MATH 6

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Chapter 1 Understanding Mathematical expressions

Lesson 1 Mathematical Expression A mathematical expression is made up of one or more operations that result in a number. The mathematical expression may be a sum, difference, product, or quotient or may be a combination of these. Example of mathematical expressions: 5+7 3+9 addition expression 15 – 3 20 – 8 subtraction expression 3x4 2x6 multiplication expression 36 ÷ 3 60 ÷ 5 division expression They all represent the same number. What number is this? Practice Exercise 1.1 Match each mathematical expression in column A with the number it represents in column B. Write the letter of your answer in the blank. A B _____ 1.) 15 – 1 a. 1 _____ 2.) 32 ÷ 8 b. 2 _____ 3.) 10 + 6 c. 23 _____ 4.) 20 – 15 d. 4 _____ 5.) 4 x 9 e. 5 _____ 6.) 7 + 16 f. 14 _____ 7.) 100 ÷ 50 g. 16 _____ 8.) 5 x 5 h. 20 _____ 9.) 5 x 4 i. 25 _____ 10.) 8 ÷ 8 j. 36

Lesson 2 Equations and Inequalities If two mathematical expressions represent the same number, we may write them we may write them with an equality sign (=) between them. This indicates that the two expressions are equal. The expression now form an equation. An equation is a statement that two mathematical expressions are equal. Example: 15 – 3 = 5 + 7 24 ÷ 2 = 3 x 4 12 = 12 12 = 12 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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If one expression does not give the same number as another expression, we write the inequality sign ( ≠ ) between them. 14 – 4 ≠ 6 x 3 The two expressions are not equal. So we have an inequality. Inequality is a statement that two mathematical expressions arenot equal Example: 14 – 4 ≠ 6 x 3 10 ≠ 18 Practice Exercise 1.2 A. Make each equation true by writing the correct symbols of operation ( +, - , x, or ÷ ) in the circles 1. 100 10 = 25 15 2. 50 25 = 5 3 3. 2 4 = 64 8 4. 30 6 = 9 4 5. 8 2 = 10 6 B. Write the correct sign ( = or ≠ ) between the mathematical expressions. 1. 25 – 7 6 + 12 2. 13 – 3 22 ÷ 2 3. 6 x 4 28 – 4 4. 60 ÷ 5 80 ÷ 8 5. 4 + 8 9+3 Lesson 3 Evaluating Mathematical Expressions To evaluate mathematical expression means to find the value of the expression. Suppose we are asked to find the value of the following expression. 4+2x3 Should we first add 4 and 2, then multiply the sum by 3? That is 4 + 2 x 3 = 6 x 3 = 18 Or should we first multiply 2 and 3, then add the product to 4? That is 4 + 2 x3 = 4 + 2 x 3 = 4 + 6 = 10 The correct answer is 10. Similarly, note the correct order of operations in the following operations in the following expressions. a. 8 – 2 x 3 b. 8 + 6 ÷ 3 8–6=2 8 + 2 = 10 To generalize, when we have several operations to evaluate in an expression, we multiply and divide first, before we add and subtract. When an expression involves addition and subtraction only, as in the expressions below, which operation do we perform first? a. 10 + 5 – 3 b. 10 – 5 + 3 15 – 3 = 12 5+3=8 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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To evaluate the expression below, we should also go from left to right, a. 4 x 6 ÷ 3 b. 10 ÷ 5 x 2 24 ÷ 3 = 8 =2x2=4 Let us have some more examples a. 4 + 22 ÷ 11 x 3 =4+2x3 divide first: 22 ÷ 11 = 2 =4+6 then multiply; 2 x 3 = 6 = 10 then add: 4 + 6 = 10 b. 10 – 2 x 6 ÷ 4 = 10 – 12 ÷ 4 = 10 – 3 =7

multiply first: 2 x 6 = 12 then divide: 12 ÷ 4= 3 finally subtract: 10 – 3 = 7

You can easily remember this order of operations by the expression: My Dear Aunt Sally where M stand for multiply, D for divide, A for add, and S for subtract (MDAS rule). Also remember that we multiply and divide from left to right. Likewise, we add and subtract from left to right. To indicate which operation to perform first, groupings of numbers are indicated using parentheses. a. (2 + 3) x 5 = 5 x 5 = 25 b. 17 – (4+3) x 2 = 17 – 7 x 2 = 17 – 14 = 3 c. (17 – 4) + 3 x 2 = 13 + 3 x 2 = 13 + 6 = 19 d. 20 ÷ ( 5 – 1 ) + 3 = 20 ÷ 4 + 3 = 5 + 3 = 8 What do you observe from the examples? In each example we first evaluated the expression inside the parentheses. Then we proceeded with the other operations based on MDAS rule. Practice Evaluate the following expressions 1. 6 – 2 + 3 = 2. 7 + 8 – 10 = 3. 3 x 5 + 2 = 4. 3 + 5 x 2 = 5. 8 ÷ 4 + 3 = 6. 18 + 6 ÷ 3 = 7. (18 + 6 ) ÷ 3 = 8. 3 x (2 + 5) ÷ 7 = 9. 2 x (6 – 3 ) + 2 = 8 10. 9 + 3 x ( 1 + 5 ) – 12 =

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Lesson 4 Exponents in Mathematical Expressions Let us now consider the following expression 3x3x3x3 3 is multiplied by itself four times. This expression can be written in exponential form as: 34 Where 3 is the base and 4 is the exponent. Similarly, these are equivalent expressions. 2 x 2 x 2 x 2 x 2 = 25 Where 2 is the base and 5 is the exponent. An exponent tells how many times a number, called the base is used as a factor. In mathematical expression that has an exponent is treated as a case of multiplication; the MDAS rule still applies. We evaluate first the exponential expressions before proceeding with the MDAS rule. a. 3 x 23 – 3 + 4 x 3 =3x8–3+4x3 = 24 – 3 + 12 = 21 + 12 = 33

evaluate 23 first: 2 x 2 x 2 = 8 apply MDAS rule

b. 4 x 32 – 2 x 23 + 3 x 42 = 4 x 9 – 2 x 8 + 3 x 14 evaluate 32, 23 and 42 first. = 36 – 16 + 48 apply MDAS rule = 20 + 48 = 68 Parentheses may also be used in mathematical expressions with exponents. 3 x (23 + 42) – 2 x 32 = 3 x (8 + 16) – 2 x 9 = 3 x 24 – 18 72 – 18 = 54 Always evaluate first the expressions inside the parentheses. Then look at those with exponents and evaluate them before applying the MDAS rule. We can now add these rules to the MDAS rule and call this new rule the PEMDAS rule, for evaluating expressions inside Parentheses and with Exponents first, then Multiply and Divide and then add and Subtract. Practice A. Write the following multiplication expressions in exponential form. Afterward identify which is the exponent and which is the base in each. 1. 3 x 3 x 3 x 3 x 3 x 3 x 3 = 37 Exponent: _____ base: _____ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2. 6 x 6 x 6 = _____ Exponent: _____ 3. 12 x 12 = _____ Exponent: _____ 4. 10 x 10 x 10 x 10 = _____ Exponent: _____ 5. 8 x 8 x 8 x 8 x 8 = _____ Exponent: _____

base: _____ base: _____ base: _____ base: _____

B. Write the multiplication expression for each exponential expression then, give the product, the base, and the exponent. Product Base Exponent 4 6. 3 = __________ ______ _____ _____ 3 7. 6 = __________ _____ _____ _____ 5 8. 2 = __________ _____ _____ _____ 3 9. 5 = __________ _____ _____ _____ 3 10. 4 = ________ _____ _____ _____ C. Evaluate each mathematical expression 11. 3 x 62 = 12. 43 ÷ 8 = 13. 3 + 52 = 14. 5 x 23 = 15. 33 – 10 = 16. 23 + 32 – 42 = 17. 3 x 23 – 2 x 32 + 5 x 42 = 18. 22 x (32 + 42) – 52 = 19. 53 – 3 x ( 22 + 32) – 42 = 20. 62 + 22 x (3 + 4) – 72 =

Lesson 5 Signs Used in Groupings In mathematical expressions, the multiplication sign (x) may be replaced by parentheses. These are some examples. a. 3 x 2 may be written as 3(2), or (3)2 or (3)(2). b. 5 x (4 + 1 ) may be written as 5 (4 + 1) or (5)(4 + 1). Hence, 3 + 4 x 5 – 2 = 3 + 4 (5) – 2 = 3 + 20 -2 = 23 – 2 = 21 Note that we get a different answer when the parentheses enclose different numbers in the expression. 3 + 4(5) – 2 = 21 Versus 3 + 3(5 – 2) = 3 + 4(3) = 3 + 12 = 15 In the second expression, we subtract 5 and 2 before multiplying by 4. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The symbol for division (÷ ) may also be replaced by a division bar, such as the one used as a symbol for grouping, as shown in the examples below. Be sure to divide the numbers first before performing the other operations. a. 15 – 30 ÷ 6 + 2 b. (15 – 3 ) ÷ (3 + 1) – 2 30

= 15 −

6

+2

=

= 15 – 5 + 2

=

12 4

15−3 3+1

− 2

− 2

= 10 + 2 =3–2 = 12 =1 This time, we must simplify the fractional expression first since it is an indicated division, then we apply the MDAS rule. 9−

12+3 4+1 15

= 9−

5

+ 2

+ 2

=9–3+2 =6+2 =8 The braces ([ ]) are also used to indicate grouping. 2 [5(3) – 2(4)] evaluate first the expressions with parentheses = 2 (15 – 8) the braces are turned into parentheses =2(7) = 14 The basic rule is to evaluate first the expressions with parentheses inside the braces. Once done, change the braces to parentheses. Evaluate further the simplified expression by applying the PEMDAS rule. Practice Evaluate each mathematical expression. 1. 72 – 6(13 – 5) = 2.

4 5 − 9(2) 2

+1=

2

3. (3 + 5) – [23 + 2(3)] = 4. 52 – 3(22) − 5.

43 25

+

2(34 ) 6

62 3(2)

=

=

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Lesson 6 Solving Word Problems There are many varied problems involving the four arithmetic operations as applied to whole numbers Problem 1 Three buses of students and 12 teachers went on a field trip. Each carried 36 students. How many people joined the field trip? Solution: Total number of people = number of students + number of teachers = 3 (36) + 12 = 108 + 12 = 120 A total of 120 people joined the field trip. Problem 2 Each student contributed P 200.00 and each teacher gave P300.00 to pay for all expenses in the trip. How much was the total amount contributed? Solution: Total amount = amount given by the students + amount given by the teachers = P200.00(108) + P300.00(12) = P 21,600.00 + P3,600.00 = P25,200.00 P25,200.00 was the total amount contributed for the field trip. Practice 1. A jumbo egg costs P6.00 each. Myrna bought 3 dozen eggs. She gave the vendor a 500-peso bill. How much change did she get? What is asked? ________________________________________ What are given? _______________________________________ Operations to be used: _________________________________ Number Sentence: _____________________________________ Solution:

Complete Answer: _____________________________________

2. In the last election, 81 people voted in a precinct that listed 104 voters. In another precinct, 74 people voted out of 107 listed voters. How many did not vote in the two precincts? What is asked? ________________________________________ What are given? _______________________________________ Operations to be used: _________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Number Sentence: _____________________________________ Solution: Complete Answer: _____________________________________

3. A distributor of bottled drinks initially had 24 cases of cola and 12 cases of orange drinks. To the first store, he delivered 4 cases of cola and 2 cases of orange drinks. To the second store, he delivered 5 cases of cola and 3 cases of orange drinks. To the third store, he delivered 7 cases of cola and 3 cases or orange drinks. How many cases of each were left in the delivery truck? What is asked? ________________________________________ What are given? _______________________________________ Operations to be used: _________________________________ Number Sentence: _____________________________________ Solution: Complete Answer: _____________________________________

4. Three bus passengers paid P180.00 each to reach town A, and four passengers paid P240.00 each to reach town B. another 5 passengers paid P300.00 each to go to a much farther town C. what was the total amount that the conductor received from these passengers? What is asked? ________________________________________ What are given? _______________________________________ Operations to be used: _________________________________ Number Sentence: _____________________________________ Solution: Complete Answer: _____________________________________ 5. One airplane flight from Manila carried 270 passengers. In Cebu, 150 of these passengers got down, and then 90 passengers got on for the final destination, Davao. How many passengers in all reached Davao? What is asked? ________________________________________ What are given? _______________________________________ Operations to be used: _________________________________ Number Sentence: _____________________________________ Solution:

Complete Answer: _____________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Chapter 2 Understanding Decimals Lesson 1 Decimals and Fractions 50

The fraction

100

is written as 0.50 in decimal notation. This is read as “ fifty

hundredths.� Fraction

Decimal

1 10 1 100 1

In Words 0.1

one tenth

0.01

one hundredth

0.001

1 000 1

0.0001

10 000

one thousandth one ten thousandth

Remember this To rename a fraction whose denominator is a power of 10, copy the numerator. Then starting from the rightmost digit of the numerator, move the decimal point to the left as many times as there are zeros in the denominator. Other examples a. b. c. d. e.

30 100 25

is written as 0.3 in decimal notation

is written as 2.5 in decimal notation

10 750

1 000 8 375

0.750 in decimal notation

10 000 375 100

is written as 0.8375 in decimal notation

is written as 3.75 in decimal notation

Practice: A. Write each fraction in decimal forms 1. Four thousandths 2. Eighteen hundredths 3. Twenty-two thousandths 4. Nine tenths 5. Seven hundredths 6. Two and fifty-three hundredths 7. One and six hundredths 8. Four and two hundred five thousandths 9. Ninety and fourteen thousandths 10. Seven hundred and seven hundredths YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Fraction _____ _____ _____ _____ _____ _____ _____ _____ _____ _____

Decimal _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ Page 9

B. Write the following fractions in decimal notations 7

11. 10 = 19

12. 2 100 = 23

13. 100 = 51

14. 1 100 = 125

15. 100 = 675

16. 1 000 = 1 482

17. 1 000 = 5 930

18. 1 10 000 = 19. 3

8 214

10 000 25 681

=

20. 10 000 =

C. Write each number in words 21. 0.59 ___________________________________ 22. 15.030 ___________________________________ 23. 1.854 ___________________________________ 24. 6.007 ___________________________________ 25. 0.023 ___________________________________ Using decimal to Represent Amounts of Money In our money system, 100 centavos (¢) = 1 peso ( ⹣). Thus a centavo is one hundredth of a peso, or 1

1¢ = 100 of a peso = ⹣0.01 These are equivalent amounts of money 25

25¢ = 100 of a peso = ⹣ 0.25 60¢ =

60 100

of a peso = ⹣ 0.60

⹣3 and 75¢ =3 and

75 100

⹣25 and 30¢ =25 and

pesos = ⹣ 3.75 30

100 50

pesos = ⹣ 25.30

⹣326 and 50¢ = 326 100 pesos = ⹣326.50 Test Use decimal to write the equivalent amounts of the following: 1. 80¢ = __________ 2. ⹣10 and 90¢ = __________ 3. ⹣50 đ?‘Žđ?‘›đ?‘‘ 25¢ = __________ 4. ⹣103 đ?‘Žđ?‘›đ?‘‘ 56¢ = ___________ 5. ⹣29 đ?‘Žđ?‘›đ?‘‘ 72¢ = __________ 6. đ?‘Ąđ?‘¤đ?‘œ đ?‘?đ?‘’đ?‘ đ?‘œđ?‘ đ?‘Žđ?‘›đ?‘‘ đ?‘“đ?‘œđ?‘&#x;đ?‘Ąđ?‘Ś đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘Łđ?‘œđ?‘ = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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7. đ?‘“đ?‘–đ?‘“đ?‘Ąđ?‘Ś − đ?‘“đ?‘œđ?‘˘đ?‘&#x; đ?‘?đ?‘’đ?‘ đ?‘œđ?‘ đ?‘Žđ?‘›đ?‘‘ đ?‘Ąđ?‘’đ?‘› đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘Łđ?‘œđ?‘ = __________ 8. đ?‘ đ?‘’đ?‘Łđ?‘’đ?‘›đ?‘Ąđ?‘Śâˆ’eight centavos = __________ 9. đ?‘ đ?‘’đ?‘Łđ?‘’đ?‘›đ?‘Ąđ?‘Ś đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą đ?‘?đ?‘’đ?‘ đ?‘œđ?‘ đ?‘Žđ?‘›đ?‘‘ đ?‘“đ?‘–đ?‘Łđ?‘’ đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘Łđ?‘œđ?‘ = __________ 10. Ten pesos and ten centavos = __________

Lesson 2 Place Value in Decimals Consider the following fractions and their equivalent decimals: 5 10 25

= 0.5

100 375

= 0.25

1 000 4 725

= 0.375

10 000

= 0.4725

hundredths

thousandths

Ten thousandths

. . . .

tenths

0 0 0 0

Decimal point

ones

What is the place value of the digit 5 in each decimal? In which decimal does the digit 5 have the greatest place value? the least place value?

5 2 3 4

5 7 7

5 2

5

The digit 5 is in the tenths place. The digit 5 is in the hundredths place. The digit 5 is in the thousandths place. The digit 5 is in the ten thousandths place. We express a decimal in words in terms of the place value of the rightmost digit. 0.5 is read as “ five tenths� 0.25 is read as “twenty-five hundredths.� 0.375 is read as “ three hundred seventy-five thousandths.� 0.4725 is read as “four thousand seven hundred twenty-five ten thousandths�

Remember this To read a decimal we must know the value of the last digit on the right. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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As with the whole numbers, we may also write decimals in expanded notation. The expanded notation is the sum of the value of each digit in the decimal number.

What part of the grid is shaded? 2 tenths and 5 hundredths 2

5

In fractional form: 10 + 100 In expanded notation: 0.2 + 0.05 In standard notation: 0.25 Read as: 25 hundredths 3 tenths, 7 hundredths, and 5 thousandths In expanded notation: 0.3 + 0.07 + 0.005 In standard notation: 0.375 Read as: 375 thousandths

Ten thousandths

thousandths

hundredths

tenths

Decimal point

ones

Test A. Indicate the place value of the underlined digit in each of the following decimals 1. 0.4635 __________ 2. 0.7643 __________ 3. 0.6784 __________ 4. 0.7895 __________ 5. 0.3456 __________ B. Write the digits of each decimal under the correct place values in the chart. Then write the decimal in expanded notation.

4.72 3.215 6.7439 2.0087 9.0703

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Lesson 3 Comparing and Ordering Decimals To compare decimal, compare the digits by place values, starting with the digits in the tenths place. The decimal that has the digit with the greater value is greater. However, if the digits are the same, continue comparing the digits in the next place value. 0.2 and 0.20 are equivalent decimals. Equivalent decimals name the same amount. When comparing decimals by digits, we may write the decimals in column, aligning the decimal points. This way we clearly see the place value where the decimals start to differ. Example Order these decimals from least to greatest: 4.8, 4.85, 4.827. In comparing decimals with different numbers of digits, we add zeros so that the decimals will have the same decimal places. So we have 4.800, 4.850, and 4.827 The ones and tenths digits are the same, so we compare the digits in the hundredths place. 4.8 4.800 4.85 4.850 4.827 4.827 Since 0 < 2 < 5, then 4.800 < 4.827 < 4.850 The numbers are arranged from least to greatest: 4.8, 4.827, 4.85 In ordering decimals, the number line can be used to compare them. Arrange these decimals from highest to lowest: 0.3, 1.7, 1.5 From highest to lowest, the numbers are 1.7, 1.5,0.3 Also 0.62 > 0.53 Practice A. Compare the following decimals in each pair. Write > = or < in the blank 1. 0.34 _____ 0.36 2. 0.440 _____ 0.44 3. 0.36 _____ 0.26 4. 0.67 _____ 0.77 5. 0.04 _____ 0.004 B. Arrange each set of decimals from the greatest to the least. 1. 0.52, 0.56, 0.57 2. 0.61, 0.651, 0.70

_____ _____ _____ _____ _____ _____

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3. 4. 5. C.

0.45, 0.54, 0.642 _____ _____ _____ 0.384, 0.37, 0.39 _____ _____ _____ 0.76, 0.688, 0.87 _____ _____ _____ Arrange each set of decimals from the least to the greatest. 1. 0.27, 0.281, 0.26 _____ _____ _____ 2. 0.63, 0.642, 0.65517 _____ _____ _____ 3. 0.710, 0.702, 0.71218 _____ _____ _____ 4. 0.34, 0.635, 0.645 _____ _____ _____ 5. 0.752, 0.753, 0.751 _____ _____ _____

Lesson 4 Rounding off Decimals

To round off decimal to a given place value, we apply the rules on rounding off whole numbers.  Note the place value to which the decimal will be rounded off.  Look at the digit next to it, going to the right.  If that digit is 5 or greater, add one to the digit that will be rounded off.  If that digit is 4 or less, retain the digit in the place value to which the decimal will be rounded off. 8.45 rounded off to the nearest tenths is 8.5 7.62 rounded off to the nearest tenths is 7.6 3.416 rounded off to the nearest hundredths is 3.42 6.3714 rounded off to the nearest thousandths 6.371 Take note the following observation: When rounding off decimals, we do not affix zeros at the end of the roundedoff decimal. The last digit we write is that in the place value to which we are rounding off. Practice Round off the following decimals to the indicated place value 1. 0.4238 to the nearest hundredths __________ 2. 0.8625 to the nearest thousandths __________ 3. 0.0635 to the nearest tenths _________ 4. 4.1562 to the nearest hundredths _________ 5. 9.4675 to the nearest thousandths __________

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Lesson 5 Adding Decimals To add decimals, align all the digits by place value. Use zeros as place holders. Then add the decimals as if adding whole numbers. Place the decimal point of the sum directly under the decimal points of the addends. Find the sum: 5.13 + 8.2651 5.31 5.1300 + 8.2651 + 8.2651 13.3951 When adding decimals, add zeros to the addends such that the addends will have the same number of decimal places. This will ensure that we add digits of the same place value. Practice: 1) 0.42 + 0.31 + 0.2 2) 2.46 + 3.32 + 4.5 3) 0.052 + 0.185 + 0.0953 4) 2.3453 + 0.46 + 1.6112 5) 3.15 + 4.083 + 5.0094 Estimating sums of decimals To estimate sums of decimals, round off each addend to the most convenient place value (to the nearest ones or to the nearest tenths). Then add the rounded-off numbers. Example 1 51.38 + 32.75 Round off to the nearest ones. 51 + 33 = 83 Round off to the nearest tenths 51.4 32.8 = 84.2 Example 2 Estimate: 5.13 + 22.76 Round off to the nearest ones 5 + 23 = 28 Round off to the nearest tenths 5.1 + 22.8 = 27.9 Practice Estimate the sum. 1. 21.79 + 3.43 = 2. 32.105 + 18.212 = 3. 14.344 + 7.58 = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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4. 81.694 + 23.298 = 5. 28.42 + 46.81 = Addition of decimals 3.7 +16.0 insert necessary zeros so that all numbers have the same number of decimals 19.7 Lesson 6 Subtracting Decimals To subtract decimals, align the digits by the same place value. Write the zeros such that the minuend and the subtrahend have the same number of decimal places. Then subtract as if subtracting whole numbers. Place the decimal point of the difference under the decimal points of the minuend and the subtrahend. Example 4.30 - 2.26 2.04 5 10 0.46 0.460 - 0.138 - 0.138 0. 322 Practice Find the difference 1. 0.9 – 0.5 = 2. 0.8 – 0.17 = 3. 0.56 – 0.34 = 4. 0.482 – 0.136 = 5. 0.24 – 0.085 = Subtracting mixed decimals Example 43.05 – 12.63 2 10 43.05 - 12.63 30.42 Practice Find the differences 1. 13.56 – 2.31 = 2. 4.8 – 1.25 = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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3. 16.20 – 3.45 = 4. 28.47 – 7.846 = 5. 49 – 28.163 =

Lesson 7 Multiplying Decimals Multiplication 4.2 has 1 decimal place X 0.8 has 1 decimal place 3.36 the number of decimal places is the total number of decimal places To multiply decimals, do the following: 1. Multiply as the whole numbers. 2. Count the number of decimal places in the factors. 3. Put the decimal point in the product so that there are as many decimal places as there in the factors. 4. If there are not enough digits in the product obtained in (1), add the necessary number of zeros to the left of that product before placing the decimal point. Example: Find 3.6 x 7.2 = 3.6 x 7.2 43.2 216 25.92 Find 0.82 x 0.07 = 0.82 x 0.07 0.0574 Since there are 4 decimal places in the factors, there must be 4 decimal places in the product. For the product to have 4 decimal places, we add zero to the left of 5 before putting the decimal point. Practice: Solve for each product a) 0.134 x 0.507

b) 3.28 x 0.48__

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c) 5.238 x 0.56__

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Solving word problems involving multiplication of Decimals Applications of multiplication of decimals include measurement of perimeters, areas, and volumes. Transactions involving money may also make use of multiplication. Example 1 Mrs. Andres bought 5 kg of sugar. If a kilogram of sugar cost P35.75, how much did Mrs. Andres pay for the sugar? Given: P35.75 – cost of 1 kg of sugar 5 kg – amount of sugar bought Solution: Total cost = cost per kg of sugar x amount of sugar bought = P35.75 x 5 = P 178.75 Mrs. Andres paid P 178.75 for the sugar. Example 2 Mr. Luna has 8.75 hectares (ha) of farm land. He sold it at P17,250.00 per hectare. How much did he get from selling the land? Given: P 17,250.00 – price per hectare of land 8.75 ha – size of the land Solution: Total price = price per hectare x number of hectares = P 17,250 x 8.75 = P150,937.50 Mr. Luna got P150,937.50 from selling his farm land. Solve each problem 1. A residential lot is in the shape of a rectangle 12.5 m wide and 24.5 m long. Find the area in square meters. 2. If the altitude of the parallel bases of a trapezoid is 10 cm and its bases are 1

8.25 cm and 12.75 cm long. Find its area. The area of a trapezoid is 2 the product of the altitude and the sum of the bases. 3. A cube 6.5 cm on a side and is made out of cardboard. Suppose the cube is unfolded, what will be (a) the perimeter (b) the total area of the polygon formed. 4. Mrs. Dulay bought a dressed chicken that weighs 1.75 kg at P115 p[er kilo, a slice of beef that weighs 1.25 kg per kilo, and a slice of pork that weighs 1.75 kg at P145 per kilo. How much was the cost of all the meat she bought? Multiplying decimals by 10, 100, and 1000 You can easily multiply a decimal by 10, 100 and 1000 by moving a decimal point to the right. 0.45 x 10 = 4.5 move 1 decimal place to the right 0.45 x 100 = 45 move 2 decimal places to the right 0.45 x 1000 = 450 move 3 decimal places to the right YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Multiplying decimals by 0.1, 0.01, and 0.001 You can also readily find the product if you multiply by 0.1, 0.01 and 0.001 by moving the decimal point to the left. 0.45 x 0.1 = 0.045 move 1 decimal place to the left 0.45 x 0.01 = 0.0045 move 2 decimal place to the left 0.45 x 0.001 = 0.00045 move 3 decimal places to the left Practice Find the product 1. 0.63 x 10 = 2. 6.2 x 100 = 3. 0.72 x 1000 = 4. 72.4 x 10 = 5. 0.042 x 100 = 6. 3.6 x 0.1 = 7. 41.6 x 0.01 = 8. 0.54 x 0.001 = 9. 64.2 x 0.001 = 10. 0.045 x 0.1 = Lesson 8 Dividing Decimals Estimating the Quotient Mila paid P 13.20 for 12 brown envelopes from the bookstore. Was the cost of one envelope more or less than P1.00? To find the answer, we inspect the dividend and the divisor. The quotient is less than 1 if the dividend is less than the divisor. The quotient is more than 1 if the dividend is greater than the divisor. In 13.2 ÷ 12, the dividend, 13.20 is greater than the divisor, 12, so the quotient is more than 1. The envelope cost more than P1.00 In estimating the quotient, we may round off the dividend and the divisor to the nearest compatible numbers so to make mental division easier Example 204.8 ÷ 47 will be rounded off 200 ÷ 50 is about 4 29.7 ÷ 6.2 will be rounded off to 30 ÷ 6 is about 5 Test State whether the quotient is less than or greater than 1, without actually dividing. 1. 8.9 ÷ 9.8 = 2. 3.953 ÷ 4.02 = 3. 145.2 ÷ 142.3 = 4. 78.2 ÷ 75.6 = 5. 103.8 ÷ 95.2 = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Dividing decimal by a whole number To divide a decimal by a whole number, divide as with whole numbers. Place the decimal point in the quotient directly above the decimal point of the dividend. Example 3.15 18 56.70 54 27 18 90 90 0 Rounding off quotients The AZ store sells three pairs of socks in a bundle at P 128.75. Find the price for each pair of socks. We round off the answer to the nearest centavos so that our answer makes sense. This means we round off to the nearest hundredths. Step 1: divide the numbers 42.91 3 128.75 12 8 6 27 27 5 3 20 18 2

Step 2: in the dividend write zero after the place to which you are rounding off to.

42.916 3 128.75 12 8 6 27 27 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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5 3 20 18 2 Step 3: round off the quotient to the designated value. 42.916 42.92 round off to hundredths Each pair of socks is about P42.92 Practice 1. 2. 3. 4. 5.

4.68 ÷ 3 = 261.2 ÷ 8 = 417.35 ÷ 5 = 9.534 ÷ 6 = 18.75 ÷ 15 =

Mental math dividing by powers of 10 Dividing by 10, 100 or 1000 Remember this 1. To mentally divide a decimal by 10, move the decimal point of the dividend 1 place to the left. 2. To mentally divide a decimal by 100, move the decimal point of the dividend 2 places to the left. 3. To mentally divide a decimal by 1000, move the decimal point of the dividend 3 places to the left. Example 0.4 ÷ 10 = 0.04 0.5 ÷ 100 = 0.005 0.6 ÷ 1000 = 0.0006

Dividing by 0.1, 0.01, or 0.001 Remember this 1. To divided mentally divide a number by 0.1, move the decimal point of the dividend 1 place to the right. 2. To mentally divide a number by 0.01, move the decimal point of the dividend 2 places to the right. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 21

3. To mentally divide the number by 0.001, move the decimal point of the dividend 3 places to the right. Recognizing the patterns makes mental division by powers of ten easy. Example 0.4 ÷ 0.1 = 4 0.5 ÷ 0.01 = 50 0.6 ÷ 0.001 = 600 Practice Divide mentally 1. 75 ÷ 10 = 2. 0.55 ÷ 10 = 3. 64 ÷ 100 = 4. 85 ÷ 10 = 5. 7.68 ÷ 1000 = 6. 0.046 ÷ 0.1 = 7. 3.4 ÷ 0.1 = 8. 8.62 ÷ 0.1 = 9. 0.047 ÷ 0.01 = 10. 0.075 ÷ 0.001 = Test Evaluate 1) 4.8 + 1.5 2) 17.7 + 4.38 3) 2.8 + 28 4) 9 + 2.7 + 8.38 5) 12.6 + 1.9 6) 4.3 – 2.6 7) 3.19 – 1.4 8) 43.7 – 5.66 9) 10 – 4.72 10) 18.5 – 7.8 11) 4.15 x 3 12) 3.6 x 0.2 13) 0.77 x 0.4 14) 0.047 x 0.5 15) 2.76 x 5 16) 8 ÷ 0.4 17) 12 ÷ 0.3 18) 7.2 ÷ 9 19) 1.44 ÷ 8 20) 3.5 ÷ 0.07

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Solve each problem 1. A landowner divided equally 0.5 ha of land among his children. What is the land area, in square meters given to each child? 2. A can of soft drink contains of 0.33 liters (L) of liquid. If six cans of soft drink can equally fill nine glasses, what part of a liter does each glass contain? 3. One American gallon (gal) contains 3.79 L of milk. If a gallon of milk will be shared by 15 pupils in a class, about how many milliliters (mL) of milk will each child receive? 4. A family consisting of a father, a mother, and their two children ate corned beef for breakfast. Three of the family members had equal shares except for the youngest child who had only half as much as the others. If they consumed 0.45 kg of corned beef, about how many grams(g) did each family member eat? Chapter 3 Understanding Fractions Lesson 1 Fractions and Decimals A fraction is an indicated division. It may denote a part of a whole. It has two parts: the numerator and the denominator. The numerator is the number written above the fraction bar. It indicates the number of equal parts taken into consideration. The denominator is wrtiien below the fraction bar. It represents the whole or the total number of equal parts.

đ?&#x;?

Each part is đ?&#x;’

đ?&#x;?

đ?&#x;?

đ?&#x;’

đ?&#x;’

one-fourths of the triangle is shaded

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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4 6

of the hexagon is shaded

Practice Write the fraction to indicate the shaded region. 1.

2. 3.

Practice Shade the region indicated by the fraction 4.

5 5

5. 4 9

Fraction Parts of a Set Fractions are also used to denote parts of a set. The set could be made up of people, living things, and nonliving objects. To find a fractional part of a set of people or objects, we divide the number of people or objects in the set by the denominator, and then multiply the quotient by the numerator. 1 3 1 2 3 4

of 36 is 12 of 12 is 6 of 100 is 75

Practice: Write the fraction for the shaded part 1.

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2. 3. Test Find the fraction in each sentence. Round off the answer to the nearest whole number 3

1. Out of 45 pupils in class, 5 passed the test. _____ pupils. 2. 3. 4.

1 4 1 3 2 5

of the 36 horses in a farm are male. _____ horses. of the 120 coconut trees in a farm have fruits. _____ coconut trees. of the 200 chickens in a farm lay eggs.. _____ chickens. 2

5. đ?‘œđ?‘“ đ?‘Ąđ?‘•đ?‘’ 42 đ?‘ đ?‘Ąđ?‘˘đ?‘‘đ?‘’đ?‘›đ?‘Ąđ?‘ đ?‘¤đ?‘•đ?‘œ đ?‘Ąđ?‘œđ?‘œđ?‘˜ đ?‘Ž đ?‘Ąđ?‘’đ?‘ đ?‘Ą đ?‘–đ?‘› đ?‘šđ?‘Žđ?‘Ąđ?‘•đ?‘’đ?‘šđ?‘Žđ?‘Ąđ?‘–đ?‘?đ?‘ , 3

passed.

_____

students. Renaming fractions as decimals and vice-versa To rename a fraction as a decimal, simply divide the numerator by the denominator. a.

1

0.5

2

2

1 2

b.

1.0 1.0 0

= 0.5

3

0.75

4

4

3.00 28 20 20 0

To rename a decimal as a fraction, express it as a fraction whose denominator is a power of 10 and move the decimal point. 2

0.2 = 10

1 decimal place

68

0.68 = 100

2 decimal places

375

0.375 = 1 000

3 decimal places

1 256

0.1256 = 10 000 4 decimal places

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Practice A. Rename the following fractions as decimals 1. 2. 3. 4. 5.

4 10 4 100 62 100 5 9 4 7

B. Rename the following decimals as fractions 6. 0.7 7. 0.602 8. 0.85 9. 0.375 10. 0.333 Lesson 2 Kinds of Fractions A fraction whose numerator is less than its denominator is a proper fraction. If the numerator is equal to or greater than the denominator, then the fraction is an improper fraction. 1

2

3

2 3 5 7

4 10

4 7

9

, , ���

, , ���

are proper fractions are improper fractions

Similar fractions have the same denominator. Dissimilar fractions have different denominators. 1 1

1

4 4 1 4

4 7

3 5

8

, , ��� , , ���

are similar fractions are dissimilar fractions

A proper fraction whose numerator is 1 is called a unit fraction. 1 1

1

2 3

4

, , ���

are unit fraction

A common fraction is a proper fraction whose numerator is greater than 1. 2 3

4

3 4

5

, , ���

are common fractions

A mixed number consists of a whole number and a fraction. 1

1

1

1

1 2 ��� 2 4 are mixed numbers. 1 2 means 1 + 2. Changing Mixed numbers into improper fractions and vice versa How to change a mixed number into improper fraction? Step 1: multiply the denominator of the fractional part by the whole number. Step 2: add the product to the numerator of the fractional part. Step 3: the sum is the numerator of the improper fraction. Step 4: the denominator remains the same. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 26

Example 3

Rename 24 into an improper fraction. Multiply the denominator by the whole number: 4 x 2 = 8 Add the product and the numerator: 8 + 3 = 11. Write the sum as the numerator and copy the denominator: 3

24 =

4đ?‘Ľ 2 +3 4

=

8+3 4

=

11 4

11 4

To rename improper fraction as a mixed number, divide the numerator by the denominator. The quotient is the whole number part. The remainder is the numerator of the fractional part. The denominator of the fractional part is the same as that of the original fraction. Example: 25 8

=

3 8

25 24 1

25 8

1

=38

Practice A. Change each mixed number into an improper fraction. 2

1. 23 = _____ 1

2. 85 = _____ 3

3. 7 4 = _____ 3

4. 27 = _____ 5

5. 18 = _____ B. Change these improper fractions into mixed numbers. 6. 7. 8. 9. 10.

14 3 28 5 65 3 72

= _____ = _____ = _____ = _____

7 114 8

= _____

Factors and Multiples A number that divides a second number with no remainder is a factor of that number. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 27

Factors of 6: 1,2,3,and 6 Factors of 12: 1,2,3,4,6, and 12 Factors of 7: 1, and 7 The common factors of 6 and 12: 1,2,3, and 6 The common factors of 6 and 7: 1 When a number has exactly two factors, itself and 1, it is called a prime number. When a number has more than two factors, it is called a composite number The greatest common factor (GCF) is the greatest factor that is common to two or more numbers. The GCF of 6 and 12 is 6. Any composite number may be written as the product of its prime factors. This process is called prime factorization. A factor tree helps in finding the prime factors of a number. 28 2 x 14 2 x 28 = 2 x 2 x 7

7

We can use the prime factorization of two numbers to find their GCF 28 = 2 x 2 x 7 24 = 2 x 2 x 2 x 3 2x2=4 The GCF of 28 and 24 is 4 A multiple of a number is the product of that number and any zero whole number. To find the multiples of a number, we simply skip count starting with the given number. Example: Multiples of 6: 6,12,18,24,30,42,48,54,60 … Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, … Common multiples of 6 and 8: 24, 48, … The least common multiples (LCM) is the least among the common multiples of the numbers. Example: what is the LCM of 2, 3, and 4? Multiples of 2: 2, 4, 6, 8, 10, 12, 16, 18, 20, 22, 24, 26,… Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, … Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, … Practice A. Construct a factor three to find the prime factors of each number. 1. 100 _____ 2. 72 _____ 3. 64 _____ 4. 54 _____ 5. 96 _____ B. Find the factors, the common factors and the GCF 6. 24 _____ 18 _____ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 28

GCF: 7. 10 12 GCF: 8. 40 72

_____ _____ _____ _____ _____ _____

C. Find the multiples, the common multiples and the LCM. 9. 2 _____ 5 _____ LCM: _____ 10. 3 _____ 7 _____ LCM: _____ Equivalent fractions Equivalent fractions are fractions that are equal to each other. They represent the same part of a whole or set. 1 2

2

=

���

4

1 3

=

2 6

To find equivalent fraction of a given fraction, we multiply both the numerator and denominator by the same whole number except zero. 3 5 3 5 3 5

3đ?‘Ľ3

=

5đ?‘Ľ3 3đ?‘Ľ4

=

5đ?‘Ľ4 3đ?‘Ľ5

=

5đ?‘Ľ5

= = =

9 15 12 20 15 25 3

Example 1: give the equivalent fractions of 5 . 3 5

9

=

15

12

=

20

���

18 30

are equivalent fractions.

Another way to form equivalent fractions is to divide both the numerator and denominator Example: find the equivalent fractions of

12 20

Factors of 12: 1, 2,3, 4, 6, 12 Factors of 20: 1, 2, 4, 5, 10, 20 Common factors other than 1: 2 and 4 12 20 12

= ,

6

20 10

12 á2 20 á2

, ���

= 3 5

6

12 á4

10

20 á4

=

3 5

are equivalent fractions.

A cross product of a pair of fractions is the product of the numerator of one fraction and the denominator of the other fraction. Example:

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 29

đ??´đ?‘&#x;đ?‘’

2

6

���

3

are equivalent?

9

3 x 6 = 18 2 3

=

6 9

2 x 9 = 18 The cross products are 2 x 9 and 3 x 6, which are both equal to 18. Thus,

2 3

���

6 9

are equivalent fractions If the cross products are equal, the fractions are equivalent. The numerator and denominator of a fraction are called the terms of the fraction. When there is a missing term in equivalent fractions, we need to know the number that was multiplied to or used to divide the given fraction. 5

đ?‘›

Example: find n such that 8 = 24 A numerator n, is missing. To find n, examine the denominators. 8 x ? = 24 think 24 á 8 = 3 5

đ?‘›

3 is the number used to multiply the terms of 8 to get 24 5

Thus 8 = 15

5đ?‘Ľ3 8đ?‘Ľ3 đ?‘›

=

24

=

đ?‘› 24

24

đ?‘› = 15 5

Therefore 8 =

15 24

Practice A. Write E if the fractions in each pair are equivalent. Write NE if they are not equivalent. 1. 2. 3. 4. 5.

3

12

���

5 4 9 10 3 4

20 24

��� ��� ���

7 1

=

52 13 4 18 14 2

���

8

=

9

=

=

=

B. Find the missing term x 6. 7. 8. 9.

3 8 9 5 1

= =

30 4 7 đ?‘Ľ

=

=

10. 7 =

đ?‘Ľ 40 81 đ?‘Ľ đ?‘Ľ 10 21 đ?‘Ľ 4 28

x = _____ x = _____ x = _____ x = _____ x = _____

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 30

Expressing fractions in simplest form A fraction is in its lowest terms or simplest form when the numerator and the denominator have no common factors other than 1. 7 9 4 5

7

7đ?‘Ľ1

is in its simplest form because 9 = 3 đ?‘Ľ 3; 7 and 9 has no common prime factor. 8

is the lowest term of 10

Example 36

Reduce 24 to its simplest form. 36

Solution: 64 =

2đ?‘Ľ2đ?‘Ľ3đ?‘Ľ3 2đ?‘Ľ2đ?‘Ľ2đ?‘Ľ2đ?‘Ľ2đ?‘Ľ2

=

9 16

Check: (36)(16) = (64)(9) 576 = 576 Practice: A. Find the common prime factors of the terms of each fraction to check if it is in its lowest terms. If not, write the lowest terms. 1. 2. 3. 4. 5.

10 18 6 25 8 36 24 52 30 49

B. Reduce the following fractions to their simplest form. 6. 7. 8. 9.

15 45 24 96 18 54 40 28 12

10. 96 Lesson 3 Comparing and Ordering fractions Comparing proper fractions 3

Which fraction is greater, 4 or

10 12

?>

To answer the question we need to change then into similar fractions. First look at the denominator. 12 is the multiple of 4: 4 x 3 = 12. Changing

3 4

into an

equivalent fraction with 12 as denominator, we have 3

= 4

3đ?‘Ľ3

= 4đ?‘Ľ3

9 12 9

10

3

Comparing two similar fractions now, since 9 < 10 then 12 < 12 and 4 < YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

10 12

Page 31

When comparing similar fractions, the fraction with the greater numerator has the greater value. Comparing mixed numbers To compare mixed numbers, change them first into improper fractions and use whichever method of comparing is convenient. 4

3

Example; which is greater 1 7 or 1 4? Solution: 4

3

Step 1: change 1 7 and 1 4 to improper fractions. 4

11

17 =

7

3

, 14 =

7 4

Step 2: cross multiply

11 7

���

7 4

Since 11 x 4 = 44, 7 x 7 = 49, and 44 < 49, Then

11 7

7

<

4

4

3

��� 1 7 < 1 4

Practice A. Using cross multiplication, determine which of the following fractions in each pair is greater. Box the greater fraction. 1. 2. 3. 4. 5.

3

3

and

7 3

9 16

and

5 7 9 10

26 5

and

7

and

13 17

and

19

11 14 16 18

B. Determine which of the fractions in each pair is greater using any method. Circle the greater fraction. 2

5

6. 2 5 ��� 2 9 6

5

4

8

3

5

7

12

7. 3 11 ��� 3 8 8. 4 5 ��� 4 11 9. 5 7 ��� 5 9 10. 6 9 ��� 6 15 Lesson 4 Estimating fractions 1

We can estimate a fraction to be closer to 0 , 2, 1 Rules on estimating the value of a fraction  When the numerator is much less than the denominator, the value of the fraction is closer to 0. Thus,

2

1

, ��� 40 10

3 25

are fractions closer

to 0 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 32



When the denominator is about half of the denominator, the value 1

of the fraction is closer to 2 . 

7

1

.

When the numerator is about equal to the denominator, the fraction 98

49

is closer to 1. Thus, 100 , 50 ��� 

19

, are all closer to 2 12 41 118 120

are all closer to 1.

We use the symbol ~ to mean closer to or approximately equal to, 98

thus we write 100 ~ 1 ���

19 41

~

1 2

Test 1

A. Fill in the blanks with 0, 2,or 1 1. 2. 3. 4. 5.

8 9 2 20 3 5 13 14 110 112

is close to _____ is close to is close to is close to is close to

_____ _____ _____ _____

B. Replace the terms of each fraction with compatible numbers to estimate 1. 2. 3. 4. 5.

32 70 3 20 126 130 21 50 30 65

đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ or _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ or _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ or _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ or _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ or _____ 1

C. Estimate each given fraction as close to 0, close to 2, or close to 1. 1. 2. 3. 4.

3 65 24 26 42 79 130 138

đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____ đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 33

3

5.

đ?‘–đ?‘ đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘Ą _____

65

Lesson 5 Addition of fractions Adding similar fractions To add similar fractions, 1. Add the numerators to get the numerator of the sum. 2. Use the common denominator as the denominator of the sum. Example: 2 7

3

5

+7=7

If the final answer is improper fraction, we change it to a whole number or a mixed number in simplest form. Adding mixed number To add mixed numbers with similar fractions, we do the following 1. Add the whole number parts. 2. Add the fractional parts. 3. Add the two sums then reduce to simplest form. Example: 1

3

1

35 + 65 = 3 + 6 + 5 +

3 5

4

4

= 9 + 5 = 95

Practice Find the sums. 1

1

1

2 2

+ 2

1. 2.

3 1

3.

5 3

4.

8 4

5.

9

+ + + +

3 3 5 5 8 7 9

= = = = =

Adding dissimilar fractions To add dissimilar fractions, change the addends into similar fractions then follow the rule on adding similar fractions. Example: 1

3

+4=? 5 1

3

The LCD of 5 and 4 is 20. 1 5 3 4

= =

20 á5 (1) 20 20 á4 (3) 20 1

Thus 5 +

3 4

= = =

4 20 15 20 4

15

20

20

+

=

19 20

Practice YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 34

Find the sums 1. 2. 3. 4. 5.

3 8 2 3 9

5 6

+

15

4

3

+

5

=

9

1

+

10 3

+

3

=

4 4

+

6

+

3

+

2 1

+

12 7 8 6

1

+

=

4

2

+

=

9

= Lesson 6 Subtraction of Fractions

Subtracting similar fractions To subtract similar fractions, we subtract the numerators. The denominators remain the same for the difference. Example 4 1 − =? 5 5 4 1 3 − = 5 5 5 Note that we change improper fractions to mixed numbers and change the fractional parts to lowest terms to get the final answer. Subtracting Mixed Numbers To subtract mixed numbers with similar fractions, change them into improper fractions, and then follow the rule on the subtraction of similar fractions. We may also subtract separately the whole numbers and the fractions parts. Example 4

3

24

45 − 25 =

5

−

13 5

=

11 5

1

= 25

When the fraction in the minuend is less than the fraction in the subtrahend, rename the minuend. 1

3

6

3

35 − 25 = 25 − 25 =

3 5

Practice 1. 2. 3. 4. 5.

5

− 6 7 10 7 8 7 9 2 3

2 6

3

−

− − −

= =

10 3 8 4 9 1 3

= = =

Find the differences 2

1. 13 −

1 3

=

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 35

4

3

1

6

1

3

2

7

2. 4 5 − 3 5 = 3. 2 7 − 1 7 = 4. 4 8 − 2 8 = 5. 3 9 − 1 9 = Subtracting dissimilar fractions Change similar fractions to similar fractions before subtraction; the denominator of the fractions should be the least common denominator. Example 2 3

−

1 2

=? 2

1

The LCM of 3 and 2 is 6 so the LCD of 3 and 2 is 6. 2 3

−

1 2

=

2(2)

1(3)

−

6

6

1

+ 6

Check:

4

=6 −

3

= 6

3 6

1

=6

4 6

Practice Subtract. Reduce answers to lowest terms. 1.

4 5

−

1 2

=

2

3

1

4 4

2. 1 3 − 3. 2 6 −

9

= =

1

2

1

3

4. 3 3 − 1 5 = 5. 4 2 − 2 8 = Lesson 7 Solving Problems Involving Addition and Subtraction of Fractions Problem: Helen had

4 5

cup of ripe mango cubes. She put

2 3

cup of these into a fruit salad she

was mixing. How much mango cubes was left? What is asked? Amount of mango cubes left What are given?

4 5

cup - original amount of mango cubes;

2 3

cup-amount of mango

cubes mixed in salad Operation to be used: Subtraction Number Sentence: 4 5

−

2 3

12

10

2

= 15 - 15 = 15

Complete answer:

2 15

cup of mango cubes was left.

Test Solve the following problems YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 36

1. An automobile gas tank is into it, making it

3 4

1 5

full of gasoline. Additional gasoline was loaded

full. What part of the full capacity of the tank was the

gasoline added to it? 2. Mr. Duque owns a 4- hectares fruit farm.

1 6

of the farm is planted with

1

rambutan trees, 3 with banana trees, and the rest with mango trees. What part of the fruit farm is planted with mango trees? Lesson 8 Multiplication of Fractions If we have the fractions đ?‘Ž

đ?‘?

x

đ?‘?

đ?‘Žđ?‘?

=

đ?‘‘

đ?‘Ž đ?‘?

đ?‘?

and đ?‘‘ , then the product is

đ?‘?đ?‘‘

If a = 4, b = 7, c = 14, and d = 8, then đ?‘Ž

x

đ?‘?

đ?‘?

4

=

đ?‘‘

7

x

14 8

To get the product we can actually multiply 4 x 14 = 56 and divide it by 7 x 8 = 56, so 56

that we get the product 56 or 1. But we can also divide out the common factors in the numerator and the denominator. Example 1: 4 7 4

14

x

8 1

=

=2

8 14

2

7 2

=1

=2 =1 Example 2: 1

6

1 5 x 10 = 5 x

10 1

= 12

Example 3: 1

1

4

9

13 x 1 8 = 3 x 4 3

x

9 8

3

8

change the mixed number to improper fraction first.

1

= 2 = 12

In multiplying fractions, look for the common factors in the numerator and denominator and cancel these out. This will make the multiplication easier. Practice: 1. 2. 3. 4.

1 3 2 5 4 7 5 9

1

x 12 = 1

x 22 = 5

x2 = 8 3

x 35 = 1

3

5. 2 3 x 24 = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 37

2

3

1

5

6. 29 x 4 = 7. 3 5 x 9 = 8. 9.

3 5 5 9

5

x6= 1

x 25 = 2

9

10. 6 3 x 10 = Lesson 9 Division of fractions To divide a fraction by another, multiply the dividend by the reciprocal of the divisor to get the quotient. đ?‘Ž đ?‘?

á

đ?‘? đ?‘‘

đ?‘Ž

đ?‘‘

=đ?‘?xđ?‘? 3

Example: 5 á

2

3

5

3

1

= 5 x 2 = 2 = 12

5

Since the multiplication is the inverse of division, the result may be checked by multiplying the quotient by the divisor to get the dividend. 3

đ?‘Ľ 2

2

6

3

= 10 or 5 5

Practice: 1.

1 3

á 1

1 10 3

2. 1 3 á 3.

4 9

4 4

á 2 12 3

2

4

3

1

3

4. 3 5 á 2 9 5. 4 5 á 3 15 6. 2 3 á 1 4 6

7. 5 á 11 7

8.

1 5

5 7

3

á 3 10 1

3

5

2

9. 6 2 á 2 4 10. 5 6 á 6 3 Lesson 10 Solving problems Involving Multiplication and Division of Fractions 3 4

of the class joined Math Club. Half of those who joined Math Club also joined

Science Club. If 15 students are in both Math Club and Science Club, how many students are in the class? Solution:

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Page 38

Given is: 15 = number of students both in Math Club and Science Club letting x be the number of students in the class, then 3 4 1 2

đ?‘Ľ = number of students whom joined Math Club 3 4

đ?‘Ľ = number of students who joined both Math Club and Science Club

Thus the equation is 1

3

2

4

đ?‘Ľ = 15

3 đ?‘Ľ = 15 8 8

3

3

8

đ?‘Ľ = 15

X=

8 3

15 (8) 3

X = 40 There are 40 students in the class. Solve the following problems 1. It is estimated that the size of the neck (circumference) is around 0ne-half that of the waist. If the waist of a woman measures 28 in., what is the size of her neck? 2. Three-fifths of the birthday cake was divided equally among six guests. The remainder of the cake was divided also equally among five guests who came late. Which of the two groups had each member received a larger share? By how much was the share larger? Chapter 4 Understanding Ratio, Proportion and Variations Lesson 1 Ratio A ratio is indicated quotient of two numbers. The ratio of a and b may be written as

đ?‘Ž đ?‘?

or a:b. Example The ratio of 2 boys to 3 girls may be written in 3 ways: 2 to 3

2:3

2 3

This means for every 2 boys there are 3 girls. The numbers 2 and 3 are called the terms of the ratio. A ratio that involves quantities with the same units of measure allows us to make three types of comparisons (a) part-to-part, (b) part-to-whole, and (c) whole-to-part. Example: there are 4 red roses and 5 white roses in a vase. The sum of all the roses is 4 + 5 = 9 roses. We now make the following comparisons. 4

a. Part-to-part: the ratio of red roses to white roses is 5 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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b. Part-to-whole: the ratio of white roses to all flowers is

5 9

9

c. Whole-to-part: the ratio of all flowers to red roses is 4

Sometimes ma whole is divided into three parts or more. We the form the ratio a:b:c where a, b, and c are parts of the whole or set. They form a partitive ratio. Example In a box there are 5 red balls, 3 blue balls, and 4 yellow balls. We then represent the contents of the box by the following partitive ratio: Red balls:blue balls: yellow balls = 5:3:4 The following comparisons can be made 5

Ratio of red balls to blue balls: 3 5

Ratio of red balls to yellow balls: 4 3

Ratio of red balls to all balls: 5+3+4 =

3 12

1

or 4

When the ratio compares quantities having different units of measure, it is called rate. We usually express a rate as a unit rate. A unit rate has 1 as its denominator. Example A car traveled 150 km in 3 hours. Its average speed (speed is expressed as a rate) Speed =

đ?‘‘đ?‘–đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘?đ?‘’ đ?‘Ąđ?‘–đ?‘šđ?‘’

=

150 đ?‘˜đ?‘š 3đ?‘•

=

50 đ?‘˜đ?‘š 1đ?‘•

The unit rate is 50 km per hour. Comparing rates In a supermarket, Manny saw these displays for prices of mangoes. A B C 5 mangoes for P75.00

3 mangoes for P48.00

10 mangoes for P155.00

Which prize display has the smallest unit price of mangoes? To find the answer, we compute the unit rate for each display. đ?‘¤đ?‘•đ?‘œđ?‘™đ?‘’ đ?‘ đ?‘Žđ?‘™đ?‘’ đ?‘?đ?‘&#x;đ?‘–đ?‘?đ?‘’

For A: đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; For B: For C:

đ?‘œđ?‘“ đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œđ?‘’đ?‘ đ?‘¤đ?‘•đ?‘œđ?‘™đ?‘’ đ?‘ đ?‘Žđ?‘™đ?‘’ đ?‘?đ?‘&#x;đ?‘–đ?‘?đ?‘’

đ?‘ƒ75.00

đ?‘ƒ15.00

= 5 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œđ?‘’đ?‘ = 1 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œ

đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œ đ?‘’đ?‘ đ?‘¤đ?‘•đ?‘œđ?‘™đ?‘’ đ?‘ đ?‘Žđ?‘™đ?‘’ đ?‘?đ?‘&#x;đ?‘–đ?‘?đ?‘’ đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œđ?‘’đ?‘

đ?‘ƒ48.00

đ?‘ƒ16.00

= 3 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œđ?‘’đ?‘ = 1 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œ = 10

đ?‘ƒ155.00 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œđ?‘’đ?‘

đ?‘ƒ15.50

= 1 đ?‘šđ?‘Žđ?‘›đ?‘”đ?‘œ

Test Write the following as ratios in their lowest terms 1. A man saves P150 for every P1,500 earned. What is the ratio of the man’s savings to his earnings? 2. In a particular subject, 16 books are available to the 40 pupils in a class. 3. Out of 15 games it played, team A won 9 of the games. What is the ratio of wins to losses? YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Write each as a rate then find the unit rate. 1. 10 postcards for P150.00 2. P600 paid for painting a wall for 3 days 3. 288 pulse beats in 4 min Lesson 2 Proportion A proportion is a statement of equality of two ratios. Properties of proportions 1. In a proportion, the product of the extremes equal to the product of the means. 2. Interchanging the first and fourth terms of a proportion still results to a proportion. 3. Interchanging the second and third terms of a proportion results to a proportion. 4. Interchanging the second term with the first term and the fourth term with the third term of a proportion results to a proportion. Finding the missing term in a proportion 1. Multiply each side of the proportion by the same number. 2. Find the cross products 3. Equate the numerators or denominators 4. Convert one ratio to a decimal Example: 5 8

=

đ?‘Ľ 72 5

(72)8 =

đ?‘Ľ 72

(72)

X = (9)(5) X = 45

Practice: Solve for the missing term in each proportion 1. 2. 3. 4. 5.

2 5 đ?‘Ľ

=

32 đ?‘Ľ 64 4 13 đ?‘Ľ 9

= = =

=

đ?‘Ľ 20 9 12 5 16 đ?‘Ľ 65 18 27

Equate the numerators or denominators 1. 2.

8 32 30 đ?‘Ľ

= =

4 đ?‘Ľ 3 8

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3. 4. 5.

4 đ?‘Ľ 6

=

15 42 55

= =

2 19 18 đ?‘Ľ 21 đ?‘Ľ

Lesson 3 Direct Proportion When two quantities a and b vary directly, their ratio is always a constant. If k đ?‘Ž represents the constant, then we have đ?‘? = đ?‘˜ or a = bk. This kind of proportion is called direct proportion. Quantity a is directly proportional to quantity b. Example The cost of x rice varies directly with the quantity y. if the unit cost is P 25 per kilogram (kg), then 50 kg of rice costs 50 x P25 = P 1,250. Using the concept of direct proportionality, x = ky đ?‘ƒ25 1 đ?‘˜đ?‘”

x 50 kg = P1,250

Practice Solve the following problems 1. In a bicycle shop, Rico can repair 6 bicycles in 30 h. (a) how many hours will it take Rico to repair 2 bicycles? (b) How many bicycles can Rico repair in 5 hours? 2. The cost per dozen of eggs is P 60. (a) How much do five dozen eggs cost? (b) What is the cost of one egg? Lesson 4 Partitive Proportion When 3 quantities x,y, and z vary directly (or more) has a constant sum k, the quantities and the sum form a partitive proportion. đ?‘Ľ+đ?‘Ś+đ?‘§ =đ?‘˜ Example: A father willed that his property worth â‚ą2,000,000 be divided among his two sons and his wife in the ratio of 2:3:5 respectively. How much will each of his sons inherit? How about his wife? How the ratio 2:3:5 called? It is called partitive ratio, and the equation that relates the quantities is called a partitive proportion. To form the partitive proportion for the above problem, add 2, 3, and 5, and then equate the ratio to the sum. 2 + 3 + 5 = 10

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The sum, which is 10, is the total number of parts into which the property is to be 2

3

divided. One of the sons will get 10 of ₱2,000,000. The other son will get 10 of 5

₱2,000,000. The wife will get 10 of ₱2,000,000. Solving each, we get 2 10 3 10 5 10

x ₱2,000,000 = ₱ 400,000 x ₱2,000,000 = ₱ 600,000 x ₱2,000,000 = ₱ 1,000,000

To check: ₱ 400,000 ₱ 600,000 + ₱ 1,000,000 ₱ 2,000,000 Finding the total number of parts in a partitive ratio is the first method in solving a problem on partitive proportion. Another way is to use algebra If we let x be a unit share of the father’s property, then 2x = share of one son 3x = share of the other son 5x = share of the wife The partitive proportion is 2x + 3x + 5x = ₱ 2,000,000 10 x = ₱ 2,000,000 x = ₱ 200,000 Thus these are the amounts of the shares: 2x = (2)( ₱ 2,000,000) = ₱ 400,000 – share of one son 3x = (3)( ₱ 2,000,000) = ₱ 600,000 – share of the other son 5x = (5)( ₱ 2,000,000) = ₱ 1,000,000 – share of the wife Exercises: Solve each problem: 1. The amounts of money from the first prize to the third prize in billiards competition have the ratio 4:2:1. If the second prize winner will get ₱ 50,000, what is the total prize money for the first and third places? 2. A piece of stick is cut into 3 parts according to the ratio 4:2:6. What is the length of each part if the length of the stick is 90 cm? 3. A businessman deposited money in the bank for his four children in the ratio 4:5:6:7 for the youngest to the oldest. If he deposited ₱ 20,000 for his youngest child, how much did he deposit in all? Lesson 5 Inverse proportion

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If one variable x varies inversely with another variable y, then xy = k, where k is the constant of the proportionality. This equation means that as x increases, y decreases, and vice versa. The equation xy = k is an inverse proportion. For example, if the area of a rectangle is constant, the length l varies inversely as the width w. đ?‘™đ?‘¤ = đ??´ Problem: A rectangle is 8 cm long and 6 cm wide. For the area to remain constant, what should be the width if the length is increased to 12 cm? Solution: the new length is 12 cm. let w be the new width. We have the following equation (8 cm)(6cm) = (12 cm)(w) = A The value of A can be computed as A = 8 cm x 6 cm = 48 cm2 Hence, (12 cm)w = 48 cm2 48 đ?‘?đ?‘š 2

w = 12 đ?‘?đ?‘š 2 w = 4 cm The width should be 4 cm Practice: Solve the problem. 1. At constant voltage V, electric current I in amperes varies inversely with the resistance R in ohms (â„Ś). If the constant voltage is 220 volts (V), and the resistance is 110 â„Ś, find the current 2. At constant temperature, the volume V of a gas varies inversely as the pressure P it exerts. If the pressure is 50 grams per square centimeter (g/cm2), find the volume of the gas. 3. If the area A is constant, one diagonal (d1) of a rhombus varies inversely as 1

the other diagonal (d2) according to the relation A = 2 đ?‘‘1 đ?‘‘2 . If the area is 2.5 cm2 and one diagonal is 5 cm, how long is the other diagonal?

Chapter 5 Understanding Percent, Probability, and the Circle graph Lesson 1 Percent When solving problems about percent, the basic equation is: percentage equals base times percent expressed as decimal. The equation can be written as an equality of two ratios or as a proportion. For example, 40% of 80 pupils is 32. 40% x 80 = (0.4)(80) = 32 Remember this The base is the whole quantity form which a percent is obtained. Percentage is the product of the percent expressed as a decimal and the base. The equation relating percentage, base, and percent is YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Percent x base = percentage Finding percentage Example: Of the 48 passengers on a bus, 25% are children. How many children are on the bus? What is asked: the number of children on the bus (the percentage) Given: the number of passenger on the bus, 48, is the base; 0.25 is the percent. Solution: Let C represent the percentage, Percentage = percent x base C = (0.25)(48) = 12 There are 12 children on the bus. Finding the percent Example: In a math test, 24 out of 32 pupils passed the test. What percent of the pupils passed the test? What is asked: the percent of pupils who passed the test? Given: 24 is the percentage; 32 is the base Solution: Percent =

đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ?‘?đ?‘Žđ?‘ đ?‘’

24

= 32 = 0.75 or 75%

75% of the pupils passed the math test. Finding the Base Example: Twenty pupils passed the test in science; this represents 80% of the pupils who took the test. How many pupils took the test? What is asked: the base (the number of pupils who took the test) Given: 20 pupils passed – percentage; 80% - percent Solution: Base =

đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ą

20

= 0.8 = 25

There were 25 pupils who took the test. 1

1

There are cases wherein percent is a mixed number, such as 72%, 129%, and 3

1504%. To change the percent into decimal Express the fractional part as a decimal. 1

72% = 7.5% 1

124 % = 12.25% 3

1504 % = 150.75% Change the whole percent into decimal.

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7.5

7.5% = 100 = 0.75% 12.25% =

12.25

= 0.1225%

100 150.75

150.75% =

100

= 1.5075%

Example: Mr. Cortez invested ₱ 50,000 on a business venture. Mr. Belo, his partner, added 3

1504% of that amount. How much did Mr. Belo invest? Solution: 3

Percent = 1504% = 150.75% = 1.5075 Percentage = 1.5075 x ₱ 50,000 = ₱ 75,375 Mr. Belo invested ₱75,375. Another example: Mr. Cruz paid ₱800 for his medicine after being given a 20% discount at the drugstore. How much was the original price of the medicine? Solution After the discount, Mr. Cruz pays only 100% - 20% = 80% of the original price. Let x = the original cost of the medicine; the base 80% of x = ₱ 800 0.80 (x) = 800 800

X = 0.80 = 1,000 Practice: Change the percents to decimals 1. 25% = _____ 2. 9% = ______ 3. 89% = _____ 4. 523% = ____ 5. 12.6% = ____ Change the fractions to percents 1. 2. 3. 4.

4 5 3

=

18 38

= =

50 126 90

=

Solve the following problems. 1. Thirty of the 48 pupils in a class submitted their home works late. What percent of the class submitted their homework late? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 2. The teacher assigned a difficult problem to class. Only 30% of the 40 pupils were able to solve it. How many were not able to solve the problem? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 3. Four of the graduating elementary pupils in a school passed the entrance test for the Philippine Science High School. These pupils represent 8% of those who took the entrance exam. How many from the school took the exam? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 4. In a farm are 16 horses, 20 carabaos, and 24 cows. a. what percent of the animals are cows? b. what percent of the animals are not horses? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________

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5. In a class of 48 pupils, 50% submitted their home works on time, 20 pupils submitted their home work late. What percent of the class did not submit their home work? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 6. On a certain day, 5 of the 40 pupils in a class were absent. What percent of the class was present? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 7. Nena spends ₹25 for her transportation to school. This represents 50% of her daily allowance. How much is Nena’s daily allowance? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 8. In a test on problem solving, 30% of the items are easy, 40% are of average difficult, and the rest are difficult items. (a) If there are 20 average items, how many problems are there in all? (b) How many are difficult? What is asked:_____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

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Complete Answer: _________________________________________________ 9. Results of a survey show that 30% of cars in a certain city are silver, 25% are black and 20% are maroon. The rest may be green, blue, beige and other colors. If there are 1000 cars in the city, how many are neither colored silver, black, nor maroon? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 10. In a rural coastal town, 40% of the males are farmers, and 25% are fisher folk. The rest are in white-collar jobs, including teachers, government employees, and business people. If there are 150 fisher folk, how many are farmers? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ Lesson 2 Other Applications of Percent Percent Discount Percent discount is the percent subtracted from the original price of an item or commodity. Example: A 10% discount is offered for an item that originally costs ₱1,000. The discount is equal to 10% of ₱1,000 = (0.10) (₱1,000) = ₱100. The discounted price is ₱1,000 minus 10% of ₱1,000. = ₱1,000 - ₱100 = ₱900 Commission Those who engage in selling real estate properties ( land and buildings), insurance and other commodities, receive commission. The commission is a certain percent of the amount of the transaction. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Example: A 5% commission is given to the real estate brokers of a certain company. If a broker is able to sell a house and lot worth â‚ą3,000,000 then his commission is 0.05 x â‚ą3,000,000 = â‚ą150,000 Tax A certain percent of a taxable amount is given to the government as tax. 1) 12% value-added tax(VAT)(by law) is charged to a customer who buys a taxable item, such as gasoline or food from a restaurant. Example: If you buy 20 L of gasoline at â‚ą44 per liter, the pre-tax cost of the gasoline is (20)(â‚ą44) = â‚ą880 The VAT is 12% of â‚ą880 (0.12)(â‚ą880) = â‚ą105.60 The total amount you pay is Pre-tax cost + VAT = â‚ą880 + â‚ą105.60 = â‚ą985.60 2) The government collects taxes on interest earned by savings from investments; such as time deposits, from banks. The amount collected is 20% of the interest earned. Example: If you made a time deposit of â‚ą1,000,000 and it earned â‚ą82,000, the government collects a 20% tax on the earned amount or interest. (0.20)(â‚ą82,000) = â‚ą16,400 So the amount your time deposit earns from the interest is only â‚ą82,000 - â‚ą16,400 = â‚ą65,600 Percent Increase or Decrease Example 1: For school year 2005 to 2006, the tuition fee in a certain school was â‚ą20,000. For school year 2006to 2007 the tuition fee was increased to â‚ą22,000. What was the increase in percent? Solution: Amount of increase = â‚ą22,000 - â‚ą20,000 = â‚ą2,000 In this case â‚ą20,000 is the base. So the percent increase is 2 000 20 000

=

1 10

= 0.10 đ?‘œđ?‘&#x; 10%

Example 2: Rounded off, the population of the Philippines in 2001 was 83 million. In 2005 it was 88 million. Compute the percent increase in population. Solution: Increase in population = 88 000 000 – 83 000 000 = 5 000 000 5 000 000

Percent increase = 83 000 000 = 0.06 or 6% YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Example: The enrolment in a certain private school for school year 2005 to 2006 was 1 250. For school year 2006 to 2007, the enrolment was only 1 000. What was the percent decrease in enrolment? Solution: Decrease in enrolment = 1 250 – 1 000 = 250 250

The base is 1 250, so the percent decrease is 1 250 = 0.2 or 20% Interest In this book, only simple interest will be taken up. In computing interest, time is another quantity considered aside from percentage, base and rate. The formula for interest (I) earned is the product of base or principal (P), rate (r), and time(t). Interest = principal x rate x time I = Prt Time depends on the unit of the rate. If the rate is percent per year, then time is in years. This is the usual unit of time in financial transactions. Example 1: If ₱10,000 (principal) is deposited in a bank that pays a 5% interest per year, how much interest will the principal earn in 4 years? Solution: The given are: P = ₱10,000, r = 0.05/year, and t = 4 years. I = Prt I = (₱15,000)(0.08)(1.5) I = ₱1,800 The investment will earn ₱1,800 Interest on loans is computed in the same way. Example 3: Mr. Valle has borrowed ₱500,000 to buy a car. The loan is being charged a 12% interest per year and is payable in 5 years. How much total interest shall Mr. Valle pay in 5 years? Solution: I = Prt = ₱500,000 x 0.12 x 5 = ₱300,000 The total amount Mr. Valle will pay is ₱500,000 + ₱300,000 = ₱800,000 Example: 1

A man invested ₱10,000 in a business that has a yearly profit equal to 124% of the capital. After 2 years, how much shall his investment earn? Solution: 1

r = 124% = 0.225 using the formula I = Prt YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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I= ₱10,000 x 0.1225 x 2 I = ₱2,450 The man’s investment will earn ₱2,450 Practice: 1. For school year 2006 to 2007, a certain private school increased the tuition fee by 10%. If the tuition for school year 2005 to 2006 was ₱24,000, what was its amount for school year 2006 to 2007? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 2. A 20% discount is given by a department store on a pair of denim pants that originally cost ₱1,800. What is the sale price of the pants? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 3. A family ate 600-peso worth of food in a restaurant. A 12% VAT was added to the bill. How much was the total bill? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 4. Mr. Chavez deposited ₱ 10,000 in a bank that gives 6% interest per year. He withdrew the total amount after 5 years. How much money did he get? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 5. Starting the year 2000, Mrs. One made yearly deposits of ₱5,000 for 3 years. The money earned interest at 7% per year. What was Mrs. Ong’s total amount of money at the end of 2005? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 6. A certain school charges ₱25,920 tuition fee per year. This amount includes an 8% increase in tuition. How much was the tuition fee last school year? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 7. The population of a certain town was 15 210 last year. This year it became 16 008. What was the percent increase in population? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 8. A real estate broker sold a house and lot for ₱ 6,000,000. He received a 1.5% commission on the sale. How much did the owner receive for selling the house and lot? What is asked: _____________________________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 9. Mr. Tan ate in a restaurant. The original bill was ₱250. Being a senior citizen, Mr. Tan was given a discount of 20%. After the discount, he was charged with 12% VAT. What was the final amount he paid for the bill? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 10. For school year 2006 to 2007, a certain private school charge ₱27,500 for tuition. This amount represents a10% increase over that of the previous year. How much was the tuition the previous year? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 11. a senior citizen was given a 20% discount on medicines but was charged with 12% VAT. The medicines were worth ₱1,200. a. How much discount did he get as a senior citizen? b. How much VAT was he charge with? c. What was the net amount that the senior citizen paid? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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12. A 4% commission was given to a real estate broker for selling real property worth â‚ą3,000,000. How much did he get as commission? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ 13. Mrs. Ramos has deposited â‚ą100,000 in a bank that gives 5% interest per year. But 20% of the interest goes to the government as tax. How much net interest will Mrs. Ramos get after 1 year? What is asked: _____________________________________________________ What are given: ____________________________________________________ Operations to be used: _______________________________________________ Number Sentence: __________________________________________________ Solution:

Complete Answer: _________________________________________________ Lesson 3 Simple Probability Consider the following statements 1. It will probably rain tomorrow. 2. Most likely, Mr. X will win in the election. 3. There is a very good chance that the weather will be fine next week. 4. There is 50% chance that it will rain tomorrow. These are all probability statements. They are educated guesses based on past experiences or previous data. Words and phrases such as probably, most likely, and chance indicate probability. How does one make an educated guess? Say, for example, you live in the western side of Luzon where the chance of rain during the months of July, August, and September is high. So if you predict that on the month of august it will probably rain, you have a good chance of being right. You are making n educated guess. But if you make the same prediction for the month of March, April, or May, then you have a good chance of being wrong. In sample statement 2, the likelihood of Mr. X winning the election may be based on the surveys. This statement is based on statistical probability. Now suppose you toss a coin. What is the chance that you will come up a head? If you throw a die, what is the probability that the face with four dots will turn up? If there are eight horses in a race, what is your chance of guessing the winner? YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Answering these questions require mathematical probability because they are based on the mathematical theory of probability. The basic definition of probability is based on the following related terms; experiment, sample space, and event. Let us now define each term. An experiment is any activity that can be done repeatedly under similar conditions some examples of experiment are rolling a die, tossing a coin and drawing a card from a deck. A sample space is the set of all possible outcomes in an experiment. For example, the sample space for tossing a coin includes a head and a tail. An outcome is also called an event. An event is a subject (or part) of a sample space. Getting an even number in rolling a die and drawing a red card from a deck of cards are events. If an event can happen in m ways and fail to happen in l ways, and if m and l ways are equally likely, then the following are true. đ?‘š 1. The ratio P = đ?‘š +đ?‘™ is the probability that m will occur 2. The ratio q =

đ?‘™ đ?‘š +đ?‘™

is the probability that m will not occur or that l will occur.

The above statements form mathematical theory of probability. The value of a probability may be anywhere from o to 1 ( or 0% to 100%). A probability of 0 would mean that an event is certainly not going to occur. A probability of 1 would mean that the event will certainly occur. The higher the probability, the more likely the event is going to occur. Example 1 Tossing of a coin If you toss a coin, it will land with either the head or the tail facing up. If you let m be the number of ways a head can come up and l that for a tail m = 1 and l = 1. đ?‘š

The probability of getting a head is P(head) = đ?‘š +đ?‘™ =

1

1

=2 1+1

And the probability of getting a tail is đ?‘™

P(not head) = q = đ?‘š +đ?‘™ =

1 1+1

=

1 2

Head is the favorable outcome because this is what you want to happen. We 1

state the probability of getting a head as 2 or 1 out of 2, and that getting a tail also 1 2

or 1 out of 2. We can say that getting a head and getting a tail are equally likely.

As shown above, a probability can be expressed as a fraction. We can also 1

represent a probability as a decimal. In our example, P = 2 can also be expressed as P = 0.5 means that there is a 50% chance of getting a head on a single toss of a coin. Example 2: Throwing a die A die is a cube with each face representing a number from 1 to 6. Suppose you want the number 4 face to come up. So you let m be the number of ways the 4 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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face can come up, which is 1: m = 1. There are 5 ways of not getting a 4 face: getting a 1,2,3,5 or 6 face. Thus l = 5. The probabilities are đ?‘™

P(4) = đ?‘š +đ?‘™ =

1 1+5 đ?‘™

=

1 6

P(not 4) = q = đ?‘š +đ?‘™ =

5 1+5

=

5 6

1

5

So you have 6 of getting a 4 face, and a 6 chance of not getting it. Take note that 1

all events are equally likely to occur; p=6 for each of the six faces of the die. Example: Horse race Assuming all the horses in the race are of equal abilities and there are eight horses participating, your probability of guessing the winner correctly is

1 8

1

P(winner) = 8 Example 4: In a box, there are six red balls and four white balls, all of the same size. If you pick one ball at random (without looking), what is the chance that it will be a red ball? In this case m= 6 and l = 4, hence đ?‘š

6

6

3

P(red) = đ?‘š +đ?‘™ = 6+4 = 10 = 5 or 0.6 đ?‘™

P (not red) = q = đ?‘š +đ?‘™ =

4

= 6+4

4 10

2

or 5 or 0.4

3

2

There is a 5 or 60% chance of getting a red ball and a 5 or 40% chance of getting a white ball. The tossing of a coin, throwing of a die, and a horse race are events. The sample space is the set of all outcomes. In the throwing of a coin, head and tail make up the sample space. In the throwing of a die, the faces 1,2,3,4,5, and 6 are the sample space; and in the horse race, horses 1 to 8 are the sample space. Remember this Probability is a number from 0 to 1 ( or 0% to 100%) that is a measure of how likely an outcome is not going to happen. A probability of 0 that an outcome is not going to occur; 1 means the outcome is certainly going to occur. An experiment is an activity that can be done repeatedly under similar conditions. A sample space of an experiment is the set of all possible outcomes, also called events. Practice Answer this: A. State whether the question or statement expresses a statistical probability or a mathematical probability. __________1. I will probably be still alive 20 years from now. __________2. The chance is 1 in 6 that a number 3 face will come up in a throw of a die. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1

__________3. The probability is 2 for a tail to come up in a single toss of a coin. __________4. There is a good chance that tomorrow will be a cloudy day. __________5. If five runners are in a 100-m sprint, what is your chance of picking the winner correctly? B. In a box of five blue balls, three red balls, and two green balls, all of the same size. 6. What is the probability of picking the blue ball? _____ 7. What is the probability of not picking the blue ball? _____ 8. What is the chance of picking the green ball? _____ 9. What is the chance of picking the red ball? _____ 10. What is the chance of not picking the red ball? _____ C. if a die is thrown, what is the probability that 11. The number 6 will come up? _____ 12. An even number will come up? _____ 13. An odd number will come up? _____ D. Suppose there are four options in a multiple-choice test and you do not know the answer to a certain item. What is the probability of 14. Correctly guessing the answer? _____ 15. Getting the wrong answer? _____ E. Ten Ping-Pong balls labeled 1 to 10 were place in a box. One ball is randomly picked. What is the probability that 16. The ball is even-numbered? _____ 17. The ball is odd-numbered? _____ 18. The number on the ball is divisible by 3? _____ 19. The number on the ball is divisible by 2? _____ 20. The number on the ball is divisible by 4? _____

Lesson 4 Circle Graph Four types of graphs – the pictograph, the bar graph, the line graph and the circle graph. A circle graph or a pie chart shows how the parts relate to the whole. It is used to compare the different parts of the whole. In a circle graph, a circle is used to represent the total or whole. The area covered by each sector or pie slice represents the area of the sectors of the whole.

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The circle graph below gives the preferred fruit-flavored ice cream of grade 6 pupils.

25% of the pupils like mango-flavored ice cream. 25% of the pupils like strawberry-flavored ice cream. 50% of the pupils like chocolate-flavored ice cream. If there are 20 pupils in the group, how many prefer strawberry ice cream? To find the answer, we multiply the total number of pupils by the percent who prefer strawberry. That is, 25

1

20 x 25% = 20 x 100 = 20 x 4 = 5 Five pupils choose strawberry ice cream. Let us examine a graph wherein the percent represented by each part is labeled Distribution of family budget

From the given circle graph, answer the questions that follow. 1. What is the large part of the budget? _____ 2. What is the least budget? _____ 3. If the budget is ₱20 000, how much will go to food? _____ 4. If the budget is ₱20 000, how much will go to savings? _____ Lesson 5 Constructing a Circle Graph The total income of a family of four is ₱20,000 per month. The table below shows how the family budget their income for a month. Construct the circle graph for the given data. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Monthly Budget for a family of 4 Item Amount allotted Food ₱10,000 Housing ₱4,000 Education ₱2,000 Clothing ₱1,000 Miscellaneous ₱2,000 Savings ₱1,000 Here are the steps in making a circle graph. 1. From the table of data, find the total. The total is ₱20,000. 2. Find what percent of the total each item (or category) in the budget is. It will also help if we write the fractional part each item represents. 10 000

Food: Housing: Education: Clothing: Miscellaneous: Savings:

20 000 4 000 20 000 2 000 20 000 1 000 20 000 2 000

= = =

=

20 000 1 000 20 000

= =

10 20 4 20 2 20 1 20 2

= 0.50 = 50% = 0.20 = 20% = 0.10 = 10%

= 0.05 = 5%

20 1 20

= 0.10 = 10% = 0.05 = 5%

4. Draw a circle into equal parts. In this case, we divide the circle into 20 parts. Thus, the items will represent the following parts. 5. Label the graph and give it a title.

monthly budget of a family of 4 savings miscellaneou 5% s 10%

clothing 5%

education 10%

food 50% housing 20%

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Food – 10 parts Housing – 4 parts Education – 2 parts Clothing – 1 part Miscellaneous – 2 parts Saving – 1 part 4. Label the graph and give it a title Monthly Budget of a Family of 4 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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You may use the protractor in constructing the circle graph. Convert the fractional part of each category into degrees by multiplying it with 3600. One circle measures 3600. Let us now consider the next example: 150 twelve-year-old children were asked for their color preferences. The data are shown below. Color Number of Fraction of the % of the Number of students total total degrees 5

Red

50

50 5 = 150 15

33%

Blue

40

40 4 = 150 15

27%

Green

30

30 3 = 150 15

20%

3600 x 15 = 720

Yellow 30

30 3 = 150 15

20%

3600 x 15 = 720

3600 x 15 = 1200

4

3600 x 15 = 960 3

3

Total = 150 Give the title of the circle graph. Do this Activity: Construct the circle graph for each of the following sets of data. Indicate the percentage for each category and give a title for the graph. 1. A grade 6 spends his monthly allowance on the following: Food - ₱400 Transportation - ₱300 School supplies - ₱100 Miscellaneous - ₱200 2. During weekdays, Danielle spends the number of hours for the activities: Sleeping time – 8 h Preparing for school – 2 h School time – 10 h Working on assignments – 3 h Watching T.V. – 1 h 4. In a camping group, 50 students are in 3, 60 students are in grade 4, 40 are in grade 5, and 50 are in grade 6. Test yourself Choose the best answer from the given points. Circle the letter of your choice. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1

1. Which decimal is equal to 52%? a. 5.5

b. 0.55

c. 0.055

d. 0.0055

3

2. The fraction 80 is equal to _____ a. 3.75% b. 37.5% 3. 125% is equal to _____ 3

a. 8

1

b. 8

4. 25% of 160 is _____ a. 20 b. 40 5. 20% of what number is32? a. 320 b. 3.2

c. 375%

d. 0.375%

c. 1.25

d. 12.5

c. 80

d. 120

c. 64

d. 160

Chapter 6 Understanding Plane Figures and Geometric Solids Lesson 1 Areas of Selected Plane Figures Area of a parallelogram A parallelogram is a plane figure that has four sides and whose opposite sides are parallel and congruent. Before we can find the area of a parallelogram, we need to know the length of its base and its altitude or the height. The base of a parallelogram can be any of its sides. Its altitude is the perpendicular distance between the chosen base and its opposite.

base altitude

h

base A parallelogram The area A of the parallelogram below is equal to b x h, where h is the altitude to the base b. Area of parallelogram = base x altitude A = bh In a square rectangle, the adjacent sides are perpendicular. Hence the perpendicular sides form a pair consisting of a base and the altitude to the base. The longer side of a rectangle is usually referred to as the length l while the shorter side is the width w. Area of a rectangle = length x width YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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A = lw The lengths of the sides of a square are equal. If s represents the length of one side of the square , then the area is the square of the side, s2. Area of square = square of the side A = s2 Example 1 Find the area of a parallelogram whose base is 15 cm and the altitude is 4 cm. Given: b = 15 cm h= 4 cm Solution: A = bh = (15 cm)(4 cm) = 60 cm2 Example 2: The area of a square garden is 64 m2. What is the length of one side of the garden? Given: A = 64 m2 Solution: A = s2 = 64 m2 S = 64đ?‘š2 S=8m The length of one side of the garden is 8 m. Area of a triangle For any triangle with base b and altitude h, the area is one-half of the product of the base and altitude 1

Area of triangle = 2 (AB)(BC) Example 1: Find the area of a triangular piece of wood whose base is 4 m and altitude is 1.5 m. Given: b = 4 m h = 1.5 m Solution: 1

A = 2bh 1

A = 2 (4 m)(1.5 m) = 3 m2 The area of the piece of wood is 3 m2. Area of a trapezoid 1

1

Area of the trapezoid = 2 b1h + 2 b2h

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1

A = 2 h ( b1 + b2) Example: Find the area of trapezoid whose altitude is 6.5 cm and the bases have lengths 7.5 cm and 12 cm. Given: h = 6.5 b1 = 7.5 cm b2 = 12 cm Solution: 1

A = 2 h ( b1 + b2) =

1 2

(6.5 cm) (7.5 cm + 12 cm)

= 63.4 cm2 Area of a circle Formula: A = đ?œ‹đ?‘&#x; 2 Where đ?œ‹ is a constant with a value approximately equal to 3.1416 Example: Find the area of a circular pizza whose diameter is 30 cm. Given: r =

đ?‘‘ 2

=

30 đ?‘?đ?‘š 2

= 15 cm

A = đ?œ‹đ?‘&#x; 2 = (3.1416)(15 cm)2 = 3.1416 (625) = 706.86 cm2 The area of the pizza is 706.86 cm2 Example 2 A circle is inscribed in a square. This means that the circle touches four sides of the square. Find the area of the circle if the side of the square is 6 cm. Given: side s of a square = 6 cm Solution: Since the circle touches the side of the square, then the diameter of the circle is equal to the side of the square. Thus we have d = 6 cm and r = 3 cm. A = đ?œ‹đ?‘&#x; 2 = (3.1416)(3 cm)2 = 28.3 cm2 The area of the circle inscribed in the square is 28.3 cm2. Practice: Solve the problem 1. Find the area of a parallelogram whose base is 250 cm and altitude is 1.5 m 2. The width of a rectangle is 3 m. what is the length if its area is 13.5 m 2? 3. A square is inscribed in a circle as shown at the right. If the diagonal of the square is 9 cm, what is the area of the circle?

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4. A trapezoid has an area of 20 cm2. One of the bases is 7 cm long. What is the length of the other base if the altitude of the trapezoid is 4 cm? 5. A semicircle whose base is 20 cm sits on the base of a triangle. If the altitude of the triangle is 8 cm, what is the area of the combined figures? Lesson 2 Parts of the geometric Solid figures Geometric solids or spatial figures are shapes that have three dimensions. A polyhedron is a geometric solid that has flat surfaces that are polygons. These surfaces are called faces. Prism has two congruent polygons that are parallel to each other. These polygons are called bases. A prism is named according to the shape of its bases. Prisms Triangular prism Cube Rectangular prism Pentagonal prism Hexagonal prism Pyramids A pyramid has only one base, which is a polygon., the pyramid is named according to the shape of its base. Triangular pyramid Square pyramid Rectangular pyramid Rectangular pyramid Pentagonal pyramid Hexagonal pyramid Cylinders A three dimensional figure that has two circular bases on parallel planes is called a cylinder. Circular cones A circular cone is a spatial figure consisting of a circular base and a curved lateral surface.

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Spheres A sphere has no vertices and no edges. All the points on the surface of a sphere have the same distance from a point inside the sphere called center. The distance from the center to any point in surface of the sphere called radius. Test Write T if the statement is true and F if it is false. 1. A cylinder and a cone are polyhedrons 2. The height of a geometric solid is the length of its altitude. 3. A rectangular prism has four rectangular sides. 4. Cylinders and cone do not have edges nor vertices. 5. A sphere has no flat surfaces. Lesson 3 Lateral Areas and Surface Areas of Geometric Solids Lateral Areas and Surface Area of a cube All the faces of a cube are squares, so the area of a face of a cube whose edge has the length s is given by the formula = s2 The lateral area is composed of the area of the four lateral sides, so Lateral area of cube = 4 s2 The surface area is the combined areas of the six faces, hence Surface area of cube = 6s2 Example: Find the lateral area and surface area of a cube whose edge is 5 cm. Solution: Area of a face = s2 = (5 cm)2 = 25 cm2 Lateral area = 4 s2 = 4 (25cm2) = 100 cm2 Surface area = 6s2 = 5 (25 cm2) = 150 cm2 Lateral area and surface area of a rectangular Prism Area of front and back faces = 2lh YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Area of left and right faces = 2wh Area of 2 bases = 2lw Lateral area of a rectangular prism = 2lh + 2wh Surface area of rectangular prism = 2lh + 2wh + 2lw Example: Find the lateral area and the surface area of a rectangular prism with the following dimensions: length l = 10 cm; width w = 6 cm; and height h = 4 cm. Solution Area of front and back faces = 2lh = 2(10 cm)(4 cm) = 80 cm2 Area of left and right faces = 2 wh =2(6 cm)(4 cm) = 48 cm2 Area of bases = 2lw = (10 cm)(6 cm) = 120 cm2 Lateral area = 2lh + 2lw = 80 cm2 + 48 cm2 = 128 cm2 Surface area = 2lh + 2wh + 2lw = 80 cm2 + 48 cm2 + 120 cm2 = 248 cm2 Lateral area and surface area of a cylinder Lateral area of cylinder = length x width = 2đ?œ‹đ?‘&#x; (h) = 2đ?œ‹đ?‘&#x;h 2 The area of a circle is đ?œ‹đ?‘&#x; so Area of 2 bases of cylinder = 2 đ?œ‹đ?‘&#x; 2 The surface area of the cylinder is then the sum of the lateral area and the area of the two bases, thus Surface area of the cylinder = 2đ?œ‹đ?‘&#x;h + 2 đ?œ‹đ?‘&#x; 2 Example Find the lateral area and the surface area of a cylinder whose radius is 5 cm and whose height is 10 cm. Solution: Lateral area = 2đ?œ‹đ?‘&#x;h = 2đ?œ‹ (5 cm)(10 cm) = 100 đ?œ‹ cm2 Area of bases = 2 đ?œ‹đ?‘&#x; 2 = 2đ?œ‹ ( 5 cm)2 = 50đ?œ‹ cm2 Surface area = 2đ?œ‹đ?‘&#x;h + 2 đ?œ‹đ?‘&#x; 2 = 100 đ?œ‹ cm2 + 50đ?œ‹ cm2 = 150 đ?œ‹ cm2 Surface area of a sphere YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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If a sphere has a radius r, then its surface area is Surface area = 4đ?œ‹đ?‘&#x; 2 Example: What is the surface area of a sphere whose radius is 6 cm? Solution: Surface area = 4đ?œ‹đ?‘&#x; 2 = 4đ?œ‹ (6 cm)2 = 4đ?œ‹ (36 cm2) = 144đ?œ‹ cm2 Practice 1. The side of a cube is 10 cm long Its surface area is _____ cm2 Its lateral area is _____ cm2 2. A shoebox is 33 cm long, 23 cm wide, and 13 cm tall. Its surface area is _____ cm2 Its lateral area is _____ cm2 3. A storeroom is in the form of a cube that 6.,5 m on one edge. What is the lateral area of the room? 4. A can of orange juice has a diameter of 10.5 cm and a height of 17 cm Its surface area is _____ cm2 Its lateral area is _____ cm2 Lesson 4 Volumes of Geometric Solids Volume of a cube In a cube the length , width, and height are all congruent. Thus its volume V is Volume of cube = (s)(s)(s) V = s3 Example: Find the volume of a cube 10 cm on a side. Given: s = 10 cm Solution: V = s3 = (10 cm)3 = 1 000 cm3 The volume of the cube is 1 000 cm3 Volume of a rectangular Prism The volume of a rectangular prism is equal to the area of its base ( length x width) multiplied by the height. Volume of rectangular prism = area of the base x height = (lw)(h) V = lwh YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Example Find the volume or capacity of a swimming pool in the shape of a rectangular solid that measures 20 m in length, 10 m in width, and 1.4 m in height. Given: l = 20 m w = 10 m h = 1.4 m solution: V = lwh = (20 m)(10 m)(1.4 m) = 280 m3 The volume of the swimming pool is 280 m3 Volume of a Pyramid 1

Volume of triangular pyramid = 6 đ?‘?đ?‘Žđ?‘• 1

Volume of a square pyramid = 3 s2h 1

Volume of a rectangular pyramid = 3 ��� Example The base of a pyramid is a right triangle with legs 4 cm and 6 cm. the altitude is 8 cm. find the volume of the pyramid. Given: legs = 4 cm and 6 cm h = 8 cm Solution: in a right triangle, either of the two legs can be taken as the side and the other is its altitude because they are perpendicular to each other. So the area of the base is 1

Base area = 2 (4 cm x 6 cm) = 12 cm2 The volume of the pyramid is 1

V = 3 base area x h 1

= 3 (12 cm2)(8 cm) = 32 cm3 Volume of a cylinder The volume of a cylinder is equal to the area of the base multiplied by the height. The base is a circle and its area is đ?œ‹đ?‘&#x; 2 Volume of a cylinder = (đ?œ‹đ?‘&#x; 2 )(h) V = đ?œ‹đ?‘&#x; 2 đ?‘• Example A can of fruit cocktail is 10 cm in diameter and a height of 11 cm. find the volume. Given: YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Diameter d = 10 cm Height h = 11 cm Solution The radius r is half the diameter 1

1

r = 2d = 2(10 cm) = 5 cm the volume of the can is V = đ?œ‹đ?‘&#x; 2 đ?‘• V= đ?œ‹ (5 cm)2 (11 cm) = đ?œ‹ (25 đ?‘?đ?‘š2 ) (11 cm) = 275 đ?œ‹ cm3 Volume of circular Formula 1

Volume of circular cone = 3 (đ?œ‹đ?‘&#x; 2 đ?‘•) Example Find the volume of a cone whose base diameter is 12 cm and height is 10 cm. Solution First we find the radius 1

1

2

2

r = d = (12 cm) = 6 cm the volume is then 1

V = 3 đ?œ‹( 6 cm)2 (10cm) V = 120 đ?œ‹ đ?‘?đ?‘š3 Volume of a Sphere The volume of a sphere with radius r is 4

V = 3 đ?œ‹đ?‘&#x; 3 Example: Find the volume of a ball whose diameter is 24 cm. Given: diameter d = 24 cm Solution: The radius is 1

r = 2 24 đ?‘?đ?‘š = 12 cm the volume of the ball is then 4

V = 3 đ?œ‹ (12 đ?‘?đ?‘š)3 = 2 304 đ?œ‹ đ?‘?đ?‘š3 Practice 1. A rectangular solid measured 3 cm by 4 cm by 5 cm. find its volume. 2. The area of the base of a prism is 48 cm 2. If its altitude is 12 cm, find its volume. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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3. The diameter of a cylinder is 12 cm and its height is 10 cm. find the volume of the cylinder. Chapter 7 Metric Conversion and Electric and Water Meter Readings Lesson 1 Systems of Measurement English System of measurement Length: 1 yard (yd) = 3 feet 1 foot = 12 inches 1 mile (mi) = 5 280 ft Mass: 1 pound (lb) = 16 ounces (oz.) Capacity 1 gallon (gal) = 4 quarts (qt) 1 qt = 2 pints (pt) The unit of area is the acre. 1 acre = 43 560 square feet (ft2) The metric system Base units include those of meter for length, gram for mass, and liter for capacity, or volume. Each prefix represents a power of 10. Some of these prefixes are: 1

Milli for 1 000 1

Centi for 100 1

Deci for 10 Kilo for 1 000 The following conversions for units of length 1 kilometer (km) = 1 000 m 1 m = 10 decimeters ( dm) 1 m = 100 centimeters ( cm) 1 m = 1000 millimeters ( mm) The base unit of mass is the gram (g) 1 000 g = 1 kilogram (kg) 1 kg = 2.205 pounds (lb) 1 ton = 1 000 kg The base unit of capacity is the liter (L) 1 L = 10 deciliters (dL) 1 L = 100 centiliters (cL) 1 L = 1 000 milliliter (mL) A unit area in the metric system is the hectare (ha) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1 ha = 10 000 m2 Unit of metric system can be converted into those of the English system and vice versa. For units of length: 1.61 km = 1 mi 1 m = 3.28 ft 1 m = 39.37 in. 2.54 cm = 1 in. For units of area: 0.404 ha = 1 acre For units of capacity: 1 gal = 3.79 L (in U.S.) For units of mass 1 kg = 2.205 lb 1 oz = 28.3 g Example 1 Yao Ming, an NBA player from China, is 7 ft and 6 in. tall. What is his height in meters? Solution: 7 ft 6 in. = 7.5 ft 1 m = 3.18 ft 1đ?‘š

So 7.5 ft x 3.28 �� = 2.29 m Yao Ming’s height in meters is 2.29 m. he cannot pass through as normal-sized door without bending. Most doors are about 2 m high. Example 2: Manny Pacquiao fights in the featherweight division of boxing. The upper weight (or mass) limit for featherweight boxers is 132 lb. what is this mass in kilograms? Solution 1 kg = 2.205 lb 1�

132 lb x 2.205 đ?‘™đ?‘? = 59.9 kg The upper weight limit for the featherweight division is 59.9 kg. Test Solve the following problems 1

1. A piece of paper is measured to be 8 2 in. by 11 in. (a) what are these dimensions in centimeters? (b) Find the area in square inches (in 2) and in square centimeters (cm2). What are asked? ______________________________________________ What are given? ______________________________________________ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Operations to be Used: _________________________________________ Number Sentences: ____________________________________________ Solution:

Complete Answers: ___________________________________________ 2. Mrs. Villa bought a gallon of ice cream for P480. What was the cost of the ice cream per quart? What are asked? ______________________________________________ What are given? ______________________________________________ Operations to be Used: _________________________________________ Number Sentences: ____________________________________________ Solution: Complete Answers: ___________________________________________

3. A can of corned beef is labeled 16 oz. what is its weight in grams? What are asked? ______________________________________________ What are given? ______________________________________________ Operations to be Used: _________________________________________ Number Sentences: ____________________________________________ Solution:

Complete Answers: ___________________________________________ 4. A bed measures 80 in. by 60 in. find its dimensions in centimeters. What are asked? ______________________________________________ What are given? ______________________________________________ Operations to be Used: _________________________________________ Number Sentences: ____________________________________________ Solution:

Complete Answers: ___________________________________________ Lesson 2 Electric Meter Reading The unit for measuring electrical energy consumed is the kilowatt-hour (KWxh). One kilowatt-hour is the total electrical energy consumed in 1 hour by an appliance whose power rating is 1 kilowatt (KW). YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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To compute the electrical energy consumed by an appliance, multiply its power rating by the length of time in hours it is used. Example: Four fluorescent lamps rated at 32 W each were used for 10 h. how many kilowatt-hours of electrical energy were used by the lamps. Solution: Electrical energy = number of lamps x power rating x number of hours used = 4 x 32 W x 10 h = 128 W x h x

1 đ?‘˜đ?‘Š 1 000 đ?‘Š

= 0.128 kWx h The four lamps used 0.128 kW x h of electrical energy.

Lesson 3 Water Meter Reading Computing the cost of water consumed Here is one scheme: 1. For consuming 20 m3 or less, a flat rate of P100 is charged. 2. The consumer must pay P15 for every cubic meter in excess of 20 m 3 but below 30 m3. 3. A consumer pays P 20 per cubic meter for water consumed in excess of 30 m3 All water consumption is rounded off to the nearest cubic meter. Example: 1. A household that consumes 18 m3 pays the flat rate of P100 only. 2. A household that consumes 22 m3 pays P100 P100 for the 1st 20 m3 + P 15 (2) P30 for the 2 m3 in excess of 20 m3 P 130 3. A household that consumes 35 m3 pays P100 P100 for the first 20 m3 P15(10) P150 for the 10 m3 in excess of 20 m3 P20(5) P100 for the 5 m3 in excess of 30 m3 P350 total amount Test Solve the following problems. 1. In the house of Mr. Lopez, the water meter reading on August 3 was 1 642.536 m3, while on September 3 it was 1 571.742 m 3. How many cubic meters of water consumed by Mr. Lopez’s household? 2. The water meter readings ion the house of Mrs. Ramos were: July 5, 2006 – 1 056.237 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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August 5, 2006 – 1 080.122 What was the water consumption for the month of July? How much did Mrs. Ramos pay for the month of July?

Chapter 8 Introduction to Algebra Lesson 1 The meaning of Integers The numbers ‌, -3, -2, -1, 0, +1, +2, +3, ‌ are integers. Integers may be positive, zero or negative. Whole numbers greater than 0 are positive integers like, +1, +2, +3 and so on are positive integers. We can write positive integers with or without the positive (+) sign. -1, -2, -3, and so on are negative integers. On the number line, negative integers are to the left of 0. Negative are less than 0. The absolute value of an integer is its distance from 0 on the number line. Distance is considered positive. The symbol for the absolute value of an integer đ?‘Ž is đ?‘Ž . What is the absolute value of 6? of -6? 6 =6 −6 = 6 The above equations tell us that the distance of 6 from 0 is the same distance of 6 from 0. -2 and +2 are opposites The opposite of +15 is -15 In the real world the following are represented by positive integers:  A gain of weight  A deposit in the bank  An increase in temperature  A move to the right These are represented by negative integers  A withdrawal from the bank  Weight loss  A decrease in temperature  A move to the left Practice A. Name the integer just before each integer on the number line 1. +6 2. -18 3. +33 B. Name the integer just after each integer on the number line. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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4. -15 5. +29 6. -43 C. Write the opposite of each integer 7. 1 8. -10 9. 100 D. Give the absolute value of each integer 10. 14 = 11. −5 = 12. −20 = E. Complete each statement 13. The opposite of a negative number is _____. 14. The opposite of the opposite number is _____. 15. Two integers that are three units from 0 are ___and ___. Comparing and Ordering Integers All positive integers are greater than 0. All negative integers are less than 0. All positive integers are greater than negative integers. To order integers we may also use the number line. Locate each integer on the number line then  Start from the left to arrange the integers from least to greatest.  Start from the right to arrange the integers from greatest to least. Example Arrange from greatest to least: 1, -5, -2 The order of the integers from greatest to least: 1,-2, -5. Practice Arrange in order from least to greatest 1. 7,-2,-4 __________ 2. -15,2,-3 __________ 3. 8, -15, 20 __________ 4. -1, -8, -10 __________ 5. 5, -3, -2 __________ 6. 0, -4, -8, 4 __________ 7. -2, 2, -5, 5 __________ 8. -6, -10, -8, 3 __________ 9. 6, -10, 8, 3 __________ 10. -6, 10, 8, -3 __________

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Lesson 2 Mathematical Operations involving Integers In adding integers or signed numbers,  When the addends have the same sign, get their sum and affix the common sign.  When the addends have different signs, get their difference and affix the sign of the greater number. Examples (+5)+ (+2) = + 7 (- 5) + (-2) = -7 (+5) + (-2) = +3 (-5) + (+2) = - 3 Practice Find the sum of each set of integers 1. 5 + (-2) = 2. – 13 + 7 = 3. -1 + (-8) = 4. -4 + 4 = 5. 12 + (-12) = 6. 1 + (-2) + 3 = 7. – 8 + 5 + (- 2) = 8. 10 + (-3) + (-7) = 9. -1 + (-4) + 3 = 10. 2 + (-3) + (-2) + 5 Subtracting Integers Each integer has an opposite. The opposite of +4 is -4, the opposite of +7 is – 7, and the opposite of -10 is +10. Opposites are the same distance from 0 on the number line, but are on opposite sides of 0. The opposite of a number is also called its additive inverse. The additive inverse of + 5 is -5 The additive inverse of 12 is -12 Rules in subtracting integers: 1. Change the sign of subtraction to addition. 2. Change the second number (subtrahend) to its opposite or additive inverse. 3. Perform the operation. Example: 8 – (+2) = 8 + (-2) = 6 8 – (-2) = 8 + (+2) = 10 -8 – (+2) = -10 Practice Perform the indicated operations. 1. 13 – 7 = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2. 8 – ( -2) = 3. -6 – 4 = 4. -5 – (-8) = 3 5. 25 – ( - 9) = 6. 14 – (-2) + 7 = 7. 2 + (-5) – (-3) = 8. -7 – 1 + (-8) 9. -1 + (-4) – (-2) = 10. 5 + ( -4) -4 = Multiplying Integers Rules on multiplication of integers: 1. The product to two integers with the same sign is positive. 2. The product of two integers with opposite signs is negative. 3. When multiplying three or more factors, apply the associative property of multiplication, then use the rules in multiplying integers. Examples a. (-2) (+4) (-3) = [(-2)(4)][(3)] = (-8)(-3) = + 24 b. [(-2)(-3)](+4) = (+6) (+4) = +24 c. [(-2)][(+4)(-3)] = (-2)(-12) = +24 Practice Evaluate each expression. 1. 7(-3) = 2. (-2)(4) = 3. 5(12) = 4. (-1)(-8) = 5. 3(-2)(5) = 6. (-4)(-5)(-6) = 7. (-3)(7)(-1) = 8. (-12)(9)(0) = 9. (-3)[(-2) + (-4)] = 10. 5 [(-4)(-1) – (-2)3] =

Dividing integers Rules in dividing integers 1. The quotient of two integers with the same sign is positive. 2. The quotient of two integers with different signs is negative. Example: a. b. c.

+8

= -4 because (-4)(-2) = +8

−2 −10 +5 −12 −4

= -2 because (-2)(5) = -10 = +3 because (+3)(-4) = -12

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d.

+15 +3

= +5 because (+5)(+3) = +15

Practice Divide 1. 25 ÷ (-5) = 2. -36 ÷ 4 = 3. -56 ÷(-7) = 4. 72 ÷ (-9) = 5. -42 ÷ (-6) = 6. -98 ÷ 2 = 7. 84 ÷ (-3) = 8. 35 ÷ 7 = 9. -81 ÷ 3 = 10. 40 ÷ (-8) = Lesson 3 Algebraic Expressions Suppose that in writing expressions, we use letter to represent any number. 3y + 5 2x – 3 These expressions are known as algebraic expressions. In the expression 3y + 5, y is called the variable, 3 is the coefficient of the variable, and 5 is the constant. An algebraic expression is a combination of numbers and variables and at least one operation A variable is a letter that represents a number in an expression. The coefficient of a variable is the number that is multiplied by the variable. The coefficient of a variable may be an integer or a fraction. A constant is a number not multiplied with a variable. How do we evaluate an algebraic expression? In 3y + 5, if y = 2, then (3)(2) + 5 = 11. In 2x – 3, if x = 7, then (2)(7) – 3 = 11. Exponents tell how many times a number is taken as a factor. Thus 2 3 = 2x 2 x 2 = 8 and 42 = 4 x 4 = 16 What does x2 mean? It means x times x or x multiplied by itself. We read x2 as “ x squared” or “x to the power 2” X2 = (x)(x) if x is 2 then x2 = 22 = (2)(2) = 4 X3 = (x)(x)(x) if x is 3 then x3 = 33 = (3)(3)(3) = 27 3x2 = (3)(x)(x) if x is 4, then 3x2 = 3(42) = (3)(4)(4) = 48 If an expression is enclosed in parentheses and then raised to a certain power, then all members of the expression are raised to that power. Example (3x)2 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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(3x)2= 32x2 = (3) ( 3)(x)(x) = 9x2 Practice A. Evaluate the following algebraic expressions when x = 2 and y = -1 1. 3x + 5 2. 8 + 4y 3. 9x – 2 4. 10 – 2y 5. 3x + 2y 1

1

2

3

B. Evaluate each algebraic expression when x = and y = 1. 2. 3. 4. 5.

2x + 1 6y – 2 1 + x2 1 -18y2 2x – 6y Lesson 4 Simple Algebraic Equations

An equation is an equality of two expressions. These are equations: 1. 4 + 1 = 3 + 2 2. 3x = 12 3. 2y + 3 = 11 The equal sign tells us that the expression at he left side of the equation has the same value as the expression at the right. In the case of equation 1, it is obvious that 4 + 1 = 5 and 3 + 2 also equals 5; hence the two expressions are equal. In equation 2, 3x is equal to 12. What must be the value of x to make the equation true? let us make a guess; if x = 4 then 3 (4) = 12; therefore the value of x must be 4. In the case of equation 3, if y = 4, then 2y + 3 = (2)(4) + 3 = 11, therefore the value of y must be 4. These examples lead us to the following: to solve an algebraic equation means to find the value of the variable so that the equation will remain true. Let us look again equation 2: 3x = 12. What value of x must be multiplied by 3 to get x? The answer is 4. How did we get 4? If we divide both sides of the equation by 3, we get 3đ?‘Ľ 3

=

12 3

x=4 The solution to the equation is the value of the variable or the unknown: x=4 Let us now solve the equation 3 following the steps. Step 1: Subtract 3 from both sides of the equation. 2y + 3 – 3 = 11 – 3 2y = 8 Step 2: Divide both sides of the equation by 2 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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2đ?‘Ś 2

=

8 2

y=4 Observe that whatever operation we do to the left side of the equation must also be done to the right side. In equation 2, we divide both sides of the equation by 3. In equation 3, we subtracted 3 from both sides of the equation, and then divide both sides by 2. Why do we need to do these? The main purpose is to isolate the variable and find its value. Example 1 2x + 5 = 13 2x + 5 – 5 = 13 – 5 2x = 8 2� 2

=

8 2

X=4 Let us generalize our procedure for solving algebraic equations. 1. Think of how we can isolate the unknown variable. It is customary that the unknown should appear at the left side of the equation. 2. Whatever operation you do to the left side of the equation, you must do the same to the side. all operations must be geared toward isolating the variable Example 2 Solve for x in this equation: 4x + 3 = 23 – x 4x + 3 = 23 – x 4x + 3 – 3 = 23 – 3 – x subtract 3 from each side 4x = 20 – x 4x + x = 20 – x + x add x to both sides 5x = 20 5� 5

=

20 5

divide both sides by 5

X=4 To check, substitute the variables in the original equation with the value you get. Check: 4x + 3 = 23 – x (4)(4) + 3 = 23 – 4 16 + 3 = 19 19 = 19 Practice Solve for the variable in the following equations. 1. 6x + 2 = 8 2. 3x – 4 = 5 3. X + 2 = 4 4. Y + 5 = 12 5. Y – 2 = 6 6. 5x + 7 = 2 7. 3y + 3 = y + 9 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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8. 7x – 9 = 11 – 3x 9. 2y – 5 = 10 –y 10. x +2 = 3 - x Solving Word Problems Using Algebraic Equations The following problems involve whole numbers only. Each may require several operations and setting up of more than one algebraic equation, each involving only one unknown. Example 1: Three friends staying in the same apartment contributed the same amount for groceries. After spending P 1,250, P 250 was left. How much did each one contribute? Given: P1,250 = amount spent P 250 = amount left Solution: Letting x be the amount each friend contributed, then 3x = total amount contributed by the three friends Setting up an equation, we get Total amount – amount spent = amount left 3x – 1 250 = 250 3x = 250 + 1250 = 1 500 3� 3

=

1 500 3

x = 500 Each friend contributed P500. Example 2: Mrs. Chua had a certain amount of money in an envelope. Mr. Chua doubled the amount. Their children John and Kathy got P750 each for their weekly allowances. P500 was left inside the envelope. What was the original amount in the envelope? Given: (2)(P750) = P 1 500 = amount given to the children P500 = amount left Solution: Letting x be the original amount in the envelope, then 2x = the amount when Mr. Chua doubled it Doubled amount of money – amount given = amount left 2x – 1 500 = 500 2x - 1 500 + 1 500 = 500 + 1 500 2� 2

=

2 000 2

x = 1 000 The original amount of money in the envelope was P 1,000 Practice YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Solve the following problems 1. Mila is 11 years old. Her mother is 4 times as old. How old is Mila’s mother. What is asked?: _______________________________________________ What given?: _________________________________________________ Operations to be Used: _________________________________________ Number Sentence: _____________________________________________ Solution: Complete Answer: _____________________________________________ 2. Mr. Chavez bought 4 boxes of chocolates. He gave 9 pieces each to her 2 daughters and 12 pieces to his wife. Six pieces were left. How many chocolates did each box contain? What is asked?: _______________________________________________ What given?: _________________________________________________ Operations to be Used: _________________________________________ Number Sentence: _____________________________________________ Solution: Complete Answer: _____________________________________________ 3. Mr. Lune deposited a certain amount in a bank. He withdrew P 20,000 to buy a refrigerator. Later he deposited P 10,000 and again got P7,000 to buy a mobile phone. He had P8,000 left in the bank. How much was his original deposit? What is asked?: _______________________________________________ What given?: _________________________________________________ Operations to be Used: _________________________________________ Number Sentence: _____________________________________________ Solution: Complete Answer: _____________________________________________ 4. Mrs. David is on diet to lose weight. After the first week of diet, she lost 1.5 kg. Then on the second week she lost 12 kg. But on the third week she gained 0.5 kg. She now weighs 62 kg. What was her original weight? What is asked?: _______________________________________________ What given?: _________________________________________________ Operations to be Used: _________________________________________ Number Sentence: _____________________________________________ Solution: Complete Answer: _____________________________________________

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Test Choose the best answer from the given options. 1. The sum of – 9 and – 3 is _____. a. – 6 b. – 12 c. + 12 d. + 6 2. 7 minus – 3 is equal to _____. a. – 4 b. – 10 c. +4 d. +10 3. The additive inverse of +8 is _____. a. +

1 8

b. +8 c. – 8 d. −

1 8

4. The product of – 3 and – 5 is _____. a. – 15 b. + 15 c. + 15 or – 15 5. (- 2 )(+3)(- 6) equals _____. a. – 24 b. + 24 c. – 36 d. +36 6.

+12 −4

equals _____.

a. – 3 b. + 3 c. – 48 d. + 48 7. The opposite of the opposite of a negative number is _____. a. Positive b. Negative c. Zero 8. The sum of a number and its opposite is _____. a. Positive b. Negative c. Zero 9. Which of the following are opposites? 1

a. – 5, 5 b. – 2, - 2 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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c. 3, - 3 d. 0, 1 10. Which of the following is not represented by a positive integer? a. 14 steps forward b. 1 250 feet above sea level c. A withdrawal of P250 d. A price increase of P50 11. Which integers are arranged in the correct order on the number line? a. 0, 1, - 2, 3, - 4 b. 12, -12, 0 c. – 3, 2, 1, 0 d. – 5, - 4, - 2, 10 12. Which of these statements is true? a. – 4 > 3 b. – 3 < - 4 c. 4 > - 3 d. 3 < -4 13. Which is not true? a. 10 = 10 b. −10 = 10 c. − 10 = −10 d. − −10 = 10 14. 7 + (-3) is equal to _____. a. 1 b. 7 c. 10 d. 4 15. −2 − +5 + −1 + +3 is equal to _____. a. – 11 b. 11 c. 5 d. – 5 16. 4 (- 2)(3)( - 1) is equal to _____. a. 24 b. – 24 c. 12 d. – 12 17. [4 + ( - 2)](- 1) is equal to _____. a. – 6 b. 1 c. – 2 d. 2 18. – 2 [5 + (- 6)] ÷ 2 is equal to _____. a. 1 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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b. – 1 3

c. − 2 d. 11 19. In the expression 5x – 3, the coefficient of the variable is _____. a. 5 b. X c. 3 d. – 3 20. When x = - 2, the expression 3 – 2x is equal to _____. a. 1 b. – 1 c. – 7 d. 7 21. The expression 6y + 5x is equal to – 1 when _____. a. X = 1 and y = 1 b. X = 1 and y = - 1 c. X = -1 and y = 1 d. X = - 1 and y = -1 1

22. When x = -2, the expression 2 � 3 is equal to _____. a. 4 b. 8 c. – 4 d. – 8 23. The equation 2x – 8 = 12 is true when x is _____. a. – 10 b. 2 c. – 2 d. 10 24. If 2x + 3 = x – 1, the value of x is _____. a. 2 b. 1 c. – 3 d. – 4 25. Which of the following statements does not describe opposites? a. They have the same absolute value. b. They have the same distance from 1. c. They have different signs. d. They are the same distance from 0.

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