Alpine-G5-Textbook-Maths-FY

Recall In the chapter on division, we have learnt how to find factors of a number. We also learnt to find the H.C.F. of the given numbers. Let us solve the following to recall the concept of H.C.F. Find the H.C.F. of these numbers. a) 36, 48 b) 26, 65 c) 16, 48 d) 20, 60 e) 11, 44 & Remembering and Understanding We have seen that 1 , 2 , 7 , 10 … are all equivalent fractions. However, the fraction 1 is 3 6 21 30 3 said to be in the lowest terms. It is because its numerator and denominator do not have any common factors other than 1. A fraction can be reduced to its lowest terms using either division or H.C.F. Reducing a fraction using division Example 7: Reduce the following fractions to their lowest terms. a) 36 b) 26 48 65 Solution: a) 36 = 36 ÷ 2 = 18 ÷ 2 = 9 ÷ 3 = 3 48 48 ÷ 2 24 ÷ 2 12 ÷ 3 4 Therefore, when reduced to its lowest terms, 36 becomes 3 . 48 4 b) 26 = 26 ÷13 = 22 65 65 ÷13 55 Therefore, when reduced to its lowest terms, 26 becomes 2 . 65 5 Reducing a fraction using H.C.F. We use the concept of H.C.F. to reduce a fraction to its lowest terms. Example 8: Reduce the following fractions to their lowest terms. a) 36 b) 26 48 65 Fractions - I 97 181010044-Alpine-G5-Textbook-Maths-FY.pdf 101 03-Feb-18 2:37:01 PM

Solution: a) Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Common factors of 36 and 48: 1, 2, 3, 4, 6, 12 The H.C.F. of 36 and 48 is 12. 36 = 36 ÷12 = 3 48 48 ÷12 4 Therefore, when reduced to its lowest terms, 36 becomes 3 . 48 4 b) Factors of 26: 1, 2, 13, 26 Factors of 65: 1, 5, 13, 65 Common factors of 26 and 65: 1, 13 The H.C.F. of 26 and 65 is 13. 26 = 26 ÷13 = 2 65 65 ÷13 5 Therefore, when reduced to its lowest terms, 26 becomes 2 . 65 5 Application Let us solve a few real-life examples that involve reducing fractions to their lowest terms. Example 9: Jai ate 4 of a watermelon and Vijay ate 16 of another watermelon of the 16 32 same size. Did they eat the same quantity of watermelon? If not, who ate Solution: more? Fraction of watermelon Jai ate = 4 16 Fraction of watermelon Vijay ate = 16 32 To compare the fractions, we must reduce them to their lowest terms so that we get like fractions. 4 4÷4 1 [H.C.F. of 4 and 16 is 4.] 16 = 16 ÷ 4 = 4 16 = 16 ÷ 8 = 2 [Using division method] 32 32 ÷ 8 4 Clearly, 1 < 2. So, 1 < 2 . 44 Therefore, Vijay ate more. 98 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 102

Example 10: Suraj and Puja were painting the walls of their room. Suraj painted 21 of the 35 wall in an hour and Puja painted 24 of the wall in the same time. Who is more Solution: efficient? 30 Part of the wall painted by Suraj in an hour = 21 = 3 35 5 (H.C.F. of 21 and 35 is 7.) Example 11: Part of the wall painted by Puja in an hour = 24 = 4 30 5 Solution: (H.C.F. of 24 and 30 is 5.) Clearly, 4 is greater than 3 . 55 Therefore, Puja does more work than Suraj in the same time. So, Puja is more efficient. Malik saves ` 550 from his monthly salary of ` 5500. Akhil saves ` 300 from his monthly salary of ` 4500. What fraction of their salary did each of them save? Fraction of salary saved by Malik = 550 = 550 ÷ 10 = 55 ÷ 55 = 1 5500 5500 ÷ 10 550 ÷ 55 10 Fraction of salary saved by Akhil = 300 = 300 ÷ 100 = 3 ÷ 3 = 1 4500 4500 ÷ 100 45 ÷ 3 15 Therefore, Malik saved 1 of his salary and Akhil saved 1 of his salary. 10 15 Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples on reducing fractions to their lowest terms. Example 12: A circular disc is divided into equal parts. Some parts of the circular disc are painted in different colours as shown in the figure. Write the fraction of each colour in its lowest terms. Solution: Total number of equal parts on the disc is 16. The number of parts painted yellow is 3. Fraction = Number of parts painted yellow = 3 Total number of equal parts 16 (The numerator and the denominator do not have any common factor other than 1. So, the fraction cannot be reduced any further.) Fractions - I 99 181010044-Alpine-G5-Textbook-Maths-FY.pdf 103 03-Feb-18 2:37:01 PM

The fraction of the disc that is painted white = Number of parts painted white =6 = 6÷2 = 3 Total number of equal parts 16 16 ÷ 2 8 (H.C.F. of 6 and 16 is 2.) The fraction of the disc that is painted red = Number of parts painted red = 4 = 4 ÷ 4 = 1 Total number of equal parts 16 16 ÷ 4 4 (H.C.F. of 4 and 16 is 4.) The fraction of the disc that is painted blue = Number of parts painted blue = 3 Total number of equal parts 16 (The numerator and the denominator do not have any common factor other than 1. So, the fraction cannot be reduced any further.) Example 13: Meena used 250 g sugar for a pudding of 1000 g. What is the fraction of sugar in the pudding? Solution: Quantity of sugar = 250 g Quantity of pudding = 1000 g Fraction of sugar in the pudding = 250 = 250 ÷ 10 = 25 ÷ 25 = 1 1000 1000 ÷ 10 100 ÷ 25 4 Therefore, sugar forms 1 of the weight of the pudding. 4 Concept 9.3: Compare Unlike Fractions Think Pooja has two circular discs coloured in green, red and white as shown. She wants to know if the parts coloured in red and green are the same. Do you know how Pooja can find it? 100 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 104

Recall In class 4, we have learnt what like and unlike fractions are. Let us recall the same. Fractions such as 1 , 2 and 3 that have the same denominator are called like fractions. 88 8 Fractions such as 1 , 3 and 3 , that have different denominators are called unlike fractions. 87 11 Let us answer the following to recall like and unlike fractions. Identify the like and unlike fractions from the following: a) 3 , 3 , 1 , 5 , 6 , 1 , 4 b) 2, 1, 1 , 7 , 5, 4 c) 5 , 4 , 7 , 5 , 11 , 3 7 5 7 7 7 4 11 22 22 12 14 15 22 15 15 26 24 15 15 & Remembering and Understanding We know how to compare like fractions. To compare two or more fractions, their denominators should be the same. Let us now learn to compare unlike fractions. Steps to compare unlike fractions: 1) Find the L.C.M. of the denominators of the given unlike fractions. Using L.C.M, convert the given unlike fractions into equivalent fractions having the same denominator. 2) Compare their numerators and find which is greater than the other. The fraction with the greater numerator is greater. Example 14: Compare these unlike fractions. a) 3 , 4 b) 3 , 1 c) 1 , 3 7 11 57 48 181010044-Alpine-G5-Textbook-Maths-FY.pdf 105 Fractions - I 101 03-Feb-18 2:37:01 PM

Solution: Solved Solve these Steps 3, 4 3, 1 1, 3 7 11 57 48 Step 1: Write like fractions L.C.M. of 7 and 11 is 77. equivalent to the given So, the equivalent fractions, using the least fractions are common multiple of their 3 = 3 ×11 = 33 and denominators. 7 7 ×11 77 4 = 4 × 7 = 28 . 11 11× 7 77 Step 2: Compare their 33 > 28 numerators and find which is So, 33 > 28 . greater or lesser. 77 77 Thus, 3 > 4 . 5 11 Example 15: Compare these unlike fractions. a) 1 , 2 b) 5 , 1 c) 1 , 6 24 63 4 12 Solution: a) 1 , 2 24 The L.C.M. of 2 and 4 is 4. So, equivalent fraction of 1 = 1×2 = 2 2 2×2 4 Since the numerators are equal, 2 = 2 44 Therefore, the given fractions are equal. b) 5 , 1 63 The L.C.M. of 6 and 3 is 6. So, 1 = 1×2 = 2. 3 3×2 6 Since 5 > 2, 5 > 2 . 66 Therefore, 5 > 1 . 63 102 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 106

c) 1 , 6 4 12 The L.C.M. of 4 and 12 is 12. So, 1= 1×3 = 3. 4 4×3 12 Since 3 < 6, 3 < 6 . 12 12 Therefore, 1 < 6 . 4 12 Application Let us see some real-life situations where we compare unlike fractions. Example 16: Esha ate 1 of an apple in the morning and 2 of the apple in the evening. 43 When did she eat a larger part of the apple? Solution: Fraction of the apple Esha ate in the morning = 1 4 Fraction of the apple she ate in the evening = 2 3 To find when she ate a larger part we must compare the two fractions. Step 1: Write like fractions equivalent to 1 and 2 with the least common multiple of 4 4 3 and 3 as their denominator. The least common multiple of 4 and 3 is 12. So, the required like fractions are: Step 2: 1 = 1×3 3 and 2 = 2×4 =8 4 4×3 = 12 3 3×4 12 Compare the numerators of the equivalent fractions. Example 17: Since 8 > 3, 8 > 3 . 12 12 Hence, 2 > 1 . 34 Clearly, Esha ate the larger part of the apple in the evening. Kumar saves 1 of his salary and Pavan saves 2 of his salary. If they earn the 46 same amount every month, then who saves a lesser amount? 181010044-Alpine-G5-Textbook-Maths-FY.pdf 107 Fractions - I 103 03-Feb-18 2:37:01 PM

Solution: To find who saves lesser, we must find the lesser of the given fractions. The L.C.M. of 4 and 6 is 12. Equivalent fractions of 1 and 2 are 3 and 4 . 4 6 12 12 Since 3 < 4, 3 < 4 . 12 12 Hence, 1 < 2 . 46 Therefore, Kumar saves lesser amount than Pavan. Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples using comparison of unlike fractions. Example 18: Colour each figure to represent the given fraction and compare them. 2 2 9 7 Solution: 2 9 2 7 Clearly, the part of the figure represented by 2 is greater than that 7 represented by 2 . Hence, 2 is greater than 2 . 97 9 Let us try to arrange some unlike fractions in the ascending and descending orders. Example 19: Arrange 2 , 1 , 2 , 3 and 1 in the ascending order. 3254 6 Solution: Write equivalent fractions of the given unlike fractions. The L.C.M. of the denominators 2, 3, 4, 5 and 6 is 60. So, the fractions equivalent to 2 , 1 , 2 , 3 and 1 with the L.C.M. as their 3254 6 denominator will be: 104 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 108

2 = 2× 20 = 40 1 = 1× 30 = 30 , 2 = 2 ×12 = 24 , 3 = 3 ×15 = 45 3 3× 20 60 , 2 2× 30 60 5 5 ×12 60 4 4 ×15 60 and 1 = 1×10 = 10 . 6 6×10 60 Comparing the numerators, 10 < 24 < 30 < 40 < 45. So, 10 < 24 < 30 < 40 < 45 . 60 60 60 60 60 Therefore, the required ascending order is 1 , 2 , 1 , 2 , 3 . 65234 Example 20: Arrange 2 , 1 , 1 , 5 , and 3 in the descending order. 7 4 8 14 16 Solution: Write equivalent fractions of the given unlike fractions. The L.C.M. of the denominators 7, 4, 8, 14 and 16 is 112. So, the fractions equivalent to 2 , 1 , 1 , 5 , and 3 with the L.C.M. as the 7 4 8 14 16 denominator will be: 2 2×16 32 1 1× 28 28 1 1×14 14 5 5× 8 40 7 = 7×16 = 112 , 4 = 4× 28 = 112 , 8 = 8×14 = 112 , 14 = 14× 8 = 112 and 3 = 3×7 = 21 . 16 16×7 112 Comparing the numerators, 40 > 32 > 28 > 21 > 14. 40 32 28 21 14 So, 112 > 112 > 112 > 112 > 112 . Therefore, the required descending order is 5 , 2 , 1 , 3 , 1 . 14 7 4 16 8 Concept 9.4: Add and Subtract Unlike Fractions Think Pooja has a round cardboard with some of its portions coloured. She knows that the fractions that represent the coloured portions are unlike. She wondered how to find the part of the cardboard that is coloured and how much of it is uncoloured. How do you think Pooja can find that? 181010044-Alpine-G5-Textbook-Maths-FY.pdf 109 Fractions - I 105 03-Feb-18 2:37:01 PM

Recall We have already learnt to compare fractions. Let us compare the following to revise the same. a) 5 and 1 b) 3 and 2 c) 1 and 2 d) 4 and 3 e) 1 and 3 77 45 88 24 6 27 27 & Remembering and Understanding Unlike fractions can be added or subtracted by first making the denominators equal and then adding up or subtracting the numerators. Let us understand the addition and subtraction of unlike fractions through some numerical examples. Solve: a) 3 + 1 b) 7 + 2 Example 21: 15 10 13 39 c) 22 + 7 100 10 Solution: a) 3 + 1 = 6 + 3 15 10 30 30 [L.C.M. of 15 and 10 is 30.] 6+3 9 3 = 30 = 30 = 10 [H.C.F. of 9 and 30 is 3.] 7 2 21 2 21+ 2 23 b) 13 + 39 = 39 + 39 = 39 = 39 [L.C.M. of 13 and 39 is 39.] c) 22 + 7 = 22 + 70 22 + 70 92 23 == = 100 10 100 100 100 100 25 [ The L.C.M. of 100 and 10 is 100 and the H.C.F. of 92 and 100 is 4.] Example 22: Solve: a) 8 – 4 b) 17 – 5 c) 14 17 9 11 30 24 – 25 50 Solution: a) 8 – 4 = 88 – 36 [L.C.M. of 9 and 11 is 99.] 9 11 99 99 88 - 36 52 == 99 99 b) 17 – 5 = 68 – 25 [L.C.M. of 24 and 30 is 120.] 30 24 120 120 106 181010044-Alpine-G5-Textbook-Maths-FY.pdf 110 03-Feb-18 2:37:01 PM

68 - 25 43 == 120 120 c) 14 17 = 28 – 17 [L. C. M. of 25 and 50 is 50.] – 25 50 50 50 28 -17 11 = = 50 50 Application In some real-life situations, we use the addition or subtraction of unlike fractions. Let us solve a few such examples. Example 23: The figure shows the coloured portion of two strips of paper. Find the total part that is coloured in both the strips. What part of the strips is not coloured? Solution: Total number of parts of the first strip = 9 Part of the first strip coloured = 2 9 Total number of parts of the second strip = 7 Part of the second strip coloured = 4 n 7 Total coloured part of the strips = 2 + 4 97 14 36 [L.C.M. of 9 and 7 is 63.] = + 63 63 = 14 + 36 50 = 63 63 50 Part of the strip that is not coloured is 2 - 63 [Since 9 + 7 = 1 + 1 = 2.] 9 7 63 63 50 63 + 63 - 50 126 - 50 76 = + – = == 63 63 63 63 63 63 Example 24: Manasa ate a quarter of a chocolate bar and her sister ate two-thirds of it. How much chocolate did they eat in all? How much chocolate is remaining? 181010044-Alpine-G5-Textbook-Maths-FY.pdf 111 Fractions - I 107 03-Feb-18 2:37:01 PM

Solution: Part of the chocolate eaten by Manasa = 1 4 Part of the chocolate eaten by Manasa’s sister = 2 3 Total chocolate eaten by Manasa and her sister = 1 + 2 43 3 + 8 = 3 + 8 = 11 12 12 12 12 [L.C.M. of 4 and 3 is 12.] Therefore, the part of the chocolate eaten by both Manasa and her sister = 11 12 Remaining part of the chocolate = 1 – 11 = 12 – 11 = 12 - 11 = 1 12 12 12 12 12 Higher Order Thinking Skills (H.O.T.S.) Let us see some more examples of addition and subtraction of unlike fractions. Example 25: In a town, 5 of the population were men, 1 were women and 1 were 8 46 children. What part of the population was a) men and women? b) men and children? c) women and children? Solution: Part of the population of the town that was men = 5 8 Part of the population of the town that was women = 1 4 Part of the population of the town that was children = 1 6 Part of the population that was men and women = 5 + 1 = 5 + 2 = 7 8 4 8 8 8 Part of the population that was men and children = 5 + 1 = 15 + 4 = 19 8 6 24 24 24 Part of the population that was women and children = 1 + 1 = 3 + 2 = 5 4 6 12 12 12 108 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 112

Example 26: In a school, 1 of the students were from the primary school, 1 were from the 35 middle school and the remaining were from the high school. What fraction of the strength of the school was from the high school? Solution: Strength of the school that was from the primary school = 1 3 Strength of the school that was from the middle school = 1 5 Strength of the school that was from the high school  51 =   185 = = 1 – 1 + 1 – 5+3 = 1 – 15 -8 = 15 - 8 = 7 3 15 15 15 15 15 Therefore, 7 of the total strength were high school students. 15 Drill Time Concept 9.1: Equivalence of Fractions 1) Check if the fractions are equivalent. a) 5 and 5 b) 3 and 14 c) 8 and 24 d) 3 and 9 e) 4 and 5 8 21 21 35 23 46 27 81 25 50 Concept 9.2: Fraction in its Lowest Terms 2) Reduce these fractions using H.C.F. d) 12 e) 12 a) 24 b) 36 c) 42 36 30 48 60 70 d) 6 e) 3 3) Reduce these fractions using division. 24 27 a) 36 b) 42 c) 26 72 84 91 Concept 9.3: Compare Unlike Fractions 4) Compare the following unlike fractions: a) 3 , 2 b) 3 , 4 c) 8 , 7 d) 5 , 3 e) 11, 5 7 14 21 42 9 18 11 7 48 181010044-Alpine-G5-Textbook-Maths-FY.pdf 113 Fractions - I 109 03-Feb-18 2:37:01 PM

Concept 9.4: Add and Subtract Unlike Fractions 5) Solve: a) 3 + 5 b) 4 + 3 c) 4 + 1 d) 19 + 5 e) 2 + 6 4 13 14 12 15 10 100 10 16 30 6) Solve: a) 4 – 3 b) 14 – 3 c) 13 – 14 d) 3 – 4 e) 15 – 16 9 11 30 24 30 60 15 30 20 40 7) Word problems a) U sha played the keyboard for 7 of an hour and did her homework for 5 of 30 12 an hour. Did she spend the same amount of time for both the activities? b) William ate 3 of a chocolate bar and Wasim ate 1 of the chocolate. Did they 16 4 eat the same part of the chocolate? Who ate less? c) M ani and Roja were painting a rectangular cardboard each. Mani painted 15 of the cardboard in an hour and Roja painted 18 of the cardboard in the 25 30 same time. Who is more efficient? d) Sudheer saves ` 360 per month from his salary of ` 3600. Hari saves ` 200 per month from his salary of ` 2400. What fraction of their salary did each of them save? e) P avani used 450 cm of satin ribbon from a bundle of satin ribbon of length 3000 cm. What fraction of the satin ribbon is used? 110 03-Feb-18 2:37:01 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 114

Chapter Fractions - II 10 Let Us Learn About • the terms ‘mixed’, ‘proper’ and ‘improper’ fractions. • a dding and subtracting mixed fractions. • m ultiplying and dividing fractions by fractions. • finding the reciprocals of fractions. Concept 10.1: Add and Subtract Mixed Fractions Think Pooja has learnt addition and subtraction of unlike fractions. She has also learnt the conversion of improper fractions to mixed fractions and vice-versa. She was curious to know if she could add and subtract improper fractions and mixed fractions too. How do you think Pooja can add or subtract mixed fractions? Recall We have learnt about the types of fractions. Let us recall them here. 1) A fraction whose numerator is greater than the denominator is called an improper fraction. 2) A fraction whose denominator is greater than the numerator is called a proper fraction. 3) The combination of a whole number and a fraction is called a mixed fraction. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 115 111 03-Feb-18 2:37:02 PM

Let us revise the concept of fractions by solving the following: 13 8 11 5 22 17 a) 6 + 9 b) 7 + 14 c) 15 + 10 8 10 9 23 54 d) 3 – 11 e) 2 – 15 f) 6 – 5 & Remembering and Understanding A mixed fraction can be converted into an improper fraction by multiplying the whole number part by the fraction’s denominator and then adding the product to the numerator. Then we write the result on top of the denominator. The addition and subtraction of mixed fractions are similar to that of unlike fractions. Let us understand the same through the following examples. Example 1: 3 + 3 2 Add: 2 5 7 Solved Solve this Steps 23 + 32 12 1 + 15 1 57 43 Step 1: Convert all the mixed 2 3 = 2 × 5 + 3 = 13 ; fractions into improper fractions. 55 5 3 2 = 3 ×7 + 2 = 23 77 7 Step 2: Find the L.C.M. and add the 2 3 + 3 2 = 13 + 23 improper fractions. 5 75 7 [L.C.M. of 5 and 7 is 35.] = 7 ×13 + 5 × 23 35 = 91+115 = 206 35 35 112 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 116

Solved Solve this 12 1 + 15 1 Steps 23 + 32 57 43 Step 3: Find the H.C.F. of the The H.C.F. of 206 and 35 is 1. Solve this numerator and the denominator of So, we cannot reduce the 12 1 from 15 1 the sum. Then reduce the improper fraction any further. fraction to its simplest form. 43 Step 4: Convert the improper fraction 206 31 into a mixed fraction. =5 35 35 Therefore, 2 3 + 3 2 57 = 5 31 . 35 Example 2: Subtract 2 3 from 3 2 57 Steps Solved 2 3 from 3 2 Step 1: Convert all the mixed fractions into improper fractions. 57 3 2 = 3 ×7 + 2 = 23 ; 77 7 2 3 = 2 × 5 + 3 = 13 55 5 Step 2: Find the L.C.M. and 3 2 - 2 3 = 23 - 13 subtract the improper fractions. 7575 [L.C.M. of 5 and 7 is 35] = 5 × 23 − 7 ×13 = 115 − 91 = 24 35 35 35 Step 3: Find the H.C.F. of the The H.C.F. of 24 and 35 is 1. So, we numerator and the denominator cannot reduce the fraction any of the difference. Then reduce further. the proper fraction to its simplest form. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 117 Fractions - II 113 03-Feb-18 2:37:02 PM

Steps Solved Solve this 2 3 from 3 2 12 1 from 15 1 57 43 Step 4: If the difference is an 24 is a proper fraction. So, we improper fraction, convert it into 35 a mixed fraction. cannot convert it into a mixed fraction. 2 3 24 Therefore, 3 7 – 2 5 = 35 Application In some real-life situations, we use the addition or subtraction of mixed fractions. Example 3: Ajit ate 5 3 biscuits and Arun ate 8 1 biscuits. How many biscuits did they eat 54 in all? How many biscuits were remaining if the box had 20 biscuits in it? Solution: Total number of biscuits in the box = 20 Number of biscuits eaten by Ajit = 5 3 5 1 Number of biscuits eaten by Arun = 8 4 Total number of biscuits eaten by both Ajit and Arun =53 +81 = 28 + 33 = 112 +165 = 277 = 13 17 5 4 5 4 20 20 20 17 = 20 – 277 = 400 − 277 Number of biscuits remaining = 20 – 13 20 1 20 20 [L.C.M. of 1 and 20 is 20.] 123 3 = =6 20 20 Therefore, Ajit and Arun ate 13 17 biscuits and 6 3 biscuits are remaining 20 20 Example 4: Veena covered 34 2 km in 2 hours and 16 1 km in the next hour. If she has to 34 travel a total of 65 3 km, how much more distance must she cover? 5 114 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 118

Solution: Total distance to be covered by Veena = 65 3 km 5 Distance covered by her in the first 2 hours = 34 2 km 3 Distance covered by her in the next hour = 16 1 km 2 14 Total distance she travelled = 34 3 km + 16 4 km 104 km + 65 km = 416 +195 km = 611 km = 50 11 km 3 4 12 12 12 Distance yet to be covered = 65 3 km – 50 11 km 5 12 = 328 km – 611 km 5 12 = 3936 − 3055 km [L.C.M. of 5 and 12 is 60.] 60 = 881 km = 14 41 km 60 60 Therefore, the distance Veena has to cover is 14 41 km. 60 Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of addition and subtraction of mixed fractions. Example 5: By how much is 41 1 greater 2 ? than 39 65 Solution: 1 – 39 2 = 247 – 197 1235 −1182 The required number = 41 6 5 6 5 = 30 = 53 = 1 23 30 30 Therefore, 41 1 is greater than 39 2 by 1 23 . 6 5 30 Example 6: By how much is 22 3 less than 50 1 ? Solution: 47 The required number = 50 1 – 22 3 = 351 – 91 = 1404 − 637 7 474 28 = 767 = 27 11 28 28 Therefore, 22 3 is less than 50 1 by 27 11 . 4 7 28 Fractions - II 115 181010044-Alpine-G5-Textbook-Maths-FY.pdf 119 03-Feb-18 2:37:02 PM

Concept 10.2: Multiply Fractions Think Pooja and each of her 15 friends had a bar of chocolate. Each of them ate 5 of the 12 chocolate. How much of the chocolate did they eat in all? How do you think Pooja can find this? Recall Recall that when we find the fraction of a number, we multiply the number by the fraction. After multiplication, we simplify the product to its lowest terms. Similarly, we can multiply a fraction by another fraction too. • F raction in its lowest terms: A fraction is said to be in its lowest form if its numerator and denominator do not have a common factor other than 1. • Reducing or simplifying fractions: Writing fractions such that its numerator and denominator have no common factor other than 1 is called reducing or simplifying the fraction to its lowest terms. • Methods used to reduce a fraction: A fraction can be reduced to its lowest terms using 1) division 2) H.C. F. Let us revise the concept by simplifying the following fractions. a) 12 b) 16 c) 13 27 24 65 d) 17 e) 9 f) 14 23 21 42 & Remembering and Understanding Multiply fractions by whole numbers A whole number can be considered as a fraction with its denominator as 1. Multiplying a fraction by 2-digit or 3-digit numbers is the same as finding the fraction of a number. 116 181010044-Alpine-G5-Textbook-Maths-FY.pdf 120 03-Feb-18 2:37:02 PM

Example 7: Find the following: a) 23 of 90 45 b) 15 of 128 32 Solution: a) 23 of 90 = 23 × 90 = 23 × 90 45 45 45 = 2070 = 46 45 Multiplying the numbers in the numerator and then dividing is tedious. It is especially so when the numbers are large. Therefore, we shall find if any of the numbers in the numerator and the denominator have a common factor. If yes, we take the H.C.F. of the numbers. We then divide the numbers to reduce the fraction to its lowest terms. Hence, 23 of 90 = 23 × 90. Here, 45 and 90 have common factors, 3, 5, 9, 15 45 45 and 45. The H.C.F. of 45 and 90 is 45. So, divide both 45 and 90 by their H.C.F. Therefore, 23 × 90 = 23 × 90 2 [Cancelling using the H.C.F. of the numbers] 45 45 1 = 23 × 2 = 46 b) 15 of 128 = 15 × 128 32 32 The H.C.F of 32 and 128 is 32. Divide 32 and 128 by 32, and simplify the multiplication. 15 × 128 4 = 15 × 4 = 60 32 1 Multiply fractions by fractions Multiplication of two fractions is simple. If a and c are two fractions where b,d  are not equal to zero, b d then a × c = a ×c b d b ×d Product of numerators Therefore, product of the fractions = Product of denominators 181010044-Alpine-G5-Textbook-Maths-FY.pdf 121 Fractions - II 117 03-Feb-18 2:37:02 PM

To multiply mixed number, we change them into improper fractions and then proceed. Multiplying the numbers in the numerator and then dividing is tedious. It is especially so when the numbers are large. Therefore, we shall check if any of the numbers in the numerator and the denominator have a common factor. We then reduce the fractions into their lowest terms and then multiply them. Let us look at an example to understand the concept. Example 8: Solve: 23 × 15 45 46 Solution: Follow these steps to multiply the two fractions. Step 1: Check if the numerator and denominator have any common factors. Observing the given fractions, we see that, a) (23, 45) and (15, 46) do not have any common factors to be reduced. b) (23, 46) and (15, 45) have common factors. Step 2: Find the H.C.F. of the numerator and the denominator that have common factors. The H.C.F. of 23 and 46 is 23. The H.C.F. of 15 and 45 is 15. Step 3: Reduce the numerator and the denominator that have common factors using their H.C.F. 1 23 × 1 = 1×1 = 1 15 3 45 46 2 3 × 2 6 Therefore, 23 × 15 = 1 . Example 9: 45 46 6 Solve: a) 2 × 5 b) 7 × 70 c) 84 × 45 56 35 63 54 60 Solution: a) 12 × 1 = 1× 1 = 1×1 = 1 15 1 3 1× 3 3 5 63 b) 17 × 2 = 1 × 2 = 1× 2 = 2 70 1 35 63 9 1 9 1× 9 9 c) 7 84 5 = 7 × 5 = 7×5 = 7 1 54 6 5 6×5 6 16 6 × 45 60 5 118 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 122

Application Let us see some real-life examples where we can use multiplication of fractions. Example 10: Tina had 1 kg of flour. She used 1 of it for a recipe. How many grams of 6 10 flour did she use? Solution: Quantity of flour Tina had = 1 kg 6 Part of the flour used by her for a recipe = 1 10 Quantity of flour used by Tina = 1 of 1 kg = 1 × 1 kg = 1×1 kg 10 6 10 6 10 ×6 = 1 kg = 1 × 1000 g = 16.67 g 60 60 Example 11: Mohan saves one-fourth of his monthly salary of ` 5500. Arjun saves two-fifths of his monthly salary of ` 4500. Who saves more and by how much? Solution: Fraction of salary saved by Mohan = 1 of ` 5500. 4 1 = 4/ × 1375 = ` 1375 5500 1 2 × ` 4500 = 2 × ` 900 = ` 1800 Fraction of salary saved by Arjun = 5 Since ` 1800 is more than ` 1375, Arjun saves more. The difference in savings = ` 1800 – ` 1375 = ` 425 Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of multiplication of fractions. Example 12: Swetha cut a big watermelon into two equal parts. Jaya cut a part into 16 equal pieces and ate 4 of them. Vijay cut a part into 32 equal pieces and ate 16 of them. Who ate more watermelon? Solution: Each equal part of the watermelon = 1 2 Fraction of watermelon Jaya ate = 4 of 1 = 4 × 1 1× 1= 1 = 16 2 16 2 4 2 8 Fraction of watermelon Vijay ate = 16 of 1 = 16 × 1 = 1 × 1 = 1 32 2 32 2 2 2 4 181010044-Alpine-G5-Textbook-Maths-FY.pdf 123 Fractions - II 119 03-Feb-18 2:37:02 PM

Comparing the fractions, we see that 1 < 1 or 1 > 1 . 8 4 48 Therefore, Vijay ate more. Example 13: Multiply the following: a) 3 , 7 , 5 b) 1 , 6 , 11, 4 757 7 7 4 11 Solution: a) 3 × 7 × 5 = 3 7577 [Cancelling the common factors in the numerator and denominator] b) 1 × 6 × 11 × 4 = 1× 6 = 6 7 7 4 11 7 × 7 49 [Cancelling the common factors in the numerator and denominator] Concept 10.3: Reciprocals of Fractions Think A chocolate bar was shared among three boys. Pooja got one-third of it. She ate it in parts over a period of four days. If she ate an equal part every day, how much chocolate did Pooja eat in a day? Do you know how to find it? Recall Let us recall the relation between multiplication and division. Multiplication and division are inverse operations. The equation 3 × 8 = 24 has the inverse relationships: 24 ÷ 3 = 8 and 24 ÷ 8 = 3 Similar relationships exist for division. The equation 45 ÷ 9 = 5 has the inverse relationships. 5 × 9 = 45 and 9 × 5 = 45 Let us revise the concept by finding the inverse relationships of the following statements. a) 3 × 4 =12 b) 21÷ 3 = 7 c) 6 × 3 = 18 d) 42 ÷ 7 = 6 120 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 124

& Remembering and Understanding Reciprocal of a fraction A number or a fraction which when multiplied by a given number gives the product as 1 is called the reciprocal or multiplicative inverse of the given number. To find the reciprocal of a fraction, we interchange its numerator and denominator. • The reciprocal of a number is a fraction. For example, the reciprocal of 20 is 1 . 20 • The reciprocal of a unit fraction is a number, For example, the reciprocal of 1 is 7. 7 • The reciprocal of a proper fraction is an improper fraction. It can be left as it is or converted into a mixed fraction, For example, the reciprocal of 3 is 7 or 2 1 . 7 3 3 • The reciprocal of an improper fraction is a proper fraction, For example, the reciprocal of 9 is 5 . 59 • The reciprocal of a mixed fraction is a proper fraction, For example, the reciprocal of 2 3 is 8 . 8 19 Note: 1) The reciprocal of 1 is 1. 2) The reciprocal of 0 does not exist as division by zero is not defined. 3) N umbers such as 4, 6, 9 and so on are converted into improper fractions by writing them as 4 , 6 , 9 before finding their reciprocals. 111 4) Fractions are reduced to their lowest terms (if possible) before finding their reciprocals. Let us find the reciprocals of some fractions. Example 14: Find the reciprocals of these fractions. a) 8 b) 4 c) 3 d) 4 17 19 11 5 Solution: To find the reciprocal of a fraction, we interchange its numerator and denominator. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 125 Fractions - II 121 03-Feb-18 2:37:02 PM

The reciprocals of the given fractions are: a) 17 b) 19 c) 11 d) 5 84 3 4 Example 15: Find the multiplicative inverses of these fractions. a) 5 b) 7 5 c) 0 d) 1 1 9 e) 33 3 Solution: To find the multiplicative inverse of a fraction, we interchange its numerator and denominator. The multiplicative inverses of the given fractions are: a) 1 b) 9 c) no multiplicative inverse 5 68 d) 1 e) 3 100 Note: 0 has no reciprocal or multiplicative inverse because we cannot multiply any number by it to get 1. Zero multiplied by any number is zero. Therefore, 0 is the only number that does not have a multiplicative inverse. Application Divide a number by a fraction The division of a number by another means to find the number of divisors present in the 1 dividend. For example, 8 ÷ 4 means to find the number of fours in 8. Similarly, 10 ÷ means to 5 find the number of one-fifths in 10. Let us understand the division by fractions through some examples. Example 16: Divide: a) 15 ÷ 1 b) 75 ÷ 3 3 5 Solution: Follow these steps to divide the given. a) 15 ÷ 1 Step 1: 3 15 Step 2: Write the number as a fraction as 15 = 1 13 Find the reciprocal of the divisor. The reciprocal of 3 is 1 . Step 3: Multiply the dividend with the reciprocal of the divisor. 15 ÷ 1 = 15 × 3 = 45 3 1 1 1 122 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 126

Step 4: Reduce the product to its lowest terms. 45 = 45 1 Step 1: Therefore, 15 ÷ 1 = 45. Step 2: 3 Step 3: b) 75 ÷ 3 5 75 Write the number as a fraction as 75 = 1 Find the reciprocal of the divisor. The reciprocal of 3 is 5 . 53 Multiply the dividend with the reciprocal of the divisor. 75 ÷ 3 = 75 × 5 Step 4: 5 13 Reduce the product to its lowest terms. 75 × 5 13 The H.C.F. of 75 and 3 is 3. Cancelling 3 and 75 by 3, we get 25 75 5 1 × 3 = 25 × 5 = 125. 1 Note: To divide a number by a fraction is to multiply it by the reciprocal of the divisor. Divide a fraction by a number The division of a fraction by a number is similar to the division of a number by a fraction. Let us understand the division of fraction by numbers through some examples. 1 Example 17: Solve: ÷ 67 3 Solution: To divide the given, follow these steps: Steps Solved Solve this 1 3 ÷ 54 ÷ 67 5 3 Step 1: Write the number as a 67 fraction. 67 = 1 Step 2: Find the reciprocal of 67 1 the divisor. The reciprocal of 1 is 67 . Step 3: Multiply the dividend 1 11 1 by the reciprocal of the 3 ÷ 67 = 3 × 67 = 3 × 67 divisor. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 127 Fractions - II 123 03-Feb-18 2:37:02 PM

Steps Solved Solve this 1 3 Step 4: Reduce the product ÷ 54 to its lowest terms. ÷ 67 5 3 11 = 3 × 67 201 11 Therefore, 3 ÷ 67 = 201 . Divide a fraction by another fraction Division of a fraction by another fraction is similar to the division of a number by a fraction. Let us understand this through some examples. Example 18: Solve: 1 ÷ 1 3 21 Solution: To solve the given sums, follow these steps: Solved Solve this 3 210 Steps 1 ÷ 1 25 ÷ 75 3 21 Step 1: Find the reciprocal of 1 21 the divisor. The reciprocal of 21 is 1 . Step 2: Multiply the dividend by 1 1 1 21 3 ÷ 21 = 3 × 1 the reciprocal of the divisor. Step 3: Reduce the product 1 7 into its lowest terms. × 21 = 7 31 11 Therefore, 3 ÷ 21 = 7. Higher Order Thinking Skills (H.O.T.S.) Let us see some real-life examples using division of fractions. Example 19: Sakshi had 7 apples. She cut them into quarters. How many pieces did she get? 124 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 128

Solution: To find the number of pieces that Sakshi will get, we must find the number of quarters in 7. That is, we must divide the total number of apples by the size of each piece of apple. Number of quarter pieces = 7 ÷ 1 = 7 × reciprocal of 1 = 7 × 4 = 28 44 Therefore, Sakshi got 28 pieces of apple. Example 20: Nani had 3 of a kilogram of sugar. He poured it equally into 4 bowls. How 5 many grams of sugar is in each bowl? Solution: Total quantity of sugar = 3 kg 5 Number of bowls = 4 Quantity of sugar in each bowl = 3 kg ÷ 4 = 3 kg × reciprocal of 4 55 3 1 3 kg = 3 50 = 5 kg × 4 = 20 × 1000 g = 3 × 50 g = 150 g 20 1 Therefore, each bowl contains 150 g of sugar. Example 21: There is 16 litres of orange juice in a bottle. 8 litres of it is poured in each 25 25 glass. How many glasses can be filled? Solution: Total quantity of orange juice = 16 litres 25 Quantity of juice poured in each glass = 8 litres 25 Number of glasses filled with juice 16 litres ÷ 8 litres 25 25 21 16 8 16 × 25 = 2 = 25 × reciprocal of 25 = 25 1 8 1 Therefore, 2 glasses are filled. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 129 Fractions - II 125 03-Feb-18 2:37:02 PM

Drill Time Concept 10.1: Add and Subtract Mixed Fractions 1) Solve: a) 3 4 + 2 3 b) 2 1 + 7 2 c) 12 1 + 13 2 d) 10 1 2 75 85 75 + 12 33 b) 10 1 – 5 3 2) Solve: 27 1 – 2 1 c) 7 2 – 4 1 d) 12 3 – 11 2 a) 4 84 89 37 Concept 10.2: Multiply Fractions 3) Multiply fractions by whole numbers. a) 12 × 64 b) 3 × 80 c) 4 × 100 d) 3 × 49 32 8 20 7 4) Multiply fractions by fractions. a) 22 × 26 b) 4 × 16 c) 3 × 51 d) 7 × 45 13 44 12 24 17 21 15 49 Concept 10.3: Reciprocals of Fractions 5) Find the reciprocal of the following: a) 27 b) 2 1 d) 50 53 c) 5 23 2 d) 1 by 1 7 49 6) Divide: a) 16 by 1 b) 14 by 2 c) 1 by 3 4 7 42 126 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 130

Chapter Decimals - I 11 Let Us Learn About • c onverting fractions to decimals and vice versa. • decimal place value chart and expanding the decimal numbers. • equivalent, like and unlike decimals. • c onverting unlike decimals to like decimals. • adding and subtracting decimals. Concept 11.1: Like and Unlike Decimals Think The teacher asked Pooja to represent the fraction of girls, if there are 556 girls in a school of 1000 students. Pooja said: “The fraction of girls is 556 ”. 1000 The teacher asked her to represent the same in the decimal form. Do you know how to represent a fraction in its decimal form? Recall In class 4, we have learnt decimals and fractions with their conversions. Let us recall them. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 131 127 03-Feb-18 2:37:02 PM

Conversion of fractions into decimals To write the given fractions as decimals, follow these steps. Step 1: Write the whole part as it is. Step 2: Place a decimal point to its right. Step 3: Write the numerator of the proper fraction part. Conversion of decimals into fractions To convert a decimal into a fraction, follow these steps. Step 1: Write the number without the decimal point. Step 2: Count the number of decimal places (that is, the number of places to the right of the decimal number). Step 3: Write the denominator with 1 followed by as many zeroes as the number of digits after the decimal point. We observe that when a decimal number is converted into a fraction, the denominator is: • 10 if there is one digit after the decimal point. • 100 if there are two digits after the decimal point. & Remembering and Understanding We know that the first place to the right of the decimal point is called the tenths. The place to the right of the tenths is the hundredths. The place value of a number increases by ten when we move from right to left and decreases by ten when we move from left to right. Consider the following example. Example 1: Convert 1396 m into km. Solution: 1m= 1 km 1000 1396 Thus, 1396 m = 1000 km To represent 1396 km in decimal form, we get a new place value to the right 1000 of the decimal point. We get the first place after the decimal point by dividing the number by 10. It is called the tenths place. We get the second place after the decimal point by dividing the number by 128 03-Feb-18 2:37:02 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 132

100. It is called the hundredths place. Similarly, we get the third place after the decimal point by dividing the number by 1000. It is called the thousandths place. 1396 Hence, 1000 km is written as 1.396 km in the decimal form. It is read as one point three nine six kilometres. Similar to the place value chart for numbers, we have a place value chart for decimals too. Decimal place value chart We can place the decimal number 1436 Decimal Tenths Hundredths Thousandths Thousands Hundreds Tens Ones point 1 1 1 1 4 36 0 10 100 1000 Example 2: In the number 426.038, a) which digit is in the hundreds place? b) which digit is in the tenths place? c) which digit is in the hundredths place? d) which digit is in the thousandths place? e) which digit is in the ones place? Solution: a) 4 is in the hundreds place. b) 0 is in the tenths place. c) 3 is in the hundredths place. d) 8 is in the thousandths place. e) 6 is in the ones place. Example 3: Write the following numbers in the decimal place value chart. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 133 Decimals - I 129 03-Feb-18 2:37:02 PM

a) 13.457 b) 450.72 c) 2153.068 Solution: Thousands Hundreds Tens Ones Decimal Tenths Hundredths Thousandths point 1 1 1 1000 100 10 1 (.) 10 100 1000 5 7 a) 13 . 4 2 8 b) 4 50 . 7 6 c) 2 1 53 . 0 Example 4: Write the expanded form of the given decimals and then write them in words. a) 5418.264 b) 315.608 c) 46.937 Solution: a) 5418.264 = 5 thousands + 4 hundreds + 1 tens + 8 ones + 2 tenths + 6 hundredths + 4 thousandths = 5 × 1000 + 4 × 100 + 1 × 10 + 8× 1+ 2× 1 +6× 1 +4× 1 = 5000 + 400 + 10 + 8 + 2+ 6 + 4 10 100 1000 10 100 1000 = Five thousand four hundred and eighteen and two hundred and sixty-four thousandths. b) 315.608 = 3 hundreds + 1 tens + 5 ones + 6 tenths + 8 thousandths = 3 × 100 + 1 × 10 + 5 ×1+ 6 × 1 + 8 × 1 = 300 + 10 + 5 + 6 + 8 10 1000 10 1000 Here you can skip = Three hundred and fifteen and six hundredths place as it hundred and eight thousandths. contains 0 c) 46.937 = 4 tens + 6 ones + 9 tenths + 3 hundredths + 7 thousandths = 4 × 10 + 6 × 1 + 9 × 1 +3× 1 +7× 1 10 100 1000 = 40 + 6 + 9 + 3 + 7 = 46.937 But, here you 10 100 1000 cannot skip hundredths place = F orty-six and nine hundred and thirty-seven thousandths. Let us learn about equivalent decimals, like decimals and 130 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 134

unlike decimals. Decimal places: The digits in the decimal part are called decimal places. For example, 4109.34 has two decimal places; 1183.6 has only one decimal place. Equivalent decimals: The decimal numbers which have equal values are called equivalent decimals. For example, 0.3, 0.30, 0.300 are equivalent decimals. Like decimals: The decimal numbers that have the same number of decimal places are called like decimals. For example, a) 2.81, 35.94, 7.32, 145.67, 214.07 and b) 0.362, 51.093, 22.678, 8091.221, 1.003 are sets of like decimals. Unlike decimals: The decimal numbers that have different number of decimal places are called unlike decimals. For example, a) 5.11, 89.018, 3.4, 671.92 and b) 59.009, 231.8, 9.05, 12.25 are sets of unlike decimals. Convert unlike decimals to like decimals Consider a pair of unlike decimals, say 5.36 and 27.2. To find their equivalent decimals with the same number of decimal places, add as many 0s to the right of the decimal as needed. The equivalent decimal of 27.2 with two decimal places = 27.20 Thus, 5.36 and 27.20 are like decimals with two decimal places. Example 5: Convert the given unlike decimals into like decimals. a) 42.7, 53.28, 261.135, 11.01 b) 0.742, 12.06, 8.5, 17.12 c) 7.23, 2.1, 0.6, 4.382 Solution: a) Unlike decimals: 42.7,53.28, 261.135, 11.01 In these decimal numbers, the third decimal number has the highest number of decimal places = 3 181010044-Alpine-G5-Textbook-Maths-FY.pdf 135 Decimals - I 131 03-Feb-18 2:37:03 PM

S o, we have to find the equivalent decimals of the other three decimal numbers such that they have three decimal places. Thus, the required like decimals are: 42.700, 53.280, 261.135, 11.010 b) Unlike decimals: 0.742, 12.06, 8.5, 17.12 Like decimals: 0.742, 12.060, 8.500, 17.120 c) Unlike decimals: 7.23, 2.1, 0.6, 4.382 Like decimals: 7.230, 2.100, 0.600, 4.382 Application Let us see some real-life examples based on decimals. Example 6: If 502 out of 1000 students in a school are girls, then write the decimal equivalent of the fraction of girls in the school. Solution: The total number of students in the school = 1000 Number of girls = 502 The fraction of girls = 502 1000 The required decimal equivalent = 0.502 Example 7: If there is 263 cm of tape on a tape roll, how many metres of tape is on the roll? Give your answer in the decimal form. Solution: The length of the tape on the tape roll = 263 cm Converting into metres = 263 m 100 The required decimal form of the length of the tape = 2.63 m Example 8: The weight of a sugar jar is 5670 g. What is its weight in kilograms? Solution: The weight of a sugar jar = 5670 g Converting into kg = 5670 kg 1000 The required decimal form of the weight of the jar = 5.67 kg 132 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 136

Higher Order Thinking Skills (H.O.T.S.) Consider the following examples. Example 9: In a small village of 1000 people, there are 238 women and 450 men. Find the fraction of men, women and children and write them in decimal form. Write the decimals using place value chart. Solution: Total number of people in the village = 1000 Number of women = 238 Fraction of women = 238 1000 Decimal form = 0.238 Number of men = 450 Fraction of men = 450 1000 Decimal form = 0.450 Number of children = 1000 – (238 + 450) = 1000 – 688 = 312 Therefore, fraction of children = 312 1000 Decimal form = 0.312 Women Ones Decimal point Tenths Hundredths Thousandths Men 1 1 1 1 (.) 10 Children 2 100 1000 0 . 4 3 8 0 . 3 5 0 0 . 1 2 181010044-Alpine-G5-Textbook-Maths-FY.pdf 137 Decimals - I 133 03-Feb-18 2:37:03 PM

Concept 11.2: Compare and Order Decimals Think Pooja went to purchase a bag to gift her mother on her birthday. She selected two bags of prices ` 455.80 and ` 455.40. She couldn’t understand which is more expensive of the two. Do you know which is more expensive of the two? Recall We have already learnt about equivalent, like and unlike decimals. To compare two decimals, we should know the concepts of equivalent decimals and like decimals. Let us recall the concepts by answering the following: a) Write four equivalent decimals of 6.1. b) Convert these unlike decimals to like decimals: 32.5, 410.635, 6, 78.7 c) Identify the pair of like decimals: a) 39.12, 56.03 b) 0.14, 0.04 c) 6.75, 83.16 d) 7.101, 6.2 & Remembering and Understanding We know that, 1) A dding any number of 0s to the right side of the decimal point does not change its value. 2) Unlike decimals can be converted to like decimals by adding zeros at the right end. Now, let us learn to compare two decimals through a few examples. Example 10: Which is greater of the given decimals? a) 69.2 and 69.02 b) 77.10 and 77.012 c) 3.5631 and 3.61 134 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 138

Solution: a) 69.2 and 69.02 Step 1: T o compare two decimals, follow these steps: Step 2: C onvert the given decimal numbers into like decimals. Step 3: 69.20, 69.02 C ompare their integral parts. The decimal with the greater integral part is greater. Here, 69 = 69 If the integral parts are the same, then we have to compare the tenths digits. The decimal with the greater digit in the tenths place is greater. If the tenths digits are the same, compare the hundredths digits and so on. 69.20 69.02 6=6 9=9 2>0 Hence, 69.20 > 69.02 Note: Always start comparing from the largest place value in the integral part. b) 77.10 and 77.012 Step 1: nConvert the unlike decimals into like decimals: 77.100, 77.012 Step 2: Compare the integral parts: 77 = 77 Step 3: Compare the tenths digits: 1 > 0 Hence, 77.10 > 77.012. c) 3.5631 and 3.61 Step 1: Convert the unlike decimals into like decimals: 3.5631, 3.6100 Step 2: Compare their integral parts: 3 = 3 Step 3: Compare the tenths digits: 5 < 6 So, 3.5631 < 3.61000. Example 11: Which is smaller between each of the given pairs of decimal numbers? a) 367.80 and 362.801 b) 21.673 and 21.673 c) 11.729 and 11.726 181010044-Alpine-G5-Textbook-Maths-FY.pdf 139 Decimals - I 135 03-Feb-18 2:37:03 PM

Solution: a) 367.80 and 362.801 Converting the unlike decimals into like decimals: 367.800, 362.801 Comparing their integral parts, 367 > 362 Hence, 362.801 is smaller. b) 21.673 and 21.673 Given decimals are like decimals. Comparing their integral parts: 21 = 21 Compare: Tenths digits: 6 = 6, Hundredths digits: 7 = 7 Thousandths digits: 3 = 3 Hence, 21.673 = 21.673. c) 11.729 and 11.726 Given decimals are like decimals. Comparing their integral parts: 11 = 11 Compare: Tenths digits: 7 = 7 Hundredths digits: 2 = 2 Thousandths digits: 9 > 6 Hence, 11.726 < 11.729. Comparing decimal numbers helps us in arranging them in ascending and descending orders. Let us see a few examples. Example 12 : Arrange the following decimal numbers in the ascending order. a) 2.1, 2.01, 3.06, 0.831 b) 15.12, 19.18, 26.7, 1.007 c) 37.502, 36.512, 67.3, 22 Solution: a) 2.1, 2.01, 3.06, 0.831 Like decimals of the given unlike decimals: 2.100, 2.010, 3.060, and 0.831 0.831 < 2.010 < 2.100 < 3.060 Thus, the ascending order is: 0.831, 2.01, 2.1, 3.06 136 181010044-Alpine-G5-Textbook-Maths-FY.pdf 140 03-Feb-18 2:37:03 PM

b) 15.12, 19.18, 26.7, 1.007 Like decimals of the given unlike decimals: 15.120,19. 180, 26.700 and 1.007 1.007 < 15.120 < 19.180 < 26.700 Thus, the ascending order is: 1.007, 15.12, 19.18, 26.7 c) 37.502, 36.512, 67.3, 22 Like decimals of the given unlike decimals: 37.502, 36.512, 67.300, and 22.000 22.000 < 36.512 < 37.502 < 67.300 Thus, the ascending order is: 22, 36.512, 37.502, and 67.3 Example 13: Arrange these decimal numbers in the descending order. a) 43.25, 43.2, 43.21, 43.127 b) 63.901, 63.09, 63.009, 6.39 c) 11.2, 11.028, 1.127, 13.02 Solution: a) 43.25, 43.2, 43.21, 43.127 Like decimals of the given decimal numbers: 43.250, 43.200, 43.210, 43.127 43.250 > 43.210 > 43.200 > 43.127 Thus, the descending order is 43.25, 43.21, 43.2, 43.127. b) 63.901, 63.09, 63.009, 6.39 Like decimals of the given decimal numbers: 63.901, 63.090, 63.009, 6.390 63.901 > 63.090 > 63.009 > 6.390 Thus, the descending order is: 63.901, 63.09, 63.009, and 6.39 c) 11.2, 11.028, 1.127, 13.02 Like decimals of the given decimal numbers: 11.200, 11.028,1.127, 13.020 13.020 > 11.200 > 11.028 > 1.127 Thus, the descending order is: 13.02, 11.2, 11.028, and 1.127 Decimals - I 137 181010044-Alpine-G5-Textbook-Maths-FY.pdf 141 03-Feb-18 2:37:03 PM

Application Let us now see a few real-life examples involving comparison of decimals. Example 14: Ramu saves ` 361.80 while Raghu saves ` 351.90. Who saves more money? Solution: To know who saves more money, we have to find which number is greater between ` 361.80 and ` 351.90. 361.80 351.90 3=3 6>5 361.80 > 351.90 Therefore, Ramu saves more money. Example 15: Leela scores 32.5 marks in English, 48.5 in Mathematics and 32.75 in Science. Arrange her marks in the descending order and find the subject in which she scored the maximum marks. Solution: English Mathematics Science 32.5 48.5 32.75 Like decimals equivalent to the unlike decimals: 32.50, 48.50 and 32.75 Descending order: 48.50, 32.75, 32.50 48.5 is the greatest number. Therefore, Leela scored the maximum marks in Mathematics. Example 16: Ravi bought two watermelons. One of them weighs 12.352 kg and the other weighs 12.365 kg. Which watermelon is heavier? Solution: Weight of one watermelon = 12.352 kg Weight of the second watermelon = 12.365 kg 138 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 142

12.352 12.365 1=1 2=2 3=3 5<6 Hence, 12.352 < 12.365 or in other words, 12.365 > 12.352. Therefore, the 2nd watermelon is heavier. Higher Order Thinking Skills (H.O.T.S.) Let us see another real-life example of comparing and ordering decimals. Example 17: In a swimming competition, there are five competitors. Four of the swimmers have had their turns. Their scores are 9.8 s, 9.75 s, 9.79 s and 9.81 s. What score must the last swimmer get in order to win the competition? Solution: Arrange the given scores in ascending order: 9.75 < 9.79 < 9.80 < 9.81 The lowest decimal is 9.75. Therefore, the last swimmer must get a score less than 9.75 s in order to win the competition. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 143 Decimals - I 139 03-Feb-18 2:37:03 PM

Concept 11.3: Add and Subtract Decimals Think Pooja went to an ice cream parlour to purchase some ice creams. She bought strawberry for ` 25.50, vanilla for ` 15.30 and chocolate for ` 32.20. She gave ` 100 to the shopkeeper. She wanted to calculate the total price before the shopkeeper gives the bill. Since the prices are in decimals, she was unable to calculate. Do you know how to find the total cost of the ice creams that Pooja bought? How much change would she get in return? Recall Addition and subtraction of decimal numbers is similar to that of usual numbers. Let us recall the conversion of unlike decimals to like decimals. Convert the given unlike decimals into like decimals. a) 4.32, 4.031, 4.1, 7.823 b) 0.7, 0.82, 4.513, 0.72 c) 1.82, 7.01, 5.321, 0.8 d) 7.32, 7.310, 7.8, 5.2 & Remembering and Understanding Add and subtract decimal numbers with the thousandths place Addition and subtraction of decimal numbers with the thousandths place is similar to that of decimals with the hundredths place. Before adding or subtracting any decimals, convert the unlike decimals to like decimal. Write the given decimal numbers such that the digits in their same places are exactly one below the other. Note: The decimal points of the numbers must be exactly one below the other. 140 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 144

Let us see a few examples. Example 18: a) Find the sum of 173.809 and 23.617. b) Subtract 216.735 from 563.726. Solution: a) b) 12 16 11 5 2/ 6/ 12 17 3 . 8 0 9 5 6/ 3/ . 7/ 2/ 6 + 23 . 617 –2 1 6 . 7 3 5 197 . 426 346 . 991 Example 19: Solve: a) 294.631 + 306.524 b) 11.904 – 6.207 Solution: a) 1 1 1 b) 11 8 9 14 294 . 631 1/ 1/ . 9/ 0/ 4/ +3 0 6 . 5 2 4 – 6 . 20 7 601 . 155 5 . 69 7 Application Let us see a few real-life examples involving addition and subtraction of decimals. Example 20: Dolly bought 450 g of tomatoes and 150 g of green chillies to make tomato pickle. What is the total weight (in kg) of the vegetables that Dolly bought? Solution: Weight of tomatoes = 450 g Weight of green chillies = 150 g The total weight of vegetables = (450 + 150) g = 600 g 1 We know that 1 kg = 1000 g or 1 g = 1000 kg 600 g = 600 kg = 0.6 kg 1000 Therefore, Dolly bought 0.6 kg of vegetables. Example 21: Vinod purchased a shirt for ` 275.40, a pair of trousers for ` 1462.30 and a pair of shoes for ` 524.95. Find the total money spent by Vinod. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 145 Decimals - I 141 03-Feb-18 2:37:03 PM

Solution: Amount spent by Vinod on a shirt = ` 275.40 Amount spent on a pair of trousers = ` 1462.30 Amount spent on a pair of shoes = ` 524.95 Total amount spent by Vinod = ` 275.40 + ` 1462.30 + ` 524.95 = ` 2262.65 Example 22: By how much should 67.23 be decreased to get 28.59? 16 11 Solution: The required number is the difference of 67.23 5 6/ . 1/ 13 and 28.59 = 67.23 – 28.59 7/ . 2/ 6/ 3/ 8 5 −2 9 Therefore, 67.23 is to be decreased by 38.64 to 38.6 4 get 28.59. Example 23: Mrs. Roopa bought 13.75 litres of milk. She used 9.2 litres of milk for making paneer. Find the quantity of milk left. Solution: Milk bought by Mrs. Roopa = 13.75 litres Milk used to make paneer = 9.2 litres Quantity of milk left = (13.75 – 9.2) litres = 4.55 litres Example 24: Pawan needs 1.40 m of cloth for a shirt and 2.45 m of cloth for a pair of trousers. His father gave a piece of cloth which is 0.65 m less than that needed. What was the length of the cloth given by Pawan’s father? Solution: Cloth needed for a shirt = 1.40 m Cloth needed for a pair of trousers = 2.45 m Total cloth needed for the suit = 1.40 m + 2.45 m = 3.85 m The length by which the cloth was short by = 0.65 m The length of the cloth given by Pawan’s father = 3.85 m – 0.65 m = 3.20 m Therefore, 3.20 m of cloth was given by Pawan’s father. Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples involving addition and subtraction of decimals. Example 25: Subtract the sum of 6.223 and 37.512 from the sum of 42.106 and 5.07. 142 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 146

Solution: We should find (42.106 + 5.07) – (6.223 + 37.512) Step 1: Step 2: Add 42.106 and 5.07 Step 3: Add 6.223 and 37.512 Subtract the sum in step 2 from 1 the sum in step 1. 4 2.1 0 6 1 6 11 + 5.0 7 0 0 6.2 2 3 +3 7 . 5 1 2 4 /7 . /1 7 6 4 7.1 7 6 4 3.7 3 5 −4 3 . 7 3 5 0 3.4 4 1 Example 26: Pooja saw a doll in the showcase of a mall. The cost of the doll was ` 85.65. She wanted to buy it, but she fell short of ` 5.75. How much money did Pooja have? Solution: Cost of the doll = ` 85.65 Amount she fell short of = ` 5.75 Money that Pooja had = ` 85.65 – ` 5.75 = ` 79.90 Therefore, Pooja had ` 79.90 Drill Time Concept 11.1: Like and Unlike Decimals 1) Write the following numbers in the decimal place value chart. a) 42.874 b) 315.097 c) 2795.741 d) 127.243 2) Write the expanded forms of the given decimals and then write them in words. a) 3578.048 b) 450.981 c) 32.62 d) 432.789 3) Convert the given unlike decimals into like decimals. a) 52.7, 25.28, 321.265, 101.51 b) 42.52, 4.7, 32.472, 48.8 c) 1.7, 32.4, 328.732, 1.82 d) 7.42, 1.821, 7.01, 432.2 4) Word problems a) 475 out of 800 parents who attended a parent-teacher meet were women. Write the decimal equivalent of the fraction of men who attended the meet. b) The distance between two places is 3564 m. What is the distance in kilometres? Decimals - I 143 181010044-Alpine-G5-Textbook-Maths-FY.pdf 147 03-Feb-18 2:37:03 PM

Concept 11.2: Compare and Order Decimals 5) Which is greater of the given pairs of decimal numbers? a) 75.9 and 75.09 b) 32.87 and 32.087 c) 3.52 and 3.5221 d) 372.12 and 347.127 6) Which is smaller of the given decimals? a) 52.6 and 52.609 b) 87.48 and 87.489 c) 585.23 and 585.235 d) 538.72 and 583.72 7) Arrange these decimal numbers in the ascending and descending orders. a) 7.9, 7.91, 8.5, 8.2 b) 26.25, 26.85, 32.96, 12.951 c) 63.824, 63.756, 63.714, 63.823 d) 253.98, 253.74, 253.945, 253.732 8) Word problems a) Riya scores 35.5 marks in English, 58.50 in Mathematics and 58.75 in Science. Arrange her marks in the descending order and find the subject in which she scored the maximum marks. b) R avi bought two pumpkins. One of them weighs 8.556 kg and the other weighs 8.305 kg. Which pumpkin is heavier? Concept 11.3: Add and Subtract Decimals 9) Add: a) 528.364 and 974.623 b) 523.97 and 49.25 c) 23.547 and 14.974 d) 242.57 and 132.60 10) Subtract: a) 954.367 – 412.650 b) 234.45 – 142.52 c) 74.812 – 35.634 d) 732.532 – 522.147 11) Word problems a) By how much should 41.65 be increased to get 98.53? b) Saritha bought 56.6 litres of water. She used 9.2 litres of water for washing her uniform. Find the quantity of water left. 144 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 148

Chapter Decimals - II 12 Let Us Learn About • m ultiplying and dividing decimals by 1-digit and 2-digit numbers. • m ultiplying decimals by 10, 100 and 1000. • multiplying and dividing a decimal number by another decimal number. • the relationship between percentages, decimals and fractions. Concept 12.1: Multiply and Divide Decimals Think Pooja bought six different types of toys for ` 236.95 each. She calculated the total cost and paid the amount to the shopkeeper. Pooja then went to a sweet shop where 410.750 kg of a sweet was prepared. She wanted to know the number of 250 g packs that can be made from it. Do you know how to find the total cost of the toys? Can you calculate how many packs of sweets can be made? Recall Multiplication and division of decimal numbers are similar to that of usual numbers. Let us recall multiplication and division of numbers by answering the following. Solve: a) 267 × 14 b) 3218 × 34 c) 7424 × 14 d) 576 ÷ 12 e) 265 ÷ 5 f) 384 ÷ 4 145 181010044-Alpine-G5-Textbook-Maths-FY.pdf 149 03-Feb-18 2:37:03 PM

& Remembering and Understanding Multiplication of decimals is similar to multiplication of numbers. When two decimal numbers are multiplied, a) count the total number of digits after decimal point in both the numbers. Say it is ‘n’. b) multiply the two decimal numbers as usual and place the decimal point in the product after ‘n’ digits from the right. Multiply decimals by 1-digit and 2-digit numbers Let us understand the multiplication of decimals through a few examples. Example 1: Solve: a) 25.146 × 23 b) 276.32 × 6 Solution: a) 25.146 × 23 T Th Th H TO 1 1 1 46 11 23 25 1 38 20 × 58 1 + 75 4 502 9 5 7 8.3 Therefore, 25.146 × 23 = 578.358 b) 276.32 × 6 Step 1: To multiply the given numbers, follow the steps outlined here. Multiply the numbers without considering the decimal point. T Th Th H T O 4 3 11 2 7 6 32 ×6 1 6 5 7 92 146 03-Feb-18 2:37:03 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 150