study time c Strong, positive, linear

d False

Complex familiar 15 a 80

22 a

Resting heart rate (bpm)

75 70 65 60 55 50

Hours of exercise 0 1 2 3 4 5 6 7 8 9 10

b Strong, negative, and linear c Yes, the data point (9 hours of exercise, 70 bpm resting heart rate) appears to be an outlier, as it deviates from the general negative trend. 16 a The teacher

b Region 3

17 a B

b Yes

Sightings

10 9 8 7 6 5 4 3 2 1

Operations 0 1 2 3 4 5 6 7 8 9 10

b No deﬁnite relationship c No 23 a

Price 17 16 15 14

Supply (kg)

Complex unfamiliar

0 600 700 800 900 1000

18 a Explanatory variable: People’s age Response variable: Daily exercise time b Explanatory variable: Type of soap Response variable: Amount of germs

b Strong, negative, linear c When very few bananas are available to be sold.

Response variable: Petrol prices

1.05 Associations between numerical variables

19 Dot 3

Simple familiar

c Explanatory variable: Day of the week

20 Strongly negative 21 a

Time

1 a Positive

b Negative

c Neither

d Positive

2 a Strong linear

c Weak linear 3 a Positive

c No correlation

b Strong non-linear d Weak linear b Negative d Positive

4 Negative

Temperature 0 1 2 3 4 5 6 7 8 9 10 b 10

5 a Strong, negative, linear correlation b Yes, it is an outlier. The point lies far from the general trend of the rest of the data.

c Moderate, negative, linear d 8°C, 69 seconds

Answers mathspace.co

725

6 a

Complex familiar

Time (min)

9 A strong, negative, linear correlation would be expected. As a country’s income per capita increases, it is likely to have better healthcare, nutrition, and sanitation, leading to a lower child mortality rate.

8 7 6 5 4

Speed (km/h) 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70

100

11 a 50

Test 2 score

60 50

Number of passer-by 0

40 30 20 10

Test 1 score

0 10 20 30 40 50 60 70 80 90 100

b Strong, positive, linear correlation 8 a

Time to help (s)

b Strong negative correlation 7 a

10 This is not a valid conclusion. The data shows a correlation, not causation. A confounding variable could be the number of hours spent studying. Students who watch more TV may be studying less, and it is the lack of study that causes the lower exam scores.

140 130

There is a strong, positive, linear association between the number of passers-by and the time taken to help. b It suggests that the more people who are present during an emergency, the longer it takes for any individual to offer help. 12 a The teacher. There is a strong, positive correlation, suggesting that classes who did well on Q7 also did well on Q10, implying the questions test similar skills. b This outlier suggests that this speciﬁc class performed unusually. They may have understood the topic for Q7 but not for Q10, or perhaps ran out of time before reaching Q10.

120 110 100

Complex unfamiliar 90

Height (cm) 0 140 145 150 155 160 165 170 175 180 185

b There appears to be no correlation or a very weak positive correlation. c The person with a height of 175 cm and an IQ of 145

726

Mathspace QLD 12 General Mathematics 2025 mathspace.co

13 The description is incomplete because ‘no correlation’ implies no relationship. This data shows a clear non-linear relationship. A better description is: Task performance is highest at a moderate level of anxiety (around 5). Performance is low when anxiety is very low or very high. The relationship is strong but non-linear, forming an inverted U-shape.

14 The ﬂaw is assuming correlation implies causation. Spending on art does not directly prevent potholes. An alternative explanation could be a confounding variable like the city’s overall budget. In years when the city budget is large, there is more money for both art and road maintenance (leading to fewer potholes). In years when the budget is tight, both are cut, leading to less art and more potholes. 15 a Within Group A, there is a strong, positive, linear correlation. b Within Group B, there is a strong, positive, linear correlation. c Overall, the data appears to have a weak negative or no correlation. This is misleading because it hides the strong positive trends that exist within each subgroup. Combining the groups masks the true nature of the relationships.

1.06 Correlation using r-values

1 Set B b iii

3 a True

c i

d ii

b False

4 Yes. In this context, a causal link is logical. It is reasonable to assume that a fertiliser is designed to improve crop yield. 5 r = −1 6 There is a very weak or no linear relationship between the variables. 7 The conclusion is incorrect because Pearson’s r-value only measures the strength of a linear relationship. The data shows a clear and strong non-linear (quadratic) relationship, even though the linear correlation is close to zero. 8 a Yes

b No

Complex familiar 9 There is a strong, negative, linear correlation. As a woman’s height increases, the probability of being turned down for a promotion decreases. 10 a r ≈ −0.88

c Yes, it is reasonable to conclude causation. Higher altitudes are known to cause lower atmospheric pressure, which in turn causes lower average temperatures. The statistical correlation supports this established scientiﬁc principle. 11 a No. b The size of the village or its population. A larger village will have more houses (more nesting places for storks) and also a larger population (leading to more births). 12 a There is a weak, negative, linear association. b No. A weak correlation does not disprove causation. It only indicates that the linear relationship is not strong. There could be a non-linear causal relationship or other factors at play. Complex unfamiliar

Simple familiar

2 a iv

b There is a strong, negative, linear association between altitude and average maximum temperature.

13 a There is a strong, positive, linear association between distance driven and maintenance cost. b Yes, it is reasonable. Increased distance driven causes more wear and tear on a vehicle, which logically leads to higher maintenance costs. c It might suggest the relationship is not perfectly linear across the entire range. Cars driven very little may have maintenance costs that are influenced more by age (fixed costs) than by distance (variable costs), causing the trend to be flatter for lower distances. 14 The conclusion is flawed because it incorrectly assumes causation from correlation. Critically, the data shows a positive correlation, meaning more libraries are associated with more crime. This finding directly contradicts the idea that libraries reduce crime—the relationship isn’t even in the direction one might hope for. The explanation for this unexpected correlation is a classic confounding variable: the city’s population. A city with a larger population will naturally have more libraries to serve its

Answers mathspace.co

727

1 R2 = 0.64

11 The R2 value changes signiﬁcantly because the outlier (the senior manager) does not follow the general trend of the other employees, and its presence weakens the measured strength of the linear association. The higher value of R2 = 0.70 gives a more accurate understanding of the underlying positive relationship between experience and productivity for the majority of the employees.

2 r = −0.7

Chapter 1 review

residents and, separately, a higher total number of crimes. This makes the two variables appear connected, even though one does not cause the other.

1.07 Coefficient of determination Simple familiar

Correlation coefficient (r)

Coefficient of determination (R2)

0.6

0.36

−0.9

0.81

±0.5

0.25

Cardio

Weights

Total

Morning

Evening

Total

150

Public transport

Car

Bicycle

Total

× 100% = 35%

4 68% 5 R2 = 0.85 6 45%

3 a Faculty b

Complex familiar

Arts

60%

30%

10%

100%

Science

40%

45%

15%

100%

7 a R ≈ 0.999 b Approximately 99.9% of the variation in the selling price of a house can be explained by the variation in its size. 8 a r = −0.9 b There is a strong, negative, linear association between a car’s age and its resale value. 9 Model B is better. A higher R2 value means that a larger percentage of the variation in the response variable (income) is explained by the explanatory variable. Model B explains 60% of the variation, while Model A only explains 45%. Complex unfamiliar 10 a R2 = 0.0225 b The ﬂaw is claiming there is ‘no relationship’. R2 only measures the strength of the linear relationship. The low R2 value (2.25%) signiﬁes that there is a very weak linear relationship, meaning price is a very poor predictor of effectiveness in a linear model. It does not rule out the possibility of a non-linear relationship.

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Mathspace QLD 12 General Mathematics 2025 mathspace.co

c Yes, there appears to be an association. A higher percentage of Arts students use public transport (60%) compared to Science students (40%). Conversely, a higher percentage of Science students drive a car (45%) compared to Arts students (30%). 4 a Explanatory: Daily temperature, Response: Number of soft drinks sold b Explanatory: Person’s age, Response: Preferred social media platform c Explanatory: Amount of rainfall, Response: Height of corn stalks 5 The association is a moderate, negative, linear correlation. 6 No, this is a case of correlation not being causation. A third, confounding variable (the size or severity of the fire) causes both more firefighters to be sent and more damage to occur. 7 a r = −0.7

b r = 0.9

8 a The primary ﬂaw is confusing correlation with causation. The data shows a relationship, but does not prove that drinking coffee causes people to handle stress better. It’s an invalid conclusion. b A possible confounding variable is the person’s job demands. People in high-stress jobs may drink more coffee to stay alert and also report higher levels of stress. The job causes both, creating the apparent association between coffee and stress. 9 R2 = 0.852 = 0.72 10 11 45% of the variation in the response variable can be explained by the variation in the explanatory variable. 12 a 4

= 65%

Overall English pass rate =

= 56.7%

c Overall, Maths still has the higher pass rate (65% vs 56.7%). However, this type of problem can lead to a paradox. The apparent contradiction arises because the group sizes are very different between campuses and subjects. The North campus has better results overall, and a higher proportion of its students take English. The South campus has worse results, and a higher proportion of its students take Maths. This unequal distribution can skew the combined results. Although not a full reversal in this speciﬁc example, it illustrates how combining data can mask or change the trends seen in subgroups. 14 Calculation: The coeﬃcient of determination is R2 = 0.22 = 0.04.

Muscle gain (kg)

Interpretation: An r value of 0.2 signiﬁes a very weak, positive, linear correlation. The R2 value of 0.04 means that only 4% of the variation in battery life can be explained by the variation in price in a linear model.

3.5 3 2.5 2

Critique of Conclusion: The blog’s conclusion is oversimpliﬁed and potentially ﬂawed for several reasons:

1.5 1 0.5

Protein intake (g)

0 20 40 60 80 100 120 140 160 180 200

b r = 0.996, R2 = 0.991 c Approximately 99.1% of the variation in muscle mass gain can be explained by the variation in daily protein intake. 13 a North Campus: Maths pass rate =

= 90%,

= 70%, South

English pass rate =

Campus: Maths pass rate =

= 40%,

= 30%. At both

English pass rate =

campuses, Maths has the higher pass rate. b Combined Data:

Overall Maths pass rate =

Pass

Fail

Total

Maths

130

200

English

150

• Linearity: The r-value only measures the linear relationship. There could be a nonlinear relationship (e.g., cheap laptops have poor battery, mid-range have the best, and expensive high-performance laptops have poor battery again). • Confounding Variables: Price is affected by many other factors besides battery, such as processing power, screen quality, and build materials. These factors are not controlled for. A high price might be for a gaming laptop with a powerful graphics card that drains the battery quickly, not for a laptop designed for long battery life. • Conclusion Scope: While the data shows price is a poor linear predictor of battery life on its own, it doesn’t mean there is “almost no relationship”. The weak correlation is itself a ﬁnding, but it doesn’t justify ignoring price completely without considering what the price is paying for.

Answers mathspace.co

729

15 • Explanatory and Response Variables:

17 a

Explanatory Variable: Type of streetlight (Categorical: Old vs. New LED). Response Variable: Number of reported crimes per month in a speciﬁc area (Numerical).

20 15 10

• Data Collection Method: Select several similar neighbourhoods. For a baseline period (e.g., 6 months), record the monthly crime statistics for all neighbourhoods with the old streetlights. Then, install new LED streetlights in half of the neighbourhoods (the ‘treatment’ group) while leaving the old lights in the other half (the ‘control’ group). Record the monthly crime statistics for all neighbourhoods for another 6 months. • Data Representation: The data could be presented in a two-way frequency table comparing the average monthly crime rates before and after for both groups. Alternatively, a comparative bar chart could show the change in average crime rates for the control vs. treatment group.

Number of stands

Person’s age (years) 50 55 60 65 70 75 80 85 90

b The form of the relationship is linear (or approximately linear). 18

Year 11

Year 12

Total

Musician

Dancer

Total

Total musicians = 38 Percentage of musicians =

× 100% = 63.33%

2.01 Least-squares lines

• Strong Association: A strong association supporting the claim would be a signiﬁcantly larger percentage decrease in the average monthly crime rate in the neighbourhoods that received the new LED streetlights, compared to a much smaller change (or no change) in the neighbourhoods that did not.

Simple familiar 1 y = 2.5x + 59.7 2 y = −1202x + 24 298 3 y = −0.5x + 9.5 4 y = 2.1x + 61.4 5 y = 1.7x + 0.3

Exam practice 16 The correct interpretation is that “36% of the variation in the response variable ( y) can be explained by the variation in the explanatory variable (x)”. None of the options perfectly match this standard wording. Option B is the closest in intent but poorly worded. Option A incorrectly swaps x and y. Let’s re-examine QCAA wording. The intended answer is likely B, interpreting “total variation” as “variation in y” and “the linear association” as “variation in x”. Oﬃcial answer would need to be checked, but B is the most plausible choice despite its ambiguity.

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Mathspace QLD 12 General Mathematics 2025 mathspace.co

6 y = 3.1x + 57.5 7 a m ≈ 2.47

b c ≈ 18.16

c y = 2.47x + 18.16 8 y = −1.91x + 114.42 9 y = 0.55x + 1.23 10 y = −4.01x + 93.27 11 y = 5.19x − 17.02 12 y = 5.58x + 42.99 13 a y = 3.3x + 3.7 c c = 3.7

b m = 3.3

Complex familiar 14 a = 3.50, r = 0.99

2 a The slope is −1929. For every one-year increase in the car’s age, its value is predicted to decrease by $1929.

= 12.08, sx = 1.87, sy = 4.81,

b The y-intercept is 27 393. At 0 years of age (when new), the car’s predicted price is $27 393.

b y = 2.55x + 3.17 15 a = 20.71, r = 0.99

= 61.43, sx = 12.05, sy = 21.16,

3 a For each additional week of training, the time to run 1 km is predicted to decrease by 0.5 minutes.

b y = 1.74x + 24.64 c The equation from technology is y = 1.74x + 25.35. The equation from the rounded summary statistics is y = 1.74x + 24.64. There are minor differences in both the gradient and the y-intercept due to the rounding of the summary statistics in part (a). 16 a

b At 0 weeks of training, the predicted time to run 1 km is 9.5 minutes. 4 a y = 2.52x + 29.58 b For each additional month in age, the height is predicted to increase by 2.52 cm.

Number of customers

Waiting time

b y = 0.61x + 5.05 17 a Answers may vary. Possible answer: (5, 15) and (10, 25) b m = 2 c The calculated gradient (m = 2) is identical to the gradient in the given equation. Complex unfamiliar 18 r = 0.63 19 Alex’s model has a steeper gradient (1.8 > 1.19) 20 Removing the outlier would change the values of the mean and standard deviation. This would alter the calculated gradient and y-intercept, resulting in a different least-squares line.

2.02 Interpret slope and y-intercept Simple familiar 1 a y = 2.5x + 4.8 b For each additional training module completed, monthly sales are predicted to increase by $2500. c With 0 training modules completed, the predicted monthly sales are $4800.

c At 0 months of age, the predicted height is 29.58 cm. 5 a y = −3.31x + 96.39 b For each additional hour of daily screen time, the test score is predicted to decrease by 3.31 marks. c With 0 hours of screen time, the predicted test score is 96.39 marks. 6 a y = 0.15x − 0.39 b For each additional year of experience, the weekly income is predicted to increase by $150. c The y-intercept is −0.39. With 0 years of experience, the predicted weekly income is −0.39 thousands of dollars, or −$390. This is not practical as income cannot be negative. Complex familiar 7 a y = 2.21x + 2 b For each week that passes, the plant’s height is predicted to increase by 2.21 cm. c The y-intercept is 2. At week 0, the predicted height is 2 cm. This is a practical value, as it could represent the initial height of the plant when the gardener started recording. 8 a y = 5.96x − 66.27 b For every 1°C increase in temperature, the number of swimmers is predicted to increase by 5.96, or approximately 6 swimmers.

Answers mathspace.co

731

c The y-intercept is −66.27. This predicts that at 0°C there would be −66 swimmers, which is not practical. 9 a The predicted score will increase by 10.4 marks.

2.03 Residual plots Simple familiar 1 a

b The ﬁrst subject, as its gradient (5.2) is steeper than the second subject’s gradient (3.8).

105

110

112

118

120

106.5 108.0

113.0

117.5

121.0

−1.5

2.0

−1.0

0.5

−1.0

37.7

37.2

21.1

27.1

28.9

30.4

31.9

33.4

34.9

8.8

6.8

−10.8

−6.3

9.1

38.5

55.1

46.2

40.1

50.5

54.0

47.3

−1.2

−1.6

1.5

1.1

0.7

2.5

3.1

4.5

5.0

6.2

−10.5

−12.0

−15.8

−17.2

−20.0

−11.0

−11.5

−16.0

−17.5

−19.5

0.5

−0.5

0.2

0.3

−0.5

Complex unfamiliar

Residuals

10 a The y-intercept is 21 000.

b The predicted value of a car when it is new (0 years old) is $21 000. 11 a Sarah: y = 1.80x + 0.60. Mike: y = 1.19x + 4.10. b Sarah’s performance. Her gradient of 1.80 means her score is predicted to increase by 1.8 marks for each extra hour of study, which is higher than Mike’s predicted 1.19 marks per extra hour. c Mike has the better predicted score with no study. His y-intercept of 4.10 predicts he would score about 4.1 marks with no study, whereas Sarah’s y-intercept of 0.60 predicts she would score less than one mark. 12 a For each additional square metre of ﬂoor area, the house price is predicted to increase by $2500.

Residuals c

Residuals d

b The predicted price for a house with 0 square metres of ﬂoor area is −$50 000. This is not practical as a house cannot have zero or negative area, nor a negative price. 13 a Answers may vary. Possible answer: (4, 16) and (12, 28) b m = 1.5 c The calculated gradient (m = 1.5) is identical to the gradient in the given equation.

732

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Residuals 2 a

−8

Residual

7 a

Residual

12.8

−32.7 −33.3

0.6

−5

11.4

−30.3 −30.3

11.1

−29.5 −29.7

0.2

6.5

−21.5

−20.1

−1.4

−5

−28.5 −27.4

−1.1

110

120

−10

x y

45.5

48.2

50.0

53.8

56.1

46.0

47.5

51.2

52.9

55.5

Residuals −0.5

−1.2

0.7

0.9

Residual 10 5 0

0.6

Temperature (°C) 20

−5

3 The residual is 2.

−10

4 Above 5 Negative 6 a

−1

Residual

Residual 1

c The linear model is not appropriate. The residual plot shows a clear, curved (parabolic) pattern, which indicates the original data has a non-linear relationship. 8 Plots A and D are appropriate. Their residuals are scattered randomly above and below the horizontal axis with no clear pattern. Plot B shows a distinct U-shaped curve, and Plot C shows increasing variance (a fan or cone shape), both of which indicate a linear model is not appropriate. Complex familiar

Practice sessions 1

−1

Residual 10 5 0

x 5

−5

c The linear model is a good ﬁt. The residuals are all very small (between 0 and −1) and are scattered randomly enough around the horizontal axis, indicating a suitable linear model.

−10

10 a The gradient is m = −2. The y-intercept is c = 32. The equation is = −2x + 32. b The residual is −3.

Answers mathspace.co

733

11 a

A linear model is suitable because there is no obvious pattern in the residual plot and the residuals are relatively small.

Residual 2 0

x 2

−2

b No, a linear model is not appropriate because the residual plot shows a clear U-shaped (parabolic) pattern, indicating the original data is non-linear. 12 a Sum = −1.5 + 1.0 − 0.5 + 1.0 = 0 b The least-squares line is calculated by minimising the sum of the squared residuals, which positions the line to be as close as possible to all points. This process naturally balances the positive and negative residuals, causing their sum to be zero. 13

20 18 16 14 12 10 8 6 4 2

15 The residual for the outlier will be a large negative value. On the residual plot, this will appear as a point far below the horizontal axis. This single, large residual can make an otherwise good model appear unsuitable, as it violates the condition that residuals should be relatively small. It can draw attention away from the random scatter of the other points.

2.04 Predictions Simple familiar 1 a y = 3.77 2 B = −17.9662 3 a x = 3.15

b x = −92.74

4 v = −63.88

5 7 hours 6 6.7 years (approx) 7 a Interpolation

b Extrapolation

8 Prediction: y = 15.26. The data is strongly correlated and the prediction is an interpolation, so the prediction is reliable.

x 0 1 2 3 4 5 6 7 8 9 10 11

Complex familiar

Complex unfamiliar 14 The least-squares regression line is approximately = 0.58x − 1.15. The residual plot is shown:

9 Prediction: y = 90.45. Although the data is perfectly correlated, the prediction is an extrapolation, so it is unreliable. 10 a y = 0.025x − 550.108 b $1449.89

Residual

c The prediction is reliable as it is an interpolation and the correlation is strong.

x 10 12 14 16 18 20 22

−1

11 a The least-squares line is y = 0.85x − 1.28. The predicted score is y = 49.72%. b The prediction is very reliable. The correlation is very strong and the prediction is an interpolation as 60 is within the data range of [56.3, 79.0]. 12 a y = 0.05x + 0.23 b 1.23

734

b y = 67.8912

Mathspace QLD 12 General Mathematics 2025 mathspace.co

c The prediction is unreliable. The correlation for the data used is moderate (r ≈ 0.69), and the prediction is an extrapolation as 20 is far outside the data range of [3, 10]. 13 a W = −60 tonnes per hectare. b No, the prediction is not reasonable. It is impossible to produce a negative amount of wheat. Complex unfamiliar 14 a The relationship between speed and fuel consumption is often non-linear (U-shaped). Team A’s model likely represents the range where efficiency improves with speed (consumption decreases), while Team B’s model represents the range where high speeds lead to lower efficiency (consumption increases). A linear model can be a good approximation for a small segment of a curve. b Team B’s model is appropriate as 80 km/h is within its data range, making the prediction an interpolation. Using Team B’s model: y = 5.2 litres/100 km. 15 a The prediction is not valid. It is an extrapolation because 150 deliveries is far outside the model’s data range of 20 − 80. The linear trend is unlikely to continue, making the prediction unreliable. b For a very high number of deliveries, factors like driver fatigue, increased traffic congestion, or logistical complexities could cause the time per delivery to increase, making the overall time grow more rapidly than the linear model predicts. 16 The prediction is highly unreliable. The model is based on data from the first 60 days of the year, so predicting for day 250 is a major extrapolation. The model also assumes a linear trend, but rainfall is a seasonal phenomenon and is highly unlikely to follow a linear pattern throughout the entire year.

2.05 Association and causation Simple familiar 1 i D

ii B

iii C

iv A

2 a False

b True

c True

d False

3 a Causation

b Association

c Causation

d Association

4 i B

ii A

5 B 6 a No, this association does not prove causation. An observed correlation shows that two variables are related, but it does not explain why. This is an observational study, which cannot rule out other external factors that might be influencing both lunch habits and academic performance. b A likely non-causal explanation is a third variable, such as parental involvement or socioeconomic status. Parents who are highly involved may be more likely to pack a lunch for their child. These same parents are also more likely to provide homework assistance and a supportive learning environment at home, which directly contributes to higher test scores. In this case, parental involvement is the common cause for both outcomes. 7 a No, this association does not prove that bicycle lanes cause a reduction in heart disease. The data comes from an observational study, which shows a correlation but cannot confirm a direct cause-and-effect relationship. It is impossible to rule out other factors that might be responsible for both the presence of bike lanes and the lower rates of heart disease. b A likely non-causal explanation is a third variable such as the suburb’s socioeconomic status or health consciousness. More affluent or health-conscious communities may have residents who are already more active, eat healthier, and have better access to healthcare, leading to less heart disease. These same communities are also more likely to advocate for and secure funding for amenities like bicycle lanes. Therefore, the

Answers mathspace.co

735

community’s overall lifestyle is the common cause for both outcomes. 8 a No, this does not mean attending tutorials causes happiness. This is an observed correlation, not proof of causation. There could be other underlying factors that lead a student to both attend tutorials and report higher levels of happiness. Without a controlled experiment, a cause-and-effect relationship cannot be confirmed. b A possible third variable is student motivation and social engagement. A student who is highly motivated, organised, and socially integrated is more likely to take advantage of academic resources like tutorials. These same traits—motivation, organisation, and a strong social network—are also known to be strong predictors of personal happiness and well-being. The underlying personality and engagement level of the student could be the common cause for both behaviours. 9 a No, this does not mean that owning an electric car causes this dietary choice. The observation shows a correlation between two lifestyle choices, but it does not establish that one causes the other. It is far more likely that a third, unmeasured factor influences a person to adopt both habits. b The most likely non-causal explanation is a third variable related to a person’s values, specifically environmental consciousness. A person who is concerned about their environmental impact is likely to make multiple lifestyle changes to align with that value. They might choose to buy an electric car to reduce their carbon emissions from transport and also choose to eat more plant-based meals to reduce the environmental footprint of their diet. The underlying value of environmentalism is the common cause driving both choices. Complex familiar 10 The manager is confusing association with causation. The likely hidden variable is the rain. Rain causes the floors to be wet (requiring signs) and also causes people to need

736

Mathspace QLD 12 General Mathematics 2025 mathspace.co

umbrellas. The signs don’t cause the umbrella sales. 11 Not necessarily. While libraries support education, it’s more likely that this is an association with a hidden variable. Suburbs with higher-educated populations may place a greater value on education and therefore demand and support the funding of more libraries. The community’s existing value for education is the likely reason for both factors. 12 It is not correct to conclude causation. A possible non-causal explanation is that dog owners often engage in more physical activity, such as daily walks. This regular exercise is a well- known factor that lowers blood pressure. The dog ownership is associated with the healthier habit. 13 a There is a negative association. As the age of the car increases, its resale value tends to decrease. b This conclusion is too simplistic because other variables also affect a car’s value. For example, the number of kilometres driven is another important variable. A 5-year-old car with very low kilometres could be worth more than a 4-year-old car with very high kilometres. 14 The major ﬂaw is that there are two “hidden variables” that were changed at the same time: the fertiliser and the amount of sunlight. The researcher cannot reliably conclude the fertiliser caused the growth because it might have been the extra sun. A better study would have all 20 plants in the same sunny location, with 10 getting the fertiliser and 10 not getting it. Complex unfamiliar 15 a The course directly teaches valuable communication and presentation skills. Employees who learn these skills are better at their jobs, perform better in meetings, and are more conﬁdent, which directly causes management to recognise their potential and promote them. b This is likely just an association. A possible non-causal explanation is that the employees who are already the most ambitious and

motivated are the ones who choose to take an optional course. Their underlying ambition is the “hidden variable” that leads to them both taking the course and working hard to earn a promotion. 16 To test for causation, a researcher should conduct a controlled experiment. The key steps would be: • Random Assignment: Randomly assign students to two groups. One group (the experimental group) gets standing desks, and the other (the control group) uses regular desks. • Comparison: After a set period, compare the academic performance of the two groups. If the group using standing desks shows a signiﬁcantly greater improvement in grades, it provides strong evidence that the desks are the cause. 17 The claim is misleading because it implies causation from an association. There are several other possible explanations. 1. A hidden variable: Professional or more dedicated athletes might consume energy drinks and also train much harder, have better coaches, and be more talented. Their dedication is the real reason they win more. 2. Reverse Causation: Athletes who win more often might get more sponsorship money, which they then use to buy products like energy drinks.

2.06 Practical applications of bivariate data Simple familiar 1 a v

b iv

e ii

f iii

c vi

d i

2 a Variable 1: Car colour (Categorical). Variable 2: Manufacturer (Categorical). b Variable 1: Hours slept (Numerical). Variable 2: Test score (Numerical). c Variable 1: Screen time (Numerical). Variable 2: Age group (Categorical). 3 a Strong, negative association b Weak, positive association

c Strong, positive association d Moderate, negative association 4 a 4.5

b 55

5 The predicted score is 91. 6 a Explanatory: Number of cats a person owns. Response: Amount of money spent on pet food. b Explanatory: Number of hours a student studies. Response: Score on a maths test. c Explanatory: Number of missed school days. Response: Final exam grade. 7 a This is not a causal relationship. The third variable is smoking. People who smoke are more likely to carry lighters, and smoking is a cause of lung cancer. b A causal link is plausible here. The availability of playgrounds provides the opportunity and encouragement for students to be more active. 8 a Group 1: 20%

b Group 2: 30%

9 No, this does not imply causation. A third variable, such as warm weather or the summer season, is likely causing both an increase in people buying ice cream and an increase in people swimming, which leads to more shark attacks. 10 a Scatterplot

b Two-way table

11 a The predicted value is −$7500. b This prediction is an extrapolation, as it is far outside the range of the original data (1 − 8 years). The linear model may not hold for older cars, and a car cannot have a negative value. Complex familiar 12 a There is a strong, negative, linear association between the age of the car and its value. b For every one-year increase in age, the car’s value is predicted to decrease by $2800. c $18 200 d Yes, this association is likely causal. The aging of a car leads to wear and tear, outdated features, and higher mileage, all of which directly cause its monetary value to

Answers mathspace.co

737

decrease. This process is known as depreciation.

• Percentage Analysis: To compare the locations, calculate row percentages:

13 a Concentration level (High or Low) b

16 • The explanatory variable is the study location and the response variable is the test result.

High concentration (%)

Low concentration (%)

Ate breakfast

83.3%

16.7%

Skipped breakfast

50.0%

c There is an association between eating breakfast and concentration. Students who ate breakfast were far more likely to report high concentration (83.3%) compared to students who skipped breakfast (50%). 14 a There is a strong, negative, linear association between screen time and sleep duration. b For each additional hour of screen time, the predicted amount of sleep a teenager gets decreases by 0.6 hours (36 minutes). c Predicted sleep = −0.6(5) + 9.2 = 6.2 hours.

Passed (%)

Failed (%)

Library

80.0%

20.0%

Home

60.0%

40.0%

• Description of Relationship: There is a clear association between study location and test results. A higher percentage of students who studied at the library passed the test (80%) compared to those who studied at home (60%). • Causality Discussion: We cannot conclude that the library causes a higher pass rate. It is likely that the library offers a quieter, more focused environment. However, it’s also possible that more motivated students are the ones who make the effort to go to the library. The location itself may not be the direct cause. 17 •

Car resale value vs. Age 28 000

15 • Graphical Analysis: A scatterplot of the data reveals a strong, positive, linear association between hours studied and exam score. As study time increases, exam scores tend to increase in a predictable, straight-line pattern.

26 000

• Numerical Analysis: The correlation coeﬃcient is approximately r ≈ 0.98, conﬁrming the very strong linear relationship. The equation of the least-squares regression line is approximately Exam Score = 7.55 × Hours + 39.66 • Interpretation: The slope of 7.55 means that for each additional hour a student studies, their predicted exam score increases by about 7.55 points. The y-intercept of 39.66 is the predicted score for a student who does zero hours of studying. • Causality: The relationship between studying and exam scores is very likely causal. Effective studying directly leads to improved knowledge and better performance.

738

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Resale value $

Complex unfamiliar

24 000 22 000 20 000 18 000 16 000 14 000

Age in years • Form, Direction, and Strength: A scatterplot shows a strong, negative, linear association between the age of a car and its resale value. As cars get older, their value consistently decreases. • Outliers/Clusters: There are no obvious outliers or clusters in this dataset; the points form a single, coherent group. • Correlation and Regression: The correlation coeﬃcient is approximately r ≈ −0.996, indicating an extremely strong negative linear relationship. The equation of the least- squares regression line is

approximately Resale Value = −2033.8 × Age + 29 899. • Interpretation: The slope of = −2033.8 means that for each additional year of age, the car’s predicted resale value decreases by about $2033.80. The y-intercept of $29 899 represents the predicted value of a brand new (0-year-old) car. • Causality: This association is causal. The process of aging causes wear and tear, mechanical issues become more likely, and models become outdated. This process, called depreciation, directly causes the car’s value to fall.

Chapter 2 review 1 y = 14.27x − 248.68 2 a For every one-year increase in the van’s age, its value is predicted to decrease by $3500. b At 0 years of age (when the van is new), its predicted value is $40 000. 3 a i Percentage = 25 × 2.5 + 5 = 67.5%. ii Interpolation, because 2.5 is within the range of the original data. b It would have been charged for 1 hour. 4 y = 2.7x + 9.5 5 a Residual = Actual − Predicted = 82 − 85 = −3 b Below the line 6 Yes, a linear model is appropriate. The residuals are scattered randomly above and below the horizontal axis with no discernible pattern, which indicates that a linear model is a good ﬁt. 7 No, this is an association, not causation. A third variable, such as hot and sunny weather, is the likely cause for the increase in sales of both sunglasses and bottled water. 8 A linear model is not appropriate because the residual plot shows a clear curved pattern, not a random scatter of points. While this curve may not be a perfect U-shape or parabola, the important thing is that there is a pattern at all. Any clear pattern in the residuals tells us that the relationship in the original data is non-linear, so a linear model is not a good fit.

9 The formula for the gradient (m) is The gradient m is 10.8. So, 10.8 = r × gives 10.8 = r × 12. Therefore, r =

. . This = 0.9.

10 The newspaper’s claim is not valid because it mistakes association for causation. A possible confounding variable is the parents’ level of education or socioeconomic status. Parents who are highly educated are more likely to own many books and also more likely to create a home environment that values and supports academic achievement (e.g., helping with homework, encouraging reading). This home environment is the likely cause of the better grades, not just the presence of books. 11 It is possible for both models to be accurate because the relationship between age and value is likely non-linear (curved). The value of a phone depreciates (loses value) very quickly in the first few years, hence the steeper slope in Model A. The rate of depreciation then slows down for older models, resulting in a less steep slope in Model B. This suggests that a single linear model would not be appropriate for the entire 5-year period, and the true relationship is a curve that is steeper at the beginning and flattens out over time. 12 Adding a point that lies exactly on the leastsquares line will not change the equation of the line ( y = 1.5x + 10). The line is already the best fit for the original points, and the new point has a residual of zero, so it doesn’t pull the line in any new direction. However, the addition of a ‘perfectly fitting’ point will strengthen the overall linear trend, causing the correlation coefficient, r, to move closer to 1 (or −1 if the trend was negative). In this case, since the slope is positive, r will increase and get closer to 1. Exam practice 13 D 14 a Pearson’s correlation coefficient is r = 0.986. This value indicates a very strong, positive, linear association between the air temperature and the number of vehicles parked.

Answers mathspace.co

739

b A confounding variable is a third variable that is associated with both the explanatory and response variables, providing an alternative explanation for their observed relationship. c Time of day, t, is a confounding variable because it inﬂuences both other variables. Temperature increases during the day, and the number of parked cars increases as commuters arrive for work. The time of day (e.g., midday) is the likely cause of both high temperatures and a full car park, not one causing the other. 15 The predicted number of ﬁsh caught is approximately 415.

d The lowest season is usually September, except for 2015 when the low season was November. Complex familiar 11 a

Fines recorded

50 45 40 35 30 25 20 15

Time Period 1 2 3 4 5 6 7 8 9 10 11 12

3.01 Read and interpret time series graphs

12 a

Simple familiar 1 a Time

b Increasing

Houses

105 100

2 a 2005 b The amount of sea ice is decreasing.

c September

3 a True

b False

c True

d True

e True

f True

g False

h True

4 a False

b True

c False

5 a Both are increasing

Time 1 2 3 4 5 6 7 8 9 10 11

b Yes c Decreasing

b 2007

d 3

6 May, June, August

13 a i

Data 25

7 a Remains constant b May 19th

c 7 days

8 a Increasing

b 4

9 a i Increasing

ii 3

b i Decreasing

ii 4

c i Increasing

ii 5

d i Increasing

ii 6

e i Decreasing

ii 7

10 a Increasing b 6 c January

740

b Increasing

Mathspace QLD 12 General Mathematics 2025 mathspace.co

10 5

Time period 0

ii 3 seasons per cycle iii Increasing iv Period 4

b i

15 a

130 Data 120 110 100 90 80 70 60 50 2

Profit (in thousands) 60 50 40 30 20

Time period 4

Quarter

1 2 3 4 5 6 7 8 9 10 11 12

ii 4 seasons per cycle iii Decreasing

b The underlying trend is stable or constant.

iv Period 11

c The data shows a clear seasonal pattern with 4 seasons per cycle. Proﬁts consistently peak in the ﬁrst quarter (Summer) and are lowest in the third quarter (Winter). This is an expected pattern for a beachside cafe, as business would be highest during the warm summer months and lowest during the cold winter months.

c i 82

Hours

80 78 76 74

16 a Decreasing

72 70

Months J

A Ju O ‘17

A Ju O ‘18

A Ju O ‘19

ii 4 seasons per cycle iii Remaining stable iv No outlier

17 The point for May in the second year would be signiﬁcantly lower than the point for May in the ﬁrst year, and it would likely break the established seasonal pattern by creating a sharp dip between April and June of that year.

d i

Complex unfamiliar

Number of bags (‘00s)

18 a

b 2013

c 2008

Profit (in thousands of dollars) 85 80

75 8

70 7

Months D J D J D J D J D J D J D J D ‘05 ‘06 ‘07 ‘08 ‘09 ‘10 ‘11 ‘12

ii 2 seasons per cycle

65 60 55

iii Decreasing

Year

iv Sudden drop June 2008 but not outlier 14 a Decreasing c Month 14

b 4

2010 2011 2012 2013 2014 2015 2016

b Increasing c 51 19 a The data shows a seasonal pattern with 4 quarters per cycle. Visitor numbers

Answers mathspace.co

741

consistently peak in the third quarter (July-September) and are lowest in the ﬁrst quarter (January-March) each year. b The seasonal pattern is likely driven by Australia’s climate and school holiday schedules. The peak in Quarter 3 (July-September) corresponds to the winter months in southern Australia, a time when many tourists travel north to escape the cold weather. The lowest numbers in Quarter 1 (January-March) align with the tropical wet season in North Queensland, which is typically hot, humid, and rainy, making it a less popular time for tourism.

b The 5-point moving average best smooths the data. 7

Time period

Raw data

3-moving median

5-moving median

13.8

−

10.9

13.4

−

13.4

19.7

13.4

12.6

10.7

−

9.3

−

3.02 Moving averages Simple familiar 1 a True

b False

c False

d True

e True 2 A 3-point moving average would be most appropriate. The span of the moving average should match the length of the seasonal cycle to effectively smooth out the seasonal ﬂuctuations. 3 a b

ii 3

b i 3

ii 3

9 a = AVERAGE(B2:B4) or = (B2+B3+B4)/3 b = AVERAGE(B2:B6) or = (B2+B3+B4+B5+B6)/5 c i = AVERAGE(B4:B6) ii = AVERAGE(B4:B8) Complex familiar 10 a

= 48 = 61.8

4 a i a = 88.6, b = 96.3, c = 102.1, d = 88.8 ii 5-point moving average b i a = 40.7, b = 40.3, c = 45.2 ii 3-point moving average 5 a 5-point moving average. It best smooths the data. b 3-point moving average. It best smooths the data, removes the most seasonality, and matches the number of points per cycle in the raw data. c 5 point moving average. It best smooths the data, removes the most seasonality, and matches the number of points per cycle in the raw data. d 5-point moving average. It removes the most seasonality. 6 a a = 163.0, b = 163.0, c = 157.9

742

8 a i 3

Mathspace QLD 12 General Mathematics 2025 mathspace.co

= 123.3 mm

b The data points are 78, 210, 82. In order: 78, 82, 210. The median is 82 mm. c The moving median is a better representation. The raw data for April (210 mm) is a clear outlier. The moving average is heavily inﬂuenced by this outlier, giving a high value of 123.3 mm. The moving median is not affected by the extreme value and gives a value of 82 mm, which is more consistent with the surrounding months and better represents the underlying trend. 11 a a = 13.0, b = 15.7, c = 18.7, d = 20.5 b i 0.401 ii 0.9922 c 3-point moving average 12 a a = 29.7, b = 31.1 b i 0.3372 ii 0.4491 c 5-point moving median data

d 89.05

Complex unfamiliar 13 a i 20°C

ii 24°C

9 0.40

b i 24°C

ii 20°C

10 a 0.67

b $1 675 000

11 a i a = 0.22

ii b = 0.89

14 Number of births

b c = 703.19

c 1.29

d 1044.74

22 20

12 a i 152%

18 16

Month

ii 129%

b b = 1809.32 c 0.85 13 a 1.25

3.03 Deseasonalise data

b $135 000

c $126 500 Simple familiar

14 a 1.28

1 a Heater sales in winter tend to be 30% more than the yearly average. b Wednesday earnings tend to be 19% less than the average weekly earnings. 2 Divide by 1.6

b $235 955.06

c $198 000 15 a i E4

ii J4

iii G18

b =AVERAGE(D11:D17) c =D18/E18*100 d =(F4+F11+F18)/3

3 a Friday

e =D18/(J2/100)

b Thursday

f =SUM(J2:J8)

c 62 cars

g =AVERAGE(J2:J8)

d 98 cars e To see the underlying trend more clearly 4 a i 10.00

ii 13.33

iii 17.67 b i 210.00%

iii 30.33

iv 37.00

b i 100.00%

d A ‘peak’ corresponds to the season with highest seasonal index and a ‘trough’ corresponds to a season with the lowest seasonal index.

ii 101.41%

iii 56.76% c i 0.9150

ii 0.9882

17 a January – April

iii 1.0968

c 1157 customers

6 a i 0.80 ii 0.97

iii 1.45

b 71.25 c 61.38 b 19.48

e 0.55

f 9.09

8 a i a = 1.06 c 1.05

c 1.69

d 12.43

18 a i 23.33

ii 31.67 iv 30.33 ii 113.67%

iii 184.64% c i 0.6078

ii b = 0.80

b May - August d 2479 customers

iii 27.33 b i 102.87%

7 a 0.77

b c = 63.70

ii Lower iv Lower

c Multiply the deseasonalised ﬁgure by the seasonal index

iii 73.58% ii 23.67

16 a i Higher iii Higher

b Divide the raw data by the seasonal index.

ii 82.50%

5 a i 28.00

Complex familiar

ii 0.8407

iii 1.5515 d Around 39% less than the average sales over the three days.

Answers mathspace.co

743

19 a The number of seasons in a cycle, which is 4.

3.04 Time series data in tables

b 0.9 c Quarter 3 because the seasonal index is the highest above 1 which means sales are above average by the most.

Simple familiar 1 a Q4

b Q1

c 30.72%

2 a 86.26%

b 61.53%

c 67.46%

e 1322.86

3 a 0.74

b 32

c 43

f 1114.20

4 a 25.6

b y = 26.9

20 1.9

5 a i A = 240

ii B = 210.00

21 a 1.04

iii C = 80.00

iv D = 240.63

d 1322.50

b There tends to be approximately 31% more waiters employed on average in Quarter 1 compared to the yearly average. c

b E = 1.049 c Increasing 6 a A = 195

b B = 167.50

c C = 83.58% Quarter Quarter Quarter Quarter 1 2 3 4

Deseasonalised No. waiters

150

138

146

157

22 a t = 14 b Week 3 in February c =AVERAGE(F11:F14) d =F12/G11*100 e Lower f U se the formula =AVERAGE(K4:N4) to check that the average is 100. Or Use the formula =SUM(K4:N4) to check that the sum is 400.

e E = 126.28% 7 a 96.62%

b Wednesday

c Lower

d 64

8 a 0.5

b Higher

c Lower

d 46

9 a 134% b Thursday and Friday c Monday, Tuesday and Wednesday d 102 10 a 0.9 b January, September, October, November, December c March, April, May, June, July, August

g Week 4

d 139

h Lower

e November

i 1416.75 j 1848 Complex unfamiliar 23 a Q1 : $15 100; Q2 : $15 300; Q3 : $15 800; Q4 : $16 500

11 a i A = 800

ii B = 1300.00

iii C = 70.00

iv D = 1100.00

b E = 80.00 12 a i 255

ii 102.14

b $18 000

b Below

c $15 300

c 0.83

d Sales in Quarter 2 are typically 10% higher than the yearly average sales.

d Decreasing

Mathspace QLD 12 General Mathematics 2025 mathspace.co

iii 68.99

v 0.63

Complex familiar 13 a i A = 1367.5

ii B = 1112

iii C = 96.873

iv D = 1329

v E = 66.646

744

d D = 180.68

iv 101

b Increasing c The seasonal index for January is 131.42, which is signiﬁcantly higher than 100. This indicates that customer numbers in January are typically over 31% higher than the yearly average, which would justify the need for more staff to provide adequate service. 14 a i A = 52

ii B = 55.56%

iii C = 81.25

iv D = 53.47%

d The trend is initially slightly increasing from t = 3 to t = 5, then decreases until t = 7, and then begins to increase again. There is no single, steady trend in the ﬁrst three years. e 40

Data

35 30 25

150 No. orders 140 130 120 110 100 90 80 70 60 50 40 30 20 10

15 10 5

Time 0 1

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

3.05 Make predictions using time series data t

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Simple familiar 1 92.8

c t = 15

2 a 66.5

d Monday

b 123.7

15 a $167 500 b $150 750 c Extrapolation, because the time period t = 19 is outside the range of the original data used to create the model. 16 a y = 0.64t + 21.78 b The number of deseasonalised visitors is increasing by an average of 0.64 thousand (or 640 visitors) each quarter.

c The prediction for time period 16 is more reliable. 3 a September b y = −1.9524t + 245.7857 c $224.31 4 a 102.8595 Complex familiar

c 31.38 thousand (or 31 380 visitors)

5 a y = 55.159t + 1121.150

d 18.828 thousand (or 18 828 visitors)

b 2050 customers 6 a 4510 cars

Complex unfamiliar 17 a i A = 27.33

ii B = 24.67

iii C = 23.63

iv D = 44

b To smooth out seasonal or cyclical ﬂuctuations in the data to help identify the underlying long-term trend. c The data is collected quarterly, which means it has a 4-point cycle (4 quarters in a year). A 4-point centred moving average is speciﬁcally designed to smooth out the effects of this four-period seasonality.

b 79.2018

b 4961 cars

7 a i 0.9718

ii 0.8915

iii 1.0581

iv 1.0847

b A = 228.0, B = 251.0 c y = 6.4536t + 195.9607 d $309 774 e Reliable because the prediction was made within one cycle of the data.

Answers mathspace.co

745

8 a Wednesday

Complex familiar

b X = 107.44, Y = 131.30

5 a A = 68

c y = 1.64t + 98.69

c C = 72.7%

d 130.20 cents e Reliable due to the prediction being made within one cycle of the available data. 9 a Q1: 0.94, Q2: 0.85, Q3: 1.17, Q4: 1.03

d D = 51

e E = 1.270 6 85.04% 7 a i X = 1122.4

ii Y = 989.6

b Q1: 233.11, Q2: 234.26, Q3: 231.01, Q4: 232.16

b 817

c $223 931.68

c Unreliable because the prediction was made beyond one cycle of the data.

Complex unfamiliar

8 a Q1: 0.6798, Q2: 1.0525, Q3: 0.9396, Q4: 1.3281

10 a $167 500

b Y1: 73.6, 76.0, 74.5, 75.3

b $150 750 c Extrapolation, because the time period t = 19 is outside the range of the original data used to create the model. 11 a y = 0.63t + 21.91 b The number of deseasonalised visitors is increasing by an average of 0.63 thousand (or 630 visitors) each quarter.

Y2: 88.3, 90.3, 90.5, 90.4 Y3: 110.3, 104.5, 106.4, 105.4 c 123 206 users d Reliable because the prediction was made within one cycle of the data Complex unfamiliar

c 31.36 thousand (or 31 360 visitors)

9 168 pods

d 18.816 thousand (or 18 816 visitors)

10 292 heaters

12 5380 air conditioners

11 $43.18

13 2316 seedlings

12 677 wetsuits

3.06 Further applications

Chapter 3 review 1 a The underlying trend is increasing.

Simple familiar 1 There is an upward trend over time. There are 5 seasons in a cycle. 2 a Sunday b 3-point moving average 3 a 16.2

b 16

4 a 425.33

b 120.19%

746

b B = 68.33

Day

Raw data

Deseasonalised data

Fri

180

498

Sat

530

441

Sun

675

470

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b Yes, the data appears to be seasonal. The cycle length is approximately 4 months, with a peak every third month of a quarter followed by a small drop. c An outlier occurs in Month 17. A possible reason could be a technical issue causing a service outage, or a competitor launching a major promotion. 2 a 92 coffees b A moving median would be more appropriate. The median is less affected by extreme values (outliers) than the mean. Using the mean would result in a smoothed value that is artificially inflated by the 250 sales on the fifth day, whereas the median would provide a more representative centre value of the period.

3 a The seasonal index for Quarter 4 is 1.30. b Deseasonalised sales = $250 000 c Sales in Quarter 2 are typically 30% below the yearly average sales (1 − 0.70 = 0.30).

Exam practice 7 The correct answer is Option D. 8 a Year 2022: Mean =

4 a Year 1 mean:

= 38.75

Year 2 mean:

= 41.75

Proportions of yearly mean:

Year 3 mean:

= 45.75

Mean =

Year 1 Q1:

= 0.903

Year 2 Q1:

= 0.910

Year 3 Q1:

= 0.874

= 445

= 415

Year 2023:

Proportions (e.g., Q1):

Season

2022 Proportion 2023 Proportion

Autumn

= 0.921

= 0.940

Winter

= 0.067

= 0.048

Seasonal Index (e.g., Q1):

Spring

= 0.461

= 0.361

Seasonal index Q1 =

Summer

= 2.551

= 2.651

= 0.896

Therefore, the ﬁnal indices are:

Seasonal indices (average of proportions):

Seasonal index Q1 = 0.896

SI (Autumn) =

= 0.93

Seasonal index Q3 = 1.029

SI (Winter) =

= 0.06

Seasonal index Q4= 0.657

SI (Spring) =

= 0.41

Seasonal index Q2 = 1.418

b The deseasonalised visitor number is 46 thousand. 5 a =AVERAGE(B2:B4) b =AVERAGE(B4:B8) 6 a Forecast = $66 000

SI (Summer) =

= 2.60

b Deseasonalised Autumn:

≈ 141.30

b Predicted actual profit = Forecast × SI = $66 000 × 1.10 = $72 600

Winter:

≈ 142.16

c Residual = Actual − Predicted = $72 000 − $72 450 = −$450

Spring:

≈ 138.39

Summer:

≈ 140.43

The season with the highest seasonally adjusted rainfall is Winter. 9 Step 1: Calculate seasonal indices Year 2021 mean =

= 242

Year 2022 mean =

= 215

Answers mathspace.co

747

Proportions:

Season

2021 Prop.

2022 Prop.

Seasonal index

Autumn

= 1.178

= 1.284

= 1.231

Winter

= 0.116

= 0.102

= 0.109

Spring

= 0.806

= 0.791

= 0.798

Summer

= 1.901

= 1.823

= 1.862

Step 2: Deseasonalise the data

Year

Autumn

Winter

Spring

Summer

2021

= 231.52

= 256.88

= 244.36

= 247.05

2022

= 224.21

= 201.83

= 213.03

= 210.53

4.01 Sequences

11 a i tn = 1.025tn − 1, t1 = 1300

ii tn = 1300 × (1.025)n − 1

Simple familiar 1 a 5

b 0.2

b i tn = tn − 1 + 77, t1 = 44

c 6

ii tn = 44 + 77 (n − 1)

2 3

12 642.5

3 a i T he sequence starts at −1 and increases by 2 each term.

Complex familiar

ii 9

13 503 316 390

b i T he sequence starts at 64 and each term is subsequently halved. ii 1 4 a No

b Yes

c Yes

d No

14 a The sequence deﬁned by An + 1 = An + 25, A1 = 12 increases faster. b The sequence deﬁned by An + 1 = 3.5An, A1 = 5 increases faster.

5 a tn = n2

b 400

15 1654.4

6 a tn = tn − 1 + 5

b tn = tn − 1 + tn − 2

16 a t10 = 39

c tn = −tn − 1 + 7 7 a 1, 3, 9, 27, 81

d tn = tn − 1 − b −8, −2, 4, 10, 16

c 0, 3, 0, 3, 0 8 a tn = tn − 1 + tn − 2, with t1 = 11 and t2 = 22 b Yes

10 a 3, 8, 13, 18, 23

748

17 a t13 = 4096 c S20 = 1 048 575

c No

d Yes

b 3, 6, 11, 18, 27

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b 103 d 16 b 16 384 d 23

Complex unfamiliar 18 a t5 = 20

b 11, 22, 33, 55, 88 9 a No

c S10 = 210

b 15th batch

c S10 = 215 19 a 25 litres c 387.5 litres

b 9th day d 6 days

20 a The number of seats in row n is found by adding n − 1 to the number of seats in the previous row. The rule is tn = tn − 1 + (n − 1), with t1 = 15. b The 10th row has 60 seats. c The total seating capacity is 466 seats. 21 a Plan A: An = An − 1 + 10, with A1 = 50. Plan B: Bn = 2 × Bn − 1, with B1 = 10. b In the 5th month, Plan B’s cost ($160) ﬁrst exceeds Plan A’s cost ($90). c Plan A has the lower total cost. The total cost for Plan A is $1260, while the total cost for Plan B is $40 950.

4.02 Arithmetic sequences

Complex unfamiliar 19 −35 20 n = 31 21 a x = 4 b The ﬁrst 5 terms are: 10, 20, 30, 40, 50.

4.03 Arithmetic sequences with tables and graphs Simple familiar 1 The gradient of the line is equal to the common difference, d, of the arithmetic sequence.

4 a 7, 9, 11, 13

b 10

b a, a + d, a + 2d, a + 3d b No

c No

d No

4 a −4

b −50

5 a 8, 12, 16, 20, 24

b −5, 0, 5, 10, 15

6 a 4, 9, 14

b −12, −16, −20

7 tn + 1 = tn + 35, t1 = 60 8 a 13, 18, 23, 28

4 2

5 a b 5

b tn = 17 − 7n

c −46

b 44

9 11, 21, 31, 41

11 a t1 = 4, d = 5

2 a −10, −6, −2, 2

10 a −7

b tn = tn − 1 + 2, t1 = 3

3 tn = 8 + 5(n − 1)

b −10, −14

1 a 20, 24

3 a Yes

b x=2

2 a 3, 5, 7, 9, 11

Simple familiar

18 a x = 8

b 3 6 tn = 5 + 3 (n − 1)

12 57

7 This is a model of linear decay because the common difference (−3) is negative.

Complex familiar

8 a tn = −8 + 5(n − 1) b

13 d = 6 14 a d = 5

b 4

15 a The ﬁrst term is 3 and the common difference is 6. b 177 16 a n ≤ 10 17 n = 32

−8

−3

9 a i d = 3 ii tn = 5 + 3 (n − 1) iii

b 1

Answers mathspace.co

749

d i d = 2.5

ii tn = 5 + 2.5 (n − 1)

iii

30 25 20

7.5

12.5

27.5

10 5

0 1 2 3 4 5 6 7 8 9 10

v Yes

b i d = −4 ii tn = 50 − 4 (n − 1) iii

0 1 2 3 4 5 6 7 8 9 10

v Yes iv

t 50 n

Complex familiar

45 40 35 30 25 20 15 10 5

10 a i

iv 53 b i

c i d = 4 ii tn = −46 + 4 (n − 1)

−3

ii d = 5

−46

−42

−38

−34

−10

n 1 2 3 4 5 6 7 8 9 10 t −5 n −10 −15 −20 −25 −30 −35 −40 −45 −50

v Yes

iii tn = −3 + 5(n − 1) iv 82 c i

Mathspace QLD 12 General Mathematics 2025 mathspace.co

ii 2 iii tn = 3 + 2 (n − 1) iv 25 d i

−4

−7

−10

−13

ii −3 iii tn = −4 − 3 (n − 1) iv −34

750

iii tn = 1 + 4(n − 1)

v Yes

ii 4

0 1 2 3 4 5 6 7 8 9 10

iii

b Vn = 1440 − 40n

11 y = 3x + 9

13 a $1440

12 a The ﬁrst term C1 is the joining fee plus one week, which is $65. The common difference is $15. The rule is Cn = 65 + 15(n − 1), which simpliﬁes to Cn = 50 + 15n.

14 a tn + 1 = tn + 4, with t1 = 28 b 80 minutes 15 a $2965 b tn = tn − 1 − 5, t0 = 3000

b 30 weeks.

Complex unfamiliar

13 n = 127

16 8750 phones

Complex unfamiliar

17 n = 22

14 n = 102 15 a tn = 2 + 5(n − 1)

b t30 = 147

16 An = −3 − 5 (n − 1)

18 a The distance to the kth ball is 2 + 3(k − 1). The total distance run is twice this: 2(2 + 3(k − 1)) = 6k − 2 metres. b 70 m

4.04 Application of arithmetic sequences

4.05 Geometric sequences Simple familiar Simple familiar 1 a

b 32 minutes

2 a $25

b $1200

3 a 50 m

b 500 m

4 a $40

b $860

5 a $400

b $1700

c 512, −4096

b tn = tn − 1 + 1.25, t0 = 1.5 b Vn = 15 000 − 1200n

c $4200 8 a $36 000

ii 2

iii 2

iv 128

b i 2

ii 2

iii 2

iv −64

c i −3

ii −3

iii −3

iv 162

d i

iii

3 a 4

b −8

4 7, −14, 28, −56 5 700 000, 728 000, 757 120, 787 404.8

b $7200

6 0.5, 2, 8, 32

c $7200

7 tn = 5tn − 1, t1 = 9

9 a $160 b Vn = Vn − 1 − 10, V0 = 160 10 a

2 a i 2

6 a $14 7 a $15 000

b −768, 3072

1 a 54, 162

8 a tn = 3n − 1

230

260

290

320

b 4 782 969

9 By choosing any term after the ﬁrst and dividing it by the previous term. 10

b 30

11 tn = −0.3 × 5n − 1

Complex familiar 11 a $175

Complex familiar b $6750

12 a There is a common difference of 21 cm between each distance. b 254 cm

12 t1 = 2 13 a r = 5

b tn + 1 = 5tn, t1 = 7

14 a r = ±4

b tn = 4tn − 1, t1 = 6

Answers mathspace.co

751

15 a The common ratio is 2 and the ﬁrst term is 2.

iii

n−1

b tn = 2 × 2

tn 25

16 a r = 4

b t1 = 3

n−1

17 tn = 5 × 6

Complex unfamiliar

18 x = 20, y = −1280

19 Yes. The ratio of consecutive terms is

and

. Since there is

a common ratio of r2, the new sequence is geometric. 20

b i

−12

−48

6144

ii r = −2

21 B = 5 × 2h

4.06 Geometric sequences with tables and graphs

iii

Simple familiar

2 a 3, 5, 7, 9, 11 b Arithmetic, because the points lie on a straight line, indicating a constant common difference. 3 tn = 5 × 2n − 1

30 20 10

−10 −20 −30 −40 −50

1 The points form an exponential curve, which is characteristic of a geometric sequence.

iv an exponential curve

, 5, 15

n 1

2 3 4 5 6 7 8

iv a zig-zag pattern 6 tn = 5 × 3n − 1 7 This is a model of exponential decay because the common ratio (0.8) is between 0 and 1.

8 151 655.234

35 30 25

9 Vn + 1 = 0.95Vn, V0 = 42.54

10 B = 5 × 2h

Complex familiar

10 5

11 a tn = 1 × 5n − 1 or tn = 5n − 1

5 a i

1536

b 78 125

ii r = 2

752

Mathspace QLD 12 General Mathematics 2025 mathspace.co

12 a

13 a r = ±4

b tn = 4tn − 1, t1 = 6

14 a

b 187 500

6 a $3.74

Complex unfamiliar 15 a From the graph, t1 = 200 and r = The rule is Bn = 200 × (1.2)n − 1.

= 1.2.

b Ali’s investment: A5 = 250 × (1.1)4 ≈ $366.03

7 0.66 m 8 1.9

Ben’s investment: B5 = 200 × (1.2)4 = $414.72

9 721 nurses

Ben’s investment is worth more.

10 $5640.21

c The 4th year is the ﬁrst year in which Ben’s investment value exceeds Ali’s. 16 a The original sequence is tn = 4 × 3n − 1. The new sequence is An = (tn)2 = (4 × 3n − 1)2 = 16 × (32)n − 1 = 16 × 9n − 1. Yes, it is a geometric sequence with A1 = 16 and common ratio r = 9.

b $3.94

Complex familiar 11 a Vn = 50 000(0.85)n

b 6 years

12 a $59 559.81

b $56 723.63 n

13 a Bn = 100(0.8)

b 5 hours

14 a 57 344 bacteria

b 896 bacteria

b The terms are 14, 22, 46, … The ratio of the ﬁrst two terms is

≈ 1.57. The ratio of the

next two terms is

≈ 2.09. Since the ratios

are not constant, Bn is not a geometric sequence. 17 a From the graph, D1 = 80 cm. The common = −0.5. The rule is

ratio is r =

Dn = −0.5 × Dn − 1, with D1 = 80. b Using the rule Dn = 80 × (−0.5)n − 1, we must find the first swing n where ∣Dn∣ < 5. For n = 5, the absolute displacement is ∣D5∣ = ∣80 × (−0.5)4∣ = 5. This is not less than 5. For n = 6, the absolute displacement is ∣D6∣ = ∣80 × (−0.5)5∣ = 2.5. This is the first value less than 5. So, the answer is the 6th swing. 18 The 10th year

Complex unfamiliar 15 a Sn = 200 × 4n − 1

b 13 107 200

16 $426 615.23 17 a Plan A (Arithmetic): 15 skips. Plan B (Geometric): 16 skips. b The Beginner plan (Plan A) is more realistic. It represents linear growth, which is sustainable. The Intermediate plan (Plan B) represents exponential growth, which would become impossibly diﬃcult very quickly.

Chapter 4 review 1 t5 = −14 2 1, 3, 5, 11 3 The ﬁrst term t1 = 5 and the common difference d = 4. 4 tn = 50 + (n − 1)(−4) or tn = 54 − 4n.

4.07 Further applications of growth and decay

5 a tn = 2 + 3(n − 1) or tn = 3n − 1

Simple familiar

6 k = 4.5

1 a $2048

b $536 870 912

b The 18th term (t18 = 53)

7 tn = tn − 1 + 3, with t1 = −5

2 16 000

8 $65

3 a V1 = $6000, V2 = $4500, V3 = $3375

9 The ﬁrst term t1 = 100 and the common ratio r = 0.5

b 42.19% 4 a The common ratio is 1 − 0.0685 = 0.9315.

10 a tn = 80 × (0.5)n − 1 b The 8th term (t8 = 0.625)

b $7 714 786 n

5 a Pn = P0 (0.9)

b 12 595

11 The common ratio is 3.

Answers mathspace.co

753

12 $21 870

7 a Yes

13 764 ﬁsh

8 Stavanger, Uxbridge, El Tigre

14 After 10 years, Option A is worth $7500 and Option B is worth $7401.22. Therefore, Option A is worth $98.78 more than Option B.

9 a Nelson

b Port Douglas

10 a Eden

b Nagele

11 a Bangkok

b Suva

15 When n = 5, A5 = 30 and B5 = 48. This is the ﬁrst time Bn > An. Exam practice 16 A 17 a 14 seats b The completed table shows 12, 14, 16, and 18 seats. The total is 12 + 14 + 16 + 18 = 60 seats.

12 a Yes

b No

b Yes

c Yes

c No

d No

13 a 35°30′ S

b 51°15′ N

14 a 145.5° E

b 112.25° W

15 a Latitude: 59°N

b Latitude: 19°N

Longitude: 126°E Longitude: 126°E c Latitude: 39°N

d Latitude: 39°N

Longitude: 146°E Longitude: 106°E

18 The calculated total cost of $1733.84 falls within the estimated range of $1500 to $2000. Therefore, the estimate is reasonable.

Longitude: 145°E Longitude: 145°E

5.01 Latitude and longitude

Longitude: 175°E Longitude: 115°E

b Latitude: 68°S d Latitude: 38°S b Latitude: 34°S

Longitude: 170°W Longitude: 170°W

1 Longitude b Small circle

c Great circle 3 A great circle must have its centre at the centre of the Earth. The centre of the circle formed by the 45°S line of latitude is not at the Earth’s centre, so it is a small circle. 4 a A

c Latitude: 38°S 17 a Latitude: 4°S

Simple familiar

2 a Great circle

16 a Latitude: 8°S

b J

c (60° N, 120° E)

d 45°

e 180°

f D

g J 5 Adelaide: (35° South, 139° East)

c Latitude: 19°S

d Latitude: 19°S

Longitude: 155°W Longitude: 175°E 18 a Latitude: 9°N

b Latitude: 69°S

Longitude: 18°E Longitude: 18°E c Latitude: 30°S

d Latitude: 30°S

Longitude: 57°E Longitude: 21°W 19 a (28°S, 155°E)

b (21°N, 134°E)

c (18°N, 144°E) 20 a 9°

b City A

21 94°

La Grange: (19° South, 122° East)

Complex familiar

Alice Springs: (24° South, 134° East) Brisbane: (27° South, 153° East)

22 The coordinates in decimal degrees are (44.43°N, 26.10°E), and the city is Bucharest, Romania.

Menzies: (30° South, 121° East)

23 The new position is 5°45′ S, 160°30′ E.

Melbourne: (37° South, 145° East)

Canberra: (35° South, 149° East) Perth: (32° South, 116° East) Darwin: (12° South, 131° East) Sydney: (34° South, 151° East) 6 G

754

Mathspace QLD 12 General Mathematics 2025 mathspace.co

24 a (45° N, 75° E)

cultural boundary, central to their seasonal movements and stories.

b Part of Starting Change journey position in latitude

Change Finishing in position longitude

(45° N, 75° E)

9° N

7° W

(54° N, 68° E)

12° S

16° E

(42° N, 84° E)

c Difference in latitude: 18°, difference in longitude: 39°.

• Yidinji: The Great Dividing Range inﬂuences their territory, creating rainforest areas that support their hunting and gathering practices, with rivers like the Barron used for ﬁshing.

5.02 Distance between two places on Earth Simple familiar 1 a Small circle

25 (67°N, 81°W )

2 90.23 m

Complex unfamiliar

3 a Meridian

26 a • Warlpiri (Northern Territory): Yuendumu, a central Warlpiri community, is at approximately (−22.25°S, 131.80°E).

b Parallel

c Small circle 4 a 25° 5 a i No

b Great circle

d Half of a great circle b 3°

ii Yes

iii No

iv No

• Noongar (Western Australia): Perth, a key Noongar area, is at (−31.95°S, 115.86°E).

b 10°

• Yidinji (Queensland): Cairns, near Yidinji country, is at (−16.92°S, 145.78°E).

6 a 6400 km

b 16 532 km

7 a 9°

b 1001 km

b • Warlpiri: The Warlpiri are known for their Jukurrpa (Dreaming) stories, which map their land through songlines. For example, the Tanami Desert’s features are central to their ceremonies, as documented in Warlpiri oral traditions. • Noongar: The Noongar people’s Six Seasons calendar, based on environmental changes, guides their cultural practices. For instance, the season of Djilba (August- September) involves moving inland to hunt, tied to the Swan River. • Yidinji: The Yidinji resisted colonial expansion in the late 19th century, notably during the 1880s Cairns settlement conflicts, protecting their land near the Barron River. c • Warlpiri: The Tanami Desert defines their territory, providing resources like waterholes for survival and ceremonial sites for Jukurrpa stories, influencing their semi-nomadic lifestyle. • Noongar: The Swan River shapes their territory, providing food (e.g., fish) and a

c 1112 km

8 7006 km 9 a 111 km

b 3558 km

c 556 km

d 5226 km

10 a 7395 km

b 14 790 km

11 a Parallel

b Small circle

c 80°

d 8063 km

12 a 339 km

b 3759 km

c 9567 km

d 7284 km

e 2078 km

f 3847 km

13 2981 km 14 a 45°

b 4702 km

15 a 4880.47 km

b 852 km

Complex familiar 16 a 3788 km

b 2369 km

c 7.60 hours 17 9.48 hours 18 18.30 hours

Answers mathspace.co

755

19 2.95 hours 20 2.5° 21 a City A is closer to the prime meridian. b 8936 km

b 6 hours

c 7 hours

d 10 hours

e 8 hours 6 a 10 hours

d 40 km

7 a Dobo and Tokyo c 5:00 a.m.

22 20 015 km b 6060 km

c 8043 km

8 a 5 hours 9 A

25 12.57 m

10 B

26 18 083.0 km

11 A b 91.05 hours

Complex unfamiliar 28 Leo has the shorter total flight time. His route, traveling south along the meridian first and then east along the parallel at 18.1°S, covers 5654.4 km in approximately 6.28 hours, compared to Ava’s route, which covers 5792.6 km in approximately 6.44 hours. The shorter distance along the parallel at 18.1°S versus 12.4°S results in Leo’s faster journey. 29 The two-leg route increases the distance by approximately 3114 km (from 6850 km to 9965 km). At 900 km/h, this results in about 3.46 of additional flight time, a 45.5% increase compared to the great-circle route.

5.03 International time zones and time differences Simple familiar 1 a 5 hours

b 4 hours

c 8 hours

2 4:00 p.m. 3 a 5 hours

b 4:00 p.m.

4 a South Sandwich Islands b 2 hours c 4:00 p.m. on the 17th day of the month d Algeria e 4 hours f 1:25 on the 14th day of the month

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Mathspace QLD 12 General Mathematics 2025 mathspace.co

b 9 hours d 9:00 p.m. b 4:00 a.m.

c Sunday

24 12.57 km

27 a 31 631 km

b 7:00 a.m.

c 1:00 p.m.

c 8896 km

23 a 4448 km

5 a 7 hours

12 10:00 p.m. 13 a 8:30 a.m.

b 9:30 p.m.

14 a 10:30 p.m.

b 1:45 a.m.

15 a 10:45 p.m.

b 1:15 a.m.

16 a 16 hours

b 8:00 a.m.

c Monday 17 11:00 p.m. on the previous day 18 GMT is based on the mean solar time at the Prime Meridian, while UTC is a more precise standard based on atomic clocks. While often used interchangeably, UTC is the oﬃcial modern standard for science and international purposes. 19 The International Date Line marks the location on Earth where a new calendar day begins. When you travel from west to east across it, you gain a day (the date goes backward by one day). 20 This means the location’s local time is 6 hours behind the time at the Prime Meridian. 21 It deviates to avoid passing through countries or island groups, which would cause different parts of the same country to have different dates. Complex familiar 22 a 14:10 23 a 7:00 a.m.

b 5:20 p.m. b 6:00 a.m.

c 5:00 a.m.

24 a Annabelle

b Henry

5 a 16:15

c i 5:00 a.m.

ii 6:00 a.m.

d i 9:00 a.m.

ii 11:00 p.m.

6 15:14 7 a 1:49 a.m.

Complex unfamiliar

8 15:35

25 Wednesday

9 a 10:45 a.m.

26 22 hours

10 5:00 a.m.

27 a 12 h and 40 mins

b 10 h and 26 mins

c 13 h and 5 mins

d 15 h and 11 mins

28 100 h and 36 mins

5.04 Australian time zones and daylight savings

b 17:15

b 10:50 p.m.

b 6:13 p.m.

11 a 12:30 p.m.

b 1:30 p.m.

12 a 4:30 p.m.

b 1:30 p.m.

c 3:30 p.m.

d 3:00 p.m.

Complex familiar 13 a i 7:16 a.m.

ii 8:16 a.m.

iii 9:16 a.m.

Simple familiar

b i 1:48 a.m.

ii 3:48 a.m.

iii 4:48 a.m.

1 a 4:42 p.m.

c i 7:21 p.m.

ii 8:21 p.m.

iii 10:21 p.m.

d i 8:13 a.m.

South Western Queensland Australia Australia (QLD) (SA) (WA) 6:15

7:45

8:15

9:45

11:15

11:45

8:45

10:15

10:45

5:25

6:55

7:25

ii 10:13 a.m.

iii 11:13 a.m.

e i 5:24 a.m. ii 6:24 a.m.

iii 8:24 a.m.

f i 3:24 p.m. ii 4:24 p.m.

iii 5:24 p.m.

g i 5:11 p.m.

iii 7:11 p.m.

ii 6:11 p.m.

14 a 4:42 a.m.

b 2:42 a.m.

c 5:42 a.m.

d 3:42 a.m.

15 a 6:00 p.m.

b 5:00 p.m.

c 7:30 p.m.

16 a i 3:50 p.m. ii 10:21 p.m.

iii 10:45 a.m.

Complex unfamiliar c

AWST

ACST

AEST

3:55

5:25

5:55

5:11

6:41

7:11

10:25

11:55

12:25

5:47

7:17

7:47

2 a i 7:46 p.m.

b i 22 h and 26 mins ii 23 h and 28 mins iii 16 h and 31 mins 17 a 8:55 a.m.

ii 6:16 p.m.

Possible answer: The Gariwerd, Yawuru, and D’harawal calendars each have six seasons.

d 11:46 a.m.

b Answers may vary.

3 a October to April b April to October c Summer d Winter b No

e No

f Yes

f 6:00 p.m.

18 a Answers may vary.

ii 22:10

4 a Yes

d 9:28 a.m.

e 3:46 a.m. g 6:30 a.m.

b 5:57 p.m. c i 23:40

b 14:20

c 3:38 a.m.

c Yes

d No

Possible answer: For the Gariwerd calendar, the season of ‘Petyan’ (Season of the Wildﬂowers) is from September to mid-November. It is signalled by the blooming of wildﬂowers and is a time of eels and birds nesting.

Answers mathspace.co

757

c Answers may vary.

Complex familiar

Possible answer: This system is based on observable local ecological events rather than ﬁxed dates on a calendar. The number of seasons is different (e.g., six instead of four), and the lengths of the seasons can vary from year to year depending on when the natural signs appear. 19 a 12:25 p.m.

16 30°W 17 3:30 a.m. on Tuesday Complex unfamiliar 18 Approximately 3 hours 19 12:10 p.m. on Wednesday

b 9:55 a.m.

20 1:00 p.m. on Sunday

c 5 h and 22 mins d 2 h and 26 mins

5.06 Distance and time zone applications

e 8 h and 18 mins f 2:00 p.m. g

15 17:50

Time

Simple familiar

Michael Townsville, QLD

6:00 a.m.

1 a 12 897 km

Katrina Whyalla, SA

6:30 a.m.

Sally

4:00 a.m.

Location

Coral Bay, WA

5.05 Longitude and time difference

2 a 13 091 km b 10:45 p.m. the previous day 3 a 7006 km b 7 hours and 47 minutes c 1:15 p.m. the same day

Simple familiar

4 a ≈ 1334 km

1 2 hours

b 4 hours and 27 minutes

2 5 hours

5 2780 km

3 5 hours

6 3336 km

4 5 hours

7 3:00 p.m.

5 4 hours

8 6:00 p.m.

6 7:00 p.m. on Tuesday

9 5:00 p.m.

7 6:00 a.m.

10 a 9:30 a.m. on 15th October

8 6:45 a.m. on Saturday 9 a 9 hours

b 3:00 p.m. the same day

b 10:10 p.m.

c 1:10 a.m.

10 8:00 a.m. 11 6:40 a.m.

b 11:30 p.m. on 17th October 11 a 6 hours and 45 minutes b 9:00 a.m. on 17th November 12 C

12 90° 13 52.5° 14 105°

Complex familiar 13 1:00 p.m. 14 The time in Sydney is 11:38 a.m. on Tuesday. 15 a 9:00 a.m. on 8th January b 2:00 p.m. on 11th January

758

Mathspace QLD 12 General Mathematics 2025 mathspace.co

16 7:15 a.m. 17 Sydney’s midnight is at 8:00 a.m. on Dec 31st in New York. London’s midnight is at 7:00 p.m. on Dec 31st in New York.

Chapter 5 review 1 a 3°

b 334 km

2 9:00 a.m. 3 67.5°

Complex unfamiliar 18 a Yes, it would likely be suitable. A 9:00 a.m. call in Sydney corresponds to 7:30 a.m. in Nhulunbuy (1.5 hours behind). This time is after sunrise but still in the cooler part of the morning, falling within the ranger’s active work period before the midday heat. b A better time might be 8:00 a.m. Sydney time (6:30 a.m. Nhulunbuy time). This is very early in the work day, ensuring the ranger is available before heading out for intensive tasks. Alternatively, 4:00 p.m. Sydney time (2:30 p.m. Nhulunbuy time) would fall during the afternoon rest period, which might also be convenient for a longer discussion. 19 22° east of Greenwich 20 a They are likely in time zone UTC+6 (centred on 90°E). Their location (95°E) is 5° east of the standard meridian. Solar noon would occur 20 minutes before standard noon, at 11:40 a.m. local standard time. b The team can place a vertical stick (gnomon) in the ground. They should track the length of its shadow. Local noon is the moment the shadow is at its shortest. Trigonometrically, this corresponds to the moment the sun’s elevation angle is at its maximum, as shadow length s relates to stick height h and sun elevation angle θ by s = h/ tan(θ ). c At 95°W, they would likely be in time zone UTC−6 (centred on 90°W). Their location is 5° west of this meridian. Because the Earth rotates eastward, the sun would reach their meridian 20 minutes after it crosses the standard meridian. Thus, solar noon would occur at 12:20 p.m. local standard time. The timing relative to the clock changes because their position relative to the centre of their new time zone is different (west instead of east).

4 a 3407 km b 4 hours and 32 minutes 5 The time in Brisbane will be 1:30 p.m. The time in Perth will be 11:30 a.m. 6 The time difference is 6 hours. Departure time from City X was 7:00 a.m. 7 a 1:00 p.m. b This time (1:00 p.m.) is likely the hottest part of the day during the wet season, making it unsuitable for a long call. A better time would be earlier in the morning. For example, a 10:00 a.m. call from Canberra would be 7:00 a.m. in Broome, which is a much cooler and more appropriate time. 8 The ship travels for 108 hours. It travels 16.5 degrees east and 10.8 degrees north. The ﬁnal location is at (24.2°S, 126.5°E). The time difference is approximately 1.1 hours. The arrival time is 7:06 p.m. on Friday. 9 Ask your teacher for worked solutions. Exam practice 10 B 11 The longitude difference is 210 degrees, which corresponds to a 14-hour time difference. The local time in town B is 4:00 p.m. 12 Ask your teacher for worked solutions.

6.01 Simple interest - Review Simple familiar 1 a $600

b $3600

2 $3840 3 a $2388

b $199

4 a $457.20

b $130.65

c $207.90

d $4536.00

e $583.15

f $506.14

Answers mathspace.co

759

5 $8600

b $374.59

6 6% p.a.

2 a $4206.27

b $2116

7 3 years

3 a $2315.25

b $315.25

8 $5000

4 a 6.1%

b $15 000

c $15 915

9 a $500 b Tn + 1 = Tn + 500, where T0 = 10 000

d $49 022.89

5 a 0.02

c Tn = 10 000 + n × 500 which simplifies to Tn = 10 000 + 500n d $12 500

b 0.08

c $5412.16

d $5858.30

6 a 0.015

b $2391.24

c $2391.24

e 10 years 10 a $2200

d November 2023

7 a tn + 1 = 1.079tn, t0 = 520

b $33 000

b $704.84/m2

c 9 years

8 a tn + 1 = 1.032tn, t0 = 5700 b tn = 5700 (1.032)n

Complex familiar 11 a Simple interest

c $7568.19

b 12%

d 19 years

c $2254 12 a Simple interest

9 a i $4719

b 7%

ii $471.90

b $1290.90

c $6750

10 a tn + 1 = 1.06tn, t0 = 8000

13 a $2000

b $100 c 5% p.a. d An + 1 = An + 100, where A0 = 2000

Value at Time beginning period of time ( years) period

Complex unfamiliar

$8000

14 a A = P + 0.15P = 1.15P

$8480

$8988.80

b $6000 15 a Investment A: AA = P + 0.07P = 1.07P Investment B: AB = P + 0.105P = 1.105P b $6000

Interest earned in time period

$8480

$480

$8988.80 $508.80 $9528.13

c $1528.13 11 a tn + 1 = 1.028tn, t0 = 50 000 c $89 294.64 d 29 years 12 a $5150

Simple familiar

760

Value at end of time period

b tn = 50 000 (1.028)n

6.02 Compound interest using sequences

1 a

iii $5190.90

Value at Time beginning period of time ( years) period

Value at end of time period

Interest earned in time period

b Vn + 1 = 1.03Vn, V0 = 5000 13 a tn + 1 = 1.06tn, t0 = 1800 b $108.00 c $1908.00

$3000

$3120

$120

d $114.48

$3120

$3244.80

$124.80

e $2022.48

$3244.80

$3374.59

$129.79

Mathspace QLD 12 General Mathematics 2025 mathspace.co

f $121.35 g $2143.83

$539.33

h $343.83

Complex unfamiliar

i 19.10%

22 Bn + 1 = 1.02Bn, B0 = 3000

j $324

23 Vn + 1 = 1.04Vn, V0 = 7500

k $19.83

6.03 Compound interest using technology

Complex familiar 14 a $33 644.52

b 16.31 years Simple familiar

15 a tn + 1 = 1.00 525tn, t0 = 27 500 b tn = 27 500 × 1.00 525n

1 a $6876.05

b $5901.67

c 35 358.08 INR

c $6548.06

d $10 412.84

d 49

e $17 610.39

e 4.08 years 16 a $2081.21

2 a $14 181.06

b $2539.47

c $5751.33

d $510.36

e $2578.28

b Vn + 1 = 1.01Vn, V0 = 2000 17 a $2500

3 a $3633.39

b $1633.39

b 0.02

4 a $7153.84

b $2153.84

c 0.24

5 $111 482.72

d Vn + 1 = 1.02Vn, V0 = 2500 18 a i $12 055.06 b Vn + 1 =

6 $21 736.60

ii $12 082.69

7 9

Vn, V0 = 12 000

c 2.747%

8 $1030.38

d Lower

9 $2071.04

19 a 0.02

10 $8450.35

A 1 5

11 $555.53

Balance at end of quarter

12 $425.97

$127.35 $6494.60

14 $6527.37

Balance at Quarter beginning Interest of quarter 4

$6367.25

c Bn + 1 = 1.02Bn, B0 = 6000

13 $4.86

15 35.16%

d $7029.96

Complex familiar

e 36 quarters

16 $255.35

20 a 4.5% b Bn + 1 = 1.045Bn, B0 = 15 000 c i $19 533.90

ii $108 723.73

17 34.67% 18 17.71%

d The balance would be lower.

19 863

e The amount of interest earned would be more than doubled.

20 $269 539

21 a $16 288.95 b $16 135.95 c Investor A has the greater balance by $153.00.

21 The high-risk investment. 22 $5535.83 23 $78.23

Answers mathspace.co

761

24 $20 831.61 25 a 4.08%

b 8.16 years

Complex unfamiliar 26 31 27 22nd September 2029

13 Compounded monthly means that interest is calculated more frequently and added to the principal. In subsequent months, interest is earned on this slightly larger principal (interest on interest), leading to a higher total return and a higher effective annual rate compared to compounding only once a year.

28 No. The time it takes for an investment to double is independent of the principal amount. Using a TVM solver, setting PV = −1, FV = 2 gives the same value for N as setting PV = −1000, FV = 2000, provided the interest rate and compounding periods are the same.

Complex familiar

6.04 Effective annual interest rate

16 Lucky Lending, because their effective rate is lower.

Simple familiar 1 a $4638.77

14 a $22 767.80

b $23 302.19

c $5922.19

d $455.55

e 2.62% 15 10.57%

17 Compounded monthly, as it has a higher effective interest rate.

b $638.77

c 5.0625%

18 2% compounded daily, as it has a higher effective interest rate.

2 a i $198.84

ii 3.98%

b i $373.16

ii 4.65%

19 Plan 2 20 $19 456.66

c i $182.66

ii 4.37%

d i $153.77

ii 4.39%

e i $176.50

ii 2.31%

f i $71.19

ii 2.73%

Complex unfamiliar 21 6.167%

3 a 4.68%

b 4.07%

c 4.35%

d 3.25%

e 2.84%

f 4.81%

g 2.53%

h 4.71%

4 a 6.9%

b 1.8%

c 9.4%

d 7.3%

e 4.4%

22 Yes, the claim is correct.

6.05 Recurrence relations for present value annuities Simple familiar

5 24% b 2.67%

c 13.23%

7 a 3%

b 1%

c 12.55%

2 a ii

8 a 0.2%

b 10.95%

9 a 0.02%

b 7.57%

3 The balance after one year decreases to $13 450.

10 a 5%

b 5.0625%

c 5.0945%

d 5.1162%

e 5.1246%

f 5.1267%

d 12.36%

1 A

6 a 3.15%

4 a 1

c 1.012

b $3654.80 6 $2880

12 Bank A offers the better return due to the higher effective annual interest rate.

7 C

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b 1.005

c i

5 a An + 1 = 1.007An − 200, A0 = 4000

11 Option A is the better investment.

762

b iii

8 a False

b False

d 1.02

9 a Recurrence for $200 payment: An + 1 = 1.02 An − 200, A0 = 5000 Recurrence for $250 payment: An + 1 = 1.02 An − 250, A0 = 5000 b Balance after four months with $200 payments: $4587.84 Balance after four months with $250 payments: $4381.76 c The $250 repayment plan results in a lower outstanding balance after the fourth month. 10 a Recurrence for $100 payment: An + 1 = 1.03An − 100, A0 = 2000 Recurrence for $125 payment: An + 1 = 1.03An − 125, A0 = 2000 b The $125 repayment plan results in a lower outstanding balance after the fourth month. 11 $5570.72

20 d ≥ $710 21 3 years 22 The recurrence relation models a reducing balance loan because the payment d is typically greater than the interest generated, causing the balance An to decrease. For an interest- only loan, the payment is set to equal the interest each period (d = iAn), so the balance does not change. The formula would not apply in the same way, as the principal would remain constant. 23 a An + 1 = 1.012An − 300, A0 = 8000 b $7169.34

6.06 Formula for present value annuities Simple familiar

12 a An + 1 = 1.02An − 1500, A0 = 12 000 b $10 740 13 a $4740

b $4690

14 a An + 1 = 1.005An − 600, A0 = 15 000 b $13 417.11 15 Interest is added ﬁrst to reﬂect the balance owed before the payment reduces it, ensuring the lender earns interest on the full balance each period. 16 a The student reversed the order: interest must be added before subtracting the payment in an ordinary annuity. b $3820 17 a iii

Complex unfamiliar

b i

c ii

1 B 2 a $2723.25

b $8903.64

c $5197.66 3 a $19 389.64 b To fund annual payments of $3000 for 8 years at 5% interest, $19 389.64 must be invested today. 4 a $12 895 b To fund a scholarship that pays $2500 per year for 6 years at 4.5% interest, $12 895 must be invested today. 5 a 1%

b 3%

6 a 0.4167% b The present value of the loan is $8352.77.

Complex familiar

7 a 0.5%

b $2091.41

18 $100.45

8 a 0.4167%

b $17 119.39

19 The balance remains constant because the payment exactly offsets the interest added each period. For a $5000 loan with 1% interest, the payment is 0.01 × 5000 = $50. So, A1 = 1.01 × 5000 − 50 = 5050 − 50 = 5000, conﬁrming the balance stays at $5000.

9 a True

b False

c False

d False

10 $4807.69 11 a $9074.74 b $11 129.56 c Increasing the number of years increases the present value because more payments are included in the calculation.

Answers mathspace.co

763

12 a 0.5%

6.07 Spreadsheet & TVM solver

b $11 281.44 c Total payments are $12 000, which is greater than $11 281.44.

Simple familiar 1 A

13 $32 702.87

2 i = 0.005

14 a $4329.48

b $7721.73 c n = 10 has a higher present value because it includes more payments, increasing the total worth despite discounting. 15 $16 524.38

Period 1

$250

$32

$7782

4 a $283.07 b Ask your teacher for worked solutions. 5 a $225.79

16 $10 087.76

b Ask your teacher for worked solutions.

Complex familiar

6 a True

17 a $11 316.90

7 $10 800

b The present value is close to the total payments of $12 000 because the low interest rate of 0.1% results in minimal discounting of future payments. 18 a $25 852.85

Payment Interest Balance

b $26 296.74

c The monthly payment option has a higher present value. Although the total money paid out is almost identical, receiving the money sooner (monthly vs. annually) means each payment is discounted for less time, resulting in a slightly higher present value. 19 a Annuity A: $7863.21. Annuity B: $7721.73. b Choose Annuity A because its present value, $7863.21, is higher than Annuity B’s, $7721.73, indicating a greater current worth. 20 $386.66

b False

c False

d True

8 a i ≈ 0.004 167 b Interest = 12 000 ∗ (0.05/12) c b = 12 000 + (0.004 167 × 12 000) − d 9 $460.59 10 $1200 11

Period

Payment Interest

Balance

$20 000

$608.44

$100

$19 491.56

$608.44

$97.46

$18 980.58

$608.44

$94.90

$18 467.04

12 a $335.22 b $272.88 c The 5-year term payment is lower.

Complex unfamiliar 21 The formula fails because i = 0 results in division by zero. Instead, the present value is the sum of payments, APV = d × n, since no discounting occurs without interest. 22 The lump sum is the better option. The present value of the annual payments is calculated as:

Since this is less than the $500 000 lump sum, taking the immediate payment is ﬁnancially better.

764

Mathspace QLD 12 General Mathematics 2025 mathspace.co

13 $9741.96 14 a $359.61

b $3391.92

c $22 234.14 Complex familiar 15 The error is inputting the interest rate as a decimal 0.06 instead of a percentage 6. The TVM solver expects the annual rate as a percentage for the I% ﬁeld. 16 a 56

b $3538.55

Complex unfamiliar

3 $16 935.38

17 A TVM solver uses high precision for calculations, while a spreadsheet may round intermediate values, causing small differences in the ﬁnal balance. To minimise this, use unrounded values in spreadsheet formulas or adjust the ﬁnal payment to clear the exact remaining balance.

4 a $14 400

18 a $515.97

b $8000

c $18 000

Month

Balance

Interest Payment

$8000

$7882.00 $32.00 $150.00

$7763.53

5 A 6

$31.53

$150.00

b 78 c Increasing the payment reduces the loan term, which lowers the total interest paid because interest accrues over fewer periods. 19 The error is that the spreadsheet incorrectly applies the full annual interest rate of 4% to a single monthly period. The rate must be converted a monthly rate by dividing by 12. The correct interest for Period 1 is , resulting in a correct end-of-period balance of $15 000 + $50.00 − $355.22 = $14 694.78 20 A longer loan term means that while each monthly payment is smaller, interest is calculated on the outstanding balance for more months. This extended period allows more total interest to accrue over the life of the loan. For example, consider a $20 000 loan at 4% annual interest: 3-year term (N = 36): The payment is ≈ $590.49. Total interest paid is: ($590.49 × 36) − $20 000 = $1257.64 5-year term (N = 60): The payment drops to ≈ $368.33. Total interest paid is: ($368.33 × 60) − $20 000 = $2099.80 Even with a much lower monthly payment, the 5-year loan costs over $800 more in total interest because the balance is paid down more slowly.

6.08 Applications of present value annuities Simple familiar 1 $11 774 2 a $9659.49

b $4849.63

7 a $111 141.46

b $144 000

8 $16 800 Complex familiar 9 a $23 943.69

b $145.79

10 The required monthly payment is $443.21. 11 $563.39 12 a $43 200 13 i A

b $36 203.71 ii B

iii D

c $6996.29 iv C

14 a $2327.34 b $4501.07 c More periods mean more payments are being made, so a larger initial lump sum (present value) is required to fund them. 15 a $9400 b The formula becomes APV = d × n, as no discounting occurs without interest. 16 a $133 535.78 b Use n = 126 for 10.5 years, adjusting the ﬁnal payment if needed. 17 a Monthly payment: $920.83, Total interest: $5249.80 b Monthly payment: $674.27, Total interest: $6638.68 c Choose the 5-year term: it has a lower total interest cost, which is better for minimising the cost of borrowing. 18 a The student calculated the total repayments, d × n, instead of the present value. Their method ignored the effect of interest and did not discount the payments over time. b $17 032.13

Answers mathspace.co

765

19 a $405 468.48

b $2372.44

c $127.56

7 a $2695.35

b $970 325.21

c $490 325.21

20 The balance after 3 withdrawals is $198 494.74.

8 a 41 full payments are required.

Complex unfamiliar

Interest on this balance for one month: $310.19 × 0.007 = $2.17

b Balance after 41 payments: $310.19 21 a The balance decreases, as the payment exceeds the interest added, reducing the principal. b The balance increases, as the payment is less than the interest added, causing the debt to grow. c The balance will grow indeﬁnitely, as the payment ($100) is less than the initial interest ($150). 22 a 42 full payments are required. b The ﬁnal payment is $244.03. 23 a The monthly payment is $2045.90. b The outstanding balance is $432 709.32.

Final payment: $310.19 + $2.17 = $312.36 Exam practice 9 A0 = 6000 A1 = 1.005 × 6000 − 420 = 6030 − 420 = 5610 A2 = 1.005 × 5610 − 420 = 5638.05 − 420 = 5218.05 The closest value is Option C. 10 The monthly repayment PMT is $3543.64 11 Interest earned = $3380.79

7.01 Recurrence relation

c The new monthly payment is $2500.30. Simple familiar

Chapter 6 review 1 a $12 000

b 3.5%

b The periodic payment added at the end of each period.

c $12 854.70 2 a $18 375

c 0.005

b $18 692.73 c The compound interest investment earns $317.73 more than the simple interest investment. 3 The total value is $30 646.63. 4 Bank A: ieff =

− 1 ≈ 4.266%

Bank B: ieff =

− 1 ≈ 4.232%

5 An + 1 = 1.005An − 450, where A0 = 20 000 6 The present value of the annuity (Option 2) is ≈ $130 079.40.

This is less than the $150 000 lump sum offered in Option 1. Therefore, Option 1 (the lump sum) is ﬁnancially superior as its present value is higher.

766

d 1.005 2 $2428.80 3 a $3325.68

b $3489.54

c $3425.89

d $3640.16

4 a $5904.50

b $6363.55

5 $1503.76

Bank A offers the better return as it has a higher effective annual rate.

APV = 10 000

1 a The total amount at the beginning of the (n + 1)th period.

Mathspace QLD 12 General Mathematics 2025 mathspace.co

6 $7660.53 7 $5017.10 8 $9176.43 9 $451.50 10 $14 477.77 11 $3428.01 12 $14 003.57 13 a False

b True

c True

16 a $23 410.45

Complex familiar

b New future value is $28 092.54. The additional value is $28 092.54 − $23 410.45 = $4682.09.

14 a $753.13 b Future value: $753.76; additional interest: $0.63 15 a $2656.81

17 Plan A FV is $4988.58. Plan B FV is $5037.18. The difference is $48.60.

b $606.28

16 a An + 1 = 1.0035An + 150, with A0 = 500 b $1110.19

18 $16 339.85 19 Eli ends up with more money. Eli’s future value is $16 496.17, while Lena’s is $16 218.73

17 $2123.05 Complex unfamiliar

Complex unfamiliar

18 $2532.03

20 The new goal is $5000. The student has saved $5037.18, which is a surplus of $37.18.

19 $216.56

7.02 Future value annuity formula

21 $406 22 a $287.53

Simple familiar 1 a 0.003

b 0.00 767

2 a 0.02 075

b 0.02 875

3 a 60

b 12

c 104

b New monthly payment is $306.70. Increase is $19.17 23 $2464.97 d 10

e 117 4 $1293.04 5 $6429.80 6 $2543.20 7 $3275.08 8 $3302.85 9 $2191.78

7.03 Practical applications of annuities

10 $4134.03 11 a $2543.20

24 The term (1 + i)n calculates the future value of a single lump sum investment. In an annuity, payments are made over time. The formula must account for the fact that each payment has a different amount of time to earn interest. The −1 and division by i are part of the geometric series summation formula used to derive the annuity formula, which correctly sums the future value of each individual payment.

b $143.20

12 $75 760.86

Simple familiar 1 Future value (AFV ): $12 000

13 $136 576.44

Number of compounding periods (n): 4 × 12 = 48

14 $69 011.61

Monthly interest rate (i): 0.06 ÷ 12 = 0.005

Complex familiar 15 a $15 455.24 b The future value is $15 455.24, which is less than $16 000. Their plan is not suﬃcient.

2 a False

b True

c True

3 $206.43 4 $183.30 5 $1019.32 6 a $372.33

b $22 339.80

Answers mathspace.co

767

7 a $7609.50

b No, he has a shortfall of $390.50.

c $7 200

d $409.50

8 a $194.00

b $9312.00

c $688.00 9 a $11 203.40

b $10 000.00

c $1203.40 10 a $724.04

b $17 376.96

c $2623.04 11 a $14 846.20

b $12 000.00

c $2846.20 12 a $2665.50

b $26 655.00

Complex unfamiliar 21 The statement is false. The relationship between the number of periods and the required payment is not linear. When the investment period is doubled, each payment has more time to earn compound interest. As a result, doubling the number of periods reduces the required periodic payment by more than half. For example, to save $10 000 at 5% p.a. compounded annually, a 10-year term requires annual payments of $795.05. A 20-year term requires payments of $302.43, which is signiﬁcantly less than half of $795.05. 22 Total amount: $104 796 Total interest: $17 555.38

c $3345.00 13 $697.96

23 $60 068.23

14 $480.49

7.04 Perpetuity formula

15 $223.30

Simple familiar

Complex familiar

1 a The total amount (present value) required to sustain the perpetuity.

16 a $282.94 b The budget is suﬃcient, as $274.41 is less than $300. At $250 monthly, the future value is $13 720.38, falling short by $1746.3. 17 Sarah’s interest: $9359.42 Tom’s interest: $13 190.93 Tom earns more interest. Although both invested the same total amount ($36 000), Tom’s longer investment period and higher interest rate allowed his money to generate signiﬁcantly more compound interest.

c 0.004 d $200 000 2 a False

19 Plan A requires total payments of $22 169.60 which requires her to pay $9.40 less. Plan B requires total payments of $22 160.20. Elena should choose Plan B, as it requires her to pay $9.20 less in total to reach the same savings goal. b $947.42

c $160 000.00

d $1000.00

4 a $250 000.00

b $4000.00

5 $2000.00

7 $13 200.00 8 $12 500.00 9 $200 000.00 10 $100 000.00 11 $80 000.00 12 $571 428.57

14 $1 250 000.00 15 6%

Mathspace QLD 12 General Mathematics 2025 mathspace.co

c True b $1500.00

13 $250 000.00

768

b True

3 a $240 000.00

6 $15 000.00

18 9.03%

20 a $15 389.46

b The interest rate per compounding period.

16 0.83% 17 a 4.00%

7 4.4% per annum b 3.00%

c 2.00%

8 a 4.80% b 4.67%

Complex familiar 18 3% p.a. compounded monthly 19 a $200 000.00 b New total amount: $266 666.67; additional amount needed: $66 666.67 20 Fund A requires $240 000. Fund B requires $300 000. The difference is $60 000. 21 The required amount is $400 000. The fund is insuﬃcient, with a shortfall of $50 000. One strategy is to secure additional funding through donations. 22 a 4.80%

c Option A requires a higher interest rate (4.80% versus 4.67%). This is because Option A pays out more money annually ($2400 × 12 = $28 800) than Option B ($28 000). To generate a larger income from the same principal, a higher rate of return is required. 9 $19 450.67 10 $43 572.30 11 $157 317.07 12 a $500 000

b $3673.26

Exam practice

b 5.00% c Option B requires a higher interest rate (5.00%) to be ﬁnancially viable compared to Option A (4.80%). 23 a = 1.03, b = 480, c = 16 000

13 C 14 a 1.3%

b $75 000

15 First, calculate the value of the retiree’s current investment:

Complex unfamiliar 24 $9268.46

Next, calculate the principal required for the

25 $80 000.00

desired perpetuity: A =

Chapter 7 review 1 a An + 1 = 1.003An + 120, with A0 = 1000 b $1494.22 2 a $10 253.35

b $1253.35

3 $443.39 4 a $18 149.45 b No, he will not reach his goal. He will have a shortfall of $850.55.

= $150 000.

Advice: The retiree’s investment of $120 915.22 is not large enough to ﬁnance the perpetuity, as they need $150 000. There is a shortfall of $29 084.78.

8.01 Introduction to graphs and networks Simple familiar 1 a 5 vertices, 3 edges

b 6 vertices, 9 edges

5 Olivia’s FV: $28 465.60, Olivia’s interest: $4465.60

2 a deg(A) = 3

b deg(B) = 2

c deg(C) = 3

d deg(D) = 4

Noah’s FV: $33 318.24, Noah’s interest: $4518.24

e deg(E) = 2

Noah has the better investment strategy. Despite a lower interest rate, his larger monthly contributions result in a higher future value and more total interest earned. 6 a $3500

b $400 000

3 a Simple

b Not simple

c Simple

d Not simple

4 a Complete

b Not complete

5 a Bipartite

b Not bipartite

6 Yes

Answers mathspace.co

769

7 a 7

b 63

8 a An edge that starts and ends at the same vertex. b Two or more edges that connect the same pair of vertices.

8.02 Network representations Simple familiar 1

Aaron

Bianca

Rochelle

Mario

9 10 edges Complex familiar 10 a No, because it has a loop and multiple edges. b No, because it has an isolated vertex. c Vertex A has the highest degree, which is 5. 11 a Directed

b Undirected

c Directed 12 The total weight is 25. 3

Complex unfamiliar 13 a The sum of these degrees is 2 + 2 + 1 + 3 = 8 Twice the number of edges is 2 × (number of edges) = 2 × 4 = 8 The sum of the degrees (8) is equal to twice the number of edges (8). Therefore, the Handshaking Lemma holds true for this graph. b No. The sum of degrees would be 7 × 3 = 21, which is an odd number. The Handshaking Lemma requires the sum of degrees to be an even number, so this graph is impossible.

14 A

B Q D

6 a Directed, because the matrix is not symmetrical. b Yes, because there is a non-zero entry (‘1’) on the main diagonal. c Yes, because there is an entry (‘2’) that is greater than 1.

770

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex familiar

7 a A

C 3 f=5 4 e = 10

5 v=6 6 No, since v + f − e = 5 − 7 + 3 = 1 ≠ 2 8 a Simple, Complete

b Simple, Bipartite

9 The sum of degrees is twice the number of edges. Total degree is 10, so there are 7 edges. Because the graph is undirected, the matrix must be symmetric. Therefore, x = 1, y = 1, and z = 1.

Complex familiar 7 a

Complex unfamiliar

b Ali received the most messages (3 + 8 = 11). This is the sum of column ‘Ali’. c

10 a Cat sent the most messages (8 + 1 = 9). This is the sum of row ‘Cat’.

b v = 5, e = 7, f = 4 c 5 + 4 − 7 = 2 8 a e = 8 b f = 5 9 v = 4, f = 4, e = 6, so 4 + 4 − 6 = 2

11 a The graph has a loop at vertex A, so it is not a simple graph. b The matrix has an edge between C and D (edge C → D is 1), but the graph does not show this edge. c

Complex unfamiliar 10 a 3f = 2e since each face has 3 edges, and each edge is shared by two faces. b e = 3v − 6 from substituting v+f−e=2

= 3 into

11 v = 6, e = 9, f = 5. Since 4f ≤ 2e gives 4 × 5 ≤ 2 × 9, or 20 ≤ 18, which is false, the graph cannot be planar.

8.04 Walks, trails and paths 8.03 Planar graph and Euler’s formula Simple familiar Simple familiar 1 v = 4, e = 5, f = 3

1 a iii

b iv

c i

d ii

e v

Answers mathspace.co

771

2 a Cycle (a closed path, which is also a trail and a walk). b Trail (vertex C is repeated, but no edges are). It is also a walk. c Path (no repeated vertices or edges). It is also a trail and a walk. 3 The shortest path is A − C − B − D with a length of 8. 4 The shortest path is X − W − Y − Z with length 8. 5 a P − Q − R − P b S − P − R − Q − P 6 A trail is a walk where no edge is repeated. A path is a trail where no vertex is repeated (except for the start and end in a cycle). Complex familiar 7 a

Simple familiar 1 a All vertices must have an even degree. b Exactly 2 vertices must have an odd degree. 2 a Eulerian (all vertices have even degree) b Semi-Eulerian (exactly 2 vertices have odd degree) 3 A−B−C−D−E−A 4 No, a Hamiltonian path does not exist. To visit the leaf vertices A and E, any path must pass through vertex C. To visit leaf vertices B and F, it must pass through vertex D. It is impossible to construct a simple path that visits all six vertices without repeating C or D. 5 An Eulerian trail must traverse every edge exactly once. A Hamiltonian path must visit every vertex exactly once.

E 5

6 a The 2 vertices with odd degrees.

12 FW

8.05 Eulerian and Hamiltonian graphs

4 6

3 R

b Answers may vary. For example, B − A − C − E − D − B − C. Complex familiar 7 a The degrees of the vertices are: deg(A) = 3, deg(B) = 4, deg(C) = 3, and deg(D) = 4.

b Shortest time is 10 minutes via the Food court.

b No, because the graph is semi-Eulerian (it has 2 odd vertices), not Eulerian.

8 The cheapest route is A − D − E − C with a cost of 2 + 6 + 5 = 13 dollars.

c The guard must start at one of the odddegree vertices and ﬁnish at the other.

9 a A − B − C − A or D − E − F − D b A − B − C − D − E c The edge CD is the bridge.

8 Answers may vary. For example, A − B − C − D − E − A. 9 2 edges need to be duplicated, connecting B and H, and E and G.

Complex unfamiliar 10 The minimum time is 22 minutes. Example route: D − B − C − E − C − A − D 11 Add edge from C to D with weight 2. The new shortest path is A − C − D. (Note: The trivial solution of adding a direct edge from A to D is excluded).

772

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Complex unfamiliar 10 a 2 triangles joined at a single vertex.

which means 5 − 7 + 4 = 2. Therefore, the formula holds.

b A complete graph on 4 vertices (K4).

7 The shortest path is A-B-C-D with a total weight of 4 + 2 + 3 = 9. B

8 Vertex degrees are: deg(A) = 3, deg(B) = 3, deg(C) = 3, deg(D) = 3, deg(E) = 4. Since there are four vertices with an odd degree, the graph is neither Eulerian nor semi-Eulerian.

C 11 No, because the centre square in a 3 × 3 knight’s tour graph is isolated.

Chapter 8 review 1 a deg(P ) = 1, deg(Q) = 3, deg(R) = 3, deg(S) = 3 b Odd vertices: P, Q, R, S. There are no even or isolated vertices.

b No, it is not possible. There is no edge from the Gym (the last location in the path) back to the Library. Therefore, a Hamiltonian cycle does not exist from the Library. 10 a The degrees of x and y can be determined through these steps: 1. The Handshaking Lemma states the sum of degrees is 2e, so the total sum is 2 × 9 = 18.

3 Graph 1 is a complete graph. Graph 2 is a bipartite graph. 4

9 a One possible Hamiltonian path is Library-Park-Cafe-School-Gym.

2. Setting up an equation with the known degrees gives 12 + x + y = 18, which simpliﬁes to x + y = 6. 3. An Eulerian graph requires all vertices to have an even degree. Therefore, x and y must be a pair of even integers that sum to 6. 4. The only possible values are 2 and 4. This makes the complete set of degrees 2, 2, 2, 4, 4, 4. b

5 Using v + f − e = 2, we have 8 − e + 5 = 2. This gives 13 − e = 2, so e = 11. There are 11 edges. 6 a

B C

E A

D E b The graph has v = 5 vertices, e = 7 edges, and f = 4 faces (including the outer face)

11 a The vertices with odd degree are A, B, C, and D. An Eulerian circuit is not possible because a graph must have all vertices of even degree to have one. b The minimum additional length is 120 m by duplicating edges AB and CD.

Answers mathspace.co

773

c The minimum total distance is the original total plus the repeated paths is 620 + 120 = 740 m.

4 Example answer:

Exam practice N

12 C 13 a The graph has v = 6, e = 7 and f = 3. Substituting into Euler’s formula gives 6 − 7 + 3 = 2, which is true. b One of the crossing edges should be redrawn so it does not cross the other edge. c

L 5 a

A 8 D

14 The earliest the guard can ﬁnish is 12:36 p.m. by following the route A → B → C → D → E → F → G → H → A.

9.01 Trees and spanning trees

Simple familiar 1 a No, because it is disconnected and has a cycle.

b No, because it has a cycle. b

c Yes

d No, because it is disconnected and has a cycle.

C 6

e No, because it has a cycle. f Yes

7 F

g No, because it has a cycle.

h Yes 2 a No 3 a No

774

5 b Yes

c Yes

d No

b Yes

Mathspace QLD 12 General Mathematics 2025 mathspace.co

E B

C 3

C A

3 D

5 E

A 11

E 10

B 6 11 edges

7 17 vertices

8 a 2 − 10

b 10

9 a Answers may vary.

Possible answer: Edges a, c, e and g

b Any one of edges a, b or c c Any one of edges a, b, c or d E

10 a

B E

C b

6 D E

A d

3 B

7 G

8 4 F

2 E

6 11 a 4

F b 1

c 1

Answers mathspace.co

775

3 a 5

Complex familiar 12

Unknown Mabel

Unknown

9 Rebecca

April

Patrick

Todd

Connor

Leonard

Mark

Unknown

Section 1 Section 1

Chapter 1

Section 2 Section 3

Chapter 2

Section 1

c 21 million dollars 4 a 4 b

Section 2

Chapter 3

Richard

Daisy Juliet

Section 3

500

B 720

740

Section 1

640

Section 2 Chapter 4

Section 3

520

720

Section 4

Section 5

Chapter 5

Section 6

560

Section 1

c 2300 km Complex unfamiliar

5 a 5 b

14 35 metres

7 7

15 The original graph is already a tree.

Simple familiar A

b The edge from F to B 5

7 2

F D

2 A

F 2 a 13

776

1 a The edge from A to D c

9.02 Prim’s algorithm

3 E

6 b 3

Mathspace QLD 12 General Mathematics 2025 mathspace.co

c 25 days

6 a 4

E 111

D 5 9

126

100

117

160 B

134 A

c 19

7 m=1

2 D

F G

14 H

14 A

4 2

6 E

c Example answer:

5 D

2 11

5 C

G I

8 17

4 9

B 3

13 a

J E

G 3

Complex familiar 9

8 n=4

6 G

3 J

Answers mathspace.co

777

12 K

D E

17 Example answer:

9 11

G 14

Complex unfamiliar

G D

17 C

b 13 km

19 a 2.8 s

b 0.5 s faster

20 a 20 m

b 22 m

9.03 Applications of minimum spanning trees

31 B

18 a 307 km

Simple familiar 15 D B

2 16 metres

12 18

1 No. An edge with the lowest weight might connect two vertices that are already connected in the tree, which would form a cycle. It will only be included if it is the cheapest edge connecting a new vertex.

4 30 + 40 + 50 + 20 + 60 = 200 metres

A 16

5 The next edge added is FE (weight 5), then DE (weight 1). The full MST is AB, BC, AF, FE and DE. The total weight is 2 + 3 + 4 + 5 + 1 = 15.

20 D

B 11

C 19

17 J

778

6 18 metres

7 B

15 13

3 C

8 10 9 21 km 10 a $90 million

Mathspace QLD 12 General Mathematics 2025 mathspace.co

b No, the project cannot be completed within the budget. The minimum cost to connect all cities is $90 million, which is $10 million over the budget.

b $34 000

Complex familiar 11 a

B 4

c The council should choose the Cable network. Its minimum connection cost of $29 000 is less than the $34 000 cost for the Wireless relay system.

3 6 5

7 b

17 a The minimum cost is $32 000.

b The new minimum cost is $36 000. Complex unfamiliar

18 Not necessarily. If adding the new road creates a cycle, and the new road’s weight is not the smallest in that cycle, it will not be part of the MST. It will only be included if it is the cheapest way to connect two parts of the tree.

C A E

19 Not necessarily. A new, very cheap road from the new town to one of the old towns might create a cycle with the old MST roads. To break the cycle, one of the old, more expensive roads might be removed, changing the original MST structure.

c $140 000 12 a 25 hours b The new minimum spanning tree has a weight of 27 hours, so the total time increases by 2 hours. 13 a

20 Yes, the minimum spanning tree will be unique. When using an algorithm like Prim’s, there will never be a tie between which edge to add next. Since there is only one choice at every step, only one possible MST can be formed.

50 F3

P 40

25 F2

9.04 Project networks Simple familiar

b 75 m 14 a Total weight = 7 b Yes. Possible answers: AB, BC, CD and AB, AD, DC. C A

C A

1 a A, B

b A, B

c D, E, H d A, B

2 a A, B, C

b A, B, C

c E

d A, B, C

3 a C, D

b A, B

c G, H

d A, B

4 a There are no prerequisite activities for activity A.

b A c F D

15 a The student’s selection of edges forms a cycle (A − B − C − A). A spanning tree cannot contain any cycles.

d A 5 a C

b A, B

c G, H

d A, B

6 a D, E

b A, B, C

c I

d A, B, C

b The correct minimum spanning tree consists of edges AB(2), BC(3), CD(5) and DE(1). The total weight is 11. 16 a $29 000

Answers mathspace.co

779

780

Activity

Prerequisite

10 a

Activity

Duration

Prerequisite

−

A, B

C, D

F, G

Activity

Duration

Prerequisite

Activity

Duration

Prerequisite

−

C, D

B, D, E

F, G

Activity

Duration

Prerequisite

Task

Time (minutes)

Prerequisite

−

A, B

C, D

H, I

F, G

Mathspace QLD 12 General Mathematics 2025 mathspace.co

11 a

Activity

Duration

Prerequisite

Activity

Duration

Prerequisite

−

A, B

−

A, B

D, E

D, E, F

F, G

Activity

Duration

Prerequisite

−

D, E

F, G

Activity

Duration

Prerequisite

−

A, B

C, D

F, G

Complex familiar 12 a D A

Start

Finish

b A

Start

13 a

E Finish

Start

A, 1

B, 8

A, 2

C, 1

B, 9

C, 4

D, 3

E, 9

E, 4

Finish

Answers mathspace.co

781

Start

A, 1

A, 3

B, 4

Start

D, 7

E, 3

C, 3 C, 1

B, 7

F, 5

D, 4

G, 5 Finish E, 6

F, 2

d A, 9

Finish

B, 8

Start d

Start A, 7

D, 7

E, 2

C, 3

B, 4

C, 5

F, 4 G, 8 Finish

D, 5 15

C, 15

G, 25

A, 30

E, 4

F, 7 Start

D, 10

Start

A, 2

B, 7

F, 15

B, 15

Finish 14 a

C, 3

E, 5

Start A, 2

B, 8

C, 7

K, 3 F, 7 M, 4

Start

D, 8

A, 10

D, 4

E, 3

G, 1

H, 2

C, 5

I, 3

D, 7

E, 4

F, 5

E, 7

F, 2

D, 5

B, 2

Finish

G, 8

Start

N, 2 J, 4 P, 8

F, 7

Finish

L, 10 H, 3

Mathspace QLD 12 General Mathematics 2025 mathspace.co

Finish

Complex unfamiliar 17 a

G, 7

N, 2 J, 2

B, 3

782

Finish

S, 5 Q, 7

Finish

b Tasks J and L can only start after D, F, and H are complete. 18

• Through activities A and C: The total duration is 2 + 9 = 11.

C, 2 B, 4

Start

• To calculate the EST for the Finish vertex, consider the two paths leading to it:

R, 1

H, 1

O, 4

T, 3

P, 1

Finish

D, 2

9.05 Forward and backward scanning

The EST for the Finish vertex is the maximum of these path durations: max(11, 10) = 11. 6 min(20 − 8, 16 − 3) = 12

Simple familiar

7 a The Latest Start Times (LSTs) are calculated by backward scanning:

1 13 2 13 3 a 18

• Through activities B and D: The total duration is 4 + 6 = 10.

b 6

• The LST for the Finish node is equal to the ﬁnal EST, which is 12.

4 max(4 + 6, 7 + 5) = 12

• The LST at the start of activity C is 12 − 7 = 5.

5 a The earliest start times (ESTs) are calculated by forward scanning:

• The LST at the start of activity D is 12 − 8 = 4.

• The EST at the Start node is 0.

• The LST for the Start node is the minimum of the two backward paths: through activity A and activity B. This is min(5 − 5, 4 − 4) = 0.

• The EST at the start of activity C is determined by the duration of activity A: 0 + 7 = 7. • The EST at the start of activity D is determined by the duration of activity B: 0 + 3 = 3. • To calculate the EST at the start of activity E, consider the two paths leading to it: • Through activities A and C: The total duration is 7 + 4 = 11. • Through activities B and D: The total duration is 3 + 6 = 9. The EST at the start of activity E is the maximum of these path durations: max(11, 9) = 11. • The EST for the Finish node is then calculated from the start of activity E: 11 + 5 = 16. b The earliest start times (ESTs) are calculated by forward scanning: • The EST at the Start vertex is 0.

b The Latest Start Times (LSTs) are calculated by backward scanning: • The LST for the Finish node is equal to the ﬁnal EST, which is 14. • The LST at the start of activity D is 14 − 5 = 9. • The LST at the start of activity E is 14 − 7 = 7. • The LST at the start of activities B and C is the minimum of the two backward paths: through activity B and activity C. This is min(9 − 6, 7 − 4) = 3. • The LST for the Start node is found by working back from the start of activity A: 3 − 3 = 0. 8 B 9 25

• The EST at the start of activity C is determined by the duration of activity A: 0 + 2 = 2. • The EST at the start of activity D is determined by the duration of activity B: 0 + 4 = 4.

Answers mathspace.co

783

9.06 Critical path

Complex familiar 10

2 5 C, 4

A, 2 0 0

Simple familiar 9 9 F, 1

Start

D, 6

1 a 28 units of time b The critical path activities are A, C and E. The sum of their durations is 5 + 12 + 11 = 28 units of time.

10 10

B, 3

G, 4

Finish

14 14

E, 5 3 3

2 32 days 11

3 7 A, 3

0 0

3 33 units of time C, 5

Start

8 12

4 7 + 10 + 5 = 22

F, 2

D, 6 B, 1

Finish 1

E, 4

14 14

5 14 units of time 6 a 18

G, 9

b Critical path: A − B

5 5

4 4

Sum: 8 + 10 = 18

7 7 C, 3

A, 4 0 0

Finish

Start

13 13

F, 1

B, 2 2 4

D, 8

Complex familiar 8 12 days

10 12

9 a

6 7

6 9 1

0 0

7 18 hours

E, 6

Start

D, 2

1 B, 5

10 11

0 0

A, 1 E, 6 4 4

F, 4

H, 5

Finish

15 15

5 5 D, 4

13 13

b The critical path is B − D − E.

8 8

E, 6 B, 5

Finish

9 9

Start

G, 2 C, 3

C, 2

A, 6

c 15

6 6

10 21 days R, 3

P, 6 0 0

Start

9 9 T, 7

Q, 4

Finish

16 16

S, 5 4 4

Complex unfamiliar 15 The LST for vertex 2 is incorrect. It is shown as 4, but it should be min(8 − 6, 9 − 5) = min(2, 4) = 2.

11 The minimum completion time is 16 units. Path A-B-D has a length of 3 + 6 + 7 = 16, and path A-C-E has a length of 3 + 4 + 9 = 16. 12 18 units of time 13 The minimum completion time is 11 days

5 5

0 0

784

Mathspace QLD 12 General Mathematics 2025 mathspace.co

9 9

Start

16 max(5 + x, 11) 17 y = 9

C, 4

A, 5

E, 2 B, 3

14 x = 10

3 3 D, 6

Finish

c 2 units of time

Complex unfamiliar 15 13

16 If x > 7, MCT = 2x + 5. If x < 7, MCT = x + 12. If x = 7, MCT = 19 17 This is incorrect. The critical path is the longest path in terms of duration. It’s called ‘critical’ because it has no slack, and its length determines the project’s minimum completion time. The project cannot ﬁnish any faster than its longest sequence of dependent tasks. 18 No, it is not possible. The minimum completion time is the sum of the durations of all activities on the critical path. Since all durations must be positive, a single activity cannot have a duration longer than the total completion time. 19 Answers may vary. 8 8

B, 10

EST

LST

Float time

Critical Path: P − R − T 6 a Forward Scan (to determine Earliest Start Times - EST): • EST(Start) = 0

• EST(End of A) = 0 + 3 = 3 • EST(End of B) = 0 + 5 = 5

• EST(End of D) = EST(End of B) + 2 = 5 + 2 = 7

18 18

A, 8 0 0

Duration

• EST(End of C) = EST(End of A) + 4 = 3 + 4 = 7

Possible answer:

Activity

C, 4

Start

Finish 22 22 D, 7

E, 9 7

• EST(Finish) = max(EST(End of C) + 6, EST(End of D) + 5)

= max(7 + 6, 7 + 5) = max(13, 12) = 13 The minimum completion time is 13 weeks.

b Backward Scan (to determine Latest Start Times - LST):

20 Answers may vary.

• LST(Finish) = 13

Possible answer:

• LST(End of C) = LST(Finish) − 6 = 13 − 6 = 7

10 10

0 0

• LST(End of D) = LST(Finish) − 5 = 13 − 5 = 8

D, 5

A, 10 Start

Finish B, 6

C, 4

15 15

• LST(End of A) = LST(End of C) − 4 = 7 − 4 = 3 • LST(End of B) = LST(End of D) − 2 = 8 − 2 = 6

6 11

• LST(Start) = m in(LST(End of A) − 3, LST(End of B) − 5)

21 q = 8 and p ≤ 6

= min(3 − 3, 6 − 5) = min(0, 1) = 0 c

9.07 Float times Simple familiar 1 a A, C, E

b A, C, E, H

2 B 3 19 4 a 0 units of time

b 0 units of time

Activity

Duration

EST

LST

Float time

Answers mathspace.co

785

d The critical path is A − C − E.

10 − 5 = 5. Since the EST at the start of Activity C is 4, the ﬂoat time is 5 − 4 = 1 day. The mistake was probably in the backward scan calculation, resulting in an incorrect LST for the start of Activity C.

7 a 16 hours b The ﬂoat times for the activities are:

• Float(G) = 0

• Float(H) = 0

• Float(I) = 1

• Float(J ) = 0

• Float(K) = 1 c H − J

8 It means the start of painting can be delayed by up to 4 days without delaying the overall completion of the shed. 9 a Yes

b No

c Yes

Complex familiar 10 a

15 By definition, the critical path is a continuous path from Start to Finish that connects nodes where the earliest start time (EST) equals the latest start time (LST). For an activity to be on the critical path, its starting node must be one of these critical nodes. The float time for an activity is calculated as LST − EST at its starting node. Since an activity on the critical path must start at a node where LST = EST, its float time is LST − EST = 0. This signifies that there is no slack, and any delay to this activity will delay the entire project.

5 9

Complex unfamiliar A, 5

C, 4

16 The duration x must be 10 days.

13 13 0 0

Start

E, 8

Finish

21 21

17 Yes, the project will be delayed. The electrical path only has 4 days of ﬂoat. A 5-day delay will use up this ﬂoat and delay the project by 5 − 4 = 1 day. The plumbing path can absorb the delay.

D, 6

B, 7 7 7

b Float(C) = 4 days, Float(D) = 0 days 11 a

B, 4 8 8

3 3 0 0

Start

A, 3

D, 2

Finish

10 10

C, 5

b Activity B (write script) has a float time of 1 hour. 12 Yes, because the 3-hour delay is less than the 5-hour float time. 13 The float time for each activity is as follows: • Float time for Activity L: 0 units of time • Float time for Activity M: 0 units of time • Float time for Activity N: 0 units of time • Float time for Activity O: 1 unit of time • Float time for Activity P: 0 units of time Non-critical activities are those with a float time greater than zero. Therefore, the non-critical activities are M, O, and P. 14 The student likely made an error in the backward scan. The LST of the finish is 10. This means the LST at the start of Activity C is

786

Mathspace QLD 12 General Mathematics 2025 mathspace.co

18 The current critical path is A-C-F, with a duration of 3 + 7 + 5 = 15 days. The path involving Activity E is A-B-E, with a duration of 3 + 4 + 7 = 14 days. The current float for Activity E is 15 − 14 = 1 day. To increase this float to at least 3 days, the duration of the A-B-E path must be shortened by at least 2 days. We cannot shorten Activity A, as that would also shorten the critical path, changing the project’s completion time. Therefore, we must shorten Activity B. Shortening the duration of Activity B by 2 days will increase the float for the A-B-E path to 3 days without affecting the critical path’s duration. 19 Float(Y) = d1 − d2 20 The delay in marketing is the bigger problem. Task C has a float of 5 weeks, so it can handle a 2-week delay. Task B and D are on the critical path and have no float. Any delay to them will delay the entire festival.

9.08 Applications of critical path analysis Simple familiar

All ﬂoat is used which means cannot reduce further. Greatest reduction is 3 days.

1 C

16 Yes. Extend G by 3 days to save $450. This covers the $400 overrun.

2 False

17 a

15 17

3 16 days

B, 5

4 The concept that reducing a project’s completion time (crashing) typically increases its cost, requiring a balance between speed and budget. 5 $250 6 The manager must shorten both critical paths simultaneously, either by crashing a common activity or by crashing one activity on each path. 7 Activity Y 8 Activity A 9 Crash activity D for one day 10 Crash activity A and activity D, Total cost: $2000 + $1500 = $3500 11 a Critical path: B-D, Minimum completion time: 19 hours b $100

10 10

C, 14

Finish

24 24

Minimum completion time: 24 days, Critical path: A-C b $560 18 a Min time: 14 weeks, Critical path: A-C b The total additional cost to reduce the project time by 3 weeks is $4000 + $3000 = $7000. c Yes, the offer should be accepted. The path involving task D has a ﬂoat of 5 weeks. Extending the task by 3 weeks is well within this ﬂoat time and will not delay the project’s overall completion. Therefore, accepting a $1000 discount is a direct cost saving with no negative impact on the schedule.

19 a 2 days

13 Crash activity A and activity E, Total cost: $600 + $400 = $1000

b Crashing D by 3 days reduces path B-D-F to 15 days. However, path A-C-E takes 16 days, so it becomes the new critical path. The project time only drops from 18 to 16 days.

Complex familiar

Start

Complex unfamiliar

12 $880

14 a

0 0

D, 7

A, 10

B, 3 A, 10

D, 1

Finish

C, 5

b $800, Crash C by 2 days 15 The ﬂoat of path B-D is 3 days. The project can be reduced up to 3 days. • Day 1: Crash C ($250), Remaining budget $750, Time = 16 • Day 2: Crash C ($250), Remaining budget $500, Time = 15 • Day 3: Crash C ($250), Remaining budget $250, Time = 14

20 This is not the best strategy because the activity that can be shortened the most may also have a very high crash cost per day. It is more cost-effective to crash the activity with the lowest cost per day, even if it can only be shortened by a small amount. 21 To determine the total cost, the project time must be reduced hour by hour, re-evaluating the critical paths at each step. 1. Reduction 1 (from 22 hours to 21 hours): • Initially, Path 1 (P1) is the only critical path with a duration of 22 hours. • To reduce the project time by one hour, crash the cheapest activity on P1, which is Activity A.

Answers mathspace.co

787

• Cost for this hour: $100

B, 6

• New path times: P1 = 21, P2 = 21, P3 = 19. • Now, both P1 and P2 are critical. 2. Reduction 2 (from 21 hours to 20 hours): • To reduce the project time further, both critical paths (P1 and P2) must be shortened. • Crash the cheapest activity on P1 (Activity A) and the cheapest on P2 (Activity C). • Cost for this hour: $100 + $80 = $180 • New path times: P1 = 20, P2 = 20, P3 = 19.

Start

D, 5

A, 4

Finish

C, 3

6 To calculate the Earliest Start Time (EST) for the vertex where activities P and Q ﬁnish, the time taken for each incoming path must be determined: • The time to arrive via Path P is: 5 + 7 = 12

• P1 and P2 remain the critical paths.

• The time to arrive via Path Q is: 8 + 3 = 11

3. Reduction 3 (from 20 hours to 19 hours):

According to the forward scanning rule, the EST for a vertex is the maximum of the values from all incoming paths.

• Crash the same combination of activities (A and C) again. • Cost for this hour: $180 • New path times: P1 = 19, P2 = 19, P3 = 19. • All three paths are now critical. The total cost is the sum of the costs for each reduction: Sum the cost Total cost = $100 + $180 + $180 from each reduction step

= $460 Evaluate the total cost

8 This statement is incorrect. The critical path is the path with the longest total duration, not necessarily the path with the most activities. Consider a project network with two distinct paths: Activities A (10 days) and B (10 days).

1 a Yes, it is a tree because it is connected and has no cycles. b No, it is not a tree because it contains a cycle (F − G − H − F ). 2 A spanning tree for a graph with V vertices has V − 1 edges. Therefore, it will have 15 − 1 = 14 edges. 3 a 6 Indie

7 The LST for vertex 3 is calculated by subtracting the duration of activity G from the LST of vertex 4. So, LST = 22 − 6 = 16 weeks.

• Path 1 has two activities with long durations:

Chapter 9 review

Dance

Therefore, the EST is max(12, 11) = 12.

Rock

Main

7 Acoustic

b The total weight of the MST is 8 + 5 + 6 + 7 = 26. The cost is 26 × $100 = $2600. 4 a Activities C and D b Activities A and B

Total duration = 10 + 10 = 20 days. • Path 2 has three activities with short durations: 8 Activities C (5 days), D (5 days), and E (5 days). Total duration = 5 + 5 + 5 = 15 days. In this example, Path 1 is the critical path because its total duration of 20 days is longer than Path 2’s duration of 15 days, even though it has fewer activities. 9 The duration of path A-B is 7 + 9 = 16 hours. The duration of path C-D is x + 10 hours. For C-D to be the sole critical path, its duration must be greater than the duration of A-B. So, x + 10 > 16, which means x > 6. 10 The ﬂoat time for any critical activity is 0. 11 The ﬂoat time for activity C is calculated as LST(start) − EST(start) = 5 − 2 = 3.

788

Mathspace QLD 12 General Mathematics 2025 mathspace.co

12 a

2 2

C, 5

7 7

A, 2

28 28

3 3

0 0

Finish

Start B, 3

13 13

0 0

Start

B, 5

A, 3

F, 30 8 8

D, 15

End

D, 4

The only non-critical activity is D and its ﬂoat time is 12 minutes.

7 10

b Critical path: A − C − E, Minimum completion time: 13 hours

10.01 Introduction to flow networks

c Float(D) = LST(start) − EST(start) = 6 − 3 = 3 hours

Simple familiar

13 The new minimum completion time is 25 − 3 = 22 days. 14 A 5-day reduction is not possible with the given information, as the parallel path B-E has a ﬂoat of only 28 − 25 = 3 days. The maximum possible reduction is 3 days. To achieve this at minimum cost, crash the cheapest activity on the critical path, which is D. The total cost for a 3-day reduction is 3 × $250 = $750. 15 Option 2 is more cost-effective. To reduce the project time, both critical paths must be shortened. Option 1 achieves this by crashing the common activity E for a cost of $500. Option 2 achieves this by crashing one activity on each of the unique parts of the paths (A and B) for a total cost of $200 + $250 = $450. Since $450 < $500, Option 2 is cheaper. Exam practice

1 a True

b True

c False

b True

c False

3 B 4 a False

5 a Source: S b Sink: T c Intermediate: A, B d 8 6 a 12

b 8

B 8

15 T

B, 20

S 6

8 8

2 2

G, 2

D, 8

F, 4

B 9

E, 40

A, 2

d 15

C, 6

c 7

S 10

8 a

17 a

d False

2 C

16 B

Start

65 65

35 35

F, 3 3 4

0 0

E, 7

C, 20

E, 6

5 T

Finish

50 50

48 48

10 44

b The critical path is A-C-E-G. The minimum completion time is the length of this path: 2 + 6 + 40 + 2 = 50 minutes

b 15 + 6 = 21 Complex familiar 9 a 25

Depot

20 Factory

10 30

Answers mathspace.co

789

b Source: Depot, Sink: Factory 10 a

C1 D

9 The edges crossing from the source set to the sink set are S → A and B → T . The capacity is 30 + 15 = 45 tons per day. 10 a A → C, B → C, A → T

b 30 + 25 + 20 = 75

C2 b Source: D, Sink: H, Intermediate nodes: C1, C2. 11 9 + 15 = 24

11 Edges crossing are A → C and B → C. The capacity is 40 + 30 = 70 vehicles per hour. The student’s answer should include a sketch of the network.

Complex unfamiliar

12 Source is A, as it is the starting point (depot) and has only outgoing ﬂow. Sink is F, as it is the destination (highway access) and has only incoming ﬂow.

12 40 − 15 = 25

10.02 Capacity of a cut

Complex unfamiliar

Simple familiar 1 a This is a valid cut because the source S is in one set and the sink T is in another set. b S → A c 10 2 a S → A and B → T b 5 + 9 = 14 3 a S → A

b 10

4 a B → T

b 15

5 a This is a valid cut because S and T are in different sets. b Edge: S → A. Capacity: 9 6 a S → B, A → B, A → T b 15 + 5 + 8 = 28 7 The edges crossing from the source set to the sink set are S → B, C → D, and C → T . The capacity is 12 + 5 + 11 = 28. Complex familiar 8 The edges crossing from the source set to the sink set are A → C, B → C, and A → T. The edge C → T does not cross from the source set. The capacity is 10 + 12 + 15 = 37 Mbps.

b Edges: A → C, B → C, B → T . Capacity: 20 + 25 + 30 = 75 Gbps. 14 Capacity: The edges are B → T and C → T. The sum is 50 + 70 = 120 MW. Relevance: This cut represents the total maximum power that can be delivered to the city from all substations combined. It is a critical value for planners to ensure the city’s power demands can be met and to identify the final bottleneck before the end user. 15 The forward-flowing edges are S → A, B → A, and B → C. The capacity is 15 + 5 + 8 = 28. 16 The forward-flowing edges are S → B and A → T . The capacity is their sum: 15 + 18 = 33. The edge B → A, with capacity 10, is ignored because it flows backward across the cut from the sink set to the source set.

10.03 Maximum flow and minimum cut Simple familiar 1 a B

b E

2 C1, C3, and C4 b 110

4 a C1, C3, and C4

Mathspace QLD 12 General Mathematics 2025 mathspace.co

13 a Edges: S → B, A → B, A → C. Capacity: 40 + 15 + 20 = 75 Gbps.

3 a 130

790

c 60 b C1: 64, C3: 82, C4: 91

5 C (Drivers D1, D2, and D4 are all connected to Z2)

c 64 5 a 100

b 50

c 50

6 a Bipartile graph: Students to projects

6 The maximum ﬂow is 325 litres/hour.

Complex familiar

7 The maximum ﬂow is 35 litres/second.

8 The maximum ﬂow is 62 litres/hour.

9 The minimum cut has a capacity of 49. The maximum ﬂow is 4900 cars per hour. 10 The capacities of the cuts are: C1 = 90, C2 = 190, C3 = 140 and C4 = 120. Cut C1 has the smallest capacity.

S3 b S1 → P1, S2 → P1, S2 → P2, S3 → P2 7 A 8

11 a P = 170, Q = 130, R = 110, S = 130 b The maximum flow is 110 litres/min, equal to the minimum cut R. 12 a The maximum flow is 382 vehicles. b The removal of the road from Wollongong to Port Kembla does not affect the original minimum cut of the network. Since the minimum cut remains the bottleneck, the maximum flow is unchanged at 382 vehicles.

Task A

Task B











9 a

Complex unfamiliar

13 a 4.3 mA b All cuts through this network have the same capacity.

b Student S1 cannot work on Project P2. 10 D

14 23

11 B

10.04 Assignment and allocation

12 C 13 a 2 projects

Simple familiar 1 a iv

b iii

c ii

d i

2 a False

b True

c False

d True

Complex familiar 14 a

3 C 4

b Students S1 and S3

Bipartile graph: Tutors to study session Ali Math Beth

b The second student (James) can work on the third project (P3). 15 a 2 students

Science

b Students S1 and S4

Carlos

Answers mathspace.co

791

16 a Bipartile graph: Students to projects

work on. This is the ﬁrst step in using algorithms to ﬁnd the optimal assignment of students to projects.

P1 S1

20 a Bipartile graph: Students to projects P2

17 a Bipartile graph: Students to projects S1

b Yes, it is possible. For example, assign S1 to P3, S2 to P2, and S3 to P1. All projects are covered, and each student has one project. 21 a

b Bipartile graph: Students to projects

b 18 a

c Answers may vary. b

Possible answer: Allow student S1 to work on Project 1. This changes the entry M11 from 0 to 1. The new matrix is:

Complex unfamiliar 19 a

10.05 Optimisation by inspection Simple familiar b The matrix shows all possible assignments compactly. It allows for quick identiﬁcation of which students can be assigned to each project and which projects a student can

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1 a True 2 a 3! = 6 3 C

b False

c False

d True

b 5 + 3 + 2 = 10

4 a 1. A → 1, B → 2, C → 3 2. A → 1, B → 3, C → 2 3. A → 2, B → 1, C → 3 4. A → 2, B → 3, C → 1 5. A → 3, B → 1, C → 2 6. A → 3, B → 2, C → 1 b 8 + 4 + 3 = 15 c Assignment A → Task 3, B → Task 2, C → Task 1 has a total cost of 5 + 4 + 5 = 14. 5 a 20 + 22 + 17 = 59 b Assignment X → Bread, Y → Cake, Z → Pastry has a total time of 15 + 18 + 17 = 50. 6 a 10 + 11 + 9 = 30 b Assignment M1 → P3, M2 → P1, M3 → P2 has a total cost of 8 + 9 + 7 = 24. 7 a 8 + 8 + 7 = 23 b Assignment P → E3, Q → E1, R → E2 has a total eﬃciency of 9 + 7 + 9 = 25. 8 a 4! = 24 b 4 + 3 + 5 + 5 = 17 c Assignment S1 → G3, S2 → G2, S3 → G4, S4 → G1 has a total time of 3 + 3 + 3 + 3 = 12. 9 a 7 + 6 + 8 + 8 = 29 b Assignment T1 → C4, T2 → C1, T3 → C3, T4 → C2 has a total effectiveness of 9 + 8 + 8 + 9 = 34. Complex familiar

b Yes, for this specific problem the student’s claim happens to be correct. By inspection, the assignment W1 → T3, W2 → T1, W3 → T2 gives the minimum cost of 24. However, this method is not a reliable strategy in general. Complex unfamiliar 13 a Assignment 1: A → 1, B → 2, C → 3, D → 4, cost = 5 + 4 + 4 + 4 = 17 Assignment 2: A → 1, B → 2, C → 4, D → 3, cost = 5 + 4 + 5 + 3 = 17 b Having two optimal assignments gives the manager flexibility to choose based on other factors (e.g., worker preference, availability, or task priority) without increasing costs, enhancing operational efficiency.

10.06 Introduction to the Hungarian algorithm Simple familiar 1 a False

b True

c True

d False

2 a Rows represent resources (workers A, B, C), and columns represent tasks (1, 2, 3). b Each matrix entry represents the cost in dollars of assigning a speciﬁc worker (resource) to a speciﬁc task. c The goal is to assign each worker to exactly one task so that the total cost of all assignments is minimised. 3 a

10 a Minimum: A → X, B → Y, C → Z, cost = 15 + 10 + 13 = 38 Maximum: A → Z, B → X, C → Y, cost = 18 + 20 + 16 = 54 b Knowing the minimum cost helps optimise expenses, while knowing the maximum cost helps understand the worst-case scenario for budgeting or assessing task eﬃciency. 11 a 1. D1 → R1, D2 → R2, D3 → R3 2. D1 → R1, D2 → R3, D3 → R2 3. D1 → R2, D2 → R1, D3 → R3 4. D1 → R2, D2 → R3, D3 → R1 b Assignment D1 → R2, D2 → R1, D3 → R3 has a total time of 25 + 28 + 26 = 79.

4 a

c The optimal assignment is R1 → T1, R2 → T3, R3 → T2 . The minimum time is 4 + 5 + 4 = 13 hours. 5 a The optimal assignment is F1 → R2, F2 → R3, F3 → R1 . The minimum cost is 100 + 110 + 140 = $350.

12 a 8 + 9 + 7 = 24

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b Total cost = $120 + $90 + $150 = $360, which is higher than the optimal cost of $350. 6 a

b The optimal assignment is V1 → A1, V2 → A2, V3 → A3. Minimum time = 15 + 10 + 10 = 35 minutes 7 The optimal assignment is S1 → D1, S2 → D2, S3 → D3. Minimum time = 3 + 2 + 4 = 9 hours Complex familiar 8 a The optimal assignment is D1 → R2, D2 → R1, D3 → R3, D4 → R4 . b Minimum time = 2 + 6 + 1 + 4 = 13 minutes. 9 a One optimal assignment is P1 → M1, P2 → M5, P3 → M3, P4 → M4, P5 → M2. b Minimum time = 13 + 14 + 20 + 26 + 26 = 99 hours 10 a The optimal assignment is B1 → D2, B2 → D3, B3 → D1 . The total cost is $180 + $200 + $210 = $590. b The optimal assignment identiﬁes the combination of boats to docks that results in the lowest possible total cost for the marina, which is $590. Complex unfamiliar 11 The algorithm converts the problem into one of determining an assignment using only zero-cost elements. Row and column reductions create these zeros without changing the optimal solution. The line-covering step checks if an independent set of n zeros can be found. If so, this corresponds to the optimal (minimum cost) assignment in the original matrix. 12 a

b The optimal assignment is Pizza → B, Tacos → A, Burgers → C. c Minimum cost = $40 + $45 + $55 = $140

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10.07 Maximisation with the Hungarian algorithm Simple familiar 1 a iv

b iii

c ii

2 a False

b True

c False

d i

3 A possible assignment that contributes to the maximum total score. 4 8 Complex familiar 5 a Rows: workers W1 (Ms. Carter), W2 (Mr. Nguyen), W3 (Ms. Kim) Columns: tasks T1 (Task X), T2 (Task Y), T3 (Task Z) Entries: productivity scores, for example, M12 = 7 is Ms. Carter’s score on Task Y b 10 6 a

7 a

b The matrix remains:

8 a

b E1 → P2, E2 → P1, E3 → P3 c The maximum profit from the matrix is 12 + 15 + 12 = 39. Since the profit is in hundreds of dollars, the total profit is 39 × $100 = $3900. 9 a Three lines are needed (e.g., Column 1, Column 3, and Row 2). b Yes, three lines equal the matrix size 3, indicating an optimal assignment can be made by selecting one 0 per row and column. 10 a

b Three lines are needed (e.g., Row 1, Row 2, Row 3). 11 a The two optimal assignments are: W1 → T3, W2 → T2, W3 → T1 OR W1 → T1, W2 → T2, W 3 → T3 . b 7 + 9 + 7 = 23

2 a

c The optimal assignment is R1 → P1, R2 → P3, R3 → P2 . The minimum time is 4 + 5 + 4 = 13 hours.

Complex unfamiliar 12 a Largest value is 9. Cost matrix:

b Row reduced:

d An optimal assignment can be made using the zeros.

3 a The optimal assignment is F1 → R2, F2 → R3, F3 → R1 . The minimum cost is $100 + $110 + $140 = $350. b For example, F1 → R1, F2 → R2, F3 → R3 costs $120 + $90 + $150 = $360, which is higher. Complex familiar

Column reduced: 4 a Row reduced:

. After column

13 a Largest value is 9. Cost matrix:

reduction, the matrix is:

b Row reduced matrix:

b The optimal assignment is W1 → T1, W2 → T2, W3 → T3, W4 → T4 . The minimum cost is 20 + 15 + 15 + 25 = 75 minutes. 5 The optimal assignment is M1 → J1, M2 → J2, M3 → J3, M4 → J4 . The minimum cost is 50 + 40 + 40 + 50 = $180.

Row reduction creates zeros, which represent the best possible assignments relative to the other options in the same row, simplifying the problem.

10.08 Apply the Hungarian algorithm: Minimisation

6 a Row reduced:

Column reduced:

Simple familiar 1 a To determine the optimal assignment of tasks to resources to minimize total cost. b To ensure each resource is assigned exactly one task and vice versa.

b The minimum number of lines is 3. The smallest uncovered element is 6. The new matrix is:

c Subtract the smallest entry in each row from all entries in that row.

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795

c The optimal assignment is P1 → T3, P2 → T2, P3 → T1, P4 → T4 . The minimum cost is 69 + 37 + 11 + 23 = 140. 7 Add a dummy row of zeros. The optimal assignment is W1 → T1, W2 → T2, W3 → T3, Dummy → T4 . The minimum cost is 15 + 10 + 10 + 0 = 35. 8 The optimal assignment is W1 → T1, W2 → T2, W3 → T3, W4 → T4 . The minimum cost is 10 + 10 + 12 + 12 = 44.

Complex familiar 5 a Rows: machines M1, M2, M3 . Columns: jobs J1, J2, J3. Entries: eﬃciency scores b 8 6 a The largest value is 13. Subtract each entry from 13. Cost matrix:

Complex unfamiliar 9 The optimal assignment is V1 → T1, V2 → T2, V3 → T3, V4 → T4, V5 → T5 . The minimum cost is 10 + 5 + 10 + 5 + 5 = 35 minutes. 10 The algorithm creates zeros representing potential assignments with minimal relative cost after row and column reductions. Covering zeros with minimal lines ensures all assignments are considered, and adjustments maintain cost equivalence, guaranteeing the sum of selected entries is the smallest possible. 11 It signiﬁes that there are multiple optimal solutions. More than one combination of assignments will result in the same minimum total cost, giving ﬂexibility in decision-making. 12 This step is necessary to maintain the relative costs. The value was subtracted twice from these intersection elements (once for the row, once for the column), so adding it back once ensures the cost relationships in the matrix remain valid for the next iteration of the algorithm.

7 a

8 a Three lines are needed. b The number of lines is 3, which equals n. No adjustment is needed. 9 a

10 a Three lines are needed. b There are two optimal assignments: 1. M1 → J3, M2 → J2, M3 → J1 . 2. M1 → J1, M2 → J2, M3 → J3 . 11 a Subtract each entry from 10:

10.09 Solve complex assignment problems Simple familiar 1 a iv

b i

c iii

2 a True

b False

c True

d ii

3 A possible assignment that contributes to the minimum total cost. 4 9

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12 a Two lines are needed (e.g., Column 2 and Row 2). b Smallest uncovered is 1. Adjusted matrix:

13 a The optimal assignment is M1 → J4, M2 → J1, M3 → J2, M4 → J3 .

17 a Row reduced:

b Total cost = 2 + 1 + 2 + 1 = 6. Yes, this is guaranteed to be the minimum cost because the Hungarian algorithm is an optimal algorithm. 14 a Add a dummy row (D4 ) with zero costs to make the matrix square:

b The optimal assignment is V1 → S2, V2 → S1, V3 → S3. The minimum cost is $30 + $22 + $20 = $72. c Suburb S4 will not be served, as it is assigned to the dummy van. 15 a Replace the restricted assignment with a very large cost, M :

Column reduction does not change the matrix. b These steps simplify the matrix by creating zeros, which represent assignments with the lowest possible relative cost. This makes it easier to identify the set of assignments that will result in the overall minimum cost for the entire problem.

Chapter 10 review 1 a The source is P . b The sink is U . c The intermediate nodes are Q, R, S, T . 2 The edges crossing the cut from the source set to the sink set are S → B, A → B and A → T . The capacity is 18 + 5 + 18 = 41.

b The optimal assignment is A1 → D3, A2 → D1, A3 → D2. The minimum time is 12 + 8 + 9 = 29 hours. Complex unfamiliar 16 a Cost matrix:

3 The max-flow min-cut theorem states that the maximum possible flow from a source to a sink in a network is exactly equal to the capacity of the minimum cut (the cut with the smallest possible capacity). 4 a C1 = 10 + 8 = 18, C2 = 7 + 5 + 4 = 16, C3 = 12 + 6 = 18 b The maximum flow is 16. According to the max-flow min-cut theorem, the maximum flow is equal to the capacity of the minimum cut, which is C2.

Row reduced:

b An optimal assignment requires assigning each of the 5 resources to a unique task. This corresponds to selecting 5 independent zeros. To ensure such a selection is possible, the minimum number of lines required to cover all zeros must be 5.

5 1. To find the maximum flow, the minimum cut of the network must be identified by checking different possible cuts. 2. After inspection, it is found that the minimum cut capacity is 45 ML/day. There may be several cuts that have this same minimum value. For example: • One cut separates the Reservoir and all pumping stations from the Town. Its capacity is 10 + 35 = 45 ML/day. • Another cut separates the Reservoir, P1, and P2 from P3 and the Town. Its capacity is 5 + 20 + 10 = 45 ML/day.

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3. According to the max-ﬂow min-cut theorem, the maximum ﬂow through the network is equal to the capacity of the minimum cut.

10 Subtract 10 from Row 1, 9 from Row 2, and 18 from Row 3:

Therefore, the maximum ﬂow is 45 ML/day. P1 30 Reservoir

11 The largest value is 33. Subtract every element from 33:

15 P3

10 20 40 P2

Town

10 C1

6 1. First, identify the bottleneck in the original network, which is the minimum cut. This cut is formed by the edges ﬂowing directly into the sink: C → T and D → T .

12 1. Perform row reduction on the cost matrix. The result is:

2. The capacity of this minimum cut is 12 + 10 = 22 units. This means the original maximum ﬂow of the network is 22 units.

2. Next, perform column reduction. Since each column already has a zero, the matrix does not change.

3. The capacity of the edge B → D is increased. However, this edge is not part of the minimum cut (the bottleneck).

3. Check if the solution is optimal. Covering all zeros requires 3 lines. Because this equals the matrix size, an optimal assignment is possible.

4. Because the bottleneck itself has not been improved, the overall maximum ﬂow of the network cannot increase. Therefore, the new maximum ﬂow is still 22 units. 7

Chloe

Music

Ben

Food

Amy

Decor

4. The optimal assignment found from the reduced matrix is:

• A → Job 3

• B → Job 1

• C → Job 2

5. Calculate the minimum cost using the values from the original matrix: 12 + 13 + 15 = 40 The minimum cost is $40 000. 13 1. Convert the problem to a cost matrix by subtracting each value from the largest value (15):

2. Perform row reduction (mins 3, 1, 0), then column reduction (mins 0, 0, 1): 9 Assignment 1: Leo → Latte, Mia → Cappuccino. Profit = $2.50 + $2.40 = $4.90 Assignment 2: Leo → Cappuccino, Mia → Latte. Profit = $2.00 + $2.20 = $4.20 Therefore, the optimal assignment is assignment 1 with a proﬁt of $4.90.

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3. Covering zeros requires only 2 lines (e.g., C1 and C2). Since 2 < 3, the matrix must be adjusted. Smallest uncovered value is 1. The adjusted matrix is:

4. This matrix is optimal. The assignment is Ali → West, Beth → North, Carlos → South. 5. Calculate the maximum sales value from the original matrix:

2. Column reduction Subtract the smallest value (0, 16, 26) from each column of the new matrix: 3. Check for optimal solution

Max Sales = 11(Ali → West) + 13(Beth → North) + 15(Carlos → South) = 39 The maximum sales value is 39 × 10, 000 = $390, 000. 14 Set up the cost matrix, replacing ⬚ with a large number M . Applying the Hungarian algorithm leads to the optimal assignment: T1 → M2, T2 → M1, T3 → M3, T4 → M4 . This means that the minimum cost is 7 + 4 + 8 + 4 = 23 hours. 15 1. To solve this, ﬁrst add a dummy column of zeros to make the matrix a square (4 × 4). 2. Next, convert the maximisation problem into a minimisation one by subtracting all values from the largest value, which is 16. 3. Apply the Hungarian algorithm to the new cost matrix. The optimal assignment is:

• Film A → Cinema 1

• Film C → Cinema 2

• Film D → Cinema 3

4. Film B is assigned to the dummy column, which means it is the ﬁlm that will not be shown. 5. Calculate the maximum attendance using the values from the original matrix: 16 + 15 + 12 = 43 The maximum attendance is 4300 people. Exam practice 16 C 17 a Capacity of Cut X = 1200 + 1000 = 2200 vehicles per hour

Cover all zeros with the minimum number of lines. This requires only 2 lines (e.g., Column 1 and Row 1). Since 2 < 3, the matrix is not optimal and must be adjusted. 4. Adjust the matrix The smallest uncovered value is 7. Subtract 7 from all uncovered numbers and add it to the number at the intersection of the covering lines. The adjusted matrix is: 5. Find the optimal assignment

The new matrix requires 3 lines to cover all zeros, so an optimal assignment can be made by selecting a unique zero for each person: • Knox must be assigned to Swim (the only zero in his row). • This leaves Levi with only one available zero at Cycle. • Finally, Jane is assigned to Run. 6. Calculate the minimum yime Use the values from the original matrix for the optimal assignment: Minimum time = 36 (Knox → Swim) + 48 (Levi → Cycle) + 66 (Jane → Run) = 150 The minimum total time for the relay is 150 minutes.

b The max ﬂow is 2000 vehicles per hour. c The new maximum ﬂow is 500 vehicles per hour. 18 1. Row reduction Subtract the smallest value (40, 36, 25) from each row. The result is:

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