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Quick Notes 10s/Tere Tekau

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Term 1

Mathematics and Statistics for Aotearoa New Zealand Second edition

Lead author: Erin Doleman

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Rich tasks designed by Marie Hirst Dr Jo Knox

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Cultural author and reviewer: Moana Jarden-Osborne

Name: Class:

Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship and education by publishing worldwide. Oxford is a registered trademark of Oxford University Press in the UK and in certain other countries. Published in Australia by Oxford University Press Level 8, 737 Bourke Street, Docklands, Victoria 3008, Australia © Brian Murray and Oxford University Press 2027 The moral rights of the author have been asserted. First published 2025 Second edition All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, transmitted, used for text and data mining, or used for training artificial intelligence, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence, or under terms agreed with the reprographics rights organisation. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer. ISBN 9780190357368

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Edited by Maryanne Park Proofread by Casey McGrath Typeset by Newgen KnowledgeWorks Pvt. Ltd., Chennai, India Printed in New Zealand by Webstar

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Contents For the teacher...............................iv Progress Passports.........................2

Strand 4: Geometry Shapes and Spatial reasoning Topic 4.1: 2D shapes..................................................152

Year 8

Topic 4.2: Angle rules............................................... 158

Pathways

Strand 1: Number

Topic 4.3: Positions and pathways...................... 168

Number structures and operations Topic 1.1: Powers of 10................................................. 9 Topic 1.2: Exponents and factors........................... 14

Strand 5: Statistics

Topic 1.4: Rounding and estimation...................... 25

Developing knowledge from, visualising and interpreting data

Topic 1.5: Multiplying and dividing.......................30

Topic 5.1: Statistical measures...............................173

Topic 1.6: Order of operations................................ 41

Topic 5.2: Statistical investigations......................178

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Topic 1.3: Negative numbers...................................20

Topic 1.7: Fractions, decimals and percentages.................................................. 46

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Topic 1.8: Multiplying fractions................................56

Topic 1.9: Multiplying decimals................................ 61

Topic 1.10: Using fractions.........................................66

Topic 1.11: Using percentages....................................71

Strand 6: Probability

Theoretical probability Topic 6.1: Calculating probabilities..................... 188

Topic 1.12: Ratios.......................................................... 76

Experimental probability

Financial mathematics

Topic 6.2: Chance experiments..............................193

Topic 1.13: Money......................................................... 81

Strand 2: Algebra Equations and relationships Topic 2.1: Simplifying expressions..........................86 Topic 2.2: Linear equations and inequalities..... 91 Topic 2.3: Factorising expressions...................... 101 Topic 2.4: Tables and graphs................................106 Topic 2.5: Number patterns.................................... 117

Strand 3: Measurement Measuring Topic 3.1: Estimating and measuring................... 122 Topic 3.2: Converting units..................................... 127 Topic 3.3: Length and area..................................... 132 Topic 3.4: Volume......................................................142 Topic 3.5: Time............................................................ 147

Quick 10s........................................................ 198

For the teacher Kia ora! Thank you for choosing Mathematics and Statistics for Aotearoa New Zealand (Second edition). This programme has been purpose-written by experienced New Zealand educators to provide complete coverage of the New Zealand Curriculum (2025). Every component has been developed with real classrooms in mind, offering practical support for teachers, engaging learning experiences for students, differentiated activities and an easy-to-implement structure.

The three components of the programme This is your central hub for planning and implementing the programme. From here you can access: • Suggested year planners • Topic plans • Interactive teaching slides to support explicit teaching • Weekly pre- and post-topic quizzes (both online and paper) • Supporting resources such as rich tasks, activity sheets and interactives • An interactive assessment builder • Various achievement reports

Student Dashboard

This gives students easy access to digital activities and interactives that support and extend learning. Designed for independent use, it keeps students engaged both at school and at home. New features include key content videos for each topic, as well as an AI powered student tutor “Kea”.

Student Workbook

This is the core learning resource for students. It provides engaging, curriculumaligned activities organised by topic and includes all strands and elements, ensuring complete curriculum coverage.

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Teacher Dashboard

Here is an example of the workbook numbering system: Geometry | Pathways

Strand

Topic 4.3 Positions and pathways

Strand number

Topic number

Element

Topic title

Topics Our programme organises curriculum practices into clear topics. Each topic typically runs for one week (with the exception of a few double or half topics), with everything you need as a teacher provided. Each topic follows a three-stage progression: Learn, Explore, Deepen. Learn: D evelop the foundational knowledge and skills for the topic. Explore: Extend understanding by investigating different aspects of the topic and introducing new concepts and ideas. Deepen: B uild on learning through increased challenge and complexity, encouraging deeper mathematical thinking. For clarity, all topic plans, topic teaching slides and supporting workbook pages are aligned with these three stages. iv

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Strand 1: Number

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New features added to the student workbooks:

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Progress passport

Number structures and operations For me

1.1

Powers of 10

Practices (NZC 2025)

Review & date

Reading, writing comparing and ordering whole numbers and decimals using positive and negative powers of 10

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R numbers and RepresentingGwhole decimals in expanded form using powers of 10

Write in expanded form using powers of 10: 79.406 Convert 10 −6 to fraction and decimal form.

Strand 2: Algebra

Representing negative powers of 10 as a fraction and a decimal, and PA S S P vice-versa Using exponents and identifying cube roots for cube numbers up to at least 125 _

1.3

Quick 10s

1.4

Rounding and estimation

√8 = 3

Multiplying and dividing

Evaluating square and cube roots for perfect squares and cubes and using Topic a calculator to approximate them for other numbers

Use a calculator to approximate √5 . What is the prime factorisation of 28?

Use a number line to show the number that is 2 places less than zero. Write > or <: − 5 − 45 + 27 =

‾

672 × 83 = 437 ÷ 23 =

× 1

37 × 4

37 × 10

Comparing and ordering negative andb = 8 to check whether Substitute positive numbers using a number 25 − b =line 15.

Using substitution to find the value of an expression or formula

Rearranging known formulae using one or two steps Forming and solving linear inequalities and representing the solution on a number line

Using rounding, estimation and benchmarks to2.3 predict results and Factorise: 8m + 12mn Factorising to check the reasonableness of expressions calculations

Factorising simple algebraic expressions

Identifying and plotting points in the four quadrants of the coordinate plane, using ordered pairs and values from a table

Plot andbylabel the points on the Multiplying and dividing numbers coordinate plane: powers of 10

Using tables, graphs in the coordinate plane, and diagrams to recognise the relationship between the ordinal position and its corresponding element in a linear pattern, develop a rule for the pattern in words, and make conjectures about further elements in the pattern

Order of operations

Describe the linear pattern:

2.4

Evaluating Tables expressions and with integers, using the order of operations graphs

6 + [4 − (3 × 5 − 7)] = 2

What is the algebraic rule of: 1, 4, 7, 10, 13, …?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2.5

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Number patterns

Give a rule in words for how to get from one triangular number Tn to the next one, Tn+1.

Review & date

Forming and solving linear equations with rational solutions

(7, − 3), (− 10, 4), (2, 9), (− 6, − 1)

1.6

Practices (NZC 2025) Simplifying algebraic expressions involving sums, products, differences and single brackets, and collecting like terms

Multiplying and dividing whole numbers

+ 3

682.4815 × 10 6 =

For my teacher I feel

tenth, hundredth or thousandth

Find a lower and upper estimate for 3.7 × 6.

437 ÷ 23 =

For me Example

Representing composite numbers as products of their prime factors, using 2.1 Expand and simplify exponents to summarise repeated Simplifying 2x(4x − 5y) + 3x 2. factors expressions Locating negative and positive numbers on a number line

2.2 Evaluating expressions involving Solve: 4a + 5 = 8 Linearaddition and negative numbers, 5 _ equations and Solve: 8h = 4 subtraction Solve inequalities Rounding whole numbers to anythe inequality and represent solutions on a number line: specified power of 10, and the rounding 3x ≤number, 6 decimals to the nearest whole

Round 2.71828 to the nearest whole number, tenth, hundredth and thousandth.

1.5

Using radicals (√ and √ ) to represent squareEquations and cube roots and relationships

√100 =

1.2

Exponents and factors

Negative numbers

Located at the back of the workbook, these ten-question number knowledge checks draw on practices from the previous year level. They are ideal for lesson starters and provide regular opportunities to consolidate number knowledge skills.

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These checklists include illustrated examples to help students understand their goals and self-assess their confidence as they progress. Teachers can track progress, add comments or stickers and celebrate achievement. Progress Passports also support meaningful learning conversations between students, teachers and whānau.

For my teacher

Example Put these in order from smallest to largest: 100, 10 −3, 10, 10 4, 100,000, 10 −1, 0.01

Topic

11-Aug-26 14:08:33

Identifying the constant increase or decrease in a linear pattern, using variables and algebraic notation to represent the rule in an equation, and using the equation to make conjectures Investigating the patterns of triangular numbers, square numbers and cube numbers, extending the patterns, creating tables of values and plotting the values on the coordinate plane

Quick 10s/Tere Tekau Term 1

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Our goal for this programme is to support and empower confident teaching, meaningful mathematical learning and success for every student. Above all, we hope that you and your students enjoy the learning, discovery and mathematical thinking it inspires.

6.54 × 1, 000 = 52 = Is 22 a prime number? Y/N Round 337, 998 to the nearest 100, 000: ‾ 10 × 4 =

Is 504 divisible by 8? Y / N 824 − 812 = 655 ÷ 100 = _ √4 = Is 41 a prime number? Y/N

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Ngā mihi nui,

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Tēnā koe! I’m Nīkau the Northland Green Gecko (Moko kākāriki).

198

Round 292, 939 to the nearest 100, 000: ‾ 5 × 14 = Is 920 divisible by 10? Y/N 94 + 76 − 20 = 911 ÷ 10 = _ √16 = Is 4 a prime number? Y / N Round 98.67 to the nearest whole number: ‾ 11 × 13 = Is 96 divisible by 5? Y / N

4 + 10 − 32 = 4.35 × 100 = _

√9 =

Is 82 a prime number? Y/N Round 77.048 to the nearest tenth: ‾ 4×8= Is 785 divisible by 10? Y/N 478 + 654 = 1.23 × 100 = _ √36 =

/10

Is 93 a prime number? Y/N Round 225.805 to the nearest hundredth: ‾ 2 × 12 = Is 645 divisible by 9? Y / N 321 ÷ 100 = _ √64 = 62 − 13 + 73 = 280 ÷ 1, 000 = 10 2 = Is 27 a prime number? Y/N

11-Aug-26 14:08:36

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Erin Doleman Programme consultant and author

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Your teacher will tell you when it’s time to refresh your skills using these Quick 10 quizzes!

Round 178.338 to the nearest whole number: ‾ 4×6=

Quick 10s/Tere Tekau 162 − 56 + 23 =

Term887 2 ÷ 100 =

12 2 = Is 75will a prime number? Your teacher tell you when it’s time to refresh your skills using these Quick 10 quizzes! Is 616 divisible by 5? Y / N Y / N 279 − 357 = Round 397, 841 to the 9.67 × 1, 000 = nearest 10: ‾ Is 899 divisible by 6? Y / N 123 × 2 = Round 94.979 62 = 6× 9 = to the nearest hundredth: Is 941 divisible‾ by 10? Is 619 divisible by 8? Y / N 102 − 339 = Is 13 a prime number? 41 × 51 121 ÷ 53 = 124 − 977 = Y /=N Y/N Is 365 divisible by 5 ? Y / N 7 × 2 + 66 = 112 ÷ 89 = Round 94.69 to the nearest 408 + 523 = 20 721 −978 645÷=100 = whole number: 82 ÷ 2 − 21 = Simplest form of _: 80 ‾ 27 ÷ 3 _ 6 = = 12 × 7 = √121 Simplest form of _: 732 ÷ 100 = 15 Is 508 divisible by 5 ? Y /9N× 25 ÷ 5 = Is 71 a prime number? 110 ÷ 10 = 119 Simplest form of _ Y/N Is 17 a prime number? /10 /10: 136 Round 866.1 to the nearest Y/N 144 ÷ 1, 000 = whole number: Round 271.145 to the Is 14 a prime number? ‾ nearest hundredth: Y/N 85 × 29 = ‾ Round 5.35 to the nearest Is 898 divisible by 3? Y / N 50 × 79 = Year 8 Student Workbook | Mathematics and Statistics for Aotearoanumber: New Zealand (Second edition) Oxford University Press whole ‾

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11-Aug-26 18:47:06

D 605 − 619 = 278 ÷ 57 = 8÷2+4=

E 398 ÷ 63 = 12 ÷ 4 − 3 × 6 =

F 9 × 25 ÷ 5 =

90 Simplest form of _: 126 26 ÷ 100 = Is 59 a prime number? Y/N Round 960, 244 to the nearest 10: ‾ 71 × 19 = Is 467 divisible by 3? Y / N 488 + 62 =

24 Simplest form of _: 48 9.8 × 10 = Is 69 a prime number? Y/N Round 834.77 to the nearest tenth: ‾ 62 × 83 = Is 602 divisible by 5? Y / N 885 + 901 = 918 ÷ 35 =

48 Simplest form of _: 72 46 ÷ 6 = 8.14 × 1, 000 = Is 56 a prime number? Y/N Round 926, 309 to the nearest 10: ‾ 19 × 99 = Is 812, 212 divisible by 9? Y/N 603 − 763 = 991 ÷ 11 =

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Getting set up on Oxford Digital Scan this QR code (or visit www.oup.com.au/nzmaths_QR) to get started! Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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PRO

PA S S P

Strand 1: Number

Number structures and operations For me

Topic

1.1

Powers of 10

For my teacher

Example

I feel

Practices (NZC 2025)

Review & date

Reading, writing comparing and ordering whole numbers and decimals using positive and negative powers of 10

Put these in order from smallest to largest: 100, 10  −3, 10, 10  4, 100,000, 10  −1, 0.01

Representing whole numbers and decimals in expanded form using powers of 10

Write in expanded form using powers of 10: 79.406 Convert 10  −6to fraction and decimal form.

Representing negative powers of 10as a fraction and a decimal, and vice-versa Using exponents and identifying cube roots for cube numbers up to at least 125 _

√8   = 3

Use a calculator to approximate √5 . What is the prime factorisation of 28?

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Exponents and factors

1.3

Negative numbers

1.4

Rounding and estimation

Use a number line to show the number that is 2places less than zero. Write > or <: − 5 7 ‾ − 45 + 27 =

Multiplying and dividing

Round 2.71828to the nearest whole number, tenth, hundredth and thousandth. Find a lower and upper estimate for 3.7 × 6.

1.6

Order of operations

672 × 83 = 437 ÷ 23 = 682.4815 × 10  = 6

× 1

Representing composite numbers as products of their prime factors, using exponents to summarise repeated factors Locating negative and positive numbers on a number line Comparing and ordering negative and positive numbers using a number line

Rounding whole numbers to any specified power of 10, and rounding decimals to the nearest whole number, tenth, hundredth or thousandth Using rounding, estimation and benchmarks to predict results and to check the reasonableness of calculations Multiplying and dividing whole numbers

37 × 4

+ 3

37 × 10

6 + [4  2 − (3 × 5 − 7)] =

Evaluating square and cube roots for perfect squares and cubes and using a calculator to approximate them for other numbers

Evaluating expressions involving negative numbers, addition and subtraction

1.5

AF T

√ 100  =

1.2

Using radicals (√  and √ ) to represent square and cube roots

Multiplying and dividing numbers by powers of 10

Evaluating expressions with integers, using the order of operations

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

For me

Topic

Fractions, decimals and percentages

1.8

Multiplying fractions

1.9

Multiplying decimals

1.10

Using fractions

1.11

I feel

Practices (NZC 2025)

Review & date

Identifying, reading, writing and representing fractions, decimals and percentages

Represent one half as a fraction, decimal and percentage. Write these in descending order: 2 7 _ _ 0.18,    , 45,    , 0.32. 5 20

Comparing, ordering and converting between fractions, decimals and percentages Multiplying whole numbers by fractions, including by improper fractions, mixed numbers and first converting to an improper fraction

2 _ 7 × 3    = 5 2 _ 9 _ Simplest form of:     ×     5 10

Multiplying fractions and representing the answer in its simplest form

.7 × 30 = 1 2.95 × 8 = 1.1 × 2.3 =

Multiplying positive decimals

Finding a fraction of a whole number, including when the result is a mixed number or improper fraction

3 _     of 18 10 4 _    of a number is 18. What is the whole 9 amount?

Finding a whole amount when given a fraction, including when the whole set is a mixed number or improper fraction Finding percentages of whole numbers

72% of 1,200 =

Identifying percentage equivalence in calculations

Equivalent calculation of: 36%of 50 If 5%of a number is 12, then the whole amount is:

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Using percentages

Example

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1.7

For my teacher

Finding the whole (100%) when given a percentage Expressing the division of quantity into two parts as a ratio

1.12

Ratios Dividing a quantity into two parts, given the part:part or part:whole ratio

The red-to-whole ratio is: Divide 28into the part:part ratio 2 : 5.

Financial mathematics Topic

1.13

Money

For me Example If Sally can save $100per week, how much can she save each year? What is the reduced price of a $330 item with a 15% discount?

For my teacher I feel

Practices (NZC 2025)

Review & date

Creating and comparing weekly, monthly and yearly finance plans Applying percentage discounts

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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PA S S P

Strand 2: Algebra

Equations and relationships

2.1

Simplifying expressions

2.2

Linear equations and inequalities

2.3

For my teacher

Example

I feel

Forming and solving linear equations with rational solutions

Solve: 4a + 5 = 8 5 _ Solve: 8h =     4 Solve the inequality and represent the solutions on a number line: 3x ≤ 6

Factorise: 8m + 12mn

Rearranging known formulae using one or two steps Forming and solving linear inequalities and representing the solution on a number line

Factorising simple algebraic expressions

(7, − 3), (− 10, 4), (2, 9), (− 6, − 1) Describe the linear pattern:

Tables and graphs

What is the algebraic rule of: 1, 4, 7, 10, 13, …?

2.5

Number patterns

Review & date

Using substitution to find the value of an expression or formula

Substitute b = 8to check whether 25 − b = 15.

Plot and label the points on the coordinate plane:

2.4

Practices (NZC 2025) Simplifying algebraic expressions involving sums, products, differences and single brackets, and collecting like terms

Expand and simplify 2x(4x − 5y) + 3x  2.

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Factorising expressions

For me

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Topic

Give a rule in words for how to get from one triangular number Tn  to the next one, T  n+1.

Identifying and plotting points in the four quadrants of the coordinate plane, using ordered pairs and values from a table Using tables, graphs in the coordinate plane, and diagrams to recognise the relationship between the ordinal position and its corresponding element in a linear pattern, develop a rule for the pattern in words, and make conjectures about further elements in the pattern Identifying the constant increase or decrease in a linear pattern, using variables and algebraic notation to represent the rule in an equation, and using the equation to make conjectures Investigating the patterns of triangular numbers, square numbers and cube numbers, extending the patterns, creating tables of values and plotting the values on the coordinate plane

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Strand 3: Measurement

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PA S S P

Measuring For me

Topic

3.1

For my teacher

Example

I feel

Estimate the angle and then measure it using a protractor.

3.2

Converting units

Converting between metric units of area (mm  2, cm  2, m  2 and km  2) and volume (mm  3, cm  3 and m  3)

Convert 1.2 m  to cm  . 2

Convert 0.5 m  3 to cm  3and then to mL.

Converting between different volume units (cm  3, m  3, mL, L)

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Calculating the area of a parallelogram and a trapezium

Calculate the area of the shape if P = 26 cm.

3.3

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Length and area

Review & date

Estimating and measuring length, area, volume, capacity, mass (weight), temperature, time and angle, using appropriate units

Estimating and measuring

Practices (NZC 2025)

5 cm

Calculating the area of a shape, given some lengths and its perimeter, and vice versa

Find the unknown length if P = 68 mmand A = 270 mm  2. a 17 mm

Calculating lengths of quadrilaterals, given their area and other sufficient information

15 mm 20 mm

Calculating the volume of triangular prisms

3.4

Volume

5 cm

20 cm 45 cm

12 cm

Ivydale

3.5

Time

Juniper Flat

30 cm

90 cm

4 cm

13:50 arr.

13:56

Convert 2 hours and 15minutes into minutes.

Calculating the volume of composite figures made up of cubes, rectangular prisms and/or triangular prisms Reading, interpreting and using timetables, charts and results that present information about duration. Converting times to a given unit (e.g. hours and minutes to minutes)

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Strand 4: Geometry

PRO

PA S S P

Shapes and Spatial reasoning For me

Topic

For my teacher

Example

I feel

Practices (NZC 2025)

Review & date

Identify the radius of Identifying and describing the parts of a circle: the radius, diameter and circumference

4.1

8 cm

2D shapes

Transforming 2D shapes on the coordinate plane, including composite shapes, by a combination of translations, reflections, rotations and scaling by any factor

Reflect the point (1, 2)across the y-axis, then move right 1unit. Find the unknown angles:

Proving that the interior angle sum of a triangle is 180°, and generalising a rule for the interior angle sum and exterior angles for any polygon

4.2

115°

88°

127°

Angle rules

64°

Reasoning about unknown angles in situations involving internal and external angles of polygons

146°

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° °

Topic

4.3

Positions and pathways

For me Example

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Pathways

What compass direction corresponds to the true bearing 225°?

I feel

For my teacher Practices (NZC 2025)

Review & date

Using map scales, compass points, distance and turn to interpret and communicate positions and pathways in coordinate systems and grid reference systems

I came. I saw. I scaled.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Strand 5: Statistics

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PA S S P

Developing knowledge from, visualising and interpreting data For me

Topic

For my teacher

Example

I feel

Calculate the mean, median and mode(s) of:

Practices (NZC 2025)

Review & date

Calculating the mean, median and mode for numerical data

7, 9, 10, 10, 11, 14, 16 Calculate the range of the data set.

Frequency

Statistical measures

Calculating the range for numerical data

Results

5.1

10 9 8 7 6 5 4 3 2 1 0

5 6 Score

Responding to statistical questions by calculating an appropriate measure of central tendency and range for a variety of data tables and data visualisations

Identifying outliers by eye and taking them into account when using range as a measure of spread

Summarise this data into a frequency table:

Give an example of a data set for which a bar chart is not a suitable visualisation.

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20, 17, 23, 15, 18, 22, 17, 24, 18, 16, 19, 20, 14, 25, 17, 21, 18, 23, 16, 22, 19, 20, 17, 18, 21, 15, 16, 19, 18, 20

AF T

Identify the outlier in the data visualisation.

Describe the shape of this data visualisation.

5.2

Planning and collecting data in order to respond to a statistical question

For a given set of data, choosing and constructing an appropriate data visualisation according to the data type

Interpreting data visualisations, including those from contemporary media

Statistical investigations

5 6is misleading 7 8 9about 10 this 11 12 What visualisation? Population index

Population change 120

120 100

Identifying when a data visualisation cannot be interpreted accurately due to missing information

100

80 60 40

Noticing and explaining outliers in a given set of data

20 0

2019

2023 Year

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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PA S S P

Strand 6: Probability

Theoretical probability Topic

For me

For my teacher

Example

I feel

What is the probability of each outcome when flipping two fair coins?

6.1

Calculating probabilities

Practices (NZC 2025)

Review & date

Calculating probabilities for events as decimals, fractions and percentages

What is the most likely sum when rolling 2 fair six-sided die?

Comparing the likelihood of different events

Calculate the probability of the event complement to drawing a Heart from a standard deck of cards.

Calculating probabilities for complementary events

Experimental probability For me

For my teacher

Example

I feel

In pairs, flip two coins 30times and record the frequency of each outcome after 30trials.

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6.2

Chance experiments

Compare the theoretical probability of each outcome with its experimental probability. Based on your results so far, do you think your coins are fair? Compare the theoretical probability of each outcome with its updated experimental probability after completing 100 trials.

Practices (NZC 2025)

AF T

Topic

Review & date

Carrying out a chance experiment and calculating the experimental probability of each outcome Comparing experimental probability (using at least 30 trials) to theoretical probability, and explaining why they differ and how increasing the number of trials reduces this difference Carrying out chance experiments of at least 100 trials and comparing the experimental probability of each individual outcome to its theoretical probability, in order to demonstrate the Law of Large Numbers

Passport ready, I’m plotting my next point on the plane!

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.1 Powers of 10

Powers of 10 with positive exponents

base exponent expanded form

A positive exponent tells us how many times to multiply the base by itself. A power of 1 0is a number which can be written with 10as the base and an integer as the exponent. The exponent is equal to the number of zeros in the basic numeral.

exponent form

34 = 3 × 3 × 3 × 3 = 81 basic numeral

10 3 = 10 × 10 × 10 = 1,000

Learn 1

Complete this table. Basic numeral 10

Expanded form 10

Exponent form 10  1

100

AF T

10 × 10 × 10

100,000

10  4

100,000,000

D R

10 × 10 × 10 × 10 × 10 × 10

10  7

10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 2 Describe the relationship between the number of zeros in the basic numeral and the exponent in the power of 1 0.

3 Write the basic numerals for each power of 10. Make sure you use commas to make the number easier to read. a 10  10:

b 10  13:

4 Describe the basic numeral form of 10  100. This amount is called one googol.

5 An exponent of 0means you have no copies of the base to multiply. Any non-zero base with an exponent of 0 is

equal to 1. ‾

a 10  0= b Compare part a with the equations in question 1. Describe the pattern you see.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Powers of 10with negative exponents A power with a negative exponent represents repeated division. The exponent indicates how many times 1 has been divided by the base. For example, 10  −1 = 1 ÷ 10 and 10  −2 = (1 ÷ 10) ÷10. We can convert between powers of 10with negative exponents, fractions and decimals. Power with negative exponent

Fraction

Decimal

10  −1

_   1   10 _   1   100 _   1   1,000

0.1

10  −2 10  −3

0.01 0.001

Learn 1

Complete this table. Exponent

− 4

− 3

− 2

10  −1

10  0

10  1

AF T

Exponent form

− 1

_1     10

Fraction/Basic numeral

D R

2 Fill in the blanks about the table in question 1.

Moving from left-to-right, multiply by to get the next power of 10. Moving from right-to-left, divide by

to get the next power of 10.

3 Convert these powers of 10to fractions.

‾

a 10  −6=

b 10  −10=

4 Describe the fraction form of 10  −100.

5 Complete this table.

Exponent

− 5

− 4

− 3

− 2

− 1

Exponent form

10  −1

Decimal form

0.1

6 Convert these powers of 10to decimals.

‾

a 10  −6=

b 10  −10=

7 Describe the decimal form of 10  −100.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Representing numbers in expanded form Writing numbers in expanded form makes it easy to see the place value of each digit. We can write the corresponding powers of 10using exponents. 1   + 7 ×  _ 1   + 3 ×  _ 1   982 . 473= 9 × 100+ 8 × 10 + 2 × 1 + 4 ×  _                10   100   1,000 = 9 × 10  2+ 8 × 10  1+ 2 × 10  0+ 4 × 10  −1+ 7 × 10  −2+ 3 × 10  −3 When 0is a digit in a number, we leave that term out of the expanded form. 1   28.03 = 2 × 10  1 + 8 × 10  0 + 3 ×  _ 100

Explore 1

Complete this place value table. Thousands

Hundreds

Tens

Ones

Tenths

Decimal

1,000

100

0.1

Fraction

1,000

100

Exponent form

10  3

10  2

10  1

10  0

Hundredths

Thousandths

_   1   100 10  −3

form.

1   + ×  ____ 1   + ×  _ ‾ ‾ ‾ 10 ‾ 100               = × 10  1+ × 10  0+ × 10  −1+ × 10  −2 ‾ ‾ ‾ ‾

D R

× 10 + × 1 a 35.42=

AF T

2 Write these numbers in expanded form, writing powers of 10as basic numerals or fractions and then in exponent

‾ ‾ 10 ‾ 100         1     = × 10  + × 10  −1 + × 10  −2 ‾ ‾ ‾

1   +  ×  _ 1   × 10 + ×  ____ b 10.96=

1   12.05= × 10 + × 1 + ×  _ 100 ‾ ‾ ‾             = × 10  1+ × 10  0+ × 10  −2 ‾ ‾ ‾

‾‾           =           ‾‾

d 3.56=          

‾‾           =           ‾‾

e 1.934=          

79.06=           ‾‾           =           ‾‾

‾‾           =           ‾‾

g 5.012=          

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 Write these numbers in expanded form, writing the powers of 1 0in exponent form. a

2.468 =

b 35.902 =

231.497 =

d 64.096 =

e 900.009 =

15.0203 =

g 101.0101 =

h 9.83451 =

4 These numbers are in expanded form. Rewrite them as basic numerals.

‾

a 4 × 10  2 + 8 × 10  1 + 2 × 10  0 + 6 × 10  −1=    

‾ 4 × 10  4 + 2 × 10  3 + 9 × 10  1 + 1 × 10  0 + 5 × 10  −1 + 7 × 10  −2=     ‾ d 6 × 10  5 + 8 × 10  4 + 2 × 10  2 + 1 × 10  1 + 9 × 10  −1=     ‾ e 9 × 10  3 + 3 × 10  1 + 1 × 10  0 + 7 × 10  −1 + 5 × 10  −2 + 2 × 10  −3=     ‾ f 7 × 10  4 + 1 × 10  3 + 6 × 10  2 + 4 × 10  0 + 8 × 10  −2=     ‾ b 7 × 10  3 + 5 × 10  2 + 3 × 10  1 + 8 × 10  −1 + 1 × 10  −2=    

AF T

5 Some students are writing numbers in expanded form but have made mistakes. Explain to them what error they

a 5.08 = 5 × 100 + 8 × 10−1

D R

have made and write the correct expanded form for the number.

b 427.3 = 4 × 10−3 + 2 × 10−2 + 7 × 10−1 + 3 × 101

6 If a number has a 0 (zero) as one of its digits, explain how this is represented in the expanded form of the number.

Include two examples in your answer.

If you’re stuck, write the place values first.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Comparing and ordering numbers using powers of 10 We can order numbers based on the size of the number or power of 1 0. Symbol

Meaning

Example

Is greater than

10,000 > 100

Is less than

100 < 10,000

Powers of 10can be ordered by the value of their exponent. For example, we know that − 2 > − 3, so 10  −2 > 10  −3.

Deepen Write each power of 10in exponent form, and then insert the correct inequality symbol (<or >). a _   1   = 10

1   = 10   _ ‾ 1,000

b _   1   = 10

100

1   = 10  _   1   = 10   _ 1,000 100,000 ‾

e 0.001 = 10

1   = 10   _ ‾ 10,000

d 0.0001 = 10

0.1 = 10  ‾

0.01 = 10  ‾

0.01 = 10  0.0001 = 10  ‾

AF T

2 Put these powers of 10in order from smallest to largest. a 100, 10  −3, 10, 10  4, 10  −1, 0.01:

D R

b 10  2, 10  −5, 10, 0.1, 10  4, 100,000, 10  0, 0.001:

10  4, 0.001, 10  6, 10, 10  −2, 100, 0.1, 10  0:

3 Put these powers of 10in order from largest to smallest. a 10,000, 10  −4, 1,000, 10  1, 0.001, 10  0, 10  −1: b 10  2, 10  4, 1,000, 0.01, 10  −3, 10, 10  −5, 1,000,000: c

10  −3, 10  4, 1, 0.01, 10  5, 100, 10  1, 0.0001:

4 Write these numbers in expanded form using exponent form for the powers of 1 0and then determine which

number is greater in value by comparing the powers of 10. a 23.45 =

2.345 = > b 483.7 =

48.37 =

> c

384.205 =

384.025 =

> 5 True or false: A number with more digits is always greater than a number with fewer digits. Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.2 Exponents and factors Exponents, squares and cubes In mathematics we often look for shortcuts. We use multiplication as a shortcut for repeated addition. For example, instead of 3 + 3 + 3 + 3, we can write 3 × 4 = 12. We use exponents as a shortcut for repeated multiplication. Instead of 3 × 3 × 3 × 3, we can write 3 4. A number in exponent form b  nhas a base b and an exponent n . We read this as b to the power of n . The equivalent repeated multiplication is called expanded form. A square number can be written in exponent form with an exponent of 2 , and a cube number can be written with an exponent of 3. base exponent expanded form exponent form

3 =3×3×3×3 = 81 4

Exponents should be written smaller at the top-right of the base. Don’t write 34 when you mean 3 4.

basic numeral

Write each repeated multiplication in exponent form by first identifying what the base and exponent should be. Repeated multiplication 3 × 3 × 3 × 3

D R

AF T

Learn

Base

Exponent

Exponent form

2 × 2 × 2 × 2 × 2 4 × 4 × 4 8 × 8 × 8 × 8 5 × 5 × 5 × 5 × 5 7 × 7 × 7 × 7 × 7 × 7 10 × 10 × 10 × 10 2 Fill in the gaps in the table.

Exponent form

Expanded form

Basic numeral

  2 4

4 × 4

  3 3   4 2   3 5   2 6   2 9 10  3

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Square roots undo squares. Cube roots undo cubes!

Square and cube roots

The square root of a whole number, when multiplied by itself, is equal to the original number. _

The symbol for a square root is √  . For example, √ 9  = 3 because 3  2 = 9. When three copies of the cube root of a number are multiplied together, you get the original number. 3 _

3 _

The symbol for a cube root is √ . For example, √8  = 2 because 2  3 = 8.

Learn 1

Write a list of the first 12square numbers.

2 Write a list of the first 5cube numbers. 3 Evaluate these square and cube roots.

‾

‾ √8 = ‾ _ m √36 = ‾ _

e √49 = i

3 _

√100 = ‾ _ g √16 = ‾ _ k √ 64 = ‾ 3 _ o √27 = ‾

‾ √125 = ‾ _ j √ 121 = ‾ _ n √25 = ‾ 3 _

b √1 =

3 _

AF T

a √4 =

‾

d √9 =

‾ √64 = ‾ _ p √144 = ‾ _

h √81 = l

3 _

4 We can find the square root of any positive number. The square root of a square number is easy to find, but we can

find the approximate values of the square roots of other numbers. _

D R

For example, as 4 < 5 <_ 9, we know that √4 < √ 5 < √ 9 , which means 2 < √ 5 < 3. Using a calculator gives us an approximate value of √   5 ≈ 2.2. Find the whole numbers above and below each square root, and then use a calculator to find a decimal a pproximation. ‾

‾

2 < √ 8 < 3 a 4 < 8 < 9, so _

‾

√  8 ≈

‾ _ √11  ≈ ‾

< 18 < 25, so < √ 18 < ‾ ‾ ‾ _ √18  ≈ ‾ _ d < 34 < , so < √ 34 < ‾ ‾ ‾ ‾ _ √34  ≈ ‾ _

‾

b 9 < 11 < , so < √ 11 <

Use your list of square numbers in question 1.

5 We can use a similar method to question 4 to find the approximate values of any cube roots using nearby cubes.

Find the whole numbers above and below each cube root, and then use a calculator to find a decimal approximation. ‾

3 _

‾

a 1 < 5 < 8, so < √5 <

‾ 3 _ < 20 < 27, so < √20 < ‾ ‾ ‾ _ 3 √20  ≈ ‾ 3 _ d < 50 < , so < √50 < ‾ ‾ ‾ ‾ _ 3 √50  ≈ ‾ 3 _

√5   ≈ c

‾

3 _

‾

b 1 < 7 < , so < √7 < 3 _

‾

√7   ≈

Use your list of cube numbers in question 2.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Factors and primes The factors of a positive whole number are the positive whole numbers it can be divided by which leave no remainder. For example, when 2is divided by 1or 2, the remainder is 0. So, the factors of 2are 1and 2 (two factors). When 4is divided by 1or 2, the remainder is 0. So, the factors of 4are 1, 2and 4 (three factors). Whole numbers can be prime, composite or neither. A prime number has exactly two factors (1and itself). For example, 2is a prime number. A composite number has more than two factors. For example, 4is a composite number. The number 1is the only whole number that is neither a prime number nor a composite number.

Explore Write the factors of the whole numbers from 1to 20. a 1:

b 2

d 4:

e 5:

g 7:

h 8:

10:

AF T

m 13:

n 14:

o 15:

p 16:

q 17:

18:

20:

D R

k 11:

19:

12:

2 What is the only number between 6and 10that is a prime number? 3 What is the only even prime number? 4 Write a list of the prime numbers between 1and 20. 5 A factor pair is made up of two factors of a number which multiply to give the number. For example, 3 and 4are a

factor pair of 12because 3 × 4 = 12. Use factor pairs to find all factors of these numbers. a 24:

b 28:

d 60:

35:

e 80:

100:

6 Explain how you used factor pairs to find all factors of the numbers in question 5.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Prime factorisation The prime factors of a number are the factors of the number which are prime. For example, the factors of 6are 1, 2 , 3and 6 , so the prime factors of 6are 2 and 3. We can use a factor tree to identify prime factors. Values at the end of branches of the tree are the prime factors of the number at the top of the tree. 6

We can multiply the prime factors of a number together to get the prime factorisation of the number. This factorisation is unique, up to reordering the prime factors, so we usually write the prime factors in increasing order. Exponents can be used to summarise repeated prime factors. For example, we can write 2  2= 4as the prime factorisation of 4.

Deepen Fill in the blanks in these factor trees to show each composite number as the product of its prime factors.

The prime factors of 10are: and .

= × ‾ ‾ ‾ 15

The only prime factor of 9is: .

AF T

D R

= × =   2 ‾ ‾ ‾ ‾ 21

The prime factors of 15are:

The prime factors of 21are:

and .

= × ‾ ‾ ‾

The prime factors of 35are:

The prime factors of 39are:

and .

= × ‾ ‾ ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

More complex factor trees For some composite numbers we need a larger factor tree. For example, if we start with 8 , we end up with 2 × 4. But 4 is not a prime number. We need to keep going until we end up with prime numbers. This is the completed factor tree for 8, which gives the prime factorisation as 8 = 2 × 2 × 2 = 2 3. 8

2 Complete each factor tree to the prime factorisation of each starting number. a

The prime factors of 20are: and . 20= × × ‾ ‾ ‾       2 = 2  × ‾ c

and . 18= × × ‾ ‾ ‾       2 = × 3  ‾ 36

The prime factors of 18are:

AF T

D R

The prime factors of 28are:

The prime factors of 36are:

and .

28= × × ‾ ‾  ‾     = × ‾ ‾

36= × × × ‾ ‾  ‾ ‾      = × ‾ ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 Draw a factor tree to determine the prime factorisation of each number. a 40

b 56

Prime factors:

Prime factorisation:

AF T

4 Complete these two factor trees for 100. b

D R

100

Prime factors:

Prime factorisation:

5 Compare your prime factorisations of 100in question 4. Did the different factor trees affect the prime factorisation?

Could two numbers have the same prime factorisation?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.3 Negative numbers Integers on the number line

Integers are positive and negative whole numbers, and zero. We write negative integers with a negative sign, like − 10, and say negative 10. Positive integers can be written with a positive sign, like + 10, but usually we would just write 10. Zero is the only number which is neither positive nor negative. We can use number lines to represent and compare the integers.

–5

–4

–3

–2

–1

Zero sits in the middle – it doesn't pick sides!

Learn 1

Write the missing numbers on this number line. –4

–3

–1

S = 4, I = 2, A = − 2, K = − 4, W = 0

–4

–3

–2

–1

D R

–5

AF T

2 Plot these letters on the number line.

3 Use the number line to answer the questions.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

90 100

a Which two numbers are 5places from zero? b Which number is 2places less than zero? c

Which number is 2places to the left of positive 1?

d Which number is 2places to the right of positive 1? e Which number is 5places to the left of negative 1? f

Which number is 8places to the right of negative 10?

4 Write the missing numbers on this number line.

–100 –90 –80

–60 –50

–30 –20 –10

5 Using the number line in question 3, write the correct symbol (>or <) between these numbers.

‾ d 1 − 3 ‾ g − 2 − 5 ‾ a − 5 7

‾ e 5 3 ‾ h − 7 − 2 ‾ b − 3 − 5

c f i

− 10 − 7 ‾ − 4 − 3 ‾ − 9 − 8 ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Positive and negative numbers We can also represent and compare negative fractions and decimals on a number line. –

–1.75 –3

–2

1 2

5 2

0.25

–1

You can think of negative numbers as counting to the left from 0 . This helps when comparing the values of negative 1 1 _ _ fractions and decimals. For example, − 1.75 < − 1and −     < −    . 4 2

Learn 1

Plot these values on the number line with a dot. 7 , − 2 _ 10  1 , − 1 _ 1 , −  _ 1 , _ _   1 , −  _  5 , 2_  1 , −  _ 4 4 4 2 4 2 4 4

–3

–2

–1

a –4

b –3

–2

–1

D R

AF T

2 Which values are plotted on this number line? Write your answers as proper fractions or mixed numbers.

3 Explain how to use the number line to help you convert a negative mixed number to a negative improper fraction

using an example from question 2.

4 Plot these values on the number line with a dot.

0.25, − 0.5, − 1.25, − 2.75, 2.25, − 1.5, 2.5, − 0.75

–3

–2

–1

The more negative a number is, the smaller it is.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

5 Which values are plotted on this number line? Write your answers as decimals. a

–3

–2

–1

e 0

f 1

2 d

6 Use the number lines in questions 1 and 2 to compare these values, and then fill in >or <.

3 2‾4

3‾

1  a −  _ _     2‾

1  b _   2  −  _ 3 4

1  e −  _ −  _ 

1  − 3 − 3 _ 4 ‾

5  2  −  _ −  _ 4‾ 3

1  d _   1  −  _

2‾

1  h −  _  − 1 _

9 4

1  −  _  g − 2 _

7 Use the number line in question 2 to compare these values, and then fill in >or <.

‾ c − 0.4 − 2.9 ‾ e − 0.6 0.6 ‾ g − 1.8 − 2.3 ‾

‾ d − 1.2 − 1.1 ‾ f − 3.4 − 3 ‾ h − 1.5 − 2.5 ‾

− 1.7 a − 2.3

− 1 b − 0.8

4‾

8 3‾

Ascending order means lowest to highest, and descending order means highest to lowest.

8 Put these values in ascending order.

3 3 4

5 4

7 4 2

3 2 2

AF T

2 , _ a −  _    , −  _ , −  _ , _  1 : 3 4 3

2 , −  _ , _ b −  _ , _  1 , − 1 _  2 : 5 , _ 1 , − 2 _ 2 , _ −  _  3 , −  _  1 : 2 3 4 4 3

D R

7 4

2 , − 1 _ 1 , _ d −  _  1 , _  2 , −  _ : 3

2 4 3

9 Put these values in descending order. a − 1.4, 2.3, − 0.7, − 2.9, 1.1: b − 2.6, − 0.5, 1.8, − 1.2, − 2.5: c

0.4, − 1.7, − 2.8, − 0.9, 2.1:

d − 1.9, − 0.6, 2.4, − 2.3, − 2.1: 10 Explain what step you would need to do first to put these fractions in ascending order.

5 , _ 9  , _ 11  −  _  7 , −  _  2 , −  _ 6 8 10 3 12

The more negative a number is, the smaller it is.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Adding and subtracting positive integers We can use a number line to add or subtract positive integers. For example, − 3 + 2 = − 1. – 3 – 2 = –5

– 3 + 2 = –1

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

Explore 1

Use the number line to evaluate these expressions. ‾ e 2 − 5 = ‾ i 3 − 7 = ‾

a − 3 + 1 =

‾ − 8 + 7 = ‾ − 4 + 10 = ‾

b − 4 − 1 = f j

− 2 + 3 = ‾ g − 7 + 4 = ‾ k − 10 + 7 = ‾

‾ h − 2 − 6 = ‾ l − 8 + 5 = ‾

d − 7 − 2 =

3 Evaluate these expressions.

‾ d − 48 + 22 = ‾ g − 50 − 14 = ‾ j − 26 − 31 = ‾ a − 12 + 37 =

D R

AF T

2 Describe any techniques you used or patterns you noticed in question 1.

‾ e − 29 + 41 = ‾ h − 39 − 23 = ‾ k − 44 − 16 = ‾ b − 45 + 28 =

c f i l

− 33 + 19 = ‾ − 18 − 27 = ‾ − 26 − 23 = ‾ − 21 + 47 = ‾

4 Can you find a quick way to evaluate expressions which subtract a positive integer from a negative integer, like

− 15 − 17? Explain.

5 Can you find a quick way to evaluate expressions which add a positive integer to a negative integer, like − 15 + 17?

Explain.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Taking away a negative makes the answer more positive!

Adding and subtracting negative integers

Adding a negative integer is the same as subtracting the positive version of that integer. For example, − 5 + (−3) = − 5 − 3 = − 8. Subtracting a negative integer the same as adding the positive version of that integer. For example, − 5 − (−3) = − 5 + 3 = − 2. (+)= + + −(+)= − +(−)= − −(−)= +

Deepen 1

Complete this table. Expression

Equivalent expression with simplified operation

Value

5 + (− 3)

5 − 3

8 − (− 5)

8 + 5

AF T

6 + (− 8) − 9 + (− 1)

D R

− 5 − (− 3) 3 − (− 9)

2 Use this number line to evaluate the expressions.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 ‾    = ‾

a 9 + (− 2)=

‾    = ‾

e − 7 − (− 2)=

− 1 + (− 9)= ‾    = ‾

‾    = ‾

m 6 + (− 1)=

‾    = ‾

b 2 + (− 6)=

− 4 + (− 5)= ‾    = ‾

9 − (− 4)= ‾    = ‾ ‾    = ‾

n 4 − (− 3)=

7 − (− 4)= ‾    = ‾

‾    = ‾

g 10 − (− 2)=

‾    = ‾

k 3 + (− 7)=

‾    = ‾

o 8 + (− 10)=

10 ‾    = ‾

d 4 + (− 9)=

‾    = ‾

h 2 − (− 7)=

− 2 − (− 8)= ‾    = ‾

‾    = ‾

p − 9 − (− 6)=

4 When a number is added to its additive inverse, the result is 0. Fill in the additive inverses of these numbers.

‾ e 2 + = 0 ‾

a − 3 + = 0

‾ 4 + = 0 ‾

b − 5 + = 0 f

− 10 + = 0 ‾ g 20 + = 0 ‾ c

‾ h 35 + = 0 ‾

d − 25 + = 0

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.4 Rounding and estimation Rounding whole numbers to the nearest 10 Rounding a number means to replace it with an approximate value which is easier to use. To round a whole number to the nearest 10, put a box around the tens digit. •

If the ones digit is 0, 1, 2, 3 or 4, we round the number down by replacing the ones digit with a 0. 1,2 5 3 ≈ 1,2 5 0

•

If the ones digit is 5, 6, 7, 8 or 9, we round the number up by increasing the tens digit by 1and replacing the ones digit by 0. 1,2 5 7 ≈ 1,2 6 0

•

If the tens digit is a 9and you need to round up, increase it to 10and then regroup the 1to the hundreds column and leave the 0as the tens digit. 1,2 9 7 ≈ 1, 2 10 0 = 1,3 0 0

Round these whole numbers to the nearest 10. a 23 ≈

b 58 ≈

e 196 ≈

289 ≈

D R

AF T

Learn c

75 ≈

d 91 ≈

g 1,504 ≈

h 9,996 ≈

2 Complete the table with whole numbers that round up and down to the values given when rounding to the nearest 1 0.

Value

Rounds up to value

Rounds down to value

Value

800

7,310

560

440

1,020

2,470

990

Rounds up to value

Rounds down to value

4 Round these whole numbers to the nearest 10. a 29 ≈

b 99 ≈

299 ≈

d 9,939 ≈

e 49,999 ≈

599,999 ≈

5 Explain how to round a number where every digit is a 9.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

4 or less, round down. 5 or more, round up.

Rounding whole numbers to a power of 10

Numbers can be rounded to any place value. To round a whole number to a power of 1 0 (e.g. 100, 1000, 10,000 etc.), put a box around the digit in that place value column. Look at the digit to the right of the box and either round up or down.

Learn 1

Round these whole numbers to the nearest 100. a 199 ≈

b 250 ≈

349 ≈

d 951 ≈

e 1,049 ≈

5,950 ≈

g 12,499 ≈

h 99,951 ≈

10,023 ≈

2 Round these whole numbers to the nearest 1,000. a 987 ≈

b 1,499 ≈

1,501 ≈

d 2,950 ≈

e 5,499 ≈

9,501 ≈

g 14,299 ≈

h 99,950 ≈

56 ≈

3 Complete this table.

814 6,529

100

AF T

Round to the nearest

1,000

10,000

100,000

D R

Value

48,372 503,918 7,104,289 82,593,701 9,450,312 4 Round these values to the nearest hundred and the nearest thousand. a 89,942

b 457,499

Nearest hundred:

Nearest thousand:

Can you explain how to round to the nearest million?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Rounding decimals Decimals can also be rounded using the same technique as for rounding whole numbers. Rounding to one decimal place is the same as rounding to the nearest tenth, rounding to two decimal places is the same as rounding to the nearest hundredth, and rounding to three decimal places is the same as rounding to the nearest thousandth. To the nearest whole number: 7 .43 ≈ 7 .00 = 7

Always check that your rounded value has the expected number of decimal places.

To the nearest tenth: 25. 4 7 ≈ 25. 5 0 = 25.5 To the nearest hundredth: 36.2 9 5 ≈ 36.2 10 0 = 36.30 To the nearest thousandth: 4.03 9 1 ≈ 4.03 9 0 = 4.039

Explore Round these decimals to the nearest whole number. a 3.7 =

b 9.2 =

24.81 =

d 37.63 =

e 81.504 =

2 Round these decimals to the nearest tenth. a 6.58 =

218.2761 =

AF T

b 22.94 =

d 465.111 =

a 4.201 =

b 239.876 =

431.5534 =

d 788.0901 =

e 222.77466 =

a 5.1204 =

b 333.9047 =

457.88205 =

d 454.33198 =

e 569.115602 =

790.4439815 =

A Round to the nearest whole number

Round to 3 decimal places

B Round to the nearest thousandth

Round to 0 decimal places

C Round to the nearest hundredth

III

Round to 1 decimal place

D Round to the nearest tenth

Round to 2 decimal places

347.803 =

e 780.6249 =

D R

1,023.98752 =

3 Round these decimals to the nearest hundredth.

6,426.430199 =

4 Round these decimals to the nearest thousandth.

5 Match the ways we can describe rounding decimals.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

6 Complete this table by rounding the decimals.

Value

Nearest whole number

Nearest tenth

Nearest hundredth

Nearest thousandth

3.1415

3.1

3.14

3.142

2.71828 1.618034 1.414214 1.732051 0.693147 0.577216 0.915966 1.202057 7 Some students have made an error when rounding these decimals. Identify the error and show the student how they

should round these decimals to the decimal places given in brackets.

0.09 ≈ 1 (nearest whole number)

b 5.243 ≈ 5.25

AF T

(nearest tenth)

D R

a 4.793 ≈ 4.7

(nearest hundredth)

d 4.12859 ≈ 4.12959

(nearest thousandth)

8 Suppose you are finding the mass of an object using two different scales. One scale reads 1 00 g, and the other scale

reads 100.0 g. Which scale is more accurate? Explain.

9 A cat is weighed at the vet and its weight is recorded as 4.6 kg. a

What place value was the weight rounded to?

b Give three possible weights of the cat in kilograms to the nearest gram (3 decimal places).

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Rounding and estimation An estimate is an approximate value. We often use estimates because they are easier to understand or visualise. For example, you might estimate that the population of New Zealand is 5 .3million which is not the exact population. We can use rounding and estimation to predict results and check the reasonableness of calculations. For example, we know that 70 < 14.7 × 5 < 75because 14 × 5 = 70is an underestimate and 15 × 5 = 75is an overestimate. Note that to get an overestimate when dividing, we can round up the dividend and round down the divisor. To get an underestimate, we round down the dividend and round up the divisor.

Deepen 1

Find lower and upper whole number estimates for these decimal calculations. a < 3.7 × 6 <

b < 5.9 × 9 <

< 7.2 × 6.1 <

d < 11.5 × 3.3 <

e < 19.93 ÷ 4.78 <

< 98.62 ÷ 9.73 <

2 Use a calculator to find the exact values of the calculations in question 1. b

AF T

D R

3 Compare the exact values of the calculations in question 2 with the lower and upper estimates you found in question 1.

Do all the exact values lie between the estimates?

4 Without using a calculator, determine whether these calculations are reasonable or unreasonable. Explain your

thinking. a 9.7 × 5.4 = 42.38 b 11.3 × 8.1 = 91.53

5 Use a calculator to check the calculations in question 4. Was your thinking correct? a 9.7 ×5.4 =

b 11.3 ×8.1 =

6 We can also estimate calculations using known benchmarks, instead of rounding to whole numbers. For example, we

could round to the nearest 0.25as quarters and halves are common benchmarks. Use this technique to estimate the following calculations. ‾ d 0.58 × 80 ≈ ‾ g 1.28 × 600 ≈ ‾ a 0.47 × 20 ≈

‾ e 0.81 × 160 ≈ ‾ h 0.82 × 40 ≈ ‾ b 0.27 × 120 ≈

c f i

0.73 × 60 ≈ ‾ 0.23 × 400 ≈ ‾ 1.55 × 30 ≈ ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.5 Multiplying and dividing Multiplying and dividing by powers of 10 Multiplying a number by a power of 10increases the place value of each digit. 56 × 10  1= 560 56 × 10  2= 5,600 56 × 10  3= 56,000

Dividing a number by a power of 10decreases the place value of each digit.

_.6 7.26 × 10  1= 72 _6 _ 7.26 × 10  2= 72 3 _6 _0 7.26 × 10  = 7,2

340 ÷ 10  1= 34.0 340 ÷ 10  2= 3.40 340 ÷ 10  3= 0.340

9.41 ÷ 10  1= 0.941 9.41 ÷ 10  2= 0.0941 9.41 ÷ 10  3= 0.00941

Learn 1

Complete this table. ×

10  1

10  2

10  3

10  4

370

3, 700

37, 000

370, 000

87 394

AF T

3,029 3.4

D R

5.934 10.001 2 Evaluate these products. a 194 × 10  2 =

b 0.1954 × 10  7 =

239.8456 × 10  1 =

d 43,985 × 10  3 =

e 682.4815 × 10  6 =

0.00384 × 10  4 =

3 Complete this table.

10  1

10  2

10  3

10  4

353,000

35,300

3,530

353

35.3

289,345 6,450 903 298.4 3.35 4 Evaluate these quotients.

a 9,754,200 ÷ 10  6 =

b 492.11 ÷ 10  2 =

825 ÷ 10  3 =

d 82,716 ÷ 10  7 =

e 28,000 ÷ 10  4 =

0.354 ÷ 10  1 =

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number properties The associative law of multiplication states that (a × b) × c = a × (b × c), where a , b and c are real numbers. This means that how we group multiplication doesn’t affect the result, so we can group numbers together which are easier to multiply. The commutative law of multiplication states that a × b = b × a, where a and b are real numbers. This means that the order we multiply in doesn’t affect the result. For example, 2 × 3 = 3 × 2 = 6.

25 × 4 × 9 = 25 × (4 × 9) = 25 × 36 = 900 25 × 4 × 9 = (25 × 4) × 9 = 100 × 9 = 900 7 × 13 = 7 × (10 + 3) = 7 × 10 + 7 × 3 = 70 + 21 = 91

The distributive law states that you can “distribute” or split up a multiplication.

Learn 1

Use the associative law to evaluate the following in two different ways. Put a star next to the working that you find easier for each calculation.

b (25 × 3) × 4= × ‾  ‾  ‾        = = ‾ ‾ 25 × (3 × 4)= × 4 × (8 × 5)= 4 × ‾  ‾     ‾      = = ‾ ‾ c (4 × 3) × 15= d (7 × 5) × 6= × × ‾  ‾ ‾  ‾         = = ‾ ‾ 4 × (3 × 15)= × 7 × (5 × 6)= × ‾  ‾ ‾  ‾         = = ‾ ‾ 2 We can combine the associative and commutative laws to make calculations easier. Identify and label the uses of the associative (A) and commutative (C) laws in this example. a (4 × 8) × 5= × 5

D R

AF T

5 × 14 × 2= 5 × (14 × 2) ‾ = 5 × (2 × 14) ‾       14  = (5 ×  2) × ‾ = 10 × 14 ‾ = 140 ‾ 3 Use the associative and commutative laws to evaluate the following. a 20 × 19 × 5

b 25 × 7 × 5 × 4

40 × 6 × 5 × 7

4 Calculate the following using the distributive law.

‾ ‾           = + ‾ ‾ = ‾

a 5 × 17= 5 × ( + )

37 × 4= × ( + ) ‾ ‾ ‾          = + ‾ ‾ = ‾

‾ ‾ ‾          = + ‾ ‾ = ‾

× ( + ) b 24 × 7=

‾ ‾ ‾          = + ‾ ‾ = ‾

× ( + ) d 8 × 35=

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Short multiplication The distributive law tells us that we can split up a number using place value to make it easier to multiply. The short multiplication algorithm uses the distributive law to multiply a whole number by a 1 - digit number.

4 × 37= 4 × (30 + 7)

= 4 × 30 + 4 × 7         = 120 + 28

3 ×

7 4

= 148

8 ones 2 tens

7 4

12 tens +2 tens

Explore 1

Evaluate these products using short multiplication. b 46 × 9

35 × 8

e 192 × 4

383 × 9

h 2,837 × 7

3,948 × 6

d 272 × 6

g 8,271 × 8

D R

AF T

a 27 × 7

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Long multiplication

Long multiplication is an algorithm used to multiply by a number with 37 × 14= 37 × 4 + 37 × 10 more than one digit. Digits are aligned       = 148  + 370 in place value columns so the addition algorithm can be used, and we include = 518 place-holding zeros to show when we are multiplying by a power of 10.

× 1

37 × 4

+ 3

37 × 10

Explore Use long multiplication to calculate the following. a

2 5 ×

3 4

2 7

3 8

7 5

2 6

2 Calculate the following using long multiplication.

3 2 4 ×                      2     4  + 0

8 4 ×

1 9

4 7

3 9

D R

2 7

AF T

2 8 3 ×                      1     3  + 0

7 3

9 2 8  ×                     3     7  + 0

3 Use long multiplication to calculate the following. a 31 × 18

b 93 × 23

86 × 46

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Solving multiplication problems Multiplication can be used to solve many problems.

Explore 1

Anahera is saving for a new game. She saves $ 15a week for 13weeks. Does she have enough for the $ 200 game? Show how you get the answer.

2 There are 12months in a year. b How many months old are you?

Counting a year as 365days, how many days old are you?

D R

and months since you were born.

AF T

a Calculate the number of years

3 The Pōhutu Geyser erupts around 17times a day. Based on that average, how many times does it erupt in: a August?

b June and July?

4 When using short or long multiplication, you can always multiply the larger number by the smaller number. Which

number property allow you to do this? 34

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Long division The terms dividend, divisor, quotient and remainder describe the values in a division calculation. The remainder is always less than the divisor. Remainders can also be written as fractions.

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50

place value columns

quotient

dividend

– 5

5 × 1 hundred

2 Identify the greatest value less than the dividend (number being divided) which is one of the multiples from step 1 multiplied by a power of 10.

– 3

5 × 6 tens

5 × 5 ones

3 Subtract the value from step 2 from the dividend using the subtraction algorithm.

–

2 827 ÷ 5 = 165 remainder 2 = 165 5 dividend

divisor

quotient

remainder

divisor

To use the long division algorithm, follow these steps. 1 Write out the first 10multiples of the divisor (number you are dividing by).

4 Record the number of copies of the divisor that have been subtracted above the dividend.

remainder

D R

Deepen

AF T

5 Repeat steps 2 to 4 using the value from Step 3 as the new dividend.

Set up and use the long division algorithm to calculate the following. a 55 ÷ 2

b 384 ÷ 7

Multiples of 2: 2, 4, ‾ , 20 ‾‾‾

993 ÷ 4

Multiples of 7: 7, 14, ‾ , 70 ‾‾‾

Multiples of 4: 4, 8, ‾ , 40 ‾‾‾

d 2,837 ÷ 3

Multiples of 3: 3, 6, ‾ , 30 ‾‾‾

2 In your maths book, use the long division algorithm to calculate the

following. Record your answers here. You can write “r” for remainder. a 232 ÷ 4 =

b 329 ÷ 7 =

365 ÷ 5 =

d 522 ÷ 9 =

e 349 ÷ 6 =

859 ÷ 7 =

g 837 ÷ 5 =

h 557 ÷ 8 =

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Short division Short division is an alternative method of writing down the steps in long division when the divisor is a 1 -digit number. The steps are the same, but how we record them changes. Remainders can also be written as decimals. You can convert a fraction remainder to a decimal, or add extra place value columns to the right of the decimal point when using short division.

1 5

2 6 5 remainder 2 = 165 — 5

8 32 27

5 4

8 32 27 20

Deepen 1

Set up and use short division to calculate the following. b 441 ÷ 9

296 ÷ 8

D R

AF T

a 290 ÷ 5

2 Set up and use short division to calculate the following. Write the remainders as fractions.

a 2,741 ÷ 6

b 5,111 ÷ 8

64,234 ÷ 7

d 39,267 ÷ 4

e 43,483 ÷ 6

59,339 ÷ 9

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 Set up and use short division to calculate the following. Write the remainders as decimals by adding extra place

value columns to the right of the decimal point. a 1,475 ÷ 2

b 285 ÷ 4

d 743 ÷ 8

5,834 ÷ 5

275,369 ÷ 5

D R

AF T

e 14,753 ÷ 2

4 In the real world, people have to decide what to do if there is a remainder. If two people were sharing three small

pizzas, they almost certainly wouldn’t leave the remainder as 3 ÷ 2 = 1 remainder 1. They would share it and say, “We get 1_   1 each”. It would be different if $ 1.50, or three tennis balls, had to be shared between them. 2 Match these divisions with the corresponding situations. A

2 5

4— 5

4 5

4 remainder 2 5

22 marbles and 5 people

II 22 apples and 5 people

III $22 and 5 people

5 When dividing an amount of money, why do we usually show the remainder as a decimal with two decimal places?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Dividing by multi-digit numbers using long division Long division can be used when a divisor has more than 1digit. The long division algorithm relies on place value columns being aligned in all steps. Because of this, most people leave out the zeros at the end of the numbers being subtracted. The zeros are implied by the space we leave for them and we bring down the digits in the dividend as we need them. Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240

Th H

24 5

– 4

–

Deepen 1

24 × 2 ones

– –

4 4

8 1

D R

24 × 1 ten

AF T

–

24 × 2 hundreds

Write out the first 10 multiples of these numbers. a 18: b 23: c

13:

2 Set up and use the long division algorithm to find the quotient and remainder of each of these. Hint: Use your lists

of multiples from question 1. a 378 ÷ 18

b 437 ÷ 23

596 ÷ 13

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 Write out the first 10 multiples of these numbers. a 21: b 18: c

17:

d 19: e 22: f

25:

4 Calculate the following using long division. b 2,981 ÷ 18

8,291 ÷ 17

82,938 ÷ 25

d 10,142 ÷ 19

D R

AF T

a 590 ÷ 21

e 59,212 ÷ 22

5 In your maths book, use the long division algorithm to calculate the following. Record your answers here, writing

any remainders as fractions. ‾ c 973 ÷ 34 =    ‾ e 9,836 ÷ 23 =    ‾ g 83,261 ÷ 29 =    ‾ a 735 ÷ 21 =   

‾ d 4,728 ÷ 12 =    ‾ f 9,108 ÷ 27 =    ‾ h 79,546 ÷ 32 =    ‾ b 846 ÷ 28 =   

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Solving division problems The division algorithm can be used to solve many problems.

Deepen 1

A conservation group plants 735native trees around a regional park. The trees are shared equally among 5 volunteer teams to plant. How many trees does each team plant?

2 A bakery in Queenstown makes 864cheese scones for

a winter festival. The scones are packed equally into boxes of 8. How many boxes can be filled?

3 Estimation and rounding are an important part of long division, especially when large numbers are involved. For

‾

b 624 ÷ 24 ≈

532 ÷ 19 ≈ ‾

D R

a 372 ÷ 31 ≈

AF T

example: How many copies of 29are there in 175? My estimate is 6because 6 × 30 = 180. Check by multiplying: 29 × 6 = 174. So, we were close: 175 ÷ 29 = 6 remainder 1. Estimate these quotients by the rounding values in the calculation. Then use long division to find out how close your estimate is to the actual value.

4 A conservation project collects 4,560native plant

seedlings. The seedlings are packed into trays holding 24plants each. How many trays are needed?

5 A honey producer has 3,782jars of mānuka honey.

The warehouse shelves hold 17jars each. How many full shelves can be stocked?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.6 Order of operations

Order of operations with whole numbers Sometimes the order that the operations in an expression should be performed is unclear. The order of operations is a convention followed so that we agree on the value of expressions.

2nd

3rd

4th

Grouping

Exponents

Evaluate any exponents. 6 × 2  3 = 6 × 8 = 48

Multiply (and) Divide

Working from left to right, evaluate any multiplication or division operations. 5 × 4 + 12 ÷ 3 = 20 + 12 ÷ 3 = 20 + 4 = 24

Add (and) Subtract

Working from left to right, evaluate any addition or subtraction operations. 12 − 5 + 6 = 7 + 6 = 13

D R

Learn 1

Don’t let your answer slither away, use GEMA!

AF T

1st

Any expressions grouped together in a bracket are evaluated first. 5 × (4 + 6) = 5 × 10 = 50

Follow the order of operations to evaluate these expressions.    ‾ = ‾

a 6 + 21 ÷ 3=

‾    = ‾

d (3 + 13) ÷ 4=

   ‾ = ‾

b 3 × 4 − 2=

   ‾ = ‾

e 8 + 3 × 15=

‾    = ‾

45 − 9 × 4=    ‾ = ‾

  2 × 2= 4    ‾ = ‾

(21 ÷ 3)  2= ‾    = ‾

g 4  2 ÷ 2=

h 3 × (15 − 3)=

k 16 − (8 ÷ 2) × 3 + 3

n 3 + (18 ÷ 2) − 3  2

o 6 × 5 − 4 × 3 ÷ 2

  2 − 3  2 7

m (2 + 8)  2 − 7  2

14 + 3  2 − 7

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

More complex expressions We can apply the order of operations to more complex expressions.

Learn Evaluate these expressions by considering the inside pair of brackets first. Show all your working. a 6 + [4  2 − (3 × 5 − 7)]

b 3 × [(20 − 4  2) ÷2] + 5

d 20 − [3  2 − 2 × (5 − 1)]

e (4 + 2) × [10 − (3 2 − 2)]

AF T

[18 − (6 + 3)]  2 − 7

D R

44 ÷ [5 + 2  3 − (6 − 4)]

The associative law of multiplication states that (a × b) × c = a × (b × c), where a, b and c are real numbers. This means that how we group multiplication doesn’t affect the result. a Calculate these division expressions. i 100 ÷ (10 ÷ 2)

ii (100 ÷ 10) ÷ 2

b Is division associative like multiplication? Explain.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Order of operations with integers The order of operations can be applied to any expression, including ones with negative integers. Remember to simplify operations with negative numbers!

(+)= + + −(+)= − +(−)= − −(−)= +

4 × (3 − (− 2)) 2 + (− 7)= 4 × (3 − (− 2))  2 + (− 7) Grouping – Brackets = 4 × 5  2+ (− 7) Exponents                        Multiplication = 4 × 25 + (−  7)    or Division = 100 + (− 7) Addition or Subtraction = 93

Explore 1

Evaluate these expressions following the order of operations. b 5   2 − (10 − 38) + 7

12 − 20 − 3 × 5 + 3 2

e − 14 + 5 × 6 + (− 7)

− 40 + (1 − (− 2))  2

h 8   2 ÷ (12 + (− 4)) − 6  2 ÷ 3

− 28 + 3  3 − (21 − 33)

d − 11 + 36 ÷ 4

g 3 × (30 − 18) − (20 − 25)

D R

AF T

a 14 + (− 9) − 6

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3  3 ÷ 9 + (− 15) − (3 × 2 − 11)

k (42 ÷ 6) × 8 − (− 27)

(7 × 9) ÷(− 1 + 4) − (− 5 − 4)

[( − 5 − (− 23)) ÷ 2 ] × [7 − (− 1 + 4)]

[6 × 2 + (−10)]× [−12 + 9 × 2]

2 Evaluate these expressions by considering the inside pair of brackets first. b 48 ÷ [10 − (2  2 − 6)]

D R

AF T

a − 9 + 6 × [9 − (5 − 8)]

d 54 ÷ [18 ÷ (6 − (− 3))] − (− 20)

e (5 − (− 3)) × [15 ÷ (10 − 7)]

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Applying the order of operations Remember to consider the order of operations when solving worded problems. Grouped operations (brackets) Exponents Multiply and Divide Add and Subtract

Deepen 1

One morning, the minimum temperature in Ōtautahi/Christchurch was − 2°C. By midday it rose by 7°C, and then suddenly cooled by 5°Cdue to a cool wind. In the mid-afternoon sun, the temperature then rose again by 3 °C. a Write and evaluate an expression to find the temperature after the mid-afternoon increase.

AF T

b Assuming the temperature cooled for the rest of the day, what was the maximum temperature in

D R

Ōtautahi/Christchurch that day?

2 Hera is trying to save money for a new pair of sports shoes worth $ 60.

•

She has $45in cash from her birthday.

•

She owes two friends $15 each.

•

She earns $12a week in pocket money in exchange for doing additional chores. She is owed three weeks’ worth of pocket money. Her parents subtract $ 4from each week as Hera did not complete all her chores.

Write and evaluate an expression to show Hera how much money she would have if she immediately repays her two friends. Can she afford the shoes at the moment?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.7 Fractions, decimals and percentages Equivalent fractions Equivalent fractions represent the same amount. For example, _   4  , _   6  , _   8   and _   20 all represent the same amount as _  2 . 10 15 20 50 5 To find an equivalent fraction, multiply or divide the numerator and denominator by the same amount. _   2  = _   2 × 10  = _   20  5 5 × 10 50

Learn 1

Find fractions on the fraction wall that are equivalent to the following. a

_   1  = 3

_   1  = 2

_   4  = 6

_   6   = 10

_   8   = 12

_   6  = 8

_   8   = 10

_   1  = 4

_   2  = 5

D R

AF T

One whole

2 Find the equivalent fractions with the given denominators. a _   2   = _    

b _   7   = _    

_   8   = _     90 15

d _   12  = _    

e _   14  = _    

_   11  = _     24 48

15 13

130

3 Find the equivalent fractions with the given numerators. a _   5   = _   20

b _   11  = _   22

_   9   = _   36 14

d _   23  = _   69

e _   35  = _   350

_   22  = _   176 61

12 54

4 Explain how to find infinitely many fractions equivalent to _  1 .

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Simplifying fractions The simplest form of a fraction is an equivalent fraction where the highest common factor (HCF) of the numerator and denominator is 1. To simplify a fraction, there are two methods. •

Divide the numerator and denominator by common factors until their HCF is 1 . 75 75 ÷ 5 _ _    =     30 30 ÷ 5 = _  15  6 What you do to the   15 _ =   ÷ 3  top, you must also do 6÷3 to the bottom! = _  5  2 Divide the numerator and denominator by their HCF. _   75 = _  75 ÷ 15  30 30 ÷ 15   = _  5  2

•

Learn a _   8   =

_   8  = 6

b _   2  =

_   36  = 24

g _   72  =

_   156  = 130

_   72   = 216

k _   204  =

e _   25  =

AF T

Use any method to simplify these fractions. 8

D R

144

d _   20  =

h _   20  =

_   120  = 72

2 Circle the fractions that are already in simplest form. Explain how you know they are in their simplest form.

_   8   _   2   _   3  _   2   _   4  _   9   _   5   7 7 11 11 12 10 12

3 The road distance from Auckland to Wellington is approximately 650 km. Lexiarna has travelled 210 kmof this

journey and Mandy says she has completed _  21 of the same journey. Show that they both have completed the same 65 amount of the road trip.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Converting between percentages and decimals To convert a decimal to a percentage, multiply the decimal by 100%. 0.46 = 0.46 × 100% = 46% To convert a percentage to a decimal, divide the percentage value by 100. 22% = 22 ÷ 100 = 0.22

Explore 1

Percentages and decimals are different ways of writing the same amount.

Convert these decimals to percentages. ‾ 0.9 = ‾ 0.8 = ‾ 1.01 = ‾ 8.375 = ‾ 20.505 = ‾

‾ d 0.12 = ‾ f 0.15 = ‾ h 3.725 = ‾ j 10.004 = ‾ l 100.667 = ‾

c e g i k

b 0.07 =

2 Convert these percentages to decimals.

c e g i k

‾ d 1% = ‾ f 90% = ‾ h 9,000% = ‾ j 2.5% = ‾ l 0.006% = ‾ b 64% =

D R

‾ 30% = ‾ 9% = ‾ 900% = ‾ 10.5% = ‾ 11.6% = ‾

a 39% =

AF T

a 0.45 =

3 Complete the conversions.

‾ b A population increased by a factor of 1.35. Write this amount as a percentage. ‾ c An event has a probability of 0.025. Express the probability as a percentage. ‾ d A jug is filled to 125% of the recommended amount. Write 125% as a decimal. ‾ a A shirt is discounted by 20%. Write 20% as a decimal.

4 Use fractions with a denominator of 100to explain why these conversion techniques work.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Converting between fractions and percentages To convert a fraction to a percentage, find an equivalent fraction with a denominator of 1 00, and then write the numerator with a percentage symbol (%). _   9   = _   45  = 45% 20 100 To convert a percentage to a fraction, write the percentage value as the numerator with a denominator of 1 00. Simplify the fraction if possible. 96% = _   96   = _   24  100 25

Explore 1

Convert these fractions to percentages. a _   1  =

b _   1  =

4 d _   6  = 5 13 _ g     = 20

5 9 e  _  = 10 h _   16  = 25

f i

_   2  = 5 11  _  = 10 _   43  = 25

2 Convert these percentages to fractions in their simplest form.

‾ 80% = ‾

b 40% = f

5% = ‾ g 125% = ‾

‾

d 60% =

‾

AF T

‾ e 12% = ‾ a 10% =

h 345% =

numerator.

D R

3 We can convert percentages with decimal values to fractions by finding an equivalent fraction with whole number

_× 10 _ 4.5 4.5% = _  4.5   = _     =   45   = _   9   100 100 × 10 1,000 200

Use this method to convert these percentages to simplified fractions. a 0.5%

b 2.5%

10.5%

d 20.2%

e 33.875%

1.325%

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Converting between fractions and decimals To convert a decimal to a fraction, put all the digits after the decimal point over the last place value. 0.307 = _   307   because 7 is in the thousandths ( _   1  ) column. 1,000 1,000 3   because 3 is in the hundredths _ 1 24.03 = 24 _ (  100 ) column. 100 There are two main ways to convert a fraction to a decimal. •

Find an equivalent fraction with a denominator of 100and then divide the numerator by 100. _   3  = _   60  = 60 ÷ 100 = 0.6 5 100

•

Divide the numerator by the denominator using short division. 0 5

Always simplify your fractions!

Explore Convert these decimals to fractions in their simplest form. Write any values greater than 1 as mixed numbers. ‾ e 0.75 = ‾ i 5.68 = ‾ a 0.1 =

‾ 0.78 = ‾ 91.03 = ‾

b 0.25 = f j

0.4 = ‾ g 1.3 = ‾ k 11.101 = ‾

‾

d 0.05 =

AF T

‾ 18.003 = ‾

h 8.82 = l

fraction.

D R

2 Use an example to explain how you know what to put as the denominator when converting from a decimal to a

3 Convert these fractions to decimals using short division. a _   1 

b _   3 

_   7  8

d _   1 

2  e  _

5   _ 6

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Fractions, decimals and percentages Given any fraction, decimal or percentage, we can convert to find the other two forms of the value. You can use the conversion methods that you find easiest.

Explore 1

Complete this table of conversions between fractions, decimals and percentages. Fraction

Decimal

Percentage

_3     4

0.75

75%

0.2 _   3   100 0.5

AF T

_   1  4 0.15

D R

40%

_   3  8

7.5% 110% 12   _ 5 2 We can represent the value of a fraction, decimal or percentage visually.

0.25

5 1 = 20 4

25%

Fill in these diagrams so they represent the same value. a

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

a 0.95

0.625

b 0.15

D R

AF T

3 Estimate the positions of the following decimals on these number lines by drawing an arrow to the correct place.

d 0.29 0

e 0.55

0.225

4 Estimate the positions of the following fractions on these number lines by drawing an arrow to the correct place. a _   1 

b _   1 

0 c

_   2  3 0

1 f

d _   1 

e _   8  0

_   5  6 0

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Comparing and ordering fractions, decimals and percentages Fractions, decimals and percentages can be compared and ordered. It’s easier to work with values that are all in the same format. For example, convert all values to decimals and then put them in ascending or descending order.

Ascending order means smallest to largest.

Explore 1

Write these values in ascending order. 1   0.25 21% a 2.5 2 _ 10

b 30% 0.31 3% _   1 

3 12 _ c 0.12 12.5%     1.2 1,000 d 5% 0.5 _   5   500% 1,000 2 Write these values in descending order.

7 20 b _   4  0.85 90% _   3  0.7 4 5 1 _ c 12% 0.2     75% 0.5 6 1 _ d 0.125     0.2 22% _   3   2 25 a 0.18 _   2  45% _     0.32

AF T

D R

3 Anika, Bo, Caleb and Danielle are sharing a cake. Anika eats _   1  , Bo eats 12%, Caleb eats 0.125and Danielle eats _   1  .

a Who eats more cake: Anika or Danielle? How can you tell without performing any conversions or finding

equivalent fractions?

b Convert all four values to the format of your choice.

Anika: Bo: Caleb: Danielle: c

Who eats the most cake?

d Put the four friends in increasing order according to how much cake they eat.

e Complete this diagram to show how much cake is eaten by each person.

Anika’s slice is shown. Do you think there is at least half of the cake leftover?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Terminating and non-terminating decimals A decimal with a finite number of non-zero decimal places is called a terminating decimal. Some decimals have an infinite (unending) number of non-zero decimal places. These are called non-terminating decimals.

Deepen 1

Complete this table by classifying the fractions as having a terminating or non-terminating decimal equivalent. _   1  _   1  _   2  _   3  _   5  _   5  _   7   _   9   _   4   _   7   _   11   7 4 6 11 2 3 8 20 30 100 25 Terminating decimal

Non-terminating decimal

D R

AF T

Working out space

2 Compare the denominators of the fractions in the two columns of the table in question 1. Can you use these to

identify whether the decimal form of a fraction is terminating or non-terminating?

3 Use your answer to question 2 to predict whether these fractions have decimal forms that are terminating or

non-terminating. Use short division to show whether your prediction is correct. a _   1  is a terminating/non-terminating decimal.

b _   3   is a terminating/non-terminating decimal.

_   5   is a terminating/non-terminating decimal. 80

d _   5  is a terminating/non-terminating decimal.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Non-terminating decimals All fractions can be converted to decimals. The result can be: •

a terminating decimal. These decimals have a finite number of non-zero decimal places. For example, _   5   = 0.3125. 16

•

a decimal with an infinite number of non-zero decimal places. These are called non-terminating decimals. There are two types of non-terminating decimals. ◦ Repeating: These non-terminating decimals have a repeating pattern after a finite number of decimal places that continues forever. We can use dots and lines to show the repeating pattern.

Can you spot the pattern in a repeating decimal?

For example, _   7  = 2.333…= 2.3˙  and 3 1 _    = 0.142857142857…= 0.‾ 142857. 7 ◦ Non-repeating: These non-terminating decimals do not have a repeating pattern. _

For example, √ 2  = 1.414213….

Deepen 1

Classify these decimals in the table.

Non-terminating and repeating

Non-terminating and non-repeating

D R

Terminating

AF T

1.666666… 7.89 0.11121314… 1.732051… 0.0625 3.141593… 0.333333… 2.718282… 1.00025 0.272727… 1.414214… 3.5

2 Rewrite the non-terminating and repeating decimals in question 1, using dots and lines to clearly show the pattern

of the repeating digits.

3 Use short division to find the decimal form of these fractions. If there is a repeating pattern, show it clearly using

a dot or line above the repeating pattern. a _   1 

b _   2 

_   4  7

4 In practice, we usually don’t need infinitely many decimal places of a number for a calculation or measurement.

Round the decimals you found in question 3 to four decimal places. a

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.8 Multiplying fractions

No fraction? No problem! Put the whole number over 1.

Multiplying whole numbers by fractions

To multiply a fraction by a whole number, multiply the numerator by the whole number and leave the denominator unchanged. _   4   × 14 = _   4 × 14   = _   56  = _   8  21 21 21 3 Cancelling means to simplify while multiplying, instead of as the final step. 1 7=_ _   4   × 14 = _   4 × 14   = _   4 × 2 ×    4 × 2 × 17   = _   8  21 21 3×7 3 3 × 7 

Learn 1

Calculate the following. 1  = a 5 ×  _ 2

b _   2  × 8 =

‾

2  = 7 ×  _ 3

d _   3  × 9 =

‾

2 Write each product as a single fraction and then divide by common factors to simplify.

= ‾ 2 = _ e 18 ×  _   3 = ‾

b _   1  × 12= _  

1 = _ 15 ×  _   3

AF T

d _   3  × 20= _  

= ‾ _   4  × 25= _   5

= ‾ 11 = _ g 8 ×  _   4

= ‾ h _   4  × 27= _   3

= ‾

D R

1 = _ a 30 ×  _  

3 Calculate the following using cancelling.

Hint: Write the numbers in both the numerator and denominator of the combined fraction as products of their prime factors. 5 = _ a 4 ×  _  4 × 6

2 ×    = ___________  2 ×   ×

b _   5  × 10= _   × 10

= _  

× 2 × 5         = ___________     ×

×    × × _____________        =        × = _  

= _  

2   × d 60 ×  _ = ___________

e _   3  × 120= ___________   ×

_   4  × 33= ___________   × 9

×    × × = ________________   ×

× × × ×         = ____________________        ×     × × = _  

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Multiplying whole numbers by mixed numbers To multiply a whole number by a mixed number, first convert the mixed number to an improper fraction. 1 4  = 3 × _ 22 = _ 1  3 × 2  _   22  = _   3 ×     3 × 22  = _   3  1× 22  = _   22  = 7 _ 9 9 9 3×3 3 3 3  × 3

Remember to cancel common factors between the numerator and denominator!

Learn 1

Calculate the following by first converting the mixed number to an improper fraction. 1 = 5 ×  _ a 5 × 2  _ 3

= _  

2 = 7 ×  _ b 7 × 3  _

= _  

3  × 12= _ 4  _   × 12 7

5  × 11= _ d 7  _   × 11 6

‾    = ‾ = ‾

5  × 8= b 5  _

‾    = ‾ = ‾

D R

2 Calculate the following. Write your answers as whole or mixed numbers.

5  × 6= d 2  _ 9

‾    = ‾ = ‾

AF T

3 = a 4 × 1  _

‾    = ‾ = ‾

1 = e 10 × 6  _ 4

1 = 9 × 4  _ 6 ‾    = ‾ = ‾ 2  × 15= 4  _ 9 ‾    = ‾ = ‾

3 The distributive law can be used to multiply a whole number by a mixed number. For example:

2  = 3 × 1 +  _ 2  3 × 1  _ ( 7 7) 2  = 3 × 1 + 3 ×  _ 7 6 _ = 3 +     7 6 _ = 3     7 Calculate the following using the distributive law. 1  a 5 × 2 _ 2

2  b 3 × 4 _ 9

3  8 × 5 _ 4

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Multiplying fractions by fractions To multiply a fraction by a fraction, multiply the numerators and the denominators, and then simplify the resulting fraction. 4  = _ _   1  ×  _   1 × 4  = _   4   3 5 3 × 5 15

1 3

Can you find any common factors to cancel?

4 5

4 15

Explore 1

Use these diagrams to find the product of the fractions. a =

1  ×  _ 1  = _  _   2 4

2  ×  _ 1  = _  _   3 2

AF T

D R

3  = _ 1  ×  _  _   3 4

2 Evaluate the following products of fractions.

3  = _ a _   1  ×  _  

5  = _ b _   2  ×  _  

3  = _ e _   7   ×  _

2  = _ _   4  ×  _   5 7

1  = _ d _   3  ×  _  

3  = _ g _   3  ×  _

1  = _ h _   5  ×  _

1  = _ _   2  ×  _   5 3

3 Evaluate the following products of fractions. Cancel any common factors so the resulting fraction is in its simplest

form. 4  a _   5  ×  _ 9 5

2  b _   6  ×  _

5  _   3   ×  _ 10 6

9   d _   2  ×  _

10  e _   4  ×  _

10  _   7   ×  _ 12 21

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

14  g _   3   ×  _ 35

40  h _   18  ×  _

72  _   24  ×  _ 45 48

4 Evaluate the following products of fractions. Cancel any common factors so the resulting fraction is in its simplest

form. 7 = a _   3  ×  _

2 = b _   9  ×  _

14 = e _   3  ×  _

9    ‾ = ‾

22 = _   5  ×  _ 9 15    ‾ = ‾

22 = g _   9   ×  _

12 = h _   6  ×  _

17 = _   3   ×  _ 4    34 ‾ = ‾

9 = _   5  ×  _ 3 4    ‾ = ‾

10 = k _   7  ×  _

3    ‾ = ‾

5    ‾ = ‾

11 = _   4  ×  _ 7 6    ‾ = ‾ 3    ‾ = ‾

7    ‾ = ‾

5 = d _   10  ×  _ 3

6    ‾ = ‾

5    ‾ = ‾

120 = _   11  ×  _ 99    8 ‾ = ‾

5 A kaitiakitanga (guardianship of the environment) mural is being painted at a school using green and blue paint.

2  green and _ The design is approximately  _   1  blue. 3 3

D R

AF T

a If the artist has brought a total of 12litres of paint to use, how much paint is there of each colour?

b After four hours, one quarter of the design is complete for each colour. How much of each colour of paint has

already been used on the mural?

In the end, the artist uses _   9  of the green paint and _  4 of the blue paint on the mural. How much paint of each 10 5 colour is used on the mural?

6 Circle the correct word or phrase to complete each statement.

Hint: Look through the products you have already found in this topic. a The product of two proper fractions is always/sometimes/never greater than 1. b The product of two improper fractions is always/sometimes/never greater than 1. c

The product of a proper fraction and an improper fraction is always/sometimes/never greater than 1.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Multiplying fractions by mixed numbers To multiply a fraction by a mixed number, convert the mixed number to an improper fraction and then multiply as fractions. 1 3 ×   7=_ 1  = _ _   9   × 2 _   9   × _   7  = _   9 × 7   = _   3  ×   21  10 3 10 3 10 × 3 10 3  1× 10

If the question uses mixed numbers, give your answer as a proper fraction or mixed number.

Deepen 1

Calculate these products by first converting the mixed numbers to improper fractions. 1  a _   3  × 2 _

1  b _   5  × 1 _

3  _   4  × 3 _ 4 9

2  d _   7   × 4 _

1  e _   2  × 5 _

1  _   3  × 6 _ 8 3

5  g _   5   × 2 _

1  _   8   × 4 _ 15 5

D R

AF T

1   a 1 _   1  × 1 _ 2

1  b 1 _   1   × 1 _

1   _ 2   2  × 3 _ 10 5

3 10

1  × 1 _  e 2  _

3 4

1  _ 3   2  × 2 _ 2 5

4  h _   7  × 3 _

2 Calculate the following products.

1  d 3 _     × 2 _

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.9 Multiplying decimals

Multiplying decimals and whole numbers using multiplication facts Multiplication facts can help us to multiply whole numbers and decimals. If we know that 6 × 7 = 42, then... 0.6 × 70= _   6   × 70 0.6 × 7= _   6   × 7 10 10 Knowing your 6 × 7 _ _ =     =  6 × 70   multiplication facts 10 10           42 420 ÷ 10 helps when multiplying = _     = _     10 10 ÷ 10 decimals. = 4.2 = 42

Learn 1

Use your knowledge of multiplication facts to calculate the following. ‾ 0.9 × 8= _     × 10 ‾

‾ 0.3 × 120= _     × 10 ‾

b 3 × 12 =

AF T

a 9 × 8 =

   = ___________   ×      10  

D R

= _     10

= ‾

0.9 × 80= _     × 10 ‾

   = ___________   ×      10   = _     10

= ‾

1.2 × 30= _     × 10 ‾

   = ___________   ×      10  

= _     10

‾ 2 Calculate the following.

= ‾

a 0.6 × 5 =

b 0.07 × 8 =

0.009 × 11 =

d 0.04 × 300 =

e 0.08 × 4,000 =

0.02 × 600 =

g 0.007 × 20,000 =

h 0.005 × 90 =

0.009 × 3 =

3 Use the given products to calculate the following. a 17 × 3 = 51

b 25 × 25 = 625

1.7 × 30 = 0.017 × 300 = c

36 × 5 = 180

3.6 × 500 = 0.036 × 50 =

2.5 × 250 = 0.25 × 2,500 = d 92 × 8 = 746

0.92 × 800 = 0.8 × 920 =

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Multiplying decimals and whole numbers by hand Short and long multiplication can be used to multiply decimals by whole numbers. Let’s calculate 1 4.7 × 3. Multiply the values as whole numbers, ignoring the decimal point. The number of decimal places in the result is the same as the total number of decimal places in the product of two numbers. 1 2 1 4 7               14.7 × 3 = 44.1 × 3 4 4 1 Use rounding and estimation to check the decimal point is in the correct place. For example, 14.7 × 3 ≈ 15 × 3 = 45, and 44.1is close to 45.

Learn Calculate the following by first multiplying the values as whole numbers. a 11.7 × 4 =

‾

b 24.8 × 3 =

‾

e 3.37 × 29 =

AF T

D R

d 40.8 × 15 =

‾

1.95 × 7 = ‾

42.575 × 14 = ‾

2 Check your results to question 1 by rounding the decimals to the nearest whole number. You may need to do your

working on another piece of paper. ‾ = ‾ × 7 c 1.95 × 7≈    ‾  = ‾

× 4   ≈ × 4 a 11.7  

   ‾  = ‾

× 29 e 3.37 × 29≈

   ‾  = ‾

b 24.8 × 3≈ × 3

   ‾  = ‾ 42.575 × 14≈ × 14 ‾     = ‾

× 15 d 40.8 × 15≈

14.7 × 3has 1decimal place, so there is 1decimal place in the product, 44.1.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Multiplying decimals by decimals Short and long multiplication can be used to multiply decimals by decimals. Suppose we want to multiply 2 .95by 1 .5. •

Step 1: Round and estimate: 2.95 × 1.5 ≈ 3 × 2 = 6. So, if the answer is not close to 6, something has gone wrong!

•

Step 2: Count the number of decimal places: In 2.95 × 1.5, there are 3 decimal places.

•

Step 3: Ignore the decimal point and multiply the values as whole numbers.

•

Step 4: Put in the decimal point so that there are the same number of decimal places as at the beginning.

•

Step 5: CHECK! Is the answer close to the estimate? Yes! (Phew!)

2 9 5 × 1 5       1    4    7  5 + 2 9 5 0 4 4 2 5 2.95 × 1.5 = 4.425

Explore 1

Calculate the following. Estimate the product by rounding the decimals to the nearest whole number. Use your estimate to check the final result. ‾ Estimate: × = ‾ ‾ ‾

‾ Estimate: × = ‾ ‾ ‾

b 1.8 × 4.2 =

D R

AF T

a 3.4 × 2.5 =

0.7 × 6.8 = ‾ Estimate: × = ‾ ‾ ‾

‾ Estimate: × = ‾ ‾ ‾

e 3.75 × 5.1 =

‾ Estimate: × = ‾ ‾ ‾

d 1.82 × 2.6 =

3.87 × 6.2 = ‾ Estimate: × = ‾ ‾ ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2 Moana is trying to calculate 3.72 × 1.64. She has correctly calculated that 372 × 164 = 61,008.

Moana then says that 3.72 × 1.64 = 61.008. 4 1 2 3 7 2 × 1 6 4             1    4  8    8 2 2 3 2 0 + 3 7 2 0 0 6 1 0 0 8 a Estimate 3.72 × 1.64by rounding the decimals to the nearest whole number. b What has Moana done incorrectly?

Explain to Moana how to use estimation to make sure the decimal point is in the right place when multiplying decimals.

D R

AF T

4 Koa says that 7.25 × 9.8 = 71.05. Maia says that Koa must be incorrect because there are 3decimal places in

7.25 × 9.8and only 2decimal places in 71.05. a Show that Koa is correct.

b Why are there only 2decimal places in the result?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Solving problems with multiplying decimals Many problems can be solved by multiplying whole numbers and decimals. Think about which technique you should use.

Remember to estimate the result to check your answer.

•

Is there a multiplication fact that can help you?

•

Is the decimal close enough to a whole number that you can use a mental technique?

•

Is long or short multiplication the best option?

Deepen A student buys 3sketchbooks for a visual arts project. Each sketchbook costs $ 5.99. How much do the sketchbooks cost altogether?

2 A whānau is filling an 8 Ldrink

dispenser for a shared lunch. They buy 7bottles of juice, with 1.2 L in each bottle. Will the dispenser be large enough, or will there be leftover juice?

3 A supermarket is selling apples for

$4.45per kilogram. Meilin buys 2.6 kgof apples. If she pays the exact amount by card, how much does she pay?

D R

AF T

4 New carpet is being laid in a classroom breakout space which sells for $ 29.50per square metre. The space is 8.4 m 2. a How much does the new carpet

in the breakout space cost?

b It takes two workers 7.5 hours

to lay the carpet. They each earn $ 32.40per hour worked. How much do the two workers earn in total to lay the carpet?

What is the total cost of replacing the carpet in the breakout space?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.10 Using fractions

Finding a fraction of a whole number Finding a fraction of a whole number has the same result as multiplying the fraction and the whole number. 1  and _ 1 . For example, _   1  of 5 is _   5  = 2 _   1  × 5 = _   5  = 2 _ 2 2 2 2 2 2

Learn 1

Find _   1  of these numbers. 2 a 40:

b 54:

15:

1 _ _ 15 ×  2 =   2 

d 21:

2 Calculate the following.

1 = _ a 40 ×  _     2

= ‾

1 = _ b 54 ×  _     2

= ‾

1 = _ d 21 ×  _    

= ‾

AF T

3 Compare your answers for questions 1 and 2. What do you notice?

a _   1  of 20

D R

4 Calculate the following. Remember to simplify any fractions. Write any improper fraction results as mixed numbers. b _   4  of 11

_   2  of 32 3

d _   3   of 18

e _   2  of 42

_   3   of 21 14

g _   7  of 9

h _   13  of 5

_   21  of 8 12

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

The unitary method In the unitary method, we find the value of one unit and then multiply to find the number of units required. Let’s use the unitary method to find _   3   of 12. 12 10 6 1 12 _ _ _ First, find    of 1 2by calculating 12 ÷ 10 =     =    . 10 10 5 6 6 6 6 6 6 6 6 6 6 — — — — — — — — — — 3 6 18 3 5 5 5 5 5 5 5 5 5 5 _ _ _ _ Then, multiply by 3 to find     of 12: 3 ×     =     = 3   . 10 5 5 5

Learn 1

Use each diagram to find the fraction of the whole number. a _   5  of 20

7 4

b  _  of 42

20 ÷ 6 = = ‾ ‾

= 42 ÷ 4 = ‾ ‾

_   7 of 42 = = 4 ‾ ‾

AF T

_   5 of 20 = 6 ‾

Divide by the denominator, then multiply by the numerator.

2 Use the unitary method to calculate the following. a _   4  of 15

D R

9 15 ÷ 9 = , so _   1  of 15 is . 9 ‾ 4 _     of 15 is . 9

b _   5  of 33

6 33 ÷ 6 = , so _   1  of 33 is . 6 ‾ 5 _     of 33 is . 6

_   7  of 36 8 36 ÷ 8 = , so _   1  of 36 is . 8 ‾ 7 _     of 36 is . 8

d _   10  of 12

e _   7  of 18

3 18 ÷ 3 = , so _   1  of 18 is . 3 ‾ 7 _     of 18 is . 3

9 12 ÷ 9 = , so _   1  of 12 is . 9 ‾ 10 _     of 12 is . 9 _   7  of 40 4 40 ÷ 4 =

, so _   1  of 40 is . 4 ‾ 7 _     of 40 is . 4

3 How does the unitary method relate to your working in question 2?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Finding the whole amount given a fraction Sometimes you know a fraction of an amount and you need to work out what the whole amount is. We can use the unitary method to find the value of one unit and then multiply to find the whole amount. For example, suppose _  3 of an amount is 8and we need to find the whole amount. 5 There are 3fifths in 8, so 1fifth is 8 ÷ 3 = _  8 . 3 8 1 . _   40  = 13 _ One whole is 5fifths, so multiply by 5:     × 5 = _ 3 3 3 1 _ The whole amount (5fifths) is 13   . 3

8 8 8 8 8 8 — — — — — 3 3 3 3 3 40 1 — = 13— 3 3

Explore 1

Find the whole amount using the diagrams. 3 4

6 11

a _    of a number is 9

b _    of a number is 30

9 ÷ 3 = ‾

30 ÷ = ‾ ‾ 30

_ c   2 of a number is 15

The whole amount is .

D R

The whole amount is .

AF T

4 9

_ d    of a number is 18

15 ÷ = ‾ ‾

18 ÷ = ‾ ‾ 18

The whole amount is . 7 6

9 4

_ e    of a number is 30

_ f    of a number is 60

30 ÷ = ‾ ‾

60 ÷ = ‾ ‾ 60

The whole amount is . 68

The whole amount is .

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2 Find the whole amount if:

3 a  _ of a number is 17

4 b  _ of a number is 14

_   5 of a number is 11 6

d _   7 of a number is 24

e _   11 of a number is 25

5 8

_ 2   1 of a number is 15. 3

D R

decimal.

AF T

3 Gabby walks 4 km, which is _    of her usual distance. How far does Gabby usually walk? Give your answer as a

3 7

4 A school is raising money for a local charity. They have already raised $150 which is __     of their goal.

What is their goal amount?

5 11

5 A puzzle contains 45 completed pieces, which is ___    of the whole puzzle. How many pieces does the puzzle contain?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Finding the whole amount using reciprocals The reciprocal of a fraction swaps the numerator and denominator. For example, the reciprocal of _  4  is _   5  and the 5 4 6 1 _ _ reciprocal of     is    = 6. If we know a fraction of an amount is a given number, then we can multiply the number by the 6 1 reciprocal of the fraction to find the whole amount.

Deepen 1

Use the reciprocal method to find the whole amount. b _   10 of a number is 25

a _   2 of a number is 20

The reciprocal of _  2  is _   . 9

The reciprocal of _  10  is _   . 7

×   20 ×  _ = ___________

×   25 ×  _ = ___________

= =_   ‾ _   5 of a number is 16 6 The reciprocal of _  5  is _   . 6

d _   21 of a number is 12

The reciprocal of _  21  is _   . 11

AF T

=_   = ‾

× 12 ×  _ = ___________  

D R

× 16 ×  _ = ___________  

=_   = ‾

2 Sue is writing a book which is expected to be 240pages. a After one month, she has written _  2 of the book. How many pages has she written? Give your answer as a

simplified fraction.

b After two months, she realises that she has written _  11 of the expected number of pages. How many pages has

10 Sue written? Give your answer as a mixed number.

If the publisher insists on the book being 240pages, how many pages does Sue need to remove from what she has written?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.11 Using percentages Percentages of whole numbers

To find a percentage of a whole number, convert the percentage to a fraction and then multiply by the whole number. 30% of 40= 3% × 40 = _   30   × 40 100 × 40 =_  30       100 1,200  12 = _______     100  1 = 12

Use cancelling to simplify the fractions in your working.

Learn 1

Find the percentages of the whole numbers. 40% of 50

e 25% of 160

45% of 90

g 35% of 800

h 65% of 500

72% of 40

k 82% of 50

63% of 900

d 70% of 300

17% of 1,200

AF T

b 20% of 60

D R

a 80% of 10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Finding percentages using the unitary method In the unitary method, we find the value of one unit and then multiply to find the desired value. When working with percentages, we usually consider a unit to be either 10%or 1%. For example, to find 30%of 40, first find 10%of 40and then multiply by 3to get 30%. 10% of 40= 40 ÷ 10 =  4     30% of 40= 3 × 4 = 12

Learn 1

Use the unitary method to find the following percentages of whole numbers. a 90% of 80

10 % of 80= ‾ 90    9   % of 80   = × ‾ = ‾ d 4% of 500

b 40% of 530

10 % of 530= ‾ 40     4   % of 530   = × ‾ = ‾ e 22% of 4,400

80% of 340

10 % of 340= ‾ 80     8   % of 340   = × ‾ = ‾ 61% of 12,200

D R

AF T

1 % of 4,400= 1 % of 12,200= 1 % of 500= ‾ ‾ ‾ 4    4          22      61   % of 500   = × 22 % of 4,400   = × 61 % of 12,200   = × ‾ ‾ ‾ = = = ‾ ‾ ‾ 2 If you know 10%of a value, describe how you could use it to find 15%of the value.

15% of a 20 kg bag of birdseed was used. How much birdseed was used?

4 A runner has completed 60% of a 15 km race. How far have they run?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Percentage equivalence As percentages are fractions in disguise, they inherit the multiplication properties of numbers. This means we can use commutativity to make calculations easier. For example, 40%of 25is equal to 25%of 40. To find 25%of a number, you can divide the number by 4. 40% of 25= _   40   × 25 100 _ =  40 × 25   Commutativity means we 100 × 40 can multiply in any order!    =_  25       100 2 × 4 = 4 × 2 = 8 = _   25   × 40 100 = 25% of 40 = 40 ÷ 4 = 10 We can abbreviate the calculation with the following. 40% of 25= 25% of 40    = 40  ÷ 4 = 10

Explore 1

Calculate the following. b 40% of 45

D R

AF T

a 45% of 40

2 Use percentage equivalence to calculate the following. a 36% of 50

Equivalent calculation:

48% of 75 Equivalent calculation:

b 72% of 25

Equivalent calculation:

d 65% of 80

Equivalent calculation:

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Finding the whole amount given a percentage Sometimes you know a percentage of an amount and you need to work out what the whole amount is. To use the unitary method, we find the value of one unit and then multiply to find the desired value. When working with percentages, we usually consider a unit to be either 10%or 1 %, but other values are possible. For example, if 60%of a number is 42, first find 10%of the number, and then multiply by 10to get 100% (i.e. the whole amount). 60% of a number is 42 10% of the number = 42 ÷ 6 = 7 100% of the number= 7 × 10 = 70 Whole amount= 70

How does this relate to finding the whole amount given a fraction?

Deepen 1

Fill in the blanks. ‾ If 1%of a number is 7, then the whole amount is:

‾ If 5%of a number is 12, then the whole amount is:

b 5% × = 100%

7 × = . ‾ ‾ c 10% × = 100% ‾ If 10%of a number is 20, then the whole amount is:

12 × = . ‾ ‾ d 20% × = 100% ‾ If 20%of a number is 36, then the whole amount is:

AF T

a 1% × = 100%

60 × = . ‾ ‾

D R

20 × = . ‾ ‾ e 25% × = 100% ‾ If 25%of a number is 60, then the whole amount is:

36 × = . ‾ ‾ 50% × = 100% ‾ If 50%of a number is 102, then the whole amount is: 102 × = . ‾ ‾

2 Use the unitary method to find the whole amounts given these percentages.

a 70%of a number is 210

b 15%of a number is 450

22%of a number is 66

d 28%of a number is 560

e 120%of a number is 720

150%of a number is 900

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 The whole amount is not always a whole number. In your answers, write the whole amount as a fraction in its

simplest form. a 40%of a number is 35

b 60%of a number is 46

d 75%of a number is 100

35%of a number is 90

105%of a number is 90

D R

AF T

e 82%of a number is 32

4 Suppose 48%of a number is 80. Natalie’s working is shown.

8% of the number is 80 4 80% of the number is 48           20% of the number is 48 ÷ 4 = 12 The whole amount is 12 × 5 = 60 a Explain what Natalie has done incorrectly.

b Show Natalie how to correctly find the whole amount.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Number structures and operations

Topic 1.12 Ratios Ratios

A ratio is a mathematical way of describing proportion. We can describe this diagram as having 3red parts to 5blue parts. This can be written as part-to-part ratio 3 : 5.

We say the ratio is 3red parts to 5blue parts, so we write it as 3 : 5, not 5 : 3.

We could also say that there are 3red parts in a total of 8parts. This can be written as a part-to-whole ratio 3 : 8.

Learn 1

Write the ratios for these diagrams. a

Red-to-blue:

Red-to-whole:

Blue-to-whole:

AF T

Red-to-blue:

D R

Red-to-blue:

Red-to-whole:

Blue-to-whole:

2 Write the ratios for each situation. a In a bag, there are 12green marbles and 13yellow

b On a bookshelf, there are 9non-fiction books

marbles.

and 22fiction books.

Green-to-yellow:

Non-fiction-to-fiction:

Green-to-whole:

Non-fiction-to-whole:

Yellow-to-whole:

Fiction-to-whole:

3 In a bag of marbles, there are 15blue marbles and 7red marbles. Sam says that the ratio of blue to red marbles

is 7 : 15. Explain why Sam is incorrect and write the correct ratio.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Fractions, percentages and ratios Fractions, percentages and ratios can all be used to describe proportion. Fractions

Percentages

Ratios

_   3  red 8 5 _     blue 8

37.5% red

Part-to-part: 3 : 5

62.5% blue

Part-to-whole: 3 : 8 or 5 : 8

Learn 1

Describe the proportions of colours in these diagrams using fractions, percentages and ratios. a

Fractions

Percentages

_   red  _ blue

% red ‾ % blue ‾

Fractions

Percentages

_   red  _ blue

% red ‾ % blue ‾

Ratios Part-to-part: or Part-to-whole:

D R

AF T

Ratios Part-to-part: or Part-to-whole:

2 Part-to-whole ratios can be thought of as a different way of writing proportion as a fraction. This means we can

simplify ratios in the same way that we simplify fractions – by cancelling common factors until the highest common factor (HCF) is 1. a

Fraction (red):  _  = _     30 10

Fraction (red):  _  = _   28

              Fraction (blue):  _  = _     30 10 Red-to-whole: : 30 = : 10 ‾ ‾ Blue-to-whole: : 30 = : 10 ‾ ‾

               Fraction (blue):  _  = _   28

Red-to-whole: : 28 = : ‾ ‾‾ Blue-to-whole: : 28 = : ‾ ‾‾ 3 We can also simplify part-to-part ratios. Write part-to-part ratios for the diagrams in question 2 and then simplify them by cancelling the HCF of the parts. a

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Working with part-to-part ratios Given a quantity and a part-to-part ratio, we can divide the quantity into parts to match the ratio. Suppose you want to divide 30into the part-to-part ratio 2 : 3.

1 Identify the total number of parts in the ratio: 2 + 3 = 5.

2 Divide the quantity by the total number of parts: 30 ÷ 5 = 6.

6 12

3 Multiply the result from step 2 by the value of each part in the ratio: 6 × 2 = 12and 6 × 3 = 18. The part-to-part ratio 2 : 3divides 30into 12and 18.

Explore 1

Use the diagrams to divide the quantities into the given part-to-part ratios. a Divide 45into the ratio 5 : 4

b Divide 90into the ratio 1 : 2

AF T

and

Divide 60into the ratio 3 : 7 60

D R

and

d Divide 28into the ratio 2 : 5

and

e Divide 40into the ratio 3 : 5

and

Divide 80into the ratio 1 : 4 80

and

Check your answer by adding the divided values together. 12 + 18 = 30

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2 Divide the quantities into the given part-to-part ratios. a Divide 50into the ratio 3 : 2

b Divide 120 into the ratio 5 : 7

d Divide 550into the ratio 5 : 6

Divide 300 into the ratio 4 :1

3 Luke and Nikau are going on a road trip together from Auckland to Wellington. a There is approximately 9hours of driving time. They will divide this into the ratio 7 : 5, with Luke driving more

AF T

than Nikau. How many minutes are Luke and Nikau each expecting to drive on the road trip?

D R

b They budget $630for their trip, which they will split 4 : 5, with Nikau paying more to make up for driving less.

How much are they each expecting to pay for the trip?

Luke starts to feel unwell and is not able to drive as much as planned. In the end, the driving is split 2 : 3, with Nikau driving more than Luke. Assuming there was exactly 9hours of driving time, how much longer did Nikau drive than he was expecting?

d They decide to split the costs 3 : 2, with Luke paying more. They ended up spending $ 510on the road trip. How

much more or less did Luke pay than he was expecting?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Working with part-to-whole ratios Given a quantity and a part-to-whole ratio, we can divide the quantity into parts to match the ratio. For example, let's look at the part-to-whole ratio 2 : 3 and the quantity 30.

1 Identify the total number of parts in the ratio: 3.

2 Divide the quantity by the total number of parts: 30 ÷ 3 = 10.

10 20

10 10

3 Multiply the result from step 2 by the part given in the ratio: 10 × 2 = 20. 4 Identify the unknown part in the ratio: 3 − 2 = 1. 5 Multiply the result from step 2 by the unknown part in the ratio: 10 × 1 = 10. The part-to-whole ratio 2 : 3divides 30into 20and 10.

Deepen 1

Use the diagrams to divide the quantities into the given part-to-whole ratios.

and Divide 100into the ratio 2 : 5 100

and

D R

b Divide 40into the ratio 7 : 10

AF T

a Divide 60into the ratio 1 : 3

and d Divide 160into the ratio 3 : 4

160

and

2 Write the part-to-part ratios which divide the quantities in question 1 in the same way as the part-to-whole ratios. a b c d 3 The quantity 416is to be divided into the part-to-whole ratio 5 : 8. Dominic’s working is shown.

416 ÷ 13= 32 5    × 32= 160 8 × 32= 256 Explain what Dominic has done incorrectly and show him how to divide the quantity correctly.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Number | Financial mathematics

Topic 1.13 Money Discounts

A discount is when a seller reduces the price of an item, usually by a given percentage.

Learn 1

Calculate the reduced price after the discount. a

Price: $ 330 Discount: 15% Reduced price:

Price: $ 80 Discount: 20% Reduced price:

Price: $ 110 Discount: 10% Reduced price: g

AF T

Price: $ 150 Discount: 8% Reduced price:

Price: $ 190 Discount: 28% Reduced price: j

Price: $ 700 Discount: 12% Reduced price: i

Price: $ 4,895 Discount: 60% Reduced price: k

Price: $ 40 Discount: 45% Reduced price:

Price: $ 158 Discount: 35% Reduced price:

D R

Price: $ 350 Discount: 18% Reduced price: l

Price: $ 280 Discount: 22% Reduced price:

Price: $ 2,995 Discount: 32% Reduced price:

2 Calculate the full price. a Discount: 15%

b Discount: 32%

Discount amount: $18

Reduced price: $170

Full price:

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Creating a budget A budget is a finance plan about income and expenditure. Income is the money someone has or expects to get. Expenditure is how they want or need to spend the money.

You can’t spend more money than you have available!

Learn 1

Maia and Ali are on holiday and have $ 25each to spend one morning. This expenses table shows how Ali spent his money. Ali’s holiday expenses Expenses

Amount

Balance

−

$25.00

Opening balance

− $10.75

Fun palace

− $14.15

Miserable time. Maia gave me $ 5.

+ $5.00

Novelty store

− $5.10

Closing balance

−

a Complete Ali’s expenses table.

AF T

Exploring and snacks

D R

b Ali’s balance is negative. Explain what this means.

2 Imagine that you could have helped Ali prepare a budget

for the morning. How might the budget have looked so that he had money left for the afternoon? Include up to five expenses. 3 Give Ali some advice! What should he do when preparing

Ali’s holiday budget Income Money from mum

$25.00

Expenditure

a budget?

Total expenditure Money for the afternoon

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Comparing finance plans Finance plans help us to consider how we spend or save money. They can help to make sure you have enough money for essentials, like housing and food, and for larger spends, like a holiday or new car.

Explore 1

Harper is 19years old. She lives with her family and gives them money each month to cover some of her expenses. She has a part-time job and also babysits one evening per month. This is her budget for a month. Harper’s monthly budget

a Fill in Harper’s total monthly income and expenditure.

Income

b A balanced budget is one where the expenditure is less

Part-time job

$1,039.95

Babysitting

$185

than or equal to the income. Is Harper’s budget balanced? Explain.

Total income

Food

$400

Toiletries & clothes

$115

Going out

$250

Mobile phone

$29.95 $330

Car Total expenditure

AF T

Expenditure

D R

2 Many people try to budget their income and expenditure to have money leftover to put into their savings.

This money could be put towards a large purchase or used as an emergency fund. a Show that Harper can save $ 100each month.

b Suggest one way Harper could increase the amount she saves each month.

3 When you put money into a savings account at a bank, the money earns interest. This is given by the bank as a

percentage of the amount of money in the account. a How much would Harper save in one year if she keeps the $ 100leftover each month at home in a savings jar?

b Complete this table to calculate how much Harper could save if she deposits her $ 100into a savings account

earning 1%interest paid at the end of each month. Give amounts correct to the nearest cent. Month

Amount (+$100)

Interest

Amount (+Interest)

Month

$100

$101

$201

$2.01

Amount (+$100)

Interest

Amount (+Interest)

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Explain to Harper the benefit of putting her leftover money into a savings account.

4 Here are the details for two mobile phone plans.

Ignite mobile phone plan

Thins mobile phone plan

$43

Price per month

Price per year

$439

Inclusions

Unlimited calls and texts

20 GBdata per month

120 GBdata per year

a Calculate the cost per year of the Ignite mobile phone plan.

b Which mobile phone plan is cheaper per year? c

Why might someone choose the more expensive mobile phone plan?

AF T

5 Kai has just started a new job and wants to create a budget to keep track of his finances. He will take home

$1,200per week. His estimated weekly expenses are:

D R

• $630on rent • $225on food

• $120on entertainment • $80on personal items.

Each month he also spends $ 60on his mobile phone plan and $ 250on household bills. a Complete this annual budget for Kai.

When working with finance plans, we use 12months or 52weeks per year.

Kai’s annual budget Income

Amount per year

Salary Total income Expenditure

Amount per year

b Is Kai’s budget balanced? How much can

he afford to save each year?

Rent Food Entertainment Personal items Mobile phone plan Household bills Total expenditure

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Buy now, pay later services A buy now, pay later service allows you to buy a product or service and delay the payment. The consumer pays 25%of the cost at the point of purchase, and then makes three additional payments of 2 5%to make up the full amount. These payments are usually made fortnightly. No interest is charged by buy now, pay later services but there are often additional fees for things like late payments, keeping an account or processing a payment.

Deepen 1

Sally buys a new stand mixer for $ 600using a buy now, pay later service. She pays 25%of the cost in each instalment. a How much does Sally pay in each of the four instalments for the stand mixer?

b Sally pays a $ 5account-establishment fee, plus one payment of an $ 8monthly account-keeping fee. She makes

all payments on time, so there are no late fees. How much does Sally pay in total for the stand mixer?

AF T

2 Alanna buys a new laptop for $ 2,200using a buy now, pay later service. She pays 25%of the cost in each instalment. a How much does Alanna pay in each of the four instalments for the laptop?

D R

b Alanna pays a $ 12account-establishment fee. She makes all payments on time, so there are no late fees. How

much does Alanna pay in total for the laptop?

3 Brian buys an exciting new board game for $ 250using a buy now, pay later service. He expects to pay 25%of the

cost in each instalment. Here is a schedule showing the amount Brian actually paid for each instalment. Payment schedule Payment

Amount

1 (on time)

$62.50

2 (late)

$72.50

3 (late)

$72.50

4 (on time)

$62.50

Why might someone choose to use a buy now, pay later service?

Other fees: Account-keeping fees (total)

$12

a How much is the late payment fee when using this service? b How much did Brian spend in total on the board game?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Algebra | Equations and relationships

Topic 2.1 Simplifying expressions Multiplying terms Any two terms can be multiplied together. We usually remove multiplication signs when working with variables, and often choose to write variables in alphabetical order (which we can do because multiplication is commutative). Doing this makes terms easier to read and compare. As with numbers, exponents show repeated multiplication. 8 × a= 8a 7 × 3b= 7 × 3 × b = 21b        c × c= c   2 12g × 4f  2= 12 × 4 × f  2 × g = 48f  2g

The commutative law applies to both numbers and variables.

Learn 1

Simplify these expressions. ‾

‾

a 2 × a =

b 4 × b =

− 6 × c = ‾

‾

d − 10 × d =

AF T

2 Simplify the following by multiplying the constant and coefficients together.

‾ d 6 × 2r = ‾

‾ e 8 × 3s = ‾

b 5 × 4g =

D R

a 4 × 2b =

c f

3 × 9h = ‾ 9 × 2x = ‾

3 Simplify these expressions using an exponent.

‾ c 6 × f × f × f × f = ‾ a 5 × d × d =

4 Simplify these expressions. a 8b × 7a= 8 × 7 × a × b

   =    ‾

‾    =    ‾    g 9b × 3q= ‾    =    ‾    j 12k × 7s=    ‾ =    ‾ 5 Simplify these expressions.    d 11p × 4q=

‾    =    ‾

a 7r × 7r × p=   

   ‾ =    ‾

d 8l × 4t × 8l=   

‾

b − 3 × y × y × y =

‾

d − 15 × g × g × g × g × g =

‾    =    ‾    e 7g × 5= ‾    =    ‾    h 4r × 6h=    ‾ =    ‾    k 6l × 5p= ‾    =    ‾    b 5x × 6y=

‾    =    ‾

b 3g × 3a × 9g=   

‾    =    ‾

   e 10m × 2x × 10x=

9w × 6u=       ‾ =    ‾ 12w × 11j=    ‾    =    ‾ 11f × 2x=    ‾    =    ‾ 3v × 10c=       ‾ =    ‾ 11s × 11s × 5s=       ‾ =    ‾

3b × 3q × 11b=    ‾    =    ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Like and unlike terms Like terms have exactly the same variables (letters). Term

Like terms

Unlike terms

− 3a, a , 7a

3b, 6c, d , 5

− 3b 

b  , 12b 

− 2b, 5b   3

2ab

4ab, − ab, b a

6a, − b, a   2 b, 4ab  2

Explore Are the following terms like or unlike? If they are like terms, give two more examples. Terms

2a, − 5a

✓

Unlike

6b, 6c

Examples

Terms

a, − 3a

5  a  2b, 2ab −  _ 6

4x   2, 4x

− 4c  2d  3, _   9   c  2d  3 10

_   1  t  3, 9t   3 2

_   5 mn  2, − 3 m  2n  2 6

− 2z, _   z   10

Examples

4xyz, − xyz

D R

− c, c   3

4 xy, 4xy −  _ 5

Unlike

8xz, _   5 xy 4

✓

_   2 ab, − ac 3

AF T

_   2  a  2bc  2, 11abc  2 7 − ijk  2, 8i   2j  2k _   1  fgh, _   5  fgh 2 8

s t  2, s   2t

a   2bc, 8 a  2cd

_   3  fg, _   2  fg 4 5

p   3q  3r  3, _   4  pqr  3 3

10pq  2, 3pq  2

12lmn, _   11  nml 12

2 Circle the like terms in each list.

_   10 jk − i 9j _   2 ik 9 3 3 4 c 12s − 3rt  _  − 9s _   12 rst 3st 13 5 xy _ e     7x − xyz − xy _   4 xy  2 7xy 2 3 4 1 g  _ ab  2 ab b  2c  _ ab  2 3ab  2c − 5ab  2 9 2 3 1 de 8d  2 _ 2 3 i 5    de  − 2e  6de  2 −  _ 4 3 a − j

8de _   8 d − de 10cde _   1 de 4 5 7 3 d 6p 8  _ pq 5 − pqr −  _  6 4 9 5 _ _ 3 2 2 f − 7     m      n  5mn  − 2m  3 _   3 mnp 7 4 10 6 2 h 10xy  2z  _  y  2z − 9xy  2z  _ xyz 4 x  2y  2z  2 3xz 5 5 7 7 _ _ 2 2 2 2 j     − 5a  b     c 2b  a  − 6ab _   1 abc  2 4 4 2 b − 1

3 Are all constants like terms? Explain.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Remember, a term includes the sign to the left.

Collecting like terms When simplifying expressions with more than one variable, identify like terms and combine them by reading from left to right. This process is known as “collecting like terms”. − 6z + 4xy + 2x − xy + 5z= 2x + (− 6z + 5z) + (4xy − xy)     = 2x − 5z + 3xy

Explore 1

Simplify these expressions. ‾

a 12x + 3x =   

‾

d − 3 j  2 + 2 j  2=   

‾

b − 4r − r =   

‾

m 4

e 2m −  _ =   

3 k = _   1 k +  _    2 2 ‾ 1  t  3= _   1  t  3 +  _    6 3 ‾

2 Simplify these expressions by collecting like terms.

‾ c ab − 4a + 2ab + 2a =    ‾ e 12x  2 + 12x  2y − 7x  2 − 7x  2y =    ‾

‾

a 9jk + 2 − 8jk − 6 =   

b 10x + 8y − 3y + x =   

2 ‾ 1 1 _ _ 4 4 4 4 2st  − s  t −     s  t −    st  =    4 2 ‾

1  m  3= d 2m   3 + 5m  2 − 9m  2 +  _    f

a 4a + 5b − 2a + b = c

6x + 3y + 10x − z =

e 2ab + 7cd − ab + 3cd = g 5t   2 + 8t − 5 − 8t  2 + 2t = i

7s + 3st + s − 4 − st = 5 4

8 9

1  n  2 −  _ n = k 2n   3 −  _  n  2 + n +  _ 4

D R

AF T

3 Simplify these expressions by collecting like terms.

b 5m − 3n + n − 2m =

d k + 1 + 4k − 3 =

h − 5pq + 10pr + 2pq − 4pr + 8p =

9x − z  3 − 10x + 4z  3 =

2 fg + 6 = 3f − 2fg + 2g +  _ 3 6 5  _  x  2y − 2xy  2 −  _  x  2y  2 + 9xy  2 − x  2y = 7 8

4 Explain why 8abc − 7ab + 6bc cannot be simplified by collecting like terms.

5 Give a simplified expression for the perimeter of these shapes. a

2gh – 1

7f + 6

11ab 4ab 9c2

f + 3gh

12c

12 – f 12c2

6 – ba

4gh 16c + 3

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Expanding one pair of brackets The distributive law applies to numbers and variables. An expression is fully expanded and simplified if there are no brackets and all like terms have been collected. Distributive law

Number examples

Algebra examples

a(b + c) = ab + ac

5 × (10 + 3)= 5 × 10 + 5 × 3    = 50  + 15 = 65

2(x + 4)   = 2 × x +  2 × 4 = 2x + 8

a(b − c) = ab − ac

12 × (5 − 3)= 12 × 5 − 12 × 3    = 60  − 36 = 24

x × (6x − 5)= x × 6x − x × 5    = 6x  2 − 5x

Deepen 1

Fill in the blanks to expand the brackets.

‾ ‾       = ‾ c 7(c + 5)= 7 × + 7 × ‾ ‾       = ‾ e 6(8 − e)= 6 × − 6 × ‾ ‾       = ‾ 2 Expand the brackets using the distributive law.

‾ ‾       = ‾ d 8(d − 5)= 8 × − 8 × ‾ ‾       = ‾ f 8(9 + f )= 8 × + 8 × ‾ ‾       = ‾

a 6(a + 7)= 6 × + 6 ×

b 9(b − 3)= 9 × − 9 ×

D R

a 4(x + 10)=

AF T

b 7(b + 1)=

6(g − 4)=    =

d 6(x − 3)=

e 3(7 − k)=

2(7 − v)=    =

g 4(2y + 5)=

h 8(8x + 2)=

7(5x − 3)=    =

k 6(11k + 8)=

12(4 − 5x)=    =

   =    =

4(9x − 2)=    =

3 Expand the brackets using the distributive law. a x(y + 3)=

b a(b − 4)=

p(7p + 2)=    =

d m(5 − 3n)=

e k(4r + 9)=

t(12s − 11)=    =

g 2x(4x − 5y)=

h 3a(4 + b)=

5m(2m − 7n)=    =

k 12c(4c − 9d)=

11e(7e + 5f )=    =

   =    =

8p(3p + 11)=    =

   =    =

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

4 Find the areas of these shapes in expanded form. b

8 3a + 7

5a – 3

5 Expand the brackets and then simplify by collecting like terms.

‾     =     ‾     c 7(x + 2) + 5x= ‾     =     ‾     e 6(x − 7) − 2x + 1= ‾     =     ‾ g 7(2 − 3x) + 4x − 3=     ‾     =     ‾     i 2(5x − 3)= ‾     =     ‾     k 7(3x − 2) + 6x + 3= ‾     =     ‾ 6 Expand and simplify these expressions.     a 5(x + 8) − 3=

‾     =     ‾     d 5(x − 3) + x= ‾     =     ‾     f 3(5x − 2) + 6= ‾     =     ‾     h 4(x + 2) + 7x= ‾     =     ‾     j 3(7 − x)= ‾     =     ‾     l 2(7 − 5x) + x + 12= ‾     =     ‾     b 2(x − 5) + 7=

AF T

D R

a 4(x + 3) − 7 + 3(x + 7) + 4= 4 × + 4 × − 7 + 3 × + 3 × + 4

           = 4x + 3x +

= b 5(x − 3) + 6 + 4(x + 7) + 3x =

= = c

8(x − 4) + 2x + 5(x − 5) − 3x + 4 = = =

d 7x(x + 4) − 8 + x + 4x(x − 2) + 3 − 5x =

= =

7 Find the mistakes in these students’ working and show how to correctly expand and simplify the expressions. a x(7 − x) + 2x(1 − x)= 7x − x + 2x − 2x  2

     = 8x − 2x  2

b 2(x + 3) + 4(x − 1)= 2x + 6 + 4x + 4

   = 2x + 4x   + 6 + 4 = 6x + 10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Algebra | Equations and relationships

Topic 2.2 Linear equations and inequalities Substitution We can substitute (replace) the variables in an algebraic expression or formula with numbers. Expression

Values of variables

Substitution

x + 5

x = 4

4 + 5 = 9

3y − 2

y = 5

3(5) − 2 = 3 × 5 − 2 = 15 − 2 = 13

2x + 4y

x = 2, y = 3

2(2) + 4(3) = 2 × 2 + 4 × 3 = 4 + 12 = 16

A = l × w

l = 7, w = 3

A = 7 × 3 = 21

Learn 1

a 5(a − 5) =

D R

AF T

Find the value of the expression by substituting the following numbers for the variables: a = 3, b = 6, c = 2, d = 12. a a − 5= b b + 9= c 4c=          = = = d d _    = e 3a + 8= f 3b= 2          = = = b = h _ g 4c − 3d + 5= i 2 a  2 − b=   d +    a       = = = 2 Find the value of the expressions by substituting: a = 7, b = 10, c = 5and d = 8. b 8(9 − c) + 7 =

(d − c)  2 =

2(c + b) d _     =

e 11 + 3(b − 12) =

c   2(5b  2 − a) =

3 Substitute the numbers into the formulae. a A= l × w (l = 4 and w = 7)

b V= l  3 (l = 5)

= d a= 180(n − 2) (n = 5)

P= 2(l + w) (l = 5 and w = 3)         =

e V= _  1  bhl

(b = 4, h = 3 and l = 7) 2         =

(a + b ) c A= _     (a = 6, b = 5 and c = 8) 2            =

4 A plumber charges a $ 50call-out fee plus $ 70per hour worked. Suppose the plumber works h hours. a Write a formula for the cost, C of the plumber: b Use this expression to find the cost of hiring the plumber for: i

5 hours

ii 9 hours

iii 13 hours.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Checking solutions using substitution The equals sign in an equation means the left-hand side (LHS) is the same in value as the right-hand side (RHS). A solution to an equation is a value that makes the LHS equal to the RHS. Linear equations have one solution. Solutions to equations can be checked using substitution.

Use brackets when substituting a negative number.

For example, the solution to 3x = 15is x = 5because 3 × 5 = 15.

Learn 1

Use substitution to check whether the value given in brackets is the solution to the equation. a

b 25 − b= 15 (b = 8)

a − 7= 3 (a = 10)

LHS= − 7 ‾            = ‾ RHS= 3

LHS= 25 − ‾            = ‾ RHS= ‾ b = 8is/is not the solution. e 8e − 4= 5e − 3 (e = _  1 ) 3

3(f + 2)= 5(3 − f ) (f = _  1 ) 2

LHS= 4 × + 5 ‾           = ‾ RHS= ‾ c = 5is/is not the solution.

LHS= 3 × ( + 2) ‾ =           ‾   RHS= 5 × (3 − ) ‾ = ‾ f=_  1 is/is not the solution. 2

LHS= 8 × − 4 ‾ =             ‾ RHS= 5 × − 3 ‾ = ‾ e=_  1 is/is not the solution. 3

D R

LHS= − 12 + 7 × ‾ =                 ‾   RHS= 2 × ‾ = ‾ d=_  4 is/is not the solution. 5

4c + 5= 15 (c = 5)

AF T

a = 10is/is not the solution. d − 12 + 7d= 2d (d = _  4 ) 5

2 Match these equations with their solutions. A 2(4x − 3) = 8(2x − 1)

x = _  1  2

B 10 − x = 2(x + 2)

C 6x + 1 = 4x + 2

D _   1 (4x + 5) = 3 + x

II x = 2

III x = _  1 

IV x = 4

3 Sometimes we can solve an equation by guessing, checking and improving on the guess.

For example, if 3x + 1 = 100. •

Try x = 30: 3 × 30 + 1 = 91(too low)

•

Try x = 35: 3 × 35 + 1 = 106(too high)

•

Try x = 33: 3 × 33 + 1 = 100(just right!)

Use substitution to check your guess.

Therefore, x = 33. Use the guess, check and improve method to find the solutions to these equations. a 5x + 3 = 98

b 7x − 9 = 159

9c + 15 = 11c − 7

d Is the guess, check and improve method an efficient method to solve equations?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Solving one-step linear equations Equations can be solved algebraically using inverse operations. These operations keep the two sides of an equation balanced. A one-step equation has one operation applied to the variable. To solve a one-step equation, apply the inverse operation to both sides. For example, in x − 4 = 7, 4is is subtracted from x . To solve the equation, add 4to both sides. x − 4= 7 (+ 4)  x    − 4 + 4= 7  +  4 x= 11 Operation Addition (+ )

Inverse operation Subtraction (− )

Subtraction (− )

Addition (+ )

Multiplication (×)

Division (÷)

Multiplication (×)

What you do to one side, you must do to the other!

Explore For each equation, identify the operation applied to x and the corresponding inverse operation. Use the inverse operation to solve the equation. Equation

Operation

x − 2 = 7

− 2

Inverse operation

Solution

+ 2

x = 9

D R

_   k = 6 8

AF T

7g = 31 5f = 23 3 = 3 x +  _ 2

2 Solve these equations using inverse operations. a a + 7 = 5

b _   b  = 2

e _   e   = _   2 

5  = _ f −  _   7  4 2

d d + 13 = _  1 

4c = 25

1  g _   8  + g = −  _ 9

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

5 4

h 8h = _    

Solving two-step linear equations A two-step equation has two operations applied to the variable. Use the order of operations (GEMA) to determine the order they are applied to the variable.

2(x + 1)= 6 (÷ 2) 2(x + 1) _ 6 _     =     2          2    x + 1= 3 (− 1) To solve a two-step equation, apply inverse operations to both sides in the reverse order x + 1 − 1= 3 − 1 x= 2 to which they are applied to the variable. For example, in the equation 2(x + 1) = 6, 1is added to the variable x and then the result is multiplied by 2.

For example, to solve 2(x + 1) = 6, divide by 2and then subtract 1.

Explore Fill in the inverse operations to solve these equations. 5a + 6= 15 5a + 6 = 15 ‾ ‾ 5a= 9        9   _   5a = _  

a= ‾ _ c   c   + 1= _  9  4 2 _   c   + 1 = _  9  4 ‾ 2 ‾          c   _      = _  7  4 2 _   c   = _  7  4 ‾ 2 ‾ c= ‾

12 + 3b= 40 12 + 3b = 40 ‾ ‾ 3b= 28 28          _     3b = _

b= ‾ 5 = _ _ d   d +   1  6 4 5 = _ _   d +   1  6 ‾ 4 ‾              = _   d +5  3  2 d + 5 = _  3  ‾ 2 ‾ d= ‾

AF T

D R

2 Solve these equations using inverse operations.

a 4a + 5 = 8

b 3 + 5b = 6

e − 4 + 6e = 1

f _     + 1 = _   7  5 5

11 = 6 d _   d  +  _

4 + 4c = 7

1  g 3g − 1 = −  _ 2

9  h _   h  − 4 = −  _ 2

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Forming and solving linear equations Linear equations can be formed and solved for many situations.

Explore 1

Write an equation for the perimeter of each of these shapes and then solve for the unknown variable. You many need to simplify your equation by adding or subtracting like terms. a

b 2b

2a + 1

Perimeter = 30 m

3b + 4

a = ‾

Perimeter = 25 m d

5–c

d+1

b = ‾

5d – 2

AF T

6d + 3 6 + 3c

c = ‾

D R

Perimeter = 12.5 m

Perimeter = 54 m

d = ‾

2 An electrician charges a $ 60call-out fee plus $ 100per hour worked. If a customer is charged $ 510, how many hours

did the electrician work?

3 Alice, Bob and Charlie win $ 1,100in a lottery. The three friends bought their ticket together but contributed

different amounts. •

Alice paid n dollars.

•

Bob paid half as much as Alice.

•

Charlie paid a third of what Alice paid.

How much of the winnings should each friend receive if they split the money in the same proportion as how they paid for the ticket?

Alice: $

Bob: $

Charlie: $

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Rearranging formulae The subject of a formula is a single variable along one side of the equation which does not appear on the other side. We usually write the subject of a formula on the left-hand side of an equation. For example, the subject of A = bhis A . Inverse operations can be applied to make a different variable the subject of a formula. For example, to make b the subject of A = bh, divide both sides of the formula by h .

A= bh (÷ h) A _ = _ bh h h _           A = b h b= _  A h

Explore 1

Rearrange these formulae to make the variable in brackets the subject. a A = lw

b A = lw

(l )

(x)

P = 2(l + w) (l )

bh 2

d A = _     (b)

P = 4l (l )

g P = 2(l + w)

(w)

h ax + by = c

(x)

AF T

e y = mx + c

(w)

D R

2 The formula p = mvrelates an object’s momentum, p , mass, m , and velocity, v. a Rearrange the formula to make v the subject. b An object has momentum p = 40 kg m/sand mass m = 8 kg. What is the velocity of the object?

v = m/s ‾ 3 The formula d = strelates distance travelled, d , average speed, s , and time, t .

a Rearrange the formula to make s the subject. b A car travels 250 kmin 3hours. What was the average speed of the car? s = km/h c

Rearrange the formula to make t the subject.

d A plane travels 3,600 kmat an average speed of 900 km/h. How long did the flight last from take off to landing?

4 The formula 2s = t(u + v)relates the displacement (distance from starting point), s , time, t , initial velocity, u, and

final velocity, v , of an object moving with constant acceleration. a Rearrange the formula to make u the subject. b An object lands 20 mfrom its starting point after 10seconds at a final velocity of 20 m/s, having moved with

constant acceleration. What was the initial velocity of the object? c

Rearrange the formula to make v the subject.

d An object is launched with an initial velocity of 5 m/sand lands 350 maway, having moved with constant

acceleration for 20seconds. What was the final velocity of the object? 96

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Solving linear inequalities An inequality has an inequality symbol instead of an equals symbol. Symbol

Meaning

Numerical examples

Linear inequalities

Is less than

10 < 9 + 2

2x + 1 < 5

Is greater than

3 × 6 > 15

3x − 5 > 10

≤

Is less than or equal to

10 ≤ 8 + 3

4x − 2 ≤ 22

≥

Is greater than or equal to

2 × 7 ≥ 14

5x + 6 ≥ − 4

The solutions to a linear inequality are any values of x that can be substituted to make the inequality true. Linear inequalities can be solved using inverse operations in the same way as linear equations.

Deepen 1

Write two values of x which would make the inequality true. a x > 4: ,

b x < 1: ,

d x ≤ − 2: ,

x ≥ 10: ,

2 Fill in the blanks to solve the linear inequalities using inverse operations. b

(+ 2)

 x    − 2 < 4  ‾ ‾ x< ‾

3 Solve these linear inequalities.

b _   m < 2

a p + 9 ≥ 3

x + 3≤ 2

(− 3)

 x         + 3 ≤ 2  ‾ ‾ x≤ ‾

AF T

x − 2< 4

D R

7x> 21

(÷ 7)

7x > _ 21 _            x> ‾ d 3b > 9

b − 11 ≤ − 3

4 Fill in the blanks to solve the linear inequalities using inverse operations. a

4x − 3> 13

(+ 3)

4x − 3 > 13 ‾ ‾ 4x> 16 (÷ 4)                4x 16 _ _   >  

5f − 2≥ 23

3h + 9< 21

(− 9)

5f − 2 ≥ 23 ‾ ‾ 5f≥ 25 (÷ 5)                5f _ 25 _   ≥  

3h + 9 < 21 ‾ ‾ 3h< 12 (÷ 3)                _   3h < _  12

f≥ ‾

h< ‾

x> ‾ 5 Solve these linear inequalities. a 3x + 8 < 5

(+ 2)

b 9x − 1 > 26

y _     − 2 ≤ 1 6

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2 ≥ 3 d _   x −  2

Representing linear inequalities on a number line The solutions to a linear inequality can be represented on a number line. •

An open circle (◦)means that the number is not included.

•

A closed circle (• )means that the number is included. Inequality x > 0

x ≥ 0

Inequality

Number line –5 –4 –3 –2 –1 0

–5 –4 –3 –2 –1 0

x < 0

x ≤ 0

Number line –5 –4 –3 –2 –1 0

–5 –4 –3 –2 –1 0

The number lines in the tables show all real number solutions to the linear inequalities. The integer solutions to x > 0 can be shown like this. –5 –4 –3 –2 –1 0

AF T

Deepen Represent these linear inequalities on the number line. a x > 1 0

x < − 1 –5 –4 –3 –2 –1

D R

–5 –4 –3 –2 –1

b x ≥ − 2

g x < 5

–5 –4 –3 –2 –1

x ≤ 2

h x ≥ − 4 –5 –4 –3 –2 –1

–5 –4 –3 –2 –1

d x ≤ 4 x

e x > − 3

2 Add dots to the number lines in question 1 to show the integer solutions to each linear inequality. 3 What linear inequalities do these number lines represent? a

–5 –4 –3 –2 –1 0

–5 –4 –3 –2 –1

–5 –4 –3 –2 –1 0

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

4 Solve each linear inequality and then show the solutions on the number line. a x − 4 > − 5

b 3x ≤ 6

–5 –4 –3 –2 –1 c

–5 –4 –3 –2 –1

e x − 6 > − 3

5x ≤ 25

–5 –4 –3 –2 –1

h _   x < 1

g x + 7 < 5

–5 –4 –3 –2 –1

AF T

–5 –4 –3 –2 –1

d _   x < 1

x + 4 ≥ 5

–5 –4 –3 –2 –1

5 Add dots to the number lines in question 4 to show the integer solutions to each linear inequality.

D R

6 Solve each linear inequality and then show the solutions on the number line. a 3x − 1 > 5

–5 –4 –3 –2 –1 c

_   x  + 5 < 7 2

–5 –4 –3 –2 –1

–5 –4 –3 –2 –1 f

_   x  − 7 > − 6 2

–5 –4 –3 –2 –1

5 ≤ 2 h _   x + 

g 5x − 11 ≤ 4

–5 –4 –3 –2 –1

–5 –4 –3 –2 –1 d 10x − 30 ≤ 20

e 3x − 7 ≥ − 4

–5 –4 –3 –2 –1

b 6x − 4 < 2

–5 –4 –3 –2 –1

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Forming and solving linear inequalities Many problems can be solved by forming linear inequalities.

Deepen 1

Consider this triangle.

3a – 3

a The length of each side of the triangle must be greater than 0 .

a+5

Write and solve three inequalities to find the possible values of a . i

b All three inequalities in part a must be true for the

triangle to exist. Which of the three inequalities makes this possible?

d Suppose the perimeter of the triangle must be

at most 34 m. Write and solve a linear inequality to find the possible values of a .

Represent the possible values of a on this number line. Should you use an open (◦)or closed (• ) circle? –5 –4 –3 –2 –1 0

AF T

2a – 4

iii

D R

e In a different colour, represent the possible values of a from part d to the number line in part c. f

Fill in the blanks in this two-sided inequality to show the possible values of a for this triangle to exist and to have a perimeter of at most 34 m. < a ≤ ‾ ‾ 2 Three friends are saving to buy a 3D printer worth at least $ 5,000. Tia can save n dollars. a Tia’s friend Alofiana can save twice as much as she can plus $ 100. Write an expression for the amount of money

Alofiana can save.

b Their third friend, Greta, can save $ 350less than Tia. Write an expression for the amount of money Greta can

save.

Write and simplify an expression for the total amount of money the three friends can save together.

d Write an inequality for the money they can save and the cost of the 3D printer. e Solve the inequality from part d.

If Tia saves the minimum amount to make $ 5,000, how much money do Alofiana and Greta each save? Alofiana:

100

Greta:

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Algebra | Equations and relationships

Topic 2.3 Factorising expressions Dividing algebraic terms We can divide an algebraic term by any other non-zero algebraic term. 1 Write the division as a fraction.

10xy 10xy ÷ 5x = _____     5x

2 Replace the coefficients with their prime factorisation.

Writing terms in expanded form helps to identify common factors.

5×2×x×y ____________ =        5×x

3 Write the variables in expanded form. 4 Cancel any common factors.

51 × 2 × x1 × y = _____________        51 × x1

5 Write the resulting simplified term.

= 2y

Learn Divide these terms by cancelling the common numerical factors. b 6b ÷ 9

e 12m ÷ 18

14ab ÷ 21

8x ÷ 12

AF T

a 4a ÷ 8

d 10y ÷ 15

g 28xy ÷ 42

h 36qr ÷ 60

d 4x ÷ x

D R

2 Divide these terms by cancelling the common factors. a 5x ÷ x

b 8y ÷ y

e 7x ÷ 2x

8x ÷ 4x

g 6x ÷ 9x

h 18x ÷ 12x

10xy ÷ 15x

k 14xy ÷ 21x

3xy ÷ 6x

2x ÷ x

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

24xy ÷ 48x

101

Highest common factor of algebraic terms The highest common factor of two algebraic terms is the largest product of numbers and variables that is a factor of both terms. To find the highest common factor of two terms, write the terms as a product of the prime factors of the coefficient and the variables. Identify the common factors and then multiply these together to find the highest common factor. 6x  2y = 2 × 3 × x × x × y 4xy  2 = 2 × 2 × x × y × y

Find what the terms have in common.

Highest common factor = 2 × x × y = 2xy The highest common factor is often called the HCF.

Learn Identify the HCF of each pair of terms by writing them in expanded form. 5a=    ‾          = 2a    ‾ HCF=    ‾ d 9d=    ‾          = 3d    ‾ HCF=    ‾ a

18g=    ‾ 9g=               ‾ HCF=    ‾ =    ‾ 2 Identify the HCF of each pair of terms. a

D R

7b=    ‾          = 10b    ‾ HCF=    ‾ e 6e=    ‾ 9e=               ‾ HCF=    ‾ =    ‾ h 21h=    ‾ 42h=                ‾ HCF=    ‾ =    ‾ b

AF T

6a  2=    ‾ 2a  3=                ‾ HCF=    ‾ =    ‾ 3d  2=    ‾ 5d  3=                ‾ HCF=    ‾ =    ‾ ‾ 2g h  3=                ‾ HCF=    ‾ =    ‾

g 7 g  2 h=   

3b  3=    ‾ 7b  5=                ‾ HCF=    ‾ =    ‾

6xy=    ‾ 12x  2y=                ‾ HCF=    ‾ =    ‾

9 h  2 i=    ‾ 4h i  2=                ‾ HCF=    ‾ =    ‾

4c=    ‾         = 5c    ‾ HCF=    ‾ 12f=    ‾ 15f=                ‾ HCF=    ‾ =    ‾ 16i=    ‾ 32i=                ‾ HCF=    ‾ =    ‾ 2c  2=    ‾ 2c  4=                ‾ HCF=    ‾ =    ‾

9mn=    ‾ 27m  2n=                ‾ HCF=    ‾ =    ‾

10i j  3=    ‾ 5 i  2 j  2=                ‾ HCF=    ‾ =    ‾

3 Write a pair of terms with the given HCF. a HCF = 5xy: and

102

b HCF = 11rs: and

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Factorising algebraic expressions Factorising is the opposite of expanding. To factorise an algebraic expression by taking out the highest common factor (HCF): 1 Identify the HCF of all terms in the expression. 2 Write the HCF in front of a pair of brackets.

You can expand a factorised expression to check you factorised it correctly.

3 Divide the terms by the HCF. 4 Write the resulting expression inside the brackets. When we factorise out the HCF, we say that the expression has been fully factorised. 6x  −  _ 8  6x − 8= 2(  _ )    2 2  = 2(3x − 4)

Explore Factorise these expressions. a 3s + 12 = _   3s + _   12 ‾( )

‾(

= ( + ) ‾ ‾ ‾

= ( − ) ‾ ‾ ‾

   ‾ =    ‾    e 10h − 15= f 18y + 12=    ‾       ‾ =    =    ‾ ‾ 2 Expand your factorised expressions in question 1 to show that you factorised correctly. d 6v + 12=   

D R

4x − 10=       ‾ =    ‾

)

b 16c − 12 = _  16c − _   12

AF T

3 Does anything happen to the expression 7y − 8when factorised by the HCF? Explain why or why not.

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103

4 Factorise these expressions by taking out the HCF. a 7a − 14

b 15 + 25b

12x − 36

d 60d − 20

e 7m + 21n

g 8f − 12g

h 20r + 24s

4a − 6b

5 Find an expression for the unknown base lengths of these rectangles and triangles. a

Base length = ‾

Area = 8a − 12

Area = 18h + 15

Base length = ‾

D R

Area = 6c + 10

AF T

6 Factorise these expressions by taking out the HCF.

‾(

)

25ab b 10a + 25ab= _   10a +  _

= ( + ) ‾ ‾ ‾

8m + 12mn=     ‾     =     ‾

‾     =     ‾

    e 15xy + 20x=

104

‾(

9xy a 6x + 9xy= _   6x +  _

‾     =     ‾

    d 14p + 21pq=

18ab + 30abc=     ‾     =     ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Writing variables with exponents in expanded form makes it easier to find the HCF.

Factorising algebraic expressions with exponents We can factorise algebraic expressions which include exponents.

Deepen 1

Factorise these expressions. ‾(     )

3 × a × a +  _ 4 × a a 3a  2 + 4a=  _

‾(     )

5 × b × b −  _ 6 × b b 5b  2 − 6b=  ____________

= ( + ) ‾ ‾ ‾

= ( − ) ‾ ‾ ‾

3c − 4c  2=     ‾     =     ‾

‾     =     ‾

d 5d − 2d  2=    

‾     =     ‾

e 6x  2 + 12x=    

9a  2 + 18a=     ‾     =     ‾

D R

AF T

2 Expand your factorised expressions in question 1 to show that you factorised correctly.

3 A student has factorised these expressions incorrectly. Show how to correctly factorise each expression. a 3x  2 + 6x = 3x  2(1 + 2x)

b 4x + 2x  2= 2x(2 − x)

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105

Algebra | Equations and relationships

Topic 2.4 Tables and graphs Points on the coordinate plane

y-axis 5

Every point on a coordinate plane can be described as an ordered pair (x, y).

4 2nd quadrant

The origin (0, 0)is the point where the x - and y -axes intersect.

•

1st quadrant

2 1

For a point ( x, y): •

the value of xdescribes how far left or right the point is compared to the origin

–5 –4 –3 –2 –1 0 –1 3rd quadrant

the value of ydescribes how far above or below the point is compared to the origin.

–2

5 x-axis

4th quadrant

–3 –4 –5

Learn Write the points on this coordinate plane as ordered pairs. Ordered pair

Point

(2, − 1)

Ordered pair

y-axis 5

AF T

Point

D R

B M

F H

O E

–5 –4 –3 –2 –1 0 –1 Q S –2

F N

–3 –4 –5

P 1 T

2 A

5 x-axis

D J H

T 2 Sort the points from question 1 according to their location on the coordinate plane.

1st Quadrant

2nd Quadrant

3rd Quadrant

4th Quadrant

x-axis

y-axis

3 Tick which inequalities are true for points in each quadrant.

Location

x > 0

x < 0

y > 0

y < 0

1st Quadrant 2nd Quadrant 3rd Quadrant 4th Quadrant

106

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Plotting points using ordered pairs We can plot ordered pairs as points on the coordinate plane.

Learn 1

Plot and label these points on the coordinate plane.

y 10 9 8 7 6 5 4 3 2 1

(7, − 3), (− 10, 4), (2, 9), (− 6, − 1), (0, − 8), (10, 5), (− 3, 7), (8, − 10), (− 9, 2), (4, − 6), (1, 10), (− 7, − 4), (6, 3), (− 2, − 9), (9, 0), (− 8, 8), (3, − 7), (− 1, 6), (5, − 2), (0, 10)

D R

AF T

–10–9 –8 –7 –6 –5 –4 –3 –2 –1–10

2 Plot and label these points on the coordinate plane.

Connect the points in alphabetical order.

A (0, 0), B (1, 1), C (− 1, 2), D (− 2, − 1), E (2, − 2), F (3, 3), G (− 3, 4), H (− 4, − 3), I (4, − 4), J (5, 5) 3 Can you see the pattern of the plotted points in

question 2? What are the next five points to be plotted? Extend the pattern on the coordinate plane. K ( , ) L ( , ) M ( , ) N ( , ) O ( , )

1 2 3 4 5 6 7 8 9 10 x

–2 –3 –4 –5 –6 –7 –8 –9 –10 y 10 9 8 7 6 5 4 3 2 1

–10–9 –8 –7 –6 –5 –4 –3 –2 –1–10

1 2 3 4 5 6 7 8 9 10 x

–2 –3 –4 –5 –6 –7 –8 –9 –10

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107

Plotting points using tables of values A table of values can be used to represent points on the coordinate plane. Both of these tables represent the points (− 2, 4), (− 1, − 1), (0, 2), (1, 5), (2, 8). x

− 2

− 1

− 4

− 1

− 2

− 4

− 1

Learn Plot and label the points in the tables of values on the coordinate plans. a

− 3

− 2

− 1

− 5

− 4

− 3

− 2

− 1

− 3

− 2

− 1

y 5

1 –5 –4 –3 –2 –1 0 –1

–3 –4

–5 –4 –3 –2 –1 0 –1

–5

5 x

–2 –3 –4 –5

− 3

− 2.5

− 2 − 1.5 − 1

− 0.5

− 1.5

− 1

− 0.5

0.5

1.5

− 0.5

− 1

0.5

2.5

0.5

− 1

–4

–3

–2

0.5

–1

y 3

1 x

–3

–1

108

D R

–2

5 x

AF T

–2

–1

3 x

–1

–2

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Linear patterns A pattern is linear if the increase or decrease between each successive term is constant (the same). • 1, 5, 9, 13, 17, …is a linear pattern because the difference between successive terms is 4 .

+4 1

+4 5

+4 13

+4 ...

• 1, 4, 9, 16, 25, …is a non-linear pattern because the difference between successive terms changes.

+3 1

+5 4

+7 9

+9 16

+? ...

Explore 1

Identify whether these patterns are linear or non-linear. Pattern

Linear

Non-linear

10, 20, 30, 40, 50, …

AF T

2, 4, 6, 8, 12, … 15, 19.5, 24, 28.5, 33, …

5, − 2, − 9, − 16, − 23, … 1, 1, 2, 3, 5, 8, …

D R

1.4, 3.9, 5.4, 7.9, 9.4, …

2 Give two examples of a linear pattern. a

3 Determine the constant change in these linear patterns and add the next three terms. a 5, 11, 17, 23, 29, , , , …

‾ ‾ ‾ Constant change:

b 100, 80, 60, 40, 20, , , , …

7, 4, 1, − 2, − 5, , , , … ‾ ‾ ‾ Constant change:

1 , −  _ 1 , 0, _ d −  _ , −  _  1 , , , , …

‾ ‾ ‾ Constant change: 3 4

4 ‾ ‾ ‾

Constant change:

4 Draw the next two terms in this linear pattern.

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109

Plotting linear patterns

Starting from 0may seem strange but we will learn why mathematicians choose to do this soon!

The terms in a linear pattern can be numbered according to their position in the sequence. Often, terms are numbered counting from 0instead of 1. We can use t n to represent the n thterm in a linear pattern. For example, in the linear pattern 1, 5, 9, 13, 17, …, we can say that t 0  = 1, t 1  = 5, t2  = 9, t 3  = 13 and t 4  = 17. Linear patterns can be plotted on the coordinate plane using the points ( n, tn  ).

Explore 1

For each of these linear patterns, complete the table of values and plot the points on the coordinate plane. a 0, 1, 2, 3, 4, …

b 5, 4, 3, 2, 1, …

tn 6

AF T

3 2

–1 0 –1

D R

1 6 x

2 1

–1 0 –1

tn 9

1 2

–1 0 –1

6 x

d 1, 3, 5, 7, 9, …

0, 2, 4, 6, 8, … n

110

6 x

–1 0 –1

6 x

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

2 What do you notice about all the linear patterns plotted in question 1? How would you describe the points?

A conjecture is an idea in maths that we think is true, but still need to prove.

3 Fill in the blanks to complete your conjecture about plotting linear patterns on

the coordinate plane. The points in a linear pattern form a on the coordinate plane. 4 Testing a conjecture means to check whether it is true for more examples.

Test your conjecture from question 3 with four more linear patterns of your choice. a Linear pattern:

tn 6

1 1

6 x

–1 0 –1

D R

Linear pattern: n

AF T

–1 0 –1 c

b Linear pattern:

1 2

6 x

tn tn 6

d Linear pattern:

tn 6

–1 0 –1

6 x

5 Do you think your conjecture is true? Explain.

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111

Describing linear patterns When describing a linear pattern, you need to provide two pieces of information. Usually these are the starting points and the constant increase or difference. For example, to describe the linear pattern 1, 5, 9, 13, 17, …, you could say “Starting from 1, add 4to get the next term”. This is an example of a rule which tells you how to get from one term to the next.

Explore 1

For each linear pattern, complete the rule which tells you where to start and how to get from one term to the next. a 1, 2, 3, 4, 5, …Starting from , to get the next term. b 8, 7, 6, 5, 4, …Starting from , to get the next term. c

2, 4, 6, 8, 10, …Starting from , to get the next term.

d − 20, − 10, 0, 10, 20, …Starting from , to get the next term. e 12, 5, − 2, − 9, − 16, …Starting from , to get the next term. f

100, 150, 200, 250, 300, …Starting from , to get the next term.

a Starting from 2, add 5to get the next term.

AF T

2 Write the first five terms of the linear pattern corresponding to these rules.

D R

, , , , b Starting from 8, add 10to get the next term.

You can add or subtract in a linear pattern.

, , , , c

Starting from − 12, add 9to get the next term. , , , ,

d Starting from 20, subtract 5to get the next term.

, , , , e Starting from 70, subtract 12to get the next term.

, , , , f

Starting from 1000, subtract 500to get the next term. , , , ,

3 Describe these linear patterns. a

112

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Algebraic equations of linear patterns

t = number in the pattern

Linear patterns can be described algebraically by the rule t = a × n + d. We can find the value of t nfor any value of n using the equation.

+a d = t0

+a t1

+a t3

a = amount added or subtracted each time n = position in the pattern (1st, 2nd, 3rd etc.) d = starting adjustment (when n = 0)

...

By counting terms starting from 0instead of 1, the algebraic equation of the linear pattern is simpler. If we start counting at 1, the equation is t = a × (n − 1) + d.

Deepen 1

Find the algebraic equations of these linear patterns. a

Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

D R

t =   

1, 4, 7, 10, 13, … Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

iv What is the value of‾ the term with n = 10?

AF T

iv What is the value of‾ the term with n = 25?

t =   

d 5, 3, 1, − 1, − 3, …

Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

t =    ‾ iv What is the value of the term with n = 100?

t =     ‾ iv What is the value of the term with n = 100?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

113

e 2, 4, 6, 8, 10, …

Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

− 5, − 10, − 15, − 20, − 25, … Difference between terms: a = ‾ ii Value of t 0: d = ‾ iii Algebraic rule: t = a × n + d

t =    ‾ iv What is the value of the term with n = 100?

2 Use the equation of each linear pattern to write the first five terms of the pattern. a t = 2n − 1

, , , ,

AF T

b t = 3n + 1

, , , , t = − 4n ,

D R

d t = − 2x + 9

Substitute n = 0, 1, 2, 3, 4 to find the first five terms of the linear patterns.

, , , , e t = 2n +2

t = 4n −1 ,

g t = 10n

h t = 8n + 1

3 Clare says that the linear pattern described by the equation y = 6n doesn’t have a value of d. Explain to Clare why

she is incorrect.

114

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Graphs of linear patterns

t 8

The graph of a linear pattern can be obtained by substituting values of n into the equation, recording the values of t in a table of values, and plotting the points on the coordinate plane.

7 6 5

t = 2n − 1

− 1

2 1 –1 0 –1

5 n

–2

Deepen For each equation of a linear pattern, complete the table of values and plot the corresponding points on the coordinate plane. a

b t = − n + 2

t = n + 1 n

AF T

D R

t 6 5 4

1 –1 0 –1

2 1 –1 0 –1

5 n

–2 1

–3

5 n

d t = 2n − 5

t = 2n 0

t 3 2

t t 9

t 6

3 2

–1 0 –1

–2

–1 0 –1

5 n

–3 –4

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115

Equations of graphs of linear patterns The equation of the graph of a linear pattern is in the form t = a × n + d.

Deepen 1

Compare the value of d with the graph of each linear pattern in question 1 on the previous page. a Circle the corresponding point in each graph. b Fill in the blanks to complete this statement.

In the graph of a linear pattern, you can find the value of d by looking at the point on the where n = which ‾ is on the axis. t 5

2 We can draw arrows and labels on

graphs of linear patterns like this, with the arrows pointing across and then up or down towards the next point.

Add similar arrows and labels to the graphs in question 1 on the previous page.

t 8 7

+1 –1

5 n

1 +2

AF T

+1 –1

–1 0 –1

5 n

D R

–2

3 Compare the labels added to the graphs in question 1 on the previous page on the up or down arrows with the equation

of each linear pattern.

Describe how you can find the value of afrom the graph of a linear pattern. 4 Describe how you can find the equation of a linear pattern given its graph.

5 Find the equations of these linear patterns. a

t 7 6

–1 0 –1

1 –1 0 –1

5 n

Equation: 116

t 5

Equation:

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Algebra | Equations and relationships

Topic 2.5 Number patterns Triangular numbers Triangular numbers can be organised into a triangular shape. The first four triangular numbers are 1, 3, 6, 10. We often use T  n to represent the n th triangular number. So, T1 = 1 and T2 = 3.

Learn 1

Find the next two triangular numbers by drawing the next two triangles in the example sequence. The first six triangular numbers are: 1, 3, 6, 10, ,

AF T

2 Do the triangular numbers form a linear pattern? Explain.

3 Give a rule in words for how to get from one triangular number T  n to the next one, T n+1   .

D R

4 Use your rule from question 3 to find the next four triangular numbers. a T7  =

b T8  =

T9  =

d T10   =

5 Complete this table of values with the first ten triangular numbers.

Tn 6 Explain how you could find the value of T  100. How

long do you think it would take you?

7 Plot and label the first nine points in the table of values

in question 5 on the coordinate plane. Tn 50 45 40 35 30 25 20 15 10 5 0

1 2 3 4 5 6 7 8 9 10 n

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117

8 Use the plotted points in question 7 to give another reason why the triangular numbers do not form a linear pattern.

9 There is a formula for the n thtriangular number. We can find the formula using two copies of the triangle for

each term. a

How many rows of dots?

ii How many dots in each row?

‾     = ‾ How many rows of dots?

× T1  iii Total number of dots=

ii How many dots in each row?

‾     = ‾ How many rows of dots?

× T2  iii Total number of dots=

ii How many dots in each row?

‾     = ‾

× T3  iii Total number of dots=

AF T

How many rows of dots?

ii How many dots in each row?

‾     = ‾

D R

× T4  iii Total number of dots=

e What is the relationship between the number of rows of dots and the number of dots in each row?

n(n + 1) 2

10 The formula for the nth triangular number is Tn = ________    . Check that the formula gives the correct value for the

first 10 triangular numbers.

11 Use the formula to find the following triangular numbers. a T20   = 118

b T50   =

T100   =

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Square numbers Square numbers can be organised into a square shape. The first four square numbers are 1 , 4, 9, 16. We can use S n to represent the n thsquare number.

Explore 1

Find the next two square numbers by drawing the next two squares in the example sequence. The first six square numbers are: 1, 4, 9, 16, ,

2 Do the square numbers form a linear pattern? Explain.

3 What is the formula for the n thsquare number?

D R

S n = ‾

AF T

4 Use your formula from question 3 to find the next four square numbers.

‾

a S7  =

‾

b S8  =

S9  = ‾

‾

d S10   =

5 Complete this table of values about the first ten square numbers.

1   2= 1

2   2= 4

1 + 3 = 4

3   2= 9

1 + 3 + 5 = 9

Addition fact

4 5 6 7 8 9 10 6 Fill in the blank.

The n thsquare number, S  n, is equal to the sum of the first n numbers.

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7 Plot and label the first seven points (n, S n  )in the table of values in question 5 on the coordinate plane. Tn 50 45 40 35 30 25 20 15 10 5 0

7 n

AF T

8 Use the plotted points in question 7 to give another reason why the square numbers do not form a linear pattern.

9 The formula for how to get from a square number, S  n, to the next one, S n+1   is Sn + 1 = Sn + 2n + 1. Label each colour

in these diagrams with the term in the formula they correspond to (Sn, 2n or 1). b

D R

S3 = 9 S4 = 16 S 3 = S   + 2 × + ‾ ‾ 10 Colour or annotate this diagram for S  5 in a similar

way to those in question 9 and then label each colour in your diagram with the term in the formula it correspond to (Sn, 2n or 1).

S 4 = S   + × + ‾ ‾ ‾ 11 Check that the formula in question 10 works for square

numbers from S  1 to S 10   .

S5 = 25 S 5 = S   + × + ‾ ‾ ‾

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Cube numbers Cube numbers can be organised into a cube. The first three cube numbers are 1 , 8 , 2 7. We can use C n to represent the nth cube number.

Thinking in 3D is hard! Try to find some physical objects to make models of the cubes on this page.

What is the formula for the n thcube number?

C n = ‾

AF T

Deepen

D R

2 Complete this table of values with the first ten cube numbers.

Cn 3 Do the cube numbers form a linear pattern? Explain.

4 The formula for getting from one cube number, C  n, to the next one, C n+1   , is C n+1   = C n  + 3n  2 + 3n + 1.

Use this formula to find C  4. Does it match the value given by your formula in question 1?

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Measurement | Measuring

Topic 3.1 Estimating and measuring Length and area Sometimes we need to estimate the length or area of an object.

Benchmarks help us to estimate by comparing the object we are estimating with a known length or area.

Common units of length are millimetres (mm), centimetres (cm), metres (m) and kilometres (km). Common units of area are square millimetres (mm  2), square centimetres (cm  2), square metres (m  2) and square kilometres (km  2).

Learn 1

Line A

Line B is 8 cm long. a Estimate the lengths of the other two

lines without measuring them. i

Line B Line C ii Line C estimate:

Line A estimate:

b Measure Line A and Line C . Write the lengths. ii Line C :

Line A :

AF T

2 Line B is 6 cm long.

Line A

Lines A and C . i

Line A estimate:

D R

a Estimate (do not measure) the lengths of

6 cm

Line B

ii Line C estimate: Line C

b Now measure Lines A and C . i c

ii Line C :

Line A :

Did the arrows affect your estimates?

3 Estimate the areas of these rectangles using this 1cm 2square as a benchmark. a

1 cm

Area ≈ cm  2 ‾

4 Use a ruler to measure the side lengths of the rectangles in question 3. Record their area here, correct to one

decimal place. ‾

a Area = cm  2

122

‾

b Area = cm  2

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Volume and capacity Volume is a measure of the amount of space that that an object takes up. Common units for volume are cubic centimetres (cm  3) and cubic metres (m  3). Capacity is a measure the maximum amount an object can hold. Common units of capacity are millilitres (mL) and litres (L).

Learn 1

Estimate the volume of each object. a tissue box A 20 cm3

B 2,000 cm3

C 20,000 cm3

B 10,000 cm³

C 500,000 cm³

B 200,000 cm³

C 20,000,000 cm³

b classroom rubbish bin A 500 cm³ c

fridge A 200 cm³

AF T

2 Which object is closest to a volume of 1 m3? A a pencil class

B a classroom desk

C a large fridge

D a school hall

D R

3 Put these objects in order from smallest volume to largest volume.

backpack, shipping container, 6-sided die, classroom cupboard, basketball

4 Write an estimate for the capacity of these objects. There may be one you can estimate in your classroom.

Choose a suitable unit of capacity for each object. a Drink bottle:

b Lunchbox:

Rubbish bin:

d Pencil case:

e Backpack:

Glasses case:

5 Here are some glass vases which are all different shapes.

a Which vase is greatest in capacity? Explain your choice.

b Which vase has the smallest capacity? Explain your choice.

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Mass and temperature Mass is a measure of the amount that an object weighs. Common units of mass are milligrams (mg), grams (g), kilograms (kg) and tonnes (t). Temperature is a measure of how hot or cold something is. In New Zealand, we measure temperature in degrees Celsius (°C).

1tonne is the same as 1, 000 kg.

Explore 1

What is something that would have its mass shown in: a tonnes?

b grams?

d milligrams?

kilograms?

2 Finding the mass of a single sheet of paper would be difficult. Explain how you could use the information on this

note pad to work out the mass of one sheet.

AF T

NOTE PAD 100 sheets

3 Each lift has a sign that shows the mass it can carry safely.

D R

a What does the lift company think is the average mass of a person?

Do you think this is a good estimate?

— LIFT — Safe carrying capacity: 1 — tonne (8 people) 2

b Suppose the lift is in a primary school. Does this new information change your answer to part a?

4 MetService provides weather forecasts for all of New Zealand. a What does “forecast” mean?

b Find the weather forecast for where you live today. What is the predicted maximum temperature? c

Do you think today’s weather forecast is accurate?

5 A thermometer is a tool used to measure temperature. When MetService uses a thermometer to measure

air temperature, the tool has a special screen or shield that protects it from sunlight and rain. Why do you think the MetService protects thermometers this way, instead of placing them in the direct sun or rain? 124

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Angles 0

100

100 80

110

120 60 13 0 50

150 30

30 1 50

60 120

arms

vertex

180

170

170 10

20 160

160 20

180

An angle has two arms and a vertex. A protractor is used to measure angles. The unit of measurement is called a degree (°). When estimating the size of an angle, it helps to refer to standard angles like a right angle (90°) or a straight angle (180°) .

70 110

Explore 1

Classify each angle and then estimate the size of the angle in degrees. b

Angle type:

D R

AF T

Angle type:

Size of angle:

Size of angle: d

Angle type:

Size of angle:

2 Use a protractor to measure the angles in question 1. How close were your estimates? a

3 How good are you at estimating angles? a On the first line, estimate an angle of 25°. b On the second line, use a protractor to draw an angle of 25°. Was your estimate a good one?

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Time Time is measured in milliseconds (ms), seconds (s), minutes (min), hours (h) and so on. We choose an appropriate unit of time depending on how long the duration of time being measured is.

Deepen 1

Estimate the time it takes you to: a brush your teeth

b wash your hair

read a novel

d watch a movie

e travel to school

say the alphabet.

2 Compare your estimates in question 1 with some other students in your class. What is similar? What is different?

D R

AF T

Did you make any assumptions?

3 Use a stopwatch to measure the amount of time it takes you to say the alphabet.

How good was your estimate in question 1? 4 The world record in the men’s 100 msprint of 9.58

s econds was set by Usain Bolt on August 16, 2009 at the World Athletics Championships in Berlin. Work with another student to complete this activity. a One student should have a stopwatch or timer. The

other student should say start and then count out 10seconds. Say stop so the first student stops the timer. Record the time on the timer here. b Swap roles and repeat part a. Record the time on the

timer here. c

126

How accurate is the timer you are using? Could it have been used to record Usain Bolt’s 1 00 msprint time?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Measurement | Measuring

Topic 3.2 Converting units Converting units of length

× 1,000

We can convert between different units of length using conversion factors. • •

Divide by the conversion factor to move from a smaller unit to a larger unit.

× 100 m

÷ 1,000

× 10 cm

÷ 100

mm ÷ 10

Multiply by the conversion factor to move from a larger unit to small unit.

1 cm= 10 mm        1 m= 100 cm 1 km= 1, 000 m

How many millimetres are there in 1 km?

Learn 1

Complete these conversion tables. a

100 cm 7, 000 m 19, 000 m

6 km

D R

7.5 km

AF T

4 km 4 m 5.5 m

250 cm 7.1 m 820 cm

3, 500 m 4.25 km

1.56 m 75 cm

9, 750 m c

5 cm

2 What factor would you multiply by, to convert from: a kilometres to centimetres?

42 cm

b kilometres to millimetres?

90 mm 3.2 cm 75 mm 125 mm 12.4 cm 99 mm

metres to millimetres?

3 What factor would you divide by, to convert from: a millimetres to metres? b millimetres to kilometres? c

centimetres to kilometres?

4 Convert these lengths to the given units.

‾ c 890, 000 mm =    m ‾ e 0.36 km =    mm ‾ a 250 m =    mm

‾

b 4.5 km =    cm

‾ 230, 000, 000 mm =    km ‾

d 720, 000 cm =    km f

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Converting units of area

× 1,0002

We can convert units of area by squaring the length conversion factors. 1 cm  2= 10  2 mm  2 = 100 mm  2 1 m  2= 100  2 cm  2 = 10,000 cm  2 1 km  2= 1,000  2 m  2 = 1, 000, 000 m  2

km2

× 1002

× 102

cm2

÷ 1,0002

÷ 1002

mm2

÷ 102

Learn 1

Complete these conversion tables. a

km  2

m  2

m  2 3 m  2

2.5 km  2

45, 000 cm  2

3, 000, 000 m  2 0.5 m  2

0.8 km  2

120, 000 cm  2

450, 000 m  2 1.2 m  2

6 km  2

9, 500 cm  2

12, 500, 000 m  2 0.08 m  2

AF T

0.035 km  2 95, 000 m  2 cm  2

mm  2

4 cm  2 800 mm  2 0.6 cm  2

D R

cm  2

2, 750 cm  2

Make sure you use square units for area.

25, 000 mm  2 23 cm  2 3, 600 mm  2 0.09 cm  2 95 mm  2 2 Find the area of this rectangle in square metres in two ways. 500 cm 200 cm

128

Convert each length to metres and then calculate the area

ii Calculate the area in square centimetres and then

use an appropriate conversion factor

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

3 Calculate the area of each shape using the two methods in question 2. a Find the area of the square in cm 2.

60 mm i

b Find the area of the rectangle in cm 2. 1,000 mm 250 mm

Find the area of the triangle in mm  2.

60 cm

bh to find the 2 area of a triangle.

Use A =

D R

AF T

100 cm i

4 Which method out of the two used in questions 2 and 3 do you prefer? Explain.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Converting units of volume

× 1003

We can convert units of volume by cubing the length conversion factors. 1 cm  3= 10  3 mm  3 = 1,000 mm  3 1 m  3= 100  3 cm  3 = 1,000,000 cm  3

× 103

cm3 ÷ 1003

mm3

÷ 103

Explore 1

Complete these conversion tables. a

cm  3

mm  3

m  3

cm  3

2 m  3

5 cm  3

3, 000, 000 cm  3

9, 000 mm  3 0.5 m  3

0.8 cm  3

250, 000 cm  3

125, 000 mm  3 18 m  3

2.4 cm  3

15, 200, 000 cm  3

48, 000 mm  3 the volume and then using a conversion factor.

AF T

2 Find the volume of each 3D shape by i converting each length and then calculating the volume, and ii calculating

100 cm 50 cm

D R

a Find the volume of the rectangular prism in m 3.

200 cm i

b Find the volume of the cube in cm  3. 50 mm

3 Which method in question 2 did you prefer? Explain.

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Volume and capacity Volume is the amount of space an object takes up. Capacity is a measure of how much an object can hold. Liquids can be measured by capacity and by volume. There are standard conversions between units of volume and capacity.

× 1003

cm3

÷ 1003 ÷ 1,000

÷1

× 1,000

1 mL= 1 cm  3 1 L = 1, 000 cm         3 3 1 m  = 1, 000 L

×1

× 1,000 L

mL ÷ 1,000

Deepen 1

Complete the conversion table.   3 m 1 m  3

cm  3

500, 000 cm  3 250, 000 mL

AF T

100 L

0.0001 m  3

D R

10, 000 cm  3

500 mL

2 Explain how to quickly convert between cubic centimetres (cm 3) and millilitres (mL).

3 Find the capacity of each 3D shape. a

15 cm 23 cm

21 cm

6m 50 cm

3 cm 11 cm

Capacity = mL

Capacity = L

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Measurement | Measuring

Topic 3.3 Length and area Area of a parallelogram A parallelogram is a quadrilateral with two pairs of parallel sides. A rhombus is a parallelogram with four sides of equal length.

h b

The formula for the area of a parallelogram is A = bh.

A rectangle is a special parallelogram, so the area formula of a rectangle is the same as for a parallelogram.

Learn 1

Which other dimension would you need to know in order to find the area of this parallelogram? (It is not drawn to scale.)

2 cm

2 Find the area of these parallelograms.

A= m × m ‾ ‾       = m  2 ‾

10 m c

AF T

D R

3 mm 2.5 mm

4 cm 6 cm

5m e

12 mm 7m 18 mm 3 Use a ruler to measure this rhombus, then use your measurements to find its area.

4 Ellie calculated the area of this parallelogram as A = 3 cm × 4 cm = 12 cm 2. What has Ellie done incorrectly?

Does she have enough information to find the area?

3 cm 4 cm 132

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Area of a triangle The formula for the area of a triangle, A =_  1 bh, is derived 2 using the area of a parallelogram.

We can use two copies of any triangle to form a parallelogram with the same base length and height as the triangle. This means that the area of the triangle is half the area of the parallelogram.

Learn 1

Turn these triangles into parallelograms by drawing another triangle.

‾

a Aparallelogram   = cm  2

Atriangle   = cm  ‾ c Aparallelogram   = cm  2 ‾ 2

Atriangle   = cm  2 ‾

‾

b Aparallelogram   = cm  2

D R

area of the corresponding triangle.

AF T

2 Each square in the grid in question 1 is 1 cm  2. Find the size of each parallelogram using A = bh, and then find the

Atriangle   = cm  2 ‾ d Aparallelogram   = cm  2 ‾ A triangle = cm  2 ‾

3 Find the areas of these triangles. b

8m 12 mm 1.5 m

13 mm

4 Find the areas of these triangles by first converting the given lengths to the same unit. a

30 mm

10 cm

20 cm

A = cm  2 ‾

0.1 m

A = cm  2 ‾

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133

Area of a trapezium

A trapezium is a quadrilateral with one pair of parallel sides. The formula for the area of a trapezium is A =_  1 (a + b)h, where a and b are the lengths of the parallel sides, 2 and h is the perpendicular height.

h a

Learn 1

Find the areas of these trapeziums. 13 cm

16 mm

18 mm

10 cm

AF T

14 m

20 mm

22 cm

20 m

D R

f 8m

12 m

11 m

10 m

9m 12 m

22 m

2 Find the area of these trapeziums by first converting the given lengths to the same unit. a

25 cm 0.3 m

c 2m

0.8 m

280 cm 3m

0.2 m

2,000 mm

5,000 mm

A = m  2 ‾ 134

A = m  2 ‾

A = mm  2 ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Area of quadrilaterals Rectangle

Parallelogram

A = lw

A = bh

Identify the shape first and then which area formula to use.

Explore 1

Trapezium A = _  1 (a + b)h 2

Find the area of each quadrilateral. a

12 m

6 cm 8 cm

17 m

1 mm

0.4 cm

A = cm  2 ‾

D R

A = mm  2 ‾

AF T

25 mm

0.3 m

f 30 cm

A = m  2 ‾ 8m

9m 700 cm

0.005 m

A = cm  2 ‾

A = m  2 ‾

2 Does the formula A = bhalso apply to rectangles? Explain.

3 A square is both a rhombus and a rectangle. a Is this statement true? Explain.

b Does it matter whether you use the formula for the area of a rectangle or a parallelogram when finding the area

of a square? Explain. Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

135

Rearranging area of quadrilaterals formulas We can rearrange the formula for the area of a quadrilateral to find an unknown length.

Explore 1

Rearrange the formula for the area of a rectangle to find the unknown length of each of these shapes. a A = 80 m  2

b A = 21 cm  2

A = 240 m  2 c

3 cm 8m

12 m

AF T

2 When working with the area of a rectangle formula, explain why it doesn’t matter which side you choose to be the

D R

length and which side you choose to be the width.

3 Rearrange the formula for the area of a parallelogram to find the unknown length of each of these shapes. a A = 399 cm  2

b A = 13.5 m  2

A = 17 m  2 4m

21 cm 3m

A= bh

cm  2= b × cm  ‾           ‾   b= cm  2 ÷ cm ‾ ‾ b= cm ‾

m  2= b × m  ‾           ‾   b= m  2 ÷ m ‾ ‾ b= m ‾

4 When working with the area formula for a parallelogram, can you choose which length is the base and which is the

height? Explain.

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5 Fill in the blanks to make h the subject of the formula for the area of a trapezium.

A= _  1 (a + b)h 2

The subject is the variable by itself on one side of the formula.

2A =         ‾ h= ___________   2A +

6 Use the formula from question 5 to find the unknown height of each shape. a A = 110 m  2

b A = 186 cm  2 2m

A = 33 m  2

11 cm

20 cm

20 m

h 4m

7 Fill in the blanks to change the subject of the formula for the area of a trapezium. b Make b the subject:

AF T

A= _  1 (a + b)h 2 2A=    ‾ 2A            =   _    h ‾ a= _  2A − ‾

D R

a Make a the subject:

A= _  1 (a + b)h 2 2A=    ‾ 2A      =   _          h ‾ b= _  2A − ‾

8 If you don’t know the length of one of the parallel sides of a trapezium, does it matter whether you rearrange the

area formula for a or for b ? Explain. 9 Use a formula from question 7 to find the unknown length of each shape. a A = 546 cm  2

b A = 176 mm  2

26 cm

28 cm

A = 10 m  2 1m

11 mm

18 mm a 5m

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137

Calculating the area of shapes using given information In general, there are three pieces of information that we can know about 2D shapes: lengths (like sides and height), perimeter and area. We can use the area and perimeter formulae of shapes to move between the lengths, area and perimeter and find any unknown information.

How could you use the perimeter to find an unknown length that you need for the area formula?

Deepen 1

Find the area of each triangle using the given information. a P = 28 m

b P = 24 cm

P = 40 mm 14 mm

10 m

10 cm

8 cm

9 mm 11 mm

D R

AF T

2 Find the area of each quadrilateral using the given information. a P = 87 mm

b P = 26 cm

P = 48 m

20 mm 21 mm a

5 cm

27 mm

d P = 42 cm

e P = 133 cm

10 cm d 9 cm

138

35 cm

f e 0.4 m 0.3 m

0.2 m

P = 56 m

12 m f

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Calculating lengths of shapes using given information We can use the area and perimeter formulae of shapes to move between the lengths, area and perimeter and find any unknown information.

Deepen 1

For each triangle: i

use the area to find the length of the base a P = 104 cm

b P = 44 m

A = 420 cm  2

A = 52 m  2

use the perimeter to find the other unknown side length. c

P = 36 mm A = 54 mm  2

36 cm

12 mm

a 20 cm

Base = cm ‾ a = cm ‾ d P = 83 m

A = 220 m  2

D R

AF T

Base = m ‾ b = m ‾

e P = 100 cm

P = 94 m A = 330 m  2

A = 300 cm  2 43 cm

35 m

Base = mm ‾ c = mm ‾

30 cm

f 20 m

Base = m ‾ d = m ‾

30 m

Base = cm ‾ e = cm ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Base = m ‾ f = m ‾

139

2 Use the area and perimeter and given sides to find the height, h , of each triangle. a P = 77 mm,

A = 240 mm  2

22 mm h mm

b P = 61 m,

A = 200 m  2

18 m

15 mm

12 cm

14 cm

mm h = m h = ‾ ‾ 3 Find the unknown length of each rectangle using the given perimeter. a P = 226 cm

P = 32 cm, A = 30 cm  2

b P = 198 m

h = cm ‾ c

P = 232 mm

33 m

47 mm

AF T

50 cm

D R

l = cm l = m ‾ ‾ 4 Find the unknown length of each rectangle using the given area. a A = 155 mm  2

b A = 345 cm  2

l = mm ‾ c

23 cm

A = 384 m  2 12 m

31 mm

l = cm l = mm ‾ ‾ 5 Find the unknown lengths in these trapeziums. a P = 68 mm,

A = 270 mm  2

b P = 104 m,

A = 575 m  2

P = 33 m, A = 51 m  2

15 mm

17 mm

l = m ‾

30 m

28 m

25 m 20 mm

a = mm ‾ 140

35 m

b = m ‾

11 m

c = m, h = m ‾ ‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Calculating the perimeter of shapes using given information If we can find all side lengths of a shape, we can calculate its perimeter.

Deepen 1

Use the given information to calculate the perimeter of these triangles and rectangles. a A = 30 cm  2

b A = 21 cm  2

13 cm

A = 300 mm  2 63 mm

20 mm 8 cm

12 cm

9 cm 36 mm

7 cm

P = ‾ d A = 12 m  2

P = ‾ e A = 1, 600 mm  2

AF T

1.5 cm

D R

P = ‾ A = 9 cm  2

32 mm

P = ‾

2 Calculate the perimeter of these composite shapes using the given information. a Arectangle   = 8 cm  2

b Asquare   = 4 cm  2

Atriangle   = 6 cm  2

A trapezium = 9 cm  2 2 cm 3 cm 3.6 cm

5 cm

P = ‾

‾

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

P = ‾

141

Measurement | Measuring

Topic 3.4 Volume

Volume of rectangular prisms and cubes

The formula for the volume of a rectangular prism is V = lwh, where l is the length, w is the width and h is the height. As the side lengths of a cube are all equal, the volume of a cube is given by V = l  3. Before applying these formulae, make sure all your measurements have the same units.

What operation undoes cubing a number?

Learn 1

Find the volume of each rectangular prism in the given unit. a

6m 2.5 m

50 m

30 m

0.09 m 4 cm 22 mm

AF T

2,000 cm

2 Find the volume of each cube. a 1.5 m

V = cm  2 ‾

V = m  2 ‾

D R

V = m  2 ‾

c 20 mm

8 cm

V = m  3 V = cm  3 ‾ ‾ 3 Find the side length, l, of cubes with the following volumes. a V= 64 cm  3 b V= 27 m  3 l   cm   3  = √   ‾ = cm ‾ d V= 343 cm  3

_____________

l    3  = √   m  ‾ = m ‾ e V= 729 m  3

_____________

l   m   3  = √   ‾ = m ‾

l      cm   3  = √   ‾ = cm ‾ 3

142

V = mm  3 ‾ c

V= 1, 000 mm  3

3 l       3  = √   mm  ‾ = mm ‾ V= 256 mm  3

____________

______________

l      mm   3  = √   ‾ = mm ‾ 3

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Volume of triangular prisms The formula for the volume of a rectangular prism is V =_  1 bhl, where b is the base of 2 the triangular face, h is the height of the triangular face and lis the length of the prism. h

Explore 1

Calculate the volume of this triangular prism. V= _  1  × cm × cm × cm 2 ‾ ‾ ‾       = cm  3 ‾

5 cm

12 cm 4 cm

D R

AF T

2 What other measurement would be needed to be able to calculate the volume of this prism? Draw it on the prism.

8 cm 3 cm

3 Find the volume of the triangular prism in question 2 if the missing dimension is: a 1 cm

b 2 cm

5 cm

V = cm  3 ‾ 25 cm.

V = cm  3 ‾ d 10 cm

V = cm  3 ‾ e 20 cm

V = cm  3 ‾

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143

4 Calculate the volume of each triangular prism. a

8 cm

14 mm

18 m

4 cm 3 cm

6 mm

5 mm

V = cm  3 ‾

V = mm  3 ‾

5 Calculate the volume of each triangular prism, by first converting each length to centimetres (cm). b

14 m

800 m

1,200 m

1.55 m

127 mm 100 mm

V = cm  3 ‾

V = cm  3 ‾ d

141 mm

D R

1,000 m

20 m

AF T

400 mm

600 mm

4m 7m

1.82 m

1,800 mm

2.06 m

V = cm  3 ‾

144

V = cm  3 ‾

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Volume of composite shapes We can find the volume of a composite shape by dividing them into smaller shapes that we know how to find the volume of. For example, this composite shape can be divided into a rectangular prism and a triangular prism.

4m 2m

20 m 8m

Deepen 1

Find the volume of the composite shape shown in the example. Vrect.   prism = lwh

m × m × m = ‾ ‾ ‾ = m  3 ‾ Vtri.   prism = _  1  bhl 2

Keep track of which volume you are finding in your working.

D R

= m  3 + m  3 ‾ ‾ = m  3 ‾

AF T

            m × m × m          =_  1  × 2 ‾ ‾ ‾ = m  3 ‾ V= V rect.   prism + V tri. prism

2 Find the volume of these composite shapes. 20 cm a

6 cm

12 cm

4 cm

45 cm 90 cm

30 cm 5 cm

3 cm

10 cm

V = cm  3 ‾

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2 cm

15 cm

1,250 mm

14 cm

450 mm

12 cm

16 cm 455 mm

105 mm 800 mm

V = cm  3 ‾

V = mm  3 ‾

7.2 m

3m 2m

7.1 m

AF T

6.5 m

V = m  3 ‾

8 cm 8 cm

D R

3 Find the volume of these composite shapes by subtracting the volume of one 3D shape from another.

6 cm

3 cm

V = cm  3 ‾

4.3 m 8m 5m

10 m

12 m

V = m  3 ‾ 146

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Measurement | Measuring

Topic 3.5 Time

Converting units of time We can convert between different units of time using conversion factors. •

Divide by the conversion factor to move from a smaller unit to a larger unit.

•

Multiply by the conversion factor to move from a larger unit to small unit.

1 minute= 60 seconds 1 hour= 60 minutes            1 day= 24 hours 1 year= 365 days

× 365

years

× 24

days ÷ 365

Fill in the blanks to convert these times.

‾      = days ‾ 2 days= 2 × hours ‾      = hours ‾ 5 hours= 5 × minutes       ‾ = minutes ‾ 10 minutes= 10 × seconds     ‾   = seconds ‾ 1 minutes 120 seconds= 120 ×  _     

hours ÷ 24

× 60 minutes

÷ 60

seconds

÷ 60

Are you converting from a smaller unit to a larger unit, or from larger unit to a smaller unit?

Learn 1

× 60

a 3 years= 3 × days

D R

AF T

= minutes ‾ 1 g 240 hours= 240 ×  _ days      = days ‾

1 hours 600 minutes= 600 ×  _     

= hours ‾ 1 h 1, 095 days= 1, 095 ×  _ years      = years ‾

2 Convert these durations of time to minutes. a 2 hours and 15 minutes b 1 hour and 48 minutes c

3 hours and 40 minutes

d 6 hours and 23 minutes

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3 Convert these durations of time to hours and minutes. a 95 minutes

b 80 minutes

d 534 minutes

216 minutes

4 Some durations of time include fractions of a unit. Convert these durations of time, using decimals where required. a 3.5 hours to minutes

b 4.8 hours to minutes

d 6.4 hours to seconds

5.2 minutes to seconds

D R

g 36 hours to days

45 seconds to minutes

AF T

e 40 minutes to hours

h 48hours and 15minutes to days

5 When you say “half a year”, we don’t usually mean 182.5days. What unit of time do you usually use and how do

you convert between that unit and years?

Which years have more than 365days? When is the next year like this?

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Durations of time We can find the amount of time that passes between a starting time and an ending time of an event. From the starting time, count in minutes up to the next hour, then add the hours and finally the minutes until the ending time. Add these durations of time together, converting units as needed. For example, suppose we want to find the duration of time between 10:45 amand 1:20 pm. There are 15minutes between 10:45 amand 11 am, 2hours between 11 amand 1 pm, and 20minutes between 1 pmand 1:20 pm, so in total there are 15 minutes + 2 hours + 20 minutes = 2 hours and 35 minutesbetween 10:45 amand 1:20 pm.

Drawing a timeline might help to find the duration of time.

Explore Complete the table. Ending time

9:30 am

10:30 am

2:15 pm

3:45 pm

10:10 am

11:40 am

3:25 pm

5:05 pm

8:20 pm

10:05 pm

11:45 am

1:30 pm

1:35 am

3:20 pm

4:55 pm

6:40 am

0905

1155

1335

1810

0450

1735

2245

0150

Duration (hours and minutes)

AF T

Starting time

D R

2 Find the ending time if the duration is 2.5hours and the starting time is: a 11 am

b 1:30 pm

3:50 am

d 8:05 pm

e 10:50 pm

11:40 pm.

3 Find the starting time if the duration is 3hours and 15minutes and the ending time is: a 11:15 am

b 5:45 pm

12:35 pm

d 8:00 am

e 3:05 pm

1:55 am.

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Charts and timetables Information about duration can be obtained from charts and timetables.

Would you fly, take the train or drive between Auckland and Wellington?

Deepen 1

This train timetable shows some trains travelling from Ashgrove to Juniper Flat. a How long does it take for Train 8219 to

Saturday

get from Ashgrove to Juniper Flat?

SERVICE NO.

8215

8219

8221

8225

8227

from Ashgrove to Juniper Flat?

Train/Coach

TRAIN

What is the difference between the times of the shortest and longjourneys from Ashgrove to Juniper Flat?

Seating/Catering

b Which train takes the shortest time to get

ASHGROVE (Ashgrove Strand) dep.

13:00

14:00

15:00

16:00

13:08u

14:08u

15:08u

16:08u

Brookvale

d If you were going to Juniper Flat

Cedarford DUNEMARA

14:26u

16:26u

Eversholt

14:34

16:34

AF T

Train 8219 and wanted to get off Fernridge Crossing to meet friend, how long would you have wait for the next train?

Fernridge Crossing

13:42

D R

e Train 8227 leaves Ashgrove Strand

est

at 4:30 pm. It has the same travelling time and stops as Train 8225. Fill in the blanks the timetable using 24-hour time.

2 Find some information online about driving or

taking the train from Auckland to Wellington.

15:35

16:40

Grey Willow

14:44

16:44

Harbourleigh

14:46

16:46

Ivydale

13:50

14:50

15:43

16:50

JUNIPER FLAT arr.

13:56

14:54

15:47

16:54

Legend

a What is the average journey duration if

First Class available.

Catering available.

u – Stops to pick up passengers only.

you drive?

14:40

on at a to

Peak service.

arr. – Arrive. dep. – Depart.

IC – Inter-City.

Reservation required on these services.

b What is the average journey duration if

you take the train? c

Do you think the driving journey duration is accurate? Explain.

3 Find some information online about flights from Auckland to Wellington. a What is the average flight duration? b Approximately how much time do you save by flying compared to catching the train? c

Why might someone choose to take the train instead of fly? Give two reasons. I

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4 Pretend you are planning an itinerary for someone visiting New Zealand for the first time from

Melbourne, Australia. a Choose five places in New Zealand for them to visit. Make sure there are flights to and from Melbourne for at

least two of the places.

b Find a suitable flight from Melbourne to one of your places and fill in this table.

Flight number

Departure time (Melbourne)

Arrival time (New Zealand)

Flight duration

Use a map to determine the order they should visit your places in. Decide how long they should spend in each place, and what some possible activities are. The first place should be where their arrival flight lands. The last place should have a flight back to Melbourne available. Place

Name of place

How long?

Possible activities

AF T

D R

3 4 5

d Find transport between your places in New Zealand. Should they catch more flights, or use public transport

like trains, buses and ferries? Or is driving the best option? Fill in this table with your plans, including the expected duration of travel. Add in some time for stops and then find the time you should plan for them to spend travelling. Travel

Method of travel (including flight/ train/bus number)

Expected duration of travel

Time for stops

Total travel time

From 1 to 2 From 2 to 3 From 3 to 4 From 4 to 5 e Find a suitable flight from your last place back to Melbourne and fill in this table.

Flight number

Departure time (New Zealand)

Arrival time (Melbourne)

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Flight duration

151

Geometry | Shapes and Spatial reasoning

Topic 4.1 2D shapes

umference circ

Circles

diameter

The radius is the distance from the edge of a circle to the centre. The diameter is the length of a line through the centre of a circle that touches opposite points on the edge of the circle.

radius

The circumference is the distance around a circle.

Learn Name the part of the circle indicated by the arrow. a

AF T

2 For each circle, identify the diameter using the given radius. b

c 8 cm

D R

5 cm

2 mm

Diameter:

3 For each circle, identify the radius using the given diameter. a

c 12 mm

8 cm

Radius:

4 Use a compass to draw these circles. a Radius = 2 cm

152

b Diameter = 3 cm

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Transformations Translations A translation moves an object in a single direction, without rotating or changing the size of the object. Translations are described by the direction and distance of the movement.

Reflections A reflection mirrors an object over an axis of symmetry. Points on an object and its reflection are the same distance away from the axis of symmetry (mirror line).

Rotations A rotation moves a shape in a circular motion around a point known as the centre of rotation.

90° 3 units to the right

axis of symmetry

If A is the original shape, then A′ is the transformed shape.

Learn Describe the translations.

y 4

2 1

–3 –2 –1 0

3 x

0 –1

–2

y 4

AF T

2 3 B'

D R

2 C' 1 –4 –3 –2 –1 0

4 x

1 x

–3

2 Translate the points on the coordinate plane.

3 Translate the shapes on the coordinate plane.

y 4

2 1 E –3 –2 –1 0 –1 C –2

2 3 A D

B 4 x

–3

a Translate A up 3 units.

–2

–1

1 0 –1

–2

b Translate B left 4 units. c

Translate C left 2units and up 1unit.

d Translate D left 5units and up 3units.

a Translate A left 2 units.

e Translate E right 6units and up 4units.

b Translate B down 3 units.

Translate F right 5units and down 5units.

Translate C right 2units and up 1unit.

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153

4 Draw the axis of symmetry for these reflections.

y 4

2 1

–3

–4 –3 –2 –1 0 –1

–2

–1

4 x

0 –1

–2

–1

0 –1

2 x

–2

5 Reflect each shape across: i

the x -axis

ii the y -axis. a

3 2

–3

–2

–1

–3

–2

–1

6 Rotate each shape about the origin.

90° clockwise

0 –1

–2 –3

D R

–2

AF T

–1

ii 90° anticlockwise iii 180° clockwise a

–3

–2

–1

0 –1

–3

–2

–1

0 –1

–2

–3

Transformed shapes may overlap with the original shape!

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Scaling shapes Scaling an object makes it smaller or larger by a set amount called the scale factor. A point is selected to use as a reference when scaling. A scale factor between 0and 1is a reduction which decreases the size of the object, while a scale factor greater than 1 is an enlargement which increases the size of the object.

What would a scale factor equal to 1do?

Explore

Enlarge these shapes by a scale factor of 2by doubling the lengths of all the lines. Draw the scaled shape next to the original shape.

AF T

D R

2 Scale this shape by the given scale factors using the origin as the point of reference. a 2 b _   1 

y 8 7 6 5 4 3 2 1 0

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155

3 Identify the scale factors used to enlarge these shapes. a

y 6

y 9

9 x

2 1

Scale factor:

9 10 11 12 13 x

Scale factor: 4 Identify the scale factors if the shapes in question 3 were reduced. a Scale factor:

b Scale factor:

5 Consider the areas of the rectangles in question 2 in units  2. a What is the area of the original rectangle?

Divide the area of the enlarged rectangle by the area of the original rectangle. Compare this value to the scale factor of the enlarged rectangle.

AF T

b What is the area of the enlarged rectangle?

D R

d What is the area of the reduced rectangle?

e Divide the area of the reduced rectangle by the area of the original rectangle. Compare this value to the scale

factor of the reduced rectangle.

6 Create your own enlargement or reduction image. State the scale factor used. What reference point will you use?

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Combining transformations Multiple transformations can be performed on the same object in a sequence. Changing the order of any transformations performed can impact the resulting object.

Deepen 1

y 4 3 2 1

Perform these transformations on the given shape in a sequence. Show the resulting shape at each step. I

–1–10 –2 –3 –4

Scale by a factor of 2(use the origin as the reference point)

II Reflect across the x -axis III Translate 2units left and 1unit up.

2 Perform each sequence of transformations on the given shape.

1 2 3 4 5 6 7 x

What happens to each point (x, y)on the shape when you scale by a factor of 2 ?

y 4 3 2

–5 –4 –3 –2 –1 0

AF T

1 3

5 x

D R

a Reflect across the y -axis, then move right 1unit. b Move right 1unit, then reflect across the y -axis.

3 Perform each sequence of transformations on the given shape. a Scale by a factor of _  1 (use the origin as the

2 reference point), then rotate 90°clockwise

b Rotate 90°clockwise, then scale by a factor of _  1 

(use the origin as the reference point)

y 6

–4 –3 –2 –1 0

6 x

–4 –3 –2 –1 0

6 x

4 Explain why it is important to pay attention to the order transformations are applied. You might like to use some

grid paper to test more examples.

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Geometry | Shapes and Spatial reasoning

Topic 4.2 Angle rules Angles

An angle is formed at the point where two straight lines meet. Angles can be named using an angle symbol, ∠ , followed by a capital letter on one of the arms of the angle, a capital letter representing the vertex and a capital letter on the other arm of the angle. The letter for the vertex must be in the middle.

A B

For example, we can call this angle ∠ ABCor ∠ CBA.

Learn Give both possible names of these angles. a

A C

∠

∠ or ∠

H I

or ∠

∠ or ∠

AF T

2 Match the terms and definitions. A Adjacent angles

B Vertical angles

II Two angles which share a common vertex and side

D Supplementary angles

D R

C Complementary angles

Two angles which sum to 90°

III Two angles which sum to 180° IV Two angles opposite each other where two straight

lines intersect

Complementary = Corner (90°) Supplementary = Straight (180°)

3 In this diagram, find a pair of angles which are: a adjacent e

b vertical c

supplementary

d complementary. 4 Find the supplement of: a 50°

b 11°

137°.

b 34°

9°.

5 Find the complement of: a 82° 158

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6 Angles on a straight line sum to 180°. Find the unknown angles in these diagrams. Show your working. a

c a

45°

24°

140° 103°

58°

27°

7 Angles at a point sum to 360°. Find the unknown angles in these diagrams. Show your working. a

72°

134°

130°

65°

d 44°

D R

AF T

110°

130°

8 Find the unknown angles in these diagrams. Show your working. c

e d 45°

a c

i 135° g

34° f

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

67° 42° h

f g 38°

159

Parallel lines and transversals Parallel lines point in the same direction and never meet, no matter how far they are extended. Matching arrows on lines show they are parallel. A transversal is a line which crosses two or more lines. Corresponding angles

Co-interior angles

Alternate angles

a b

a + b = 180°

Corresponding angles on parallel lines are equal.

Co-interior angles on parallel lines are supplementary.

Alternate angles on parallel lines are equal.

Learn On this diagram, label: a the transversal b a pair of corresponding angles

a pair of co-interior angles

D R

AF T

d a pair of alternate angles.

2 This diagram shows a pair of parallel lines with a transversal.

Find every pair of:

H E

a corresponding angles

b co-interior angles

alternate angles.

3 Consider the diagram in question 2. a Find all angles equal to ∠ ABD.

b Find all angles equal to ∠ EFH.

Find all angles which are supplementary to ∠ABD.

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4 Describe the relationship between ∠ CBDand ∠ GFH. 5 Are there any pairs of complementary angles in the diagram in question 2? Explain.

6 For each diagram, name the relationship between the known and unknown angles, then find the size of the

unknown angle. a

c 77°

127°

a = ‾

b = ‾

7 From the information shown, calculate the sizes of the marked angles.

D R

b = ‾ f = ‾ j = ‾ n = ‾

a = 135° e = ‾ i = ‾ m = ‾

c = ‾

AF T

121°

c = ‾ g = ‾ k = ‾ o = ‾

d = ‾ h = ‾ l = ‾ p = ‾

8 If the diagram in question 7 was changed to this diagram, could you still find all the angles? Explain.

a e

i m

j n

k o

d h

l p

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161

Interior angles of triangles The interior angle sum of a triangle is 180°. We can use this information to find an unknown angle in a triangle.

We can write △ ABCfor a triangle with vertices A, Band C .

Explore 1

Find the unknown angles in these diagrams. a

72°

26° 63°

37°

a = ‾

c = ‾

AF T

46°

55°

b = ‾

82°

D R

d = ‾

49°

f = ‾

e = ‾ 2 Let’s prove that the interior angles of a triangle sum to 180°. a First, we draw this diagram by choosing one of the sides as the base of the triangle. P

How does the line passing through P and Q relate to the triangle?

b Label ∠ CABand an angle alternate to it in the diagram with single angle arcs.

As alternate angles are equal, these two angles are equal. c

Label ∠ ACBand an angle alternate to it in the diagram with double angle arcs. As alternate angles are equal, these two angles are equal.

d What do we know about the sum of ∠ PBA, ∠ABCand ∠ CBQ?

e Hence, what do we know about the interior angles of a triangle?

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Exterior angles of triangles

An exterior angle of a triangle is an angle between one of the sides and a line extended from the adjacent side. It is equal to the sum of the two opposite interior angles.

c b

d = a + b

Explore 1

Find the unknown angle in two ways. a Angles on a straight line sum to 180° 79° 56°

45°

of the two opposite interior angles

x + = 180 °

b Exterior angles are equal to the sum

x = 180° −

x = + =

= 2 Find the unknown angles. a

b 63°

109° a

77°

AF T

45°

66°

a =

41°

c =

D R

b =

3 Let’s prove that an exterior angle is equal to the sum of the

two opposite interior angles. Look at the diagram in the example and fill in the blanks.

a + b + c= ‾ c + d= ‾           a   +     b + c= c + d

a + b + c − = c + d − ‾ ‾ a + b= ‾ 4 Find the sum of the three exterior angles of these triangles. a

c 134°

139°

103°

96°

129° 78° 130°

128° 143°

Sum = d

Sum =

What do the exterior angles of a triangle add up to?

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163

Interior angles of quadrilaterals A quadrilateral is a polygon with four sides. The interior angle sum of a quadrilateral is 360°. We can use this information to find an unknown angle in a quadrilateral.

Explore 1

Find the unknown angles in these diagrams. b

c 72°

106° 95°

95°

62° b

b =

c =

d =

e = a =

AF T

2 We can use triangles to show that the interior angle sum of a quadrilateral is 3 60°.

First, divide the quadrilateral into triangles by drawing a line between two opposite vertices. a How many triangles is the quadrilateral split into?

A B

D R

b What is the interior angle sum of each triangle?

C c

Explain how you can use this information to find the interior angle sum of a quadrilateral.

3 A kite is a quadrilateral with two pairs of equal-length adjacent sides. We can show that there is a pair of equal,

opposite angles in a kite. a First, divide the kite into two triangles as shown.

b A

c d1

d2 D

Classify the two triangles by side length.

b1 c

ii What do we now know about b  1

and d1  ? iii What about b  2 and d 2  ?

b Fill in the blanks.

b = + and d = + . ‾ ‾ ‾ ‾ As d 1  = and d 2  = , we also know that d = + . ‾ ‾ ‾ ‾ Hence, b = d.

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Exterior angles of quadrilaterals

An exterior angle of a quadrilateral is an angle between one of the sides and a line extended from the adjacent side.

Convex quadrilaterals have no interior angles greater than 180°.

c d

Explore 1

Find the unknown angles in these diagrams. °

° 89°

63° 133°

57°

64°

107°

81°

° °

D R

AF T

67°

97°

135°

° °

2 Find the sum of the exterior angles of the quadrilaterals in question 1. a

3 For each vertex in a quadrilateral, there are two choices of exterior angle. Use this diagram to explain why it doesn’t

matter which one we choose. a b

e1 d c

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165

Interior angles of polygons The interior angle sum of a polygon with n sides is 180° × (n − 2).

Deepen 1

Check that the formula, sum = 180° × (n − 2), gives the correct interior angle sum for triangles and quadrilaterals. a Triangle

b Quadrilateral

2 Find the interior angle sum of these polygons using the formula, s um = 180° × (n − 2). b Hexagon

AF T

a Pentagon

Octagon

3 For each polygon, find the interior angle sum and then check that it is equal to the value given by the formula,

sum = 180° × (n − 2). b 107° 124° 80° 98°

D R

131°

Sum =

106°

74°

97° 98°

122°

106°

132°

95°

Sum =

215° 228° 94° 73°

4 The formula, sum = 180° × (n − 2), comes from splitting up a polygon into triangles. Follow these steps to

understand the formula. a Draw a polygon with at least 5 sides. b Number the vertices in order from 1to n , where n is the

number of sides of the polygon. c

Draw lines connecting vertex 1and vertices 3to n − 1to split up the polygon into triangles.

d Write an expression for the number of triangles in terms

of n . e Explain how you can use this information to show that the interior angle sum of a polygon is 1 80° × (n − 2).

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Exterior angles of polygons

An exterior angle of a polygon is an angle between one of the sides and a line extended from the adjacent side.

c e

A convex polygon has no interior angles greater than 180°. f

Deepen 1

Find the unknown angles in these diagrams. a

37° 115°

88°

127°

64° 146°

95° °

AF T

115°

52°

139°

80°

D R

2 Find the sum of the exterior angles of the polygons in question 1. b

3 The exterior angle sum of any convex polygon is 360°.

Follow these steps to understand why this is true. a Draw a convex polygon with at least 5 sides. Extend some

sides to show an exterior angle at each vertex. b Label the interior angles a  1 to a n  , where n is the number of

sides in your polygon.

Label an exterior angle adjacent to a  1 as 180° − a 1. For the remaining interior angles, label an exterior angle in a similar way.

d Fill in the blanks. You will need to use the formula for the

interior angle sum of a polygon. Sum  interior angles = a 1  + a 2 + … + an 

= ‾ Sum  exterior angles = 180° − a1  + 180° − a2  + … + 180° − an 

                       − (a  + a  + … + a  )     = 180° × 1 2 n ‾ = 180° × − ‾ ‾ = 180° × − 180° × n + 180° × 2 ‾ = ‾

What happens to exterior angles when the adjacent interior angle is greater than 180°?

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Geometry | Pathways

Topic 4.3 Positions and pathways Coordinate systems and grid references A coordinate system uses a pair of coordinates (x, y)to describe a precise location. An example of a coordinate system is the coordinate plane (as seen in Topic 2.4). Grid references refer to a wider area instead of a specific point. A rectangular grid is overlaid on a map. The axes are labelled between grid lines, so that each reference refers to a whole rectangle instead of a point on the grid. A combination of letters and numbers is often used to label the horizontal and vertical axes.

Learn 1

Can you find a map of where you live? Does it use grid references or coordinates?

On a map, grid lines help to show the position of something. On this map, the grid references show positions in squares. Dunedin is in C2.

D4?

ii E6?

4 3

If you flew from Wellington to Christchurch, in which direction would you be heading?

D R

AF T

b What is the grid reference in which there are two cities

shown?

a Which place is in:

d Which city is approximately north-west of Tauranga? 1 A

2 Here is the same map with a coordinate system instead of grid references. a Which place is closest to: i

(4, 4)?

ii (3.5, 5.5)? b Do any coordinate points show two cities? Explain.

5 4 3 2 1

3 Describe one advantage of using coordinates instead of grid references.

4 Many maps use grid references instead of coordinates. Why do you think this is the case?

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Map scales A map’s scale is the ratio between a distance on the map and the corresponding distance in the physical world. A map scale could be presented as a ratio, as in 1 cm : 1 km, which means that 1 cmon the map represents 1 kmin the physical world.

Explore 1

Suppose a map has a scale of 1 cm : 1 km. a If these are the distances on the map, what is the distance in the physical world? Give your answers in kilometres. i

2 cm:

ii 10 cm:

iii 35 mm:

iv 19 mm:

b If these are the distances in the physical world, what would the distances on the map be? i

50 km:

ii 4.8 km:

iii 2, 300 m:

iv 500 m:

2 The map scale 1 cm : 1 kmcould also be presented as 1 : 100, 000. This means that if you measure a distance on a

map, then you can multiply it by 100, 000to get the distance in the physical world. Using your understanding of converting between metric units, explain where the value 100, 000comes from in this map scale.

AF T

3 Most maps use a visual representation of a scale. This is a map of Oxford, which is about 5 0 kmnorth-west

of Christchurch.

D R ne La

PI ai M at

Ro ad ill tH rn Bu

Stree

treet

eet

ka We

Totara Dr

Weka

Hana

treet

S Rata

ai Str

Tui S Place

High

Burnett

Kowh

Ahika

venue Park A

Maia

Stree

Maaka

Aven

Petrol Station Ambulance

Rata

Park

Tawera

Olivea

oad

Hill R

Coney

Centre

Burnt

Shoppin

eet ain Str

l RD

Oxford Motel

Meyer

t Hil

Church

Oxford Squash Club

Oxford Museum

Burn

b Describe the map scale in words.

Bay Ro

Oxford School

a Circle the scale on the map.

Oxford Hospital

xte

eet

Str

Ruru P

100 m

Find Maia’s house. How do you know Maia lives less than 500 mfrom Harewood Road?

d What can be found about 200 mwest of the shopping centre?

e Find Hana and Maaka’s houses. Whose journey to school is about 1 km? Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

169

4 This is a map of most of the North Island. Kerikeri 7

Whangārei

Mangawhai Warkworth

Auckland 5 Tauranga

Hamilton

Rotorua

New Plymouth Te Ika-a-Māui 3 /North Island

Gisborne Napier

Whanganui 2

Palmerston North

b Describe the map scale in words.

Which town is approximately: i

D R

a Circle the scale on the map.

AF T

Upper Hutt Wellington B C D

Scale 100 km

100 kmsouth-west of Gisborne?

ii 65 kmnorth-west of Palmerston North? iii 150 kmsouth-east of Hamilton? d Find the approximate direct distance between Auckland and Wellington. e If you drove from Auckland to Wellington, would you drive the exact distance you gave in part d? Explain.

170

Estimate the driving distance between Auckland and Wellington by looking at the roads marked on the map. Check your answer using a map online. How good was your estimate?

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Bearings

000°

A true bearing is a way to describe precise direction using angles. True bearings are measured as the clockwise angle between north and the given direction. True bearings are always written with three digits, and sometimes have the letter Tadded to indicate that they are a true bearing.

315°

045°

090°

270°

225°

135° 180°

Deepen 1

Write the compass direction for each of these true bearings. a 000°:

b 045°:

d 135°:

090°:

225°:

AF T

e 180°:

h 315°:

D R

g 270°:

2 What compass direction would 360°be?

3 Look back at the map of most of the North Island. a Of the eight main points on a compass, estimate the closest direction to the one you would be heading in if you

were flying from: i

Rotorua to Gisborne

ii Auckland to New Plymouth

True bearings are always written with 3 digits.

iii New Plymouth to Wellington

iv Wellington to Napier.

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b A3 60°protractor would be helpful for the following. If you don’t have one, use a 1 80°protractor very carefully.

Give the true bearing to the nearest degree for the four flights in part a. i

Rotorua to Gisborne:

ii Auckland to New Plymouth:

iii New Plymouth to Wellington:

iv Wellington to Napier:

4 Explain why the 8main compass directions are not used as directions by pilots.

AF T

5 If a pilot were using the following true bearings, which town would she be flying to? a 305° from Tauranga:

D R

b 45° from Napier:

215° from Hamilton:

6 Which two towns are at a heading of 350°from Auckland?

7 Compass bearings are similar to true bearings. We identify whether the direction is closer to north or south, and then

describe how far east or west the direction is. For example, we could describe, north-east as the true bearing 045° or as the compass bearing N45°E (read as north 45° east). Complete the following table. Direction

Closer to north or south?

Compass bearing

north-east

north

N45°E

north-west south-east south-west 172

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Statistics | Developing knowledge from, visualising and interpreting data

Topic 5.1 Statistical measures Measures of central tendency

To understand numerical data, we can look for a way to describe the centre of the data. The mean is the numerical average of the values in a data set. To find the mean, add the data values together and then divide by the number of data values. The mode is the data value with the highest frequency. There can be no mode, one mode or more than one mode in a data set. The mode can be found for categorical or numerical data. There is no mode when all the values appear with the same frequency. The median is the middle value for sorted numerical data. •

If there is an odd number of data values, the median is the middle value.

•

If there is an even number of data values, the median is the average (mean) of the two middle values.

Learn Calculate the mean, median and mode(s) of these data sets. b 1, 2, 4, 6, 7, 7

AF T

a 7, 9, 10, 10, 11, 14, 16

Mean =

Median =

Mode(s) =

Mean =

Median =

D R

Mode(s) =

6, 8, 8, 10, 12, 14

Mean =

Median =

Mode(s) =

2 This frequency table shows the number of siblings each Year 7 student in a class of 2 2has.

Number of siblings 0 1 2 3 4 5

Frequency 5 9 4 3 0 1

a Find the: i

mean

ii median iii mode(s).

b Was it easier to find any of the three values in part a with the data presented in a frequency table? Explain.

Do you prefer lists of data values or frequency tables?

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3 This stem-and-leaf plot shows the time it takes 28students to get to school.

Stem Leaf 0 359 1 1239 2 5788 3 24899 4 002345 5 7899 6 02 Key: 1 ∣ 2 = 12 minutes a Find the: i

ii median

mean

iii mode(s).

b Was it easier to find any of the three values in part a with the data presented in a stem-and-leaf plot? Explain.

AF T

4 This bar graph shows the ages of 50players at a netball club.

Ages at a netball club 14 10

D R

Frequency

12 8 6 4 2 0

15 16 Age (years)

a Find the: i

ii median

mean

iii mode(s).

b Was it easier to find any of the three values in part a with the data presented in a bar graph? Explain.

5 The arrows A, B and C represent the mean, median and mode of the data. Without performing any calculations,

identify which statistical measure belongs to which arrow.

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Range and outliers The range measures the spread of a numerical data set. It is equal to the difference between maximum and minimum values of a data set. For the data set 3, 5, 5, 8, 9, 9, 10, 12, the range is 12 − 3 = 9. An outlier is a data value that is much smaller or larger than the other data values. When an outlier is present, we need to consider the impact of an outlier when using the range as a measure of the spread of the data set. For example, if we add 120to the example data set, the range becomes 120 − 3 = 117, which is not a good representation of the data set.

Explore Calculate the range of these data sets. a 7, 9, 10, 10, 11, 14, 16:

Number of siblings 0 1 2 3 4 5

b 1, 2, 4, 6, 7, 7:

Frequency 5 9 4 3 0 1

D R

14 Frequency

12 10 8

6, 8, 8, 10, 12, 14:

Stem Leaf 0 359 1 1239 2 5788 3 24899 4 002345 5 7899 6 02 Key: 1 ∣ 2 = 12 minutes

2 Elliot thinks that the range of the data set in

Ages at a netball club

AF T

question 1d is 9 − 0 = 9. Explain why Elliot is incorrect.

6 4 2 0

15 16 Age (years)

3 Here is a data set.

Frequency

Results 10 9 8 7 6 5 4 3 2 1 0

a Calculate the range. b Identify the outlier. 1

5 6 Score

Calculate the range without the outlier.

d Do you think the range with or without the outlier is a better representation of the spread of the data set? Explain.

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Using measures of central tendency and spread To answer a statistical question, we often need to calculate a measure of central tendency or spread. For example, if we are investigating the heights of students in the class, we would need to collect the data of the height of all the students and then calculate the mean, median, mode and range.

Deepen 1

A teacher surveyed his 27students about their confidence in maths. He used a scale from 1 (not very confident) to 5(very confident). The results are shown in this bar graph.

10 9 8 7 6 5 4 3 2 1 0

a Calculate the mean, median, mode

and range of the data set. i

Mean =

ii Median = iii Mode(s) = iv Range =

3 Score

AF T

Frequency

Student confidence in maths

b Based on the statistical measures you have calculated, how would you respond to the question “How confident

D R

are the students in maths?” Explain your answer.

2 This stem-and-leaf plot shows the amount of screen time each week (to the nearest hour) of 2 5Year 8 students.

Stem Leaf 0 79 1 122344556677889 2 0123344 3 4 Key: 1 ∣ 5 = 15 hours

a Calculate the mean, median, mode and range of the data set. i

Mean =

ii Median = iii Mode(s) = iv Range =

b Based on the statistical measures you have calculated, how would you respond to the question “How much

screen time do Year 8 students have each week?” Explain your answer.

176

A newspaper article uses this data to claim that students spend over a day each week on their phones. Does the data support this claim?

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3 This frequency table shows the amount of sleep students in a Year 8 class got last night (to the nearest hour).

Hours of sleep last night 3 4 5 6 7 8 9 c

Frequency

a Identify the outlier in the data set. b Calculate the mean, median, mode and range of the data set including the

1 0 0 3 7 9 5

outlier and then again excluding the outlier. Measure With outlier Without outlier

Mean

Median

Mode(s)

Range

Based on the statistical measures calculated with the outlier, how would you respond to the question “Did students in Year 8 get enough sleep last night?”

d Based on the statistical measures calculated without the outlier, how would you respond to the question “Did

e Which response is better? Explain.

D R

AF T

students in Year 8 get enough sleep last night?”

4 We can use measures of centre and spread to compare two data sets. The heights of students in two Year 8 classes

was collected and the mean, median, mode and range were found for both classes. Measure Class 8A Class 8B

Mean 161 cm 165 cm

Median 160 cm 166 cm

Mode(s) 160 cm 168 cm

Range 18 cm 29 cm

a Based on the statistical measures, which class has: i

taller students?

ii more consistent student heights? b Explain how you can tell which class has more consistent student heights.

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Statistics | Developing knowledge from, visualising and interpreting data

Topic 5.2 Statistical investigations Dot plots and bar graphs Univariate data is data about a single variable.

Dot plots and bar graphs can be used to represent univariate numerical or categorical data. They display the frequency of data values. Dot plots and bar graphs should have a title, both axes labelled and clearly labelled data values. A bar graph should also have a consistent scale starting at 0on the frequency axis.

Learn 1

Students in Year 8 were asked how many minutes they spent working on homework last week. Their responses are listed. 20, 17, 23, 15, 18, 22, 17, 24, 18, 16, 19, 20, 14, 25, 17, 21, 18, 23, 16, 22, 19, 20, 17, 18, 21, 15, 16, 19, 18, 20 a Summarise the data into this frequency table.

Data value

AF T

Tally

Frequency b Represent the data as a dot plot.

Frequency

D R

Time spent on homework by Year 8 students

Why might you choose a bar graph instead of a dot plot to represent a data set?

14 15 16 17 18 19 20 21 22 23 24 25 Time spent on homework (min) 2 Students in Year 8 were asked their favourite sport. Their responses are listed.

Rugby, Tennis, Cricket, Netball, Rugby, Other, Cricket, Rugby, Netball, Tennis, Rugby, Cricket, Other, Rugby, Netball, Rugby, Cricket, Tennis, Rugby, Other, Cricket, Netball, Rugby, Cricket, Rugby, Tennis, Other, Cricket, Rugby a Summarise the data into this frequency table.

Data value

Tally

b Represent the data as a bar graph. Favourite sports of Year 8 students

Frequency

Rugby Netball Tennis Other

Frequency

Cricket

10 8 6 4 2 0

Rugby Cricket Netball

Tennis

Other

Sport

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Other univariate data visualisations Stem-and-leaf plots can be used to represent numerical data. Each piece of data is split into two parts: the stem (the first digit(s)) and the leaf (the last digit). The place value of the stems is indicated using a key. Leaves are written in increasing order away from the stem in neat columns.

Preferred day for assessments 10% 10%

Monday 15%

Stem Leaf 1 2468 2 135 3 147

Tuesday Wednesday

35% 30%

Thursday Friday

Key: 1 ∣ 2 = 12 A pie graph can be used to represent categorical or discrete numerical data. They are useful when we are interested in the proportion of each category.

Learn 1

Consider this stem-and-leaf plot. Stem Leaf 1 3578 2 1346 3 1358

a If the key is 1 ∣ 3 = 130, list the data values and find the range.

AF T

Range =    ‾ b If the key is 1 ∣ 3 = 0.13, list the data values and find the range.

D R

Range =    ‾

2 Create a stem-and-leaf plot for this data set.

212, 215, 218, 219, 221, 224, 226, 229, 230, 233, 235, 237

3 Create a stem-and-leaf plot for this data set.

1.4, 1.6, 1.9, 2.2, 2.4, 2.7, 2.9, 3.1, 3.3, 3.6, 3.8, 3.9 Stem

Leaf

Stem Leaf

Key: ∣ = ‾ ‾ ‾

4 Complete the table and then use it to draw the corresponding pie graph. Make sure you label each sector with its

name and the corresponding percentage. Favourite pet

Frequency

Cat

Dog

Fish

Bird

Percentage proportion _   8   = _   2 = 40% 20 5

Angle 2 _     × 360° = 144° 5

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Bivariate bar graphs Paired categorical variables can be visualised through a stacked bar graph or a clustered bar graph. A stacked bar graph displays an additional categorical variable in each bar. The bars are divided into sections and a legend shows what each section represents. Stacked bar charts help us to compare totals and the proportions of the stacked variable.

100 90 80 70 60 50 40 30 20 10 0

Stacked bar graph

Clustered bar graph

Ways to get to school across a week

Ways to get to school across a week 40 35

Train Car Walk

Frequency

A clustered bar graph displays data for two-or-more categorical variables side-by-side. Clustered bar graphs help us to compare values directly between groups within the same category.

25 15 10

Bus Monday

Tuesday Wednesday Thursday

Bus Walk Car Train

5 0

Friday

Monday Tuesday Wednesday Thursday Friday

Day

These graphs represent the same data.

AF T

Explore Look at the bar graphs in the example.

D R

a Identify the two categorical variables. and b Which graph would you use to compare: i

the proportion of each way to get to school across the week

ii the total number of students travelling to school each day iii the number of students who took the bus each day? 2 This table shows the eye colours of students in Years 7 and 8.

Blue 14 12

Green 8 10

reate a stacked bar graph C to represent the data in the table. Remember to fill in the legend with the colour or pattern you use for each year level.

Brown 20 22

Hazel 8 6

a Identify the two categorical variables in the table.

and Eye colours in Years 7 and 8

45 40 35 Frequency

Eye colour Year 7 Year 8

Year 7

Year 8

25 20 15 10 5 0

Blue

Green

Brown

Hazel

Colours 180

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3 Consider again the table of eye colours of students in Years 7 and 8. Create a clustered bar graph to represent the

data in the table. Remember to fill in the legend with the colour or pattern you use for each year level. Eye colours in Years 7 and 8 25

Frequency

Year 7 Year 8

15 10 5 0

Blue

Green

Brown

Hazel

Colours 4 Stacked bar graphs often display percentage proportion, instead of raw frequency. a There are 50students in each of Years 7 and 8. Calculate the percentage proportion of each eye colour in both

year levels.

Brown Hazel

Percentage proportion

28%

Frequency

Percentage proportion Frequency Percentage proportion Frequency

Year 8 12 10

AF T

Green

Year 7 14

D R

Blue

Eye colour Frequency

Percentage proportion

b Draw a stacked bar chart with the two year levels on the horizontal axis and the eye colours as the stacked

variable with a legend. Use percentage frequency instead of frequency on the vertical axis. 5 Compare the three graphs of the

Eye colours in Years 7 and 8

same data that you have drawn in this topic. How do they impact your interpretation of the data?

Percentage frequency

100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0

Blue Green Brown Hazel

Year 7

Colours

Year 8

Don’t forget to fill in the legend!

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Time-series graphs

Number of visitors to the museum per month 7000

A time-series graph is a line graph that shows the value of a numerical variable at regular intervals over a period of time.

Number of visitors

6500

We can describe the trends in time-series data. When describing trends we look at the big picture, instead of the details. •

6000 5500 5000 4500 4000 3500 3000

An upwards trend shows data values increasing as time passes.

6 7 Month

•

A downwards trend shows data values decreasing as time passes.

•

A stationary trend shows data values not increasing or decreasing as time passes.

•

Sometimes there is no observable trend in the data.

The break in the vertical axis shows that there is a jump up Look at the time-series graph in the example. to the data values a Describe the trend from Month 2to Month 7. from 0which is not to scale.

AF T

Explore

b Describe the trend from Month 7to Month 12.

If the months correspond to the months of the year (i.e. Month 1 is January, etc), what could explain the peak in the middle of the year?

D R

2 Casey started an online store and has been tracking her sales each week. a Create a time-series graph to show Casey’s sales. Plot the points and then join successive points with a straight line.

Number of sales 6 15 22 25 24 32 38 44 53 58

Online store sales 65 60 55 50 Number of sales

Week 1 2 3 4 5 6 7 8 9 10

45 40 35 30 25 20 15 10 5 0

Week

b Describe the trend in the time-series graph. c If Casey continues to run her business, do you think her sales will always follow the same trend? Explain.

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Choosing data visualisations When choosing a data visualisation, consider whether the data is univariate or bivariate, and whether any variables are numerical or categorical. Think about the number of data values and how they are represented in a data visualisation. Also consider anyone looking at your data visualisation. Do they need to be able to calculate statistical measures, or do you want them to compare the proportions of different categories?

Explore 1

Consider the data in this frequency table. Favourite ice cream flavour Vanilla Chocolate Strawberry Cookies and cream

a Why is a dot plot not suitable to represent this data?

Frequency 82 76 41 58

b Explain why a bar graph is a suitable way to represent this data.

Full-time Part-time Casual Unemployed

D R

18–24 25 30 22 13

AF T

2 Consider the data in this table which gives the age group and employment status of 3 00people.

25–39 45 18 12 10

40–54 40 16 9 5

55+ 28 14 6 7

a Why might someone choose to represent this data as a clustered bar graph?

b Why might someone choose to represent this data as a stacked bar graph?

Which variable would you put on the horizontal axis and which as the one described by a legend? Explain.

This table gives the daily maximum temperature in ° Cin Wellington over a two-week period. Day Temperature (°C)

1 28

2 22

3 23

4 19

5 22

6 18

7 19

8 20

9 19

10 17

11 19

12 18

13 17

14 18

4 What type of data visualisation would you use for this data? What would it help you to identify?

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Interpreting data visualisations A good data visualisation should allow viewers to discern the variable(s) and who the data was collected from. We can also find other information, like the units for numerical variables, the frequency or proportion of categories and identify any patterns or trends.

Deepen 1

This bar graph shows bivariate data.

Average rainfall in March Nelson

a Identify the categorical

b Identify the numerical

variable. c

What are units for the

City/Town

variable.

Christchurch Queenstown Invercargill

numerical variable?

Dunedin

d Does it make sense to calculate

10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 Average rainfall (mm)

AF T

any statistical measures for this data? Explain.

D R

2 This line graph shows the monthly school canteen sales

of ice blocks and soup. Use the information to complete the activities. a When were the highest sales of: i

Sales of ice blocks and soup at the school canteen

= Ice block = Soup

120 110 100

ice blocks

b In which month do you think the soup

machine was broken? In which month were 115ice blocks sold?

Number of sales

ii soup?

Key

80 70 60 50 40 30

d Sales of both soup and ice blocks were low

in January. Give the most likely reason.

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

e Describe the trend of soup sales over the school year.

The canteen manager can’t decide whether to continue with soup sales in summer. What would your advice be?

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Identifying and interpreting outliers An outlier is a data value that is much smaller or larger than the other data values. Data visualisations make outliers easier to identify. When we identify an outlier in a data set, we should try to explain why it is present. Some outliers are mistakes in the data set, while others might be very rare events.

Can you have an outlier in categorical data?

Deepen 1

Consider this dot plot.

Scores recorded in an online game

a Identify the outlier.

Give two possible reasons that would explain the presence of the outlier.

AF T

D R

II 2 Consider this bar graph.

Weights of adult cats at the vet today 10

b Do you think the outlier represents a mistake in the

Frequency

a Identify the outlier.

data? Explain.

6 4 2 0

Weight (kg) c

Suppose the value of the outlier was instead 29. Give two reasons why we might choose to discard the outlier before analysing the data. I

II 3 What would an outlier look like on a time-series graph? Explain.

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Describing the shape of data visualisations With a univariate numerical data display we can describe the shape of the data. The sides of these visualisations are known as tails. If the sides taper at the same rate, the data visualisation is symmetrical. If they taper at an uneven rate, the shape is skewed. If an outlier is present, describe the shape of the rest of the data set and include that there is an outlier. Symmetrical

Positively skewed

Negatively skewed

Deepen Describe the shape of each data visualisation. Remember to mention an outlier if one is present. a

Stem Leaf 3 8 4 13579 5 1357 6 0

5 6 7 8 9 10 11 12

5 6 7 8 9 10 11 12 13 14 15 16

D R

0 1 2 3 4 5

AF T

Stem Leaf 2 2456789 3 0138 4 5 5

Key: 1 ∣ 3 = 13

Stem Leaf 3 0 4 5 6 8 7 2456789 8 012 Key: 1 ∣ 3 = 13

2 Consider the data visualisations in question 1. a Calculate the mean and median for the graph in 1a.

Mean: Median: b Calculate the mean and median for the graph in 1b, excluding the outlier.

In a symmetrical distribution with no outliers, the mean and median are approximately equal to the mode.

Mean: Median: c

Calculate the mean and median for the graph in 1c, excluding the outlier. Mean: Median:

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Misleading graphs Sometimes data visualisations are missing information which makes them misleading. When you look at a data visualisation, check for features like a title, axis labels, a legend or key and a consistent scale.

Who might want to use a misleading data display?

Deepen Describe what is misleading in each data visualisation. a Population index

120

120 100

Water consumption over time

Population change 100

Amount used

80 60 40

16 15 14 13

20 2019

2023 Year

4 Month

AF T

D R

C A

42%

28%

31%

3 Find a misleading graph in the media. You could look

2 What is misleading about this bar graph?

in a newspaper or online. Explain how the graph is misleading.

Favourite colours

Frequency

8 6 4 2 0

Red

Yellow

Green Colours

Blue

Purple

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Probability | Theoretical probability

Topic 6.1 Calculating probabilities Theoretical probability with coins Probabilities can be expressed on a scale from 0(impossible) to 1(certain). If all outcomes are equally likely, the theoretical probability of an event is: number of ways the event can happen   ________________________________             total number of possible outcomes The probability of an event A can be written as P(A). Many probability experiments are about flipping fair coins.

Learn 1

List the sample space if you flip a fair coin once. Sample space = { }

2 Find:

‾

a P(heads) =

b P(tails) =

AF T

3 Imagine flipping two fair coins. Complete this table with the possible outcomes, where H represents landing on

heads and Trepresents landing on tails.

4 How many equally likely outcomes are

H Coin 1

D R

Coin 2

(H, H)

( , )

there when flipping two fair coins?

( , )

5 What is the probability of each outcome

( , )

when flipping two fair coins?

6 Find:

‾

a P((H, H)) =

b P((T,T)) =

P((H,T) or (T, H)) = ‾

7 The complement of an event A is the event Not A, which we can write as A ′ . The probabilities of complementary

events sum to 1.

Let A be the event of flipping two fair coins and both landing on heads. a Describe the outcomes in A ′.  b Calculate P(A′)  . 8 This tree diagram shows the possible outcomes of flipping two

First coin

Second coin

fair coins. Follow each branch to find all the possible outcomes.

How many outcomes are there if you flip three fair coins?

Outcomes H

H, H

H T H T T

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Theoretical probability with dice Many probability experiments are about rolling dice. Pay attention to the number of sides on any die as this impacts the probability of each outcome.

A fair die has an equal chance of rolling any value.

Explore 1

Consider rolling a fair six-sided die numbered from 1to 6. a List the sample space if you roll a fair six-sided die numbered from 1 to 6once.

Sample space = { }

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b Here are six events. Calculate:

P(rolling a 2) = ii P(rolling a 6) = ‾ ‾ iii P(rolling an even number) = iv P(rolling a number less than 6) = ‾ ‾ v P(rolling a prime number) = vi P(rolling a number greater than or equal to 2) = ‾ ‾ c Describe the complement of each event in part b without using the word “not” and calculate the corresponding probability. Remember, probabilities of complementary events sum to 1.

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P((rolling a 2)′  ) = ‾ ii P((rolling a 6)′ ) = ‾ iii P((rolling an even number)′ ) = ‾ iv P((rolling a number less than 6)′ ) = ‾ v P((rolling a prime number)′ ) = ‾ vi P((rolling a number greater than or equal to 2)′ ) = ‾

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2 Consider rolling two fair six-sided dice numbered from 1to 6,

and recording the outcome as an order pair. Complete this table to show all possible equally likely outcomes.

Tables help us to find all outcomes in multi-stage experiments.

Second die

First die

(1, 1)

(1, 2)

(1, 3)

(1, 4)

(1, 5)

(1, 6)

(2, 1)

(3, 1)

(4, 1)

(5, 1)

(6, 1)

3 If you roll two fair six-sided dice: a how many possible outcomes are there?

b what is the probability of each outcome?

4 Use your answers to questions 2 and 3 to help you calculate the following. b P(rolling a 1 using either of the dice) = ‾ ‾ c P(the sum of the numbers rolled being equal to 7) = ‾

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a P(rolling two of the same number) =

probability. a

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5 List the outcomes which correspond to the complement of each event in question 4 and calculate the corresponding

P((rolling two of the same number)′ ) = ‾ b

P((rolling a 1 using either of the dice)′ ) = ‾ c P((the sum of the numbers rolled being equal to 7)′ ) = ‾ 6 If you were playing a game where you had to roll two fair six-sided dice and predict the outcome, what value would

you pick? Explain. Coin

Die 1

7 Consider an experiment where you flip a coin and then roll a fair

4-sided die. a Use this tree diagram to find all possible outcomes of this experiment.

Write the outcomes as ordered pairs, like in question 2.

All of the outcomes are equally likely. What is the probability of T

each outcome?

b How many possible outcomes are there? c

Outcomes (H, 1)

2 3 4

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Theoretical probability with cards Many probability experiments are about drawing cards from a standard deck of cards. Each deck has 5 2cards, broken up into 4suits of 13cards.

Deepen 1

The suits in a standard deck of cards are Hearts, Diamonds, Clubs and Spades. Consider drawing a card from a standard deck of cards. a List the sample space for the suits of a pack of cards.

Sample space = { } b Find:

P(Hearts) = ii P(Diamonds) = ‾ ‾ iii P(Clubs) = iv P(Spades) = ‾ ‾ c What is the sum of the four probabilities in part b?

AF T

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2 Each of the four suits in a standard deck of 52cards has 13cards: Ace, 2, 3, 4, 5 , 6 , 7, 8, 9, 10, Jack, Queen, King.

Consider drawing a card from a standard deck of cards. a Find:

P(Ace) = ‾ b Find: i

P(Ace of Hearts) = ‾

‾

ii P(5) =

iii P(Jack) =

‾

ii P(5 of Diamonds) =

‾

iii P(Jack of Clubs) =

3 Each suit in a standard deck of cards is either Red or Black. Diamonds and Hearts are Red, and Clubs and Spades

are Black. Consider drawing a card from a standard deck of cards. a Find:

P(Red) = ‾ b Find: i

‾

ii P(Black) =

P(Red Ace) = ii P(Black 5) = ‾ ‾ c The cards Jack, Queen and King are known as Picture cards. Find: i

‾

iii P(Red Jack) =

P(Picture card) = ‾ ii P(Red Picture card) = ‾ iii P(Picture card of Spades) = ‾ i

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4 Calculate the following probabilities using your answers from questions 1 to 3.

‾ c P((5 of Diamonds)′ ) = ‾ e P((Red Jack)′ ) = ‾

The probabilities of complementary events add to 1.

‾

a P((Hearts)′ ) =

b P((Ace)′ ) =

‾ P((Picture card)′ ) = ‾

d P((Red)′ ) = f

5 Suppose you flip a fair coin and then draw a card from a standard deck of cards. Note that we are using shorthand:

T = Tails, Hd = Heads, Ht = Hearts, D = Diamonds, C = Clubs and S = Spades. a Complete this tree diagram of possible outcomes if you record which suit the card belongs to. Coin

Suit

Outcomes Ht

b All of the outcomes in the tree diagram are

equally likely. Find:

(Hd, Ht)

P((Hd, D)) = ‾ ii P(The suit is C) = ‾ iii P((The coin lands on Hd)′ ) = ‾ i

C S Ht D

C S

Complete this table of possible outcomes if you record the value of the card, but not the suit. Coin flip T

d All of the outcomes in the table are equally likely. Find:

P((T, Jack)) = ‾ ii P(The card is 10) = ‾ iii P((The card is K)′ ) = ‾ e Draw a tree diagram to find the number of outcomes if you instead record whether the card is Red or Black. i

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AF T

A 2 3 4 5 6 Card value

7 8 9 10 J Q

Number of outcomes:

Probability of each outcome:

6 Consider the experiment from question 5. Circle the event in each pair which is more likely. Circle both events if

they are equally likely. a Tails The card is an Ace c

b (Heads, Red) Hearts

Picture card (Heads, 2 ≤ Card value ≤ 5)

d (Tails, (Picture card)′ ) (Heads, card shows a prime number) 192

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Probability | Experimental probability

Topic 6.2 Chance experiments Chance experiments with coins Some chance-based situations can only be explored through probability experiments. When running a chance experiment, we record the number of times each outcome occurs. number of    trials in which outcome occurs The experimental probability of an outcome is  __________________________________       . total number of trials Experimental probability is also known as relative frequency.

Learn You will need two coins of the same value to conduct this chance experiment. Create a copy of this table with lots of space to record your results throughout this experiment. Note that ( H,T)is the same as (T, H). Outcome

Tally

Frequency after 30 trials

Frequency after 100 trials

Two Heads ( H, H)

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One Head and one Tail (H,T) Two Tails (T,T)

In pairs, flip two coins 30times and record the frequency of each outcome after 30trials in the corresponding column of the table.

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2 Calculate the experimental probability of each outcome after 30trials.

‾

a Experimental P(H, H) =

‾

b Experimental P(H,T) =

Experimental P(T,T) = ‾

3 Compare the theoretical probability of each outcome with its experimental probability. Based on your results so far,

do you think your coins are fair? 4 Flip the coins another 70times and combine the results with your initial 30trials. Record the frequency of each

outcome after 100trials in the corresponding column of the table. 5 Calculate the experimental probability of each outcome after 100trials.

‾

a Experimental P(H, H) =

‾

b Experimental P(H,T) =

Experimental P(T,T) = ‾

6 Compare the theoretical probability of each outcome with its updated experimental probability. Based on your

updated results, do you still agree with your answer to question 3? Explain.

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7 Are there any other factors which may have affected your results?

8 Marama says that she can complete the experiment faster on her own by flipping the same coin twice. Would this

change impact the results of the experiment? Explain.

9 The Law of Large Numbers states that as the number of trials in a chance experiment increases, the e xperimental

probability will approach the experiment’s theoretical probability. This means that as you do more trials, the experimental probabilities should get closer in value to the theoretical probabilities of each outcome. Does your chance experiment agree with the Law of Large Numbers? What would happen if you did 1 , 000flips of the two coins in total?

towards a coin being fair or biased.

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10 A biased coin is not equally likely to land on Heads or Tails. We can use a chance experiment to find evidence

a Describe a chance experiment you could run to help to determine whether a coin is biased. Consider the number

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of trials you would run.

b The probability of a coin landing on Heads 5 times in a row is 0.03125. Suppose a coin is flipped 5 times and

the outcome each time is Heads. Is the coin biased? Explain.

The probability of a coin landing on Heads 50 times in a row is less than 0.0000000000001. Suppose a coin is flipped 50times and the outcome each time is Heads. Is the coin biased? Explain.

d Can you use a chance experiment to be 100%sure that a coin is biased? Explain.

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Chance experiments with dice Many chance experiments are about rolling dice. A fair die has an equally likely chance of rolling each outcome. For a biased die, the outcomes are not equally likely. A weighted die is tampered with to increase the likelihood of it rolling a specific value.

Explore You will need two six-sided dice numbered from 1to 6to conduct this chance experiment. 1

In pairs, roll the two dice and add the two rolled values together 100times. Record the outcomes in this table. Outcome (sum of dice rolls)

Tally

Frequency after 100 trials

2 3 4 5

AF T

6 7 8

D R

9 10 11 12

2 Calculate the experimental probability of each outcome in question 1. The experimental probability of the sum being: a 2 is

b 3 is

4 is

d 5 is

e 6 is

7 is

g 8 is

h 9 is

11 is

k 12 is

10 is

3 Determine the theoretical probability of each outcome in question 1. Hint: Look back at questions in Topic 6.1.

‾ e P(6) = ‾ i P(10) = ‾ a P(2) =

‾ P(7) = ‾ P(11) = ‾

b P(3) = f j

P(4) = ‾ g P(8) = ‾ k P(12) = ‾ c

‾

d P(5) =

‾

h P(9) =

4 Compare the theoretical and experimental probabilities of the outcomes, considering the Law of Large Numbers.

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

195

5 Suppose you have a six-sided die numbered from 1to 6that you suspect is a weighted die. Describe a chance

experiment you could run to collect evidence that it is a weighted die.

6 A spreadsheet was used to simulate rolling a weighted die 3, 000times.

Outcome

Frequency 50 trials

Frequency 100 trials

Frequency 3, 000 trials

322

147

133

430

885

1,083

You can use a spreadsheet to simulate chance experiments with thousands of trials.

a Complete this table of experimental probabilities. Use a calculator to find values correct to four decimal places.

Experimental probability after 50 trials

3 4

Experimental probability after 3, 000 trials

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Experimental probability after 100 trials

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Outcome

5 6 b Which experimental probability values should you use to estimate the probability of each outcome? Explain, with

reference to the Law of Large Numbers.

Estimate the theoretical probability of each outcome correct to two decimal places. Make sure the estimated probabilities sum to 1.

P(1) ≈ ii P(2) ≈ ‾ ‾ iv P(4) ≈ v P(5) ≈ ‾ ‾ d Does the experimental data suggest that the die is weighted? Explain. i

196

‾ vi P(6) ≈ ‾

iii P(3) ≈

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Comparing theoretical and experimental probability We can conduct chance experiments to compare theoretical and experimental probability.

Deepen 1

Create a spinner with 10 equal parts. You could make a physical one, using a circle like the one shown, or create one digitally. Label the parts with the numbers from 1 to 10.

2 What is the theoretical probability of each outcome of spinning your spinner? 3 Find the theoretical probability of each event. a P(even number) =

b P(odd number) =

P(prime number) =

d P(square number) =

e P(cube number) =

P(less than 5) =

4 Run a chance experiment by spinning your spinner 100 times. Record the frequency of each outcome in this table.

Outcome

Tally

Frequency

1 2

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3 4 5

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6 7 8 9 10

5 The experimental probability of spinning a: a 1 is

b 2 is

g 7 is

h 8 is

6 is

3 is

d 4 is

e 5 is

9 is

10 is

6 Calculate the experimental probability of each event in question 3. a

7 Compare your calculated experimental probabilities (in questions 4 and 5) with the corresponding theoretical

probabilities (in questions 2 and 3). What do you notice? Do you think your spinner is fair?

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197

Quick 10s/Tere Tekau Term 1 Your teacher will tell you when it’s time to refresh your skills using these Quick 10 quizzes!

.54 × 1, 000 = 6 5  2 = Is 22a prime number? Y/N Round 3 37, 998to the nearest 100, 000: ‾ 10 × 4 = Is 504 divisible by 8? Y / N 824 − 812 = 655 ÷ 100 = _ √ 4   = Is 41a prime number? Y/N

Round 2 92, 939to the nearest 100, 000: ‾ 5 × 14 = Is 920 divisible by 10? Y/N 94 + 76 − 20 = 911 ÷ 10 = _ √ 16  = Is 4a prime number? Y / N Round 9 8.67to the nearest whole number: ‾ 11 × 13 = Is 96 divisible by 5? Y / N

/10

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198

AF T

Is 93a prime number? Y/N Round 2 25.805to the nearest hundredth: ‾ 2 × 12 = Is 645 divisible by 9? Y / N 321 ÷ 100 = _ √ 64  = 62 − 13 + 73 = 280 ÷ 1, 000 = 10  2 = Is 27a prime number? Y/N

Round 178.338 to the nearest whole number: ‾ 4 × 6 = Is 616 divisible by 5? Y / N 279 − 357 = 9.67 × 1, 000 = 6  2 = Is 13a prime number? Y/N Round 9 4.69to the nearest whole number: ‾ 12 × 7 = Is 508 divisible by 5 ? Y / N

/10

C + 10 − 32 = 4 4.35 × 100 = _ √  9= Is 82a prime number? Y/N Round 7 7.048to the nearest tenth: ‾ 4 × 8 = Is 785 divisible by 10? Y/N 478 + 654 = 1.23 × 100 = _ √36  =

/10

F 62 − 56 + 23 = 1 887 ÷ 100 = 12  2 = Is 75a prime number? Y/N Round 3 97, 841to the nearest 10: ‾ 6 × 9 = Is 941 divisible by 10? Y/N 408 + 523 = 978 ÷ 100 = _ √121  =

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

/10

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/10

99.01 × 100 = _ √25  = Is 61a prime number? Y/N Round 7 4.88to the nearest tenth: ‾ 8 × 3 = Is 59 divisible by 4? Y / N 618 + 514 = 6.03 × 100 = _ √ 100  = 8 × 94 =

AF T

  2 = 7 52 × 4 = Is 83a prime number? Is 936 divisible by 3? Y / N Y/N 704 − 294 = Round 2 3.27to the nearest 560 ÷ 10 = _ 100: √16  = ‾ 7 × 12 = Is 92a prime number? Is 813 divisible by 3? Y / N Y / N 536 + 368 = Round 8 49, 873to the 4.3 × 100 = nearest 10, 000: _ ‾ √49  = 7 × 2 = Is 39a prime number? Is 170 divisible by 10? Y/N Y/N Round 2 61.8to the nearest 187 − 727 = whole number: ‾

0 × 80 = 4 Is 426 divisible by 4? Y / N 102 − 87 = 0.35 × 1, 000 = 8  2 = Is 25a prime number? Y/N Round 1 .17940to the nearest thousandth: ‾ 2 × 12 = Is 270 divisible by 9? Y / N 244 + 395 =

27 ÷ 10 = 9 11  2 = Is 53a prime number? Y/N Round 9 4.8778to the nearest tenth: ‾ 11 × 8 = Is 807 divisible by 9? Y / N 841 + 747 = 2.29 × 10 = 4  2 = Is 75a prime number? Y/N

Round 9 97, 146to the nearest 100, 000: ‾ 22 × 5 = Is 849 divisible by 6? Y / N 306 − 610 = 8.98 × 1, 000 = _ √ 81  = Is 61a prime number? Y/N Round 5 73, 299to the nearest 100: ‾ 92 × 64 = Is 879 divisible by 10? Y/N

/10

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199

Quick 10s/Tere Tekau Term 1

17 − 604 = 4 8.65 × 1, 000 = 9  2 = Is 526a prime number? Y/N Round 6 .975to the nearest hundredth: ‾ 5 × 10 = Is 1, 296 divisible by 3? Y/N 772 − 393 = 619 ÷ 1, 000 = 3  2 =

Is 47a prime number? Y/N Round 6 0.96to the nearest tenth: ‾ 14 × 74 = Is 477 divisible by 2? Y / N 332 − 158 = 5.7 ÷ 10 = _ √144  = Is 87a prime number? Y/N Round 3 68, 278to the nearest 1, 000: ‾ 3 × 8 =

Is 1, 296 divisible by 3? Y/N 203 + 478 = 8 × 10 = _ √ 81  = Is 48a prime number? Y/N Round 7 .694to the nearest hundredth: ‾ 16 × 44 = Is 927 divisible by 2? Y / N 229 − 525 = 110 ÷ 1, 000 =

/10

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P   2 = 5 Is 81a prime number? Y/N Round 5 .84to the nearest whole number: ‾ 23 × 12 = Is 228 divisible by 4? Y / N 60 − 278 = 0.012 × 10 = _ √ 4   = Is 89a prime number? Y/N Round 5 97, 928to the nearest 100, 000: ‾

/10

200

AF T

× 8 = 4 Is 671 divisible by 8? Y / N 85 + 733 = 9.315 × 100 = _ √16  = Is 89a prime number? Y/N Round 1 85.101to the nearest tenth: ‾ 13 × 9 = Is 395 divisible by 6? Y / N 389 − 336 =

250 ÷ 1, 000 = _ √  9= Is 77a prime number? Y/N Round 5 23, 969to the nearest 10, 000: ‾ 12 × 2 = Is 929 divisible by 2? Y / N 295 − 40 = 80.3 × 1, 000 = _ √ 36  = Is 83a prime number? Y/N

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Quick 10s/Tere Tekau Term 2 Your teacher will tell you when it’s time to refresh your skills using these Quick 10 quizzes!

Round 9 4.979to the nearest hundredth: ‾ 41 × 51 = Is 365 divisible by 5? Y / N 721 − 645 = 27 ÷ 3 = 9 × 25 ÷ 5 = 119 Simplest form of _    : 136 144 ÷ 1, 000 = Is 14a prime number? Y/N Round 5 .35to the nearest whole number: ‾

23 × 2 = 1 Is 619 divisible by 8? Y / N 124 − 977 = 112 ÷ 89 = 82 ÷ 2 − 21 = 6 Simplest form of _    : 15 110 ÷ 10 = Is 17a prime number? Y/N Round 2 71.145to the nearest hundredth: ‾ 50 × 79 =

Is 899 divisible by 6? Y / N 102 − 339 = 121 ÷ 53 = 7 × 2 + 66 = 20 Simplest form of  _ : 80 732 ÷ 100 = Is 71a prime number? Y/N Round 8 66.1to the nearest whole number: ‾ 85 × 29 = Is 898 divisible by 3? Y / N

D 05 − 619 = 6 278 ÷ 57 = 8 ÷ 2 + 4 =

/10

D R

/10

AF T

98 ÷ 63 = 3 12 ÷ 4 − 3 × 6 =

F 9 × 25 ÷ 5 =

90 Simplest form of _    : 126 26 ÷ 100 = Is 59a prime number? Y/N Round 9 60, 244to the nearest 10: ‾ 71 × 19 = Is 467 divisible by 3? Y / N 488 + 62 =

24 Simplest form of _    : 48 9.8 × 10 = Is 69a prime number? Y/N Round 8 34.77to the nearest tenth: ‾ 62 × 83 = Is 602 divisible by 5? Y / N 885 + 901 = 918 ÷ 35 =

48 Simplest form of  _ : 72 46 ÷ 6 = 8.14 × 1, 000 = Is 56a prime number? Y/N Round 9 26, 309to the nearest 10: ‾ 19 × 99 = Is 812, 212 divisible by 9? Y/N 603 − 763 = 991 ÷ 11 =

/10

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201

Quick 10s/Tere Tekau Term 2

− 5 + 10 × 9 = 1 250 ÷ 14 =

8 ÷ 2 × 9 − 1 =

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30 Simplest form of _    : 50 582 ÷ 100 = Is 32a prime number? Y/N Round 2 41, 633to the nearest 10: ‾ 94 × 36 = Is 439 divisible by 4? Y / N 329 − 456 = 693 ÷ 33 =

58 ÷ 28 = 2 36 × 89 = 89.7 × 1, 000 = Is 6a prime number? Y / N Round 5 9, 158to the nearest 100: ‾ 66 × 86 = 20 Simplest form of  _ : 22 513 + 388 = 215 ÷ 47 = 7 + 5 × 9 + 5 =

/10

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/10

202

39 Simplest form of  _ : 78 637 ÷ 63 = Round 7 1.279to the nearest hundredth: ‾ 112 ÷ 1, 000 = Is 87a prime number? Y/N 48 × 34 = Is 458 divisible by 4? Y / N 539 − 583 = 20 ÷ 4 + 5 × 12 =

14 Simplest form of  _ : 77 125 ÷ 17 = 12 ÷ 3 ÷ 2 + 6 = 742 ÷ 100 = Is 73a prime number? Y/N Round 1 .79to the nearest tenth: ‾ 49 × 44 = Is 589 divisible by 10? Y/N 533 − 530 = 102 ÷ 24 =

7 × 10 + 12 ÷ 4 = 25 Simplest form of  _ : 50 Is 782 divisible by 3? Y / N 582 ÷ 46 = 40 × 21 = 930 × 100 = Is 53a prime number? Y/N Round 8 26, 160to the nearest 1, 000: ‾ 17 × 85 = Is 949 divisible by 4? Y / N

/10

79 + 142 = 6 606 ÷ 79 = 49 ÷ 7 × 4 − 9 =

162 Simplest form of  _ : 180 Is 85a prime number? Y/N 1.75 × 100 = Round 9 1.342to the nearest hundredth: ‾ Is 139 divisible by 10? Y / N 40 × 43 = 949 + 592 =

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

19 Simplest form of _    : 38 0.99 × 1, 000 = Is 16a prime number? Y/N Round 3 80, 955to the nearest 10, 000: ‾ 49 × 97 = 94 − 907 = Is 907 divisible by 9? Y / N 13 × 72 =

/10

76 Simplest form of  _ : Round 5 87, 775to the 114 nearest 100: 3 − 1 ÷ 4 × 9 = ‾ Is 399 divisible by 2? Y / N 67 × 60 = 344 − 968 = 323 ÷ 75 = Is 940 divisible by 4? Y / N 93 × 9 = 930 ÷ 45 = 974 − 365 = 3 − 11 × 5 − 5 = 2.06 × 1, 000 = 30 _ Round 6 6.72to the nearest Simplest form of  75 : 10: 9.20 × 10 = ‾ 720 ÷ 10 = 53 ÷ 10 = Is 37a prime number? 8  2 = Y/N

/10

Is 97a prime number? Round 7 09, 397to the Y/N nearest 100, 000: ‾ Round 7 68.8to the nearest 18 × 57 = 10: 536 ÷ 15 = ‾ 35 × 82 = Is 876 divisible by 6? Y / N 895 ÷ 71 = 124 − 512 = Is 706 divisible by 6? Y / N 50 ÷ 5 − 6 × 11 = 12 214 + 352 = Simplest form of  _ : 120 8 × 11 + 8 × 2 = 49 × 56 = 6 Simplest form of _    : 988 ÷ 1, 000 = 8 _ 8.57 × 1, 000 = √ 81  = Is 57a prime number? Y / N

/10

AF T

84 ÷ 7 = 5 2 + 16 ÷ 8 + 2 =

D R

R Is 64a prime number? Y/N Round 7 40, 728to the nearest 10: ‾ 18 × 65 = Is 793 divisible by 9? Y / N 226 ÷ 66 = 499 + 911 = 5 + 1 − 6 × 2 = 412 ÷ 53 = 88 Simplest form of  _ : 96 946 ÷ 1, 000 =

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

/10

203

Quick 10s/Tere Tekau Term 3 Your teacher will tell you when it’s time to refresh your skills using these Quick 10 quizzes!

6 3 _     −  _  = 4 8 5 Simplest form of 2 4 ×  _  6 1 _ 39 ÷     = 13 2 Bigger: 0.59 or  _ ? 6 ‾ 8 × 8 − 10 × 7 =

Bigger: 37% or 0.04? ‾ 9 2 _     + 4 _  = 7 28 2 Simplest form of 24 ×  _ : 6 1 31 ÷  _  = 19 6.15 × 100 =

Is 186 divisible by 8? Y / N

AF T

Is 72a prime number? Y/N 417 ÷ 45 = Round 7 40, 728to the 416 × 1, 000 = nearest 10: ‾ Round 1 2.85to the nearest 51 × 51 = whole number: ‾ Is 186 divisible by 8? Y / N 9  2 = 10 ÷ 10 + 10 × 8 =

883 − 246 = Is 860 divisible by 2? Y / N 8.78 × 10 = Round 2 23.899to the nearest hundredth: ‾

/10

2 Simplest form of 4 6 ×  _ : 3 7 _ Bigger: 0.88 or    ? 8 ‾ 5 12 _     − 3 _  = 10 40 932 ÷ 67 =

1 Bigger: 0.52 or  _ ? 3 ‾ 6 Simplest form of 14 ×  _ : 7 9 4 _ _     − 1    = 7 28 1 _ 38 ÷     = 20 Is 987 divisible by 10? Y / N Is 578 divisible by 6? Y / N

204

8 + 10 − 2 × 2 =

/10

F 458 ÷ 76 = 6 × 9 + 11 − 4 =

7 × 9 × 6 + 11 =

6 Simplest form of  _ : 21 Bigger: 0.29 or 45% ‾ 3 2 _     +  _  = 4 32 2 Simplest form of 49 ×  _ : 5 1 _ 52 ÷     = 15 58 × 10 =

950 ÷ 46 =

Is 28a prime number? Y / N

936 ÷ 100 =

595 ÷ 100 =

Round 5 4.02to the nearest whole number: ‾ 8 × 3 − 7 × 2 = 24 Simplest form of _    : 56 5 2 _ _     + 2    = 2 22

4 × 84 =

/10

1 63 ÷  _  = 7 48 Simplest form of  _ : 128 12 4 _     − 1 _  = 10 60 3 Simplest form of 2 ×  _ : 4 669 ÷ 61 =

D R

/10

30 Simplest form of  _ : 120

/10

76 × 27 =

/10

Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

71 × 34 = Is 802 divisible by 9? Y / N   2 = 3

/10

J 12 × 7 − 6 × 11 = 1 Bigger: 0.07 or  _ ? 7 ‾ 6 7 _     +  _  = 5 40 5 Simplest form of 3 7 ×  _ : 6 1 _ 55 ÷     = 4 642 ÷ 27 = 117 Simplest form of _    : 143 2 6  = Is 654 divisible by 8? Y / N 0.84 × 100 =

/10

4 Simplest form of 50 ×  _ : 6 5 _ 6 _     +     = 8 40 36 Simplest form of _    : 90 281 ÷ 23 =

2 Bigger: 0.36 or  _ ? 4 ‾ 5 Simplest form of 11 ×  _ : 10 1 4 _     +  _  = 9 45 1 + 10 × 11 + 10 = 30 Simplest form of  _ : 50 294 − 459 =

751 − 841 = Is 305 divisible by 5? Y / N 2 Bigger: 7% or  _  ? 5 ‾ 2 × 11 + 12 × 4 =

5.68 × 1, 000 = 8 × 28 = _

1.062 × 10 =

√ 81  =

Round 9 08, 677to the nearest 100, 000: ‾

620 ÷ 100 =

AF T

4 + 12 × 4 − 3 = 1 35 ÷  _  = 17 7 Simplest form of 5 0 ×  _ : 9 887 × 1, 000 =

/10

1 43 ÷  _  = 19 1 Simplest form of 12 ×  _ : 6 3 _ 4 _     +     = 2 12 Bigger: 0.27 or 45%? ‾ 19 Simplest form of  _ : 57 6 − 5 × 12 − 5 =

60 Simplest form of  _ : 160 8 _ Bigger: 13% or    ? 10 ‾ 3 6 _     −  _  = 10 50 5 Simplest form of 18 ×  _ : 6 1 _ 49 ÷     = 5 8.66 × 1, 000 =

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Bigger: 0.15 or 48%? ‾ 3 Simplest form of 3 8 ×  _  8 194 ÷ 98 =

893 ÷ 65 =

√121  = 8 × 42 =

961 − 22 =

Is 257 divisible by 6? Y / N

45 × 100 =

633 − 21 =

18 × 17 =

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Quick 10s/Tere Tekau Term 3

8 2 _     −  _  = 5 10 1 Bigger: 3% or  _ ? 3 ‾ 32 Simplest form of  _ : 40 1 Simplest form of 22 ×  _ : 6 1 _ 52 ÷     = 15 8 − 6 ÷ 2 + 1 =

9 Simplest form of 10 ×  _ : 10 1 10 ÷  _  = 7 18 Simplest form of  _ : 54 5 _ Bigger: 5% or    ? 8 ‾ 3 × 6 + 12 ÷ 2 =

9.12 × 10 =

3.68 × 100 =

Is 66 divisible by 10? Y / N

Is 150 divisible by 9? Y / N

85 × 98 =

769 ÷ 38 =

90 × 54 =

  2 = 7

11  2 =

Round 3 82, 067to the nearest 100, 000: ‾

Bigger: 0.89 or 15%? ‾ 15 Simplest form of _    : 27 2 _ 7 _     +     = 7 35 2 × 11 × 10 − 11 = 3 Simplest form of 2 4 ×  _ : 4 1 _ 60 ÷     = 16 548 ÷ 10 =

830 + 918 =

P 6 × 2 ÷ 4 × 4 = _

√16  = 52.9 × 1, 000 = Is 984 divisible by 6? Y / N 3 Bigger: 0.12 or  _ ? 5 ‾ 5 2 _     −  _  = 10 120 3 Simplest form of 3 6 ×  _ : 4 1 _ 39 ÷     = 7 6 × 5 ÷ 10 − 12 = 3.66 × 100 =

12  2 =

/10

1 65 ÷  _  = 12 1 Simplest form of 41 ×  _ : 5 6 4 _ _     − 3    = 8 64 Bigger: 73% or 0.5? 36 ‾ Simplest form of  _ : 60 12 × 9 − 10 + 7 =

1 39 ÷  _  = 10 6 Simplest form of 31 ×  _ : 7 8 _ Bigger: 55% or    ? 10 ‾ 2 2 _     + 4 _  = 5 55 24 Simplest form of  _ : 36 45 ÷ 15 =

378 ÷ 77 =

2.88 × 100 =

238 − 894 = Is 794 divisible by 6? Y / N

Is 898 divisible by 10? Y/N

2.44 × 10 =

83 × 64 =

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342 + 471 =

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

Quick 10s/Tere Tekau Term 4 Your teacher will tell you when it’s time to refresh your skills using these Quick 10 quizzes!

A .2 × 28 = 0 11 _     of 32 = 2 25 % of 268 = 1 18 ÷  _  = 11 4 Simplest form of 4 5 ×  _ : 5 1 _ Bigger: 0.43 or    ? 9 ‾ 6 Simplest form of _    : 9 11 + 7 × 10 + 3 =   2 = 4

1 Bigger: 0.05 or  _ ? 4 ‾ 3 10 _     + 2 _  = 7 56 2 Simplest form of 21 ×  _ : 6 170 × 0.9 =

2 Simplest form of 40 ×  _ : 10 1 57 ÷  _  = 12 5.88 − 97.2 =

3.62 − 9.16 = 0 % of 418 = 5 2 _     of 75 = 5 693 ÷ 10 = Is 394 divisible by 2? Y / N

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12.1 − 0.4 =

90 × 8 =

D 1 Bigger: 0.86 or  _ ? 5 ‾ 6 2 _     − 5 _  = 5 50 1 Simplest form of 4 1 ×  _ : 5 1 _ 10 ÷     = 16 4.38 + 3.79 = 20 × 0.6 = 1 7 _     of 88 = 2 10 % of 114 = 640 ÷ 1, 000 = Is 807 divisible by 3? Y / N

/10

9.06 + 11.6 = 3.41 × 10 = Is 898 divisible by 9? Y / N 82 × 32 =

/10

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70 × 0.5 = 1 12 _     of 96 = 6 75 % of 157 =

0 % of 622 = 1 4 _     of 64 = 12 13 × 0.1 = 94.5 − 2.3 = 1 53 ÷  _  = 4 Bigger: 0.14 or 57%? ‾ 6 3 _     −  _  = 7 42 15 Simplest form of  _ : 60 20 ÷ 5 × 4 + 10 =

F 964 ÷ 41 = 9 × 3 − 12 − 9 = 18 Simplest form of  _ : 36 5 _ Bigger: 94% or    ? 8 ‾ 3 6 _     +  _  = 7 28 1 30 ÷  _  = 14 2.85 + 5.9 = .7 × 7 = 0 4 _     of 13 = 3 25 % of 253 =

67.3 × 10 =

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Quick 10s/Tere Tekau Term 4

0 % of 836 = 5 5 _     of 87 = 10 33.3 + 2.89 = 9 8 _     −  _  = 7 84 1 54 ÷  _  = 5 26 Simplest form of  _ : 52 8.36 × 100 =

1 Bigger: 0.33 or  _ ? 3 ‾ 6 5 _     −  _  = 4 32 2 Simplest form of 19 ×  _ : 3 1 _ 61 ÷     = 9 13.4 − 3.07 = 3 × 0.4 =

6 3 _     + 1 _  = 3 15 1 7 ÷  _  = 6 3 Simplest form of 12 ×  _ : 4 67.9 − 0.39 = 11 _     of 73 = 5 50 % of 207 =

1 % of 216 =

8.615 × 100 =

590 + 592 =

902 ÷ 41 =

538 + 198 =

21 ÷ 3 + 3 =

1.58 × 100 =

27 ÷ 9 + 6 − 4 =

√ 100  =

√49  =

J 55 Simplest form of _    : 77 Bigger: 0 .19or 23% ‾ 8 3 _     +  _  = 9 72 1 14 ÷  _  = 15 5.87 + 54.6 = 12 _     of 88 = 11 75 % of 830 = 0.4 × 8 =

168 ÷ 58 =

/10

208

/10

Bigger: 0.35 or 14%? ‾ 10 12 _     + 1 _  = 9 90 1 _ 70 ÷     = 3 9.28 + 2 = 12 _     of 98 = 3 25 % of 335 = 5.76 × 100 = 380 ÷ 26 =

822 ÷ 1, 000 =

63 × 79 =

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24 Simplest form of  _ : 40 3 1 _ _     + 2    = 3 12

/10

L 1 41 ÷  _  = 15 5.32 + 74.5 = .4 × 11 = 0 4 _     of 44 = 12 25 % of 191 = 1 4 _     − 2 _  = 4 4 1 _ 16 ÷     = 11 Is 944 divisible by 8? Y / N 50 × 57 = _

√25  =

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

29.8 − 1.04 =

0.1 × 19 = 1 8 ÷  _  = 20 8 2 _     + 3 _  = 5 50 554 − 981 = 58 × 86 =

.8 × 6 = 0 3 _     of 40 = 10 50 % of 34 = 1 57 ÷  _  = 17 939 ÷ 27 =

20 Simplest form of _    : 48 477 ÷ 1, 000 =

16 Simplest form of  _ : 192 1.83 × 10 =

Bigger: 0.17 or 25%? ‾ 10 % of 890 =

10 ÷ 8 ÷ 2 × 2 =

0 % of 74 = 2 2 _     of 79 = 3 4.57 × 1, 000 =

604 ÷ 34 =

9 − 7 + 5 × 10 =

P Is 923 divisible by 4? Y / N 24 Simplest form of _    : 132 3 Bigger: 71% or  _ ? 8 ‾ 3 1 _     +  _  = 6 24 1 65 ÷  _  = 6 20 % of 240 = 5.06 − 9.25 = .2 × 8 = 0 12 _     of 76 = 4 742 ÷ 10 =

/10

10  2 = 27 Simplest form of  _ : 45 5 _ Bigger: 0.54 or    ? 6 ‾ 6 9 _     − 3 _  = 10 20 7.05 − 5.3 = 0.3 × 12 =

/10

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/10

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7.99 + 34 =

867 ÷ 45 = 6 Bigger: 46% or  _ ? 7 ‾ 1 1 _     + 1 _  = 2 8 4.97 × 10 =

6 5 _     +  _  = 3 12 1 _     of 82 = 10 0.74 + 5.29 = 0.4 × 9 =

Is 841 divisible by 10? Y / N 2 5 % of 80 = 1 689 ÷ 85 = 2 ÷  _  = 13 75 % of 100 = Is 834 divisible by 5? Y / N 2 _ 32     of 38 = Simplest form of  _ : 10 56 1 39 ÷  _  = 0.5 × 21 = 10 1193 + 182 = 10 % of 80 =

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Year 8 Student Workbook | Mathematics and Statistics for Aotearoa New Zealand (Second edition) Oxford University Press

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Notes

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Notes

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Notes

Quick Notes 10s/Tere Tekau

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Term 1

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