5 MATHEMATICS TEXTBOOK PART-2 Name: ____________________________ Section: ________________ Roll No.: ______ School: ____________________________

Preface ClassKlap partners with schools, supporting them with learning materials and processes that are all crafted to work together as an interconnected system to drive learning. ClassKlap presents the latest version of the VISA NEP series – updated and revised after considering the perceptive feedback and comments shared by our experienced reviewers and users. Designed specifically for CBSE schools, the VISA NEP series endeavours to be faithful to the spirit of the National Curriculum Framework (NCF) 2005. Therefore, our books strive to ensure inclusiveness in terms of gender and diversity in representation, catering to the heterogeneous Indian classroom. The books are split into two parts to manage the bag weight. The aim of the NCF 2005 regarding Mathematics teaching is to develop the abilities of a student to think and reason mathematically, pursue assumptions to their logical conclusion and handle abstraction. The VISA NEP Mathematics textbooks and workbooks for CBSE schools offer the following features: S tructured as per Bloom’s taxonomy to help organise the learning process according to the different levels involved Student engagement through simple, age-appropriate language Supported learning through visually appealing images, especially for grades 1 and 2 Increasing rigour in sub-questions for every question in order to scaffold learning for students Word problems based on real-life scenarios, which help students to relate Mathematics to their everyday experiences Mental Maths to inculcate level-appropriate mental calculation skills S tepwise breakdown of solutions to provide an easier premise for learning of problem-solving skills Overall, the ClassKlap VISA NEP Mathematics textbooks, workbooks and teacher companion books aim to enhance logical reasoning and critical thinking skills that are at the heart of Mathematics teaching and learning. – The Authors

I Will Learn About I Recall Contains the list of learning objectives to be covered in the Discusses the prerequisite chapter knowledge for the concept from the previous academic I Think year/chapter/concept/term Introduces the concept and arouses curiosity among students I Remember and Understand Train My Brain Explains the elements in detail that Checks for learning to gauge the form the basis of the concept Ensures understanding level of students, that students are engaged in learning testing both skill and knowledge throughout Pin-up Note Contains key retention points concerning the concept I Apply I Explore (H.O.T.S.) Connects the concept to Encourages students to extend real-life situations by enabling the concept learnt to advanced students to apply what has been scenarios learnt through the practice questions Connect the Dots Maths Munchies A multidisciplinary section that Aims at improving speed of connects a particular topic to calculation and problem solving other subjects in order to enable with interesting facts, tips or tricks students to relate better to it Drill Time A Note to Parent Additional practice questions at Engages a parent in the the end of every chapter out-of-classroom learning of their child and conducting activities to reinforce the learnt concepts

Contents Class 5 8 Money 8.1 Unitary Method in Money............................................................................................ 1 9 Fractions - I 9.1 Equivalence of Fractions.............................................................................................. 7 79.2 Fraction in its Lowest Terms........................................................................................ 12 09.3 Compare Unlike Fractions......................................................................................... 16 +9.4 Add and Subtract Unlike Fractions........................................................................... 21 1 -10 Fractions - II 3 46 x10.1 Add and Subtract Mixed Fractions........................................................................... 27 10.2 Multiply Fractions........................................................................................................ 32 9 510.3 Reciprocals of Fractions............................................................................................. 36 8 211 Decimals - I 11.1 Like and Unlike Decimals........................................................................................... 44 11.2 Compare and Order Decimals................................................................................. 51 11.3 Add and Subtract Decimals...................................................................................... 57 12 Decimals - II 12.1 Multiply and Divide Decimals.................................................................................... 63 12.2 Percentages................................................................................................................ 71 13 Measurements 13.1 Perimeter, Area and Volume..................................................................................... 77 14 Data Handling 14.1 Circle Graphs.............................................................................................................. 89

Chapter Money 8 I Will Learn About • unitary method in money. • problems based on unitary method in money. • c hart of exchange rates. Concept 8.1: Unitary Method in Money I Think Pooja’s father gave her a bill from a supermarket. Pooja found that there are addition, subtraction and multiplication operations in that bill. She wondered if she could use division too. Can you guess where and how each operation is used in a supermarket bill? 8.1 I Recall We have already learnt about mathematical operations such as addition, subtraction, multiplication and division. We have also learnt about decimals. Let us answer the following to recall the different operations involving money. a) ` 436.25 + ` 703.75 b) ` 565 − ` 209.50 c) ` 368.80 × 36 d) ` 911.25 ÷ 27 e) ` 495 ÷ 11 1

8.1 I Remember and Understand We use multiplication to find the value of many units from the value The method of of a single unit. We use division to find the value of a single unit from finding the value of the value of many units. We can also find the value of different one unit and then number of units by following these two steps. finding the value of many units is called Step 1: Find the value of a single unit from the value of the given the Unitary Method. units. Use division to obtain the same. Step 2: Using the value obtained in Step 1, find the value of many units. We need to use multiplication for the same. Let us now see a few examples to understand the unitary method better. Example 1: The cost of 12 erasers is ` 36. What is the cost of 18 erasers? Solution: The cost of 12 erasers = ` 36 Step 1: Cost of 1 eraser = ` 36 ÷ 12 =`3 Step 2: Cost of 18 erasers = 18 × ` 3 =` 54 Therefore, 18 erasers cost ` 54. Example 2: Three pens cost ` 39. Find the cost of a dozen pens. Solution: Cost of 3 pens = ` 39 Step 1: Cost of 1 pen = ` 39 ÷ 3 = ` 13 Step 2: 1 dozen = 12 items Cost of a dozen pens = Cost of 12 pens = ` 13 × 12 = ` 156 Therefore, a dozen pens cost ` 156. Example 3: The cost of 25 notebooks is ` 525. How many notebooks can be bought for ` 1575? Solution: Cost of 25 notebooks = ` 525 2

Cost of 1 notebook = ` 525 ÷ 25 = ` 21 Amount with which notebooks are to be bought = ` 1575 Number of notebooks that can be bought = Total amount ÷ Cost of each notebook = ` 1575 ÷ ` 21 = 75 Therefore, 75 notebooks can be bought. Train My Brain Solve these: a) The cost of 12 mangoes is ` 360. Find the cost of 6 mangoes. b) The cost of 15 caps is ` 450. Find the cost of 3 caps. c) The cost of 20 chairs is ` 4000. Find the cost of 25 chairs. 8.1 I Apply The unitary method can be used to compare two or more items. Consider these examples. Example 4: Mona observed that a pack of four soaps, each of 150 g, costs ` 60. Another pack of six soaps, each of 100 g costs ` 54. Which pack should Mona buy so that she spends lesser amount of money? Solution: Weight of 4 soaps, each of 150 g = 150 g × 4 = 600 g Cost of 4 soaps = ` 60 Cost of 1 g of soap = 60 ÷ 600 = ` 1 = 1 × 100 paise = 10 paise 10 10 Weight of 6 soaps, each of 100 g = 100 g × 6 = 600 g Cost of 6 soaps = ` 54 So, cost of 1 g soap = 54 ÷ 600 = ` 9 = 9 × 100 paise = 9 paise 100 100 Since 9 paise < 10 paise, Mona has to buy the pack of 6 soaps of 100 g each to spend a lesser amount of money. Example 5: A set of eight plastic sharpeners costs ` 48.80. A set of six steel sharpeners costs ` 48.60. Which set of sharpeners is expensive? Money 3

Solution: Cost of 8 plastic sharpeners = ` 48.80 `6.10 `8.10 Cost of 1 plastic sharpener = ` 48.80 ÷ 8 )8 `48.80 )6 `48.60 Cost of 6 steel sharpeners = ` 48.60 − 48 − 48 Cost of 1 steel sharpener = ` 48.60 ÷ 6 8 6 −8 −6 Since ` 8.10 > ` 6.10, the set of steel 0 0 0 sharpeners is expensive. −0 − 0 Example 6: The cost of 17 kg of guavas is ` 552.50. 0 What is the cost of 10 kg of guavas? 32.50 Solution: Cost of 17 kg of guavas = ` 552.50 )17 552.50 Cost of 1 kg of guavas = ` 552.50 ÷ 17 −51 = ` 32.50 42 Cost of 10 kg of guavas = ` 32.50 × 10 − 34 =` 325 85 Therefore, 10 kg of guavas cost ` 325. − 85 00 − 00 00 8.1 I Explore (H.O.T.S.) Different countries of the world have different currencies. We see a chart in banks called the chart of exchange rates. It shows the money (in rupees) that we get in exchange for the money of other countries. Observe the chart given below. This gives the exchange rates as on 23rd January 2017. Country Currency Indian Rupees U.S.A. Dollar 68.12 England Pound 84.98 China Yuan 9.94 Sri Lanka Rupee (SL) 0.46 U.A.E. Dirham 18.54 Germany Euro 73.13 Let us see a few examples on the exchange of money using this chart. 4

Example 7: Rita’s aunt from the U.S. gifted her 15 US dollars. Rita used ` 300 for her dress from that amount. How much money was left with her? Give your answer in rupees. (Hint: 1 US dollar = ` 64.14, as of December 19, 2017) Solution: 1 US dollar = ` 64.14 15 US dollars = ` 64.14 × 15 = ` 962.1 Amount Rita spent for her dress = ` 300 Amount left with Rita = ` 962.1 – ` 300 = ` 662.1 Therefore, Rita has ` 662.1 left with her. Example 8: Ravi works for a Sri Lankan company for a salary of 1000 SL (Sri Lankan Rupees). Suresh works for a company in China for a salary of 1200 yuans. Who earns more in terms of Indian currency? (Hint: 1 SL = ` 0.42 and 1 yuan = ` 9.69 ) Solution: We know that 1 Sri Lankan Rupee = ` 0.42 1000 Sri Lanka Rupees = ` 0.42 × 1000 = ` 420 Also, 1 Chinese yuan = ` 9.69 1200 yuans = 1200 × ` 9.69 =` 11628 As, ` 11628 > ` 420, Suresh earns more than Ravi. Maths Munchies 213 ` 10000 is the highest denomination that RBI has printed in its history. ` 1000 and ` 10000 notes were in circulation between 1938 and 1946 but were eventually demonetised. ` 1000, ` 5000 and ` 10000 notes were reintroduced in 1954 and demonetised in 1978. Recently, in 2016, the currency notes of ` 1000 and ` 500 have been demonetised in our country. ` 500 was reintroduced with new ` 2000 and ` 200 notes. Money 5

Connect the Dots Social Studies Fun The pound is the currency used by Egypt, Falkland Islands, Gibraltar, Guernsey, Jersey, and the United Kingdom. It is called as Pound sterling. Centuries ago, it was known as Troy Pound. Science Fun In India, currency papers are made up of cotton and cotton rag. The currency is not a paper but a cloth that gives the look and feel of paper. Drill Time Concept 8.1: Unitary Method in Money Word problems a) The cost of 15 apples is ` 84. What is the cost of 20 apples? b) Eight pairs of slippers cost ` 328. How much does a dozen pair of slippers cost? c) The cost of five textbooks is ` 525. How many textbooks can be bought for ` 945? d) R ahul observed that a pack of five creams, each of 150 g, costs ` 500 and another pack of six creams each of 100 g costs ` 450. Which pack should Rahul buy so that he spends lesser amount? e) The cost of 20 kg of potatoes is ` 310. What is the cost of 12 kg of potatoes? A Note to Parent Help your child to practise the unitary method while buying monthly grocery. The unitary method is an important skill used in day-to-day life. 6

Chapter Fractions - I 9 I Will Learn About • equivalent fractions and cross- multiplying equivalent fractions. • finding the missing numerators or denominators in the fractions. • reducing fractions using division and H.C.F. • the term 'unlike fraction' and comparing unlike fractions. • adding and subtracting unlike fractions. Concept 9.1: Equivalence of Fractions I Think Pooja eats two pieces of a cake that was cut into four equal pieces. Farhan eats three pieces of cake of the same size that was cut into six equal pieces. Do they eat the same amount of cake? 9.1 I Recall In Class 4, we have learnt about equivalent fractions. Let us revise them here. Suppose a pizza is cut as shown. Rohan eats 2 of the pizza. Then the piece of pizza he gets 8 is = . 7

Suraj eats 1 of the pizza. Then the piece of pizza he gets is . 4 We see that the pieces of pizza eaten by both are of the same size. So, we say that the fractions 2 and 1 are equivalent. 84 21 We write them as 8 = 4 . Recall the following: • The multiples of a number are obtained by multiplying it by 1, 2, 3 and so on. For example, the multiples of 6 are 6, 12, 18, 24, 30, 36 and so on. • The numbers that divide a given number exactly are called its factors. For example, the factors of 6 are 1, 2, 3 and 6. • Equivalent fractions are obtained by multiplying or dividing the numerator and the denominator of the given fraction by the same number. For example, 1 , 3 , 5 , 2 and so on, are equivalent fractions. 7 21 35 14 9.1 I Remember and Understand Let us now understand the process used to check the equivalence of the given fractions. Cross-multiplication: To check if two fractions are equivalent, Fractions that denote we cross multiply them. In cross-multiplication, we multiply the the same part of a numerator of the first fraction by the denominator of the second. whole are called Then, we multiply the denominator of the first by the numerator of equivalent fractions. the second. If the cross products are equal, the given fractions are equivalent. Otherwise, they are not equivalent. Example 1: Check if these fractions are equivalent. a) 3 and 9 b) 1 and 2 Solution: 5 15 36 a) 3 and 9 5 15 8

Cross-multiplying, we get 39 5 15 T he cross products are 5 × 9 = 45 and 3 × 15 = 45. Since the cross products are equal; the given fractions are equivalent. b) 1 and 2 36 Cross-multiplying, we get 12 36 T he cross products are 1 × 6 = 6 and 3 × 2 = 6. Since the cross products are equal, the given fractions are equivalent. Example 2: Check if these fractions are equivalent. a) 1 and 2 b) 3 and 9 4 10 9 18 S olution: a) 1 and 2 4 10 Cross-multiplying, we get 12 4 10 T he cross products are 1 × 10 = 10 and 4 × 2 = 8. Since the cross products are not equal (10 ≠ 8), the given fractions are not equivalent. b) 3 and 9 9 18 Cross-multiplying, we get 39 9 18 T he cross products are 3 × 18 = 54 and 9 × 9 = 81. Since the cross products are not equal (54 ≠ 81), the given fractions are not equivalent. Train My Brain Check if the following fractions are equivalent. a) 5 and 20 b) 36 and 9 c) 27 and 11 4 16 24 8 32 6 Fractions - I 9

9.1 I Apply Let us solve a few examples based on the concept of equivalence of fractions. Example 3: Kiran played for 1 of a day and did his homework for 3 of the day. Did he 5 15 spend the same amount of time for both the activities? Solution: If Kiran spent the same amount of time for both the activities, the given fractions must be equivalent. We check the equivalence of fractions by cross-multiplication. 13 5 15 1 × 15 = 5 × 3 = 15 As the cross products are equal, the given fractions are equivalent. Therefore, Kiran spent the same amount of time for both the activities. Example 4: Clock A shows 2 of an hour and Clock B shows 3 of an hour. Are both the 12 15 clocks showing the same time? Solution: If both the clocks are showing the same time, the given fractions must be equivalent. We check the equivalence of fractions by cross-multiplication. 23 12 15 2 × 15 = 30; 12 × 3 = 36 As the cross products are not equal, the given fractions are not equivalent. Therefore, both the clocks are not showing the same time. 9.1 I Explore (H.O.T.S.) Let us now solve some examples where equivalent fractions are used in real-life situations. Example 5: If the given fractions are equivalent, find the missing numerators in the brackets. a) 15 = [ ] b) [ ] = 3 25 5 49 7 10

Solution: Given that the fractions are equivalent, we know that their cross products are equal. a) 15 × 5 = 25 × [ ] 3 × 5 × 5 = 25 × [ ] 3 × 25 = 25 × [ ] 3 = [ ] Therefore, the missing number in the brackets is 3. Example 6: b) [ ] × 7 = 49 × 3 [ ] × 7 = 7 × 7 × 3 Solution: [ ] × 7 = 7 × 21 [ ] = 21 Therefore, the missing number in the brackets is 21. If the given fractions are equivalent, find the missing denominators in the brackets. a) 14 7 b) 18 = 9 28 = [ ] [] 27 As the fractions are equivalent, their cross products are equal. a) 14 × [ ] = 28 × 7 14 × [ ] = 2 × 14 × 7 14 × [ ] = 14 × 2 × 7 [ ] = 2 × 7 [ ] = 14 Therefore, the missing number in the brackets is 14. b) 18 × 27 = [ ] × 9 9 × 2 × 27 = [ ] × 9 [ ] × 9 = 9 × 54 [ ] = 54 Therefore, the missing number in the brackets is 54. Fractions - I 11

Concept 9.2: Fraction in its Lowest Terms I Think Pooja knows the method of finding equivalent fractions by both division and multiplication. She wants to know where she could use the division method of finding equivalent fractions. Do you know where it is used? 9.2 I 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 9.2 I Remember and Understand 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 A fraction can be denominator do not have any common factors other than 1. reduced to its lowest Reducing a fraction using division terms using either division or H.C.F. 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 12

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. Solution: a) 36 b) 26 48 65 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 Train My Brain Reduce these fra62c56ti=on62s56u÷÷s11in33g=H25.C.F. 26 2. a) 110 Therebfo) re9,1w hen reducedcto) 1it0s8lowest terms, 65 5 50 117 132 becomes 9.2 I Apply 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 more? Fractions - I 13

Solution: 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. 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 efficient? 30 Solution: Part of the wall painted by Suraj in an hour = 21 = 3 35 5 (H.C.F. of 21 and 35 is 7.) Part of the wall painted by Puja in an hour = 24 = 4 30 5 (H.C.F. of 24 and 30 is 6.) Clearly, 4 is greater than 3 . 55 Therefore, Puja does more work than Suraj in the same time. So, Puja is more efficient. Example 11: Malik saves ` 550 from his monthly salary of ` 5500. Akhil saves ` 300 from his Solution: 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 14

9.2 I Explore (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.) 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 Fractions - I 15

Concept 9.3: Compare Unlike Fractions I 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? 9.3 I 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 9.3 I Remember and Understand We know how to compare like fractions. Let us now learn to compare unlike fractions. Steps to compare unlike fractions: To compare two or more fractions, their 1) Find the L.C.M. of the denominators of the given unlike denominators should fractions. Using L.C.M, convert the given unlike fractions into be the same. 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. 16

Example 14: Compare these unlike fractions. Solution: a) 3 , 4 b) 3 , 1 c) 1 , 3 7 11 57 48 Solved Solve this Steps 3, 4 3, 1 1, 3 7 11 57 48 Step 1: Write like fractions equivalent to the given L.C.M. of 7 and 11 is 77. fractions, using the least common multiple of their So, the equivalent denominators. fractions are 3 = 3 ×11 = 33 and 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 . 7 11 Example 15: Compare these unlike fractions. a) 1 , 2 b) 5 , 1 c) 1 , 6 24 63 4 12 S olution: 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 Fractions - I 17

Therefore, 5 > 1 . 63 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 Train My Brain Compare: b) 4 and 13 c) 7 and 4 a) 5 and 9 21 6 11 15 16 4 9.3 I Apply 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: 1 = 1×3 3 and 2 = 2×4 =8 4 4×3 = 3 3×4 12 12 18

Step 2: Compare the numerators of the equivalent fractions. Since 8 > 3, 8 > 3 . 12 12 Example 17: 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? 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. 9.3 I Explore (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 Fractions - I 19

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: 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 Example 20: Solution: 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 Arrange 2 , 1 , 1 , 5 , and 3 in the descending order. 7 4 8 14 16 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 20

Concept 9.4: Add and Subtract Unlike Fractions I 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? 9.4 I 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 9.4 I Remember and Understand Let us understand the addition and subtraction of unlike fractions through some numerical examples. b) 7 + 2 Unlike fractions can be Example 21: Solve: a) 3 + 1 13 39 added or subtracted by first making the 15 10 denominators equal c) 22 + 7 and then adding up or subtracting the 100 10 numerators. S olution: 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.] Fractions - I 21

Example 22: Solve: a) 8 – 4 b) 17 – 5 c) 14 17 Solution: 9 11 30 24 – 25 50 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.] [L. C. M. of 25 and 50 is 50.] 30 24 120 120 68 - 25 43 == 120 120 c) 14 – 17 = 28 17 – 25 50 50 50 28 -17 11 == 50 50 Train My Brain Solve the following: a) 13 + 1 17 39 c) 5 – 7 d) 29 – 7 46 b) + 16 24 8 13 30 4 9.4 n I Apply 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 22

Part of the second strip coloured = 4 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 2 50 126 - 50 76 =– = = 1 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? 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 9.4 I Explore (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? Fractions - I 23

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 8488 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 Example 26: 4 6 12 12 12 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 31 51 = 51+53 185 = = 1 – + 1 – = 1 – 15 -8 = 15 - 8 = 7 15 15 15 15 Therefore, 7 of the total strength were high school students. 15 Maths Munchies 213 If two fractions have the same numerator but different denominators, the fraction with the greater denominator is smaller. For example, 1 < 1 < 1 < 1 . 8532 24

Connect the Dots Social Studies Fun Did you know that the Moon is 1 the size of the 4 Earth? Interestingly, your weight on the Moon is 1 6 of that on the Earth. Try calculating what your weight on the Moon would be. English Fun From the words given below, circle the nouns and underline the adjectives. a) Like fraction b) Unlike fraction c) Equivalent fraction 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. a) 24 b) 36 c) 42 d) 12 e) 12 48 60 70 36 30 3) Reduce these fractions using division. d) 6 e) 3 24 27 a) 36 b) 42 c) 26 72 84 91 Fractions - I 25

Drill Time 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 Concept 9.4: Add and Subtract Unlike Fractions 5) Solve: b) 4 + 3 c) 4 + 1 d) 19 + 5 e) 2 + 6 a) 3 + 5 14 12 15 10 100 10 16 30 4 13 6) Solve: b) 14 – 3 c) 13 – 14 d) 3 – 4 e) 15 – 16 a) 4 – 3 30 24 30 60 15 30 20 40 9 11 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) Mani 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? A Note to Parent Write some fractions on a piece of paper. Ask your child to compare and add them. 26

Chapter Fractions - II 10 I Will 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 I 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? 10.1 I 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. 27

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 10.1 I Remember and Understand The addition and subtraction of mixed fractions are A mixed fraction can be converted similar to that of unlike fractions. Let us understand into an improper fraction by the same through the following examples. multiplying the whole number part by the fraction’s denominator and 3 2 then adding the product to the Add: 2 5 7 numerator. Then we write the result on top of the denominator. Example 1: + 3 Solved Solve this Steps 23 + 32 12 1 + 15 1 57 43 Step 1: Convert all the mixed 23 = 2×5+3 = 13 ; fractions into improper fractions. 5 5 5 3 2 = 3 ×7 + 2 = 23 77 7 Step 2: Find the L.C.M. and add 2 3 + 3 2 = 13 + 23 the improper fractions. 5 75 7 [L.C.M. of 5 and 7 is 35.] = 7 ×13 + 5 × 23 35 = 91+115 = 206 35 35 28

Steps Solved Solve this 12 1 + 15 1 Step 3: Find the H.C.F. of the 23 + 32 numerator and the denominator 57 43 of the sum. Then reduce the improper fraction to its simplest The H.C.F. of 206 and 35 is 1. Solve this form. So, we cannot reduce the 12 1 from 15 1 Step 4: Convert the improper fraction any further. fraction into a mixed fraction. 43 206 31 =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 Fractions - II 29

Steps Solved Solve this 2 3 from 3 2 12 1 from 15 1 57 43 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. 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 – 2 = 7 5 35 Train My Brain Solve the following: b) 2 1 5 11 5 –21 a) 3 1 + 2 1 + 2 c) 4 – 2 d) 2 39 93 48 48 10.1 I Apply In some real-life situations, we use the addition or subtraction of mixed fractions. Example 3: Ajit ate 5 3 1 biscuits. How many biscuits did they eat biscuits and Arun ate 8 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 30

=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 20 123 3 [L.C.M. of 1 and 20 is 20.] = =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 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 4 Total distance she travelled = 34 2 km + 16 1 km 34 104 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 10.1 I Explore (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 Fractions - II 31

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 4 7 4= 28 = 767 = 27 11 28 28 Therefore, 22 3 is less than 50 1 by 27 11 . 4 7 28 Concept 10.2: Multiply Fractions I 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? 10.2 I 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 32

d) 17 9 f) 14 23 e) 21 42 10.2 I Remember and Understand Multiply fractions by whole numbers Multiplying a fraction by 2-digit or 3-digit numbers is the same as A whole number can be considered finding the fraction of a number. as a fraction with its denominator as 1. 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 Fractions - II 33

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 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 H1.C.F. = 1 1 23 × 15 = 1×1 3 45 46 2 3 × 2 6 Therefore, 23 × 15 = 1 . 45 46 6 Example 9: Solve: a) 2 × 5 b) 7 × 70 c) 84 × 45 Solution: 56 35 63 54 60 1 a) 12 × = 1× 1 = 1×1 = 1 15 5 1 3 1× 3 3 63 34

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 1 6 45 60 5 6 Train My Brain Solve the following: a) 14 94 b) 2 14 c) 4 3 54 7 7 21 15 12 10.2 I Apply 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 of 1 kg 10 6 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. 14 =4 × 5500 1375 = ` 1375 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 Fractions - II 35

10.2 I Explore (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 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 Solution: 757 7 7 4 11 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 I 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? 36

10.3 I 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 10.3 I Remember and Understand Reciprocal of a fraction A number or a fraction which when multiplied by To find the reciprocal of a fraction, we interchange its a given number gives the numerator and denominator. product as 1 is called the • The reciprocal of a number is a fraction. reciprocal or multiplicative For example, the reciprocal of 20 is 1 . inverse of the given number. 20 1 is 7. • The reciprocal of a unit fraction is a number, For example, the reciprocal of 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) Numbers 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) F ractions are reduced to their lowest terms (if possible) before finding their reciprocals. Fractions - II 37

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. The reciprocals of the given fractions are: a) 17 b) 19 c) 11 d) 5 Example 15: 8 4 3 4 Find the multiplicative inverses of these fractions. a) 5 b) 7 5 c) 0 d) 1 e) 33 1 9 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. Train My Brain Find the reciprocal of the following: a) 3 b) 17 c) 22 7 11 10.3 I Apply Divide a number by a fraction The division of a number by another means to find the number of divisors present in the dividend. For example, 8 ÷ 4 means to find the number of fours in 8. Similarly, 10 ÷ 1 means to 5 find the number of one-fifths in 10. 38

Let us understand the division by fractions through some examples. 1 b) 75 ÷ 3 Example 16: Divide: a) 15 ÷ 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 Step 4: 3 11 1 Reduce the product to its lowest terms. 45 = 45 Step 1: 1 Step 2: Step 3: Therefore, 15 ÷ 1 = 45. 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 75 5 Reduce the product to its lowest terms. × 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 Fractions - II 39

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 is . 1 67 Step 3: Multiply the dividend 1 ÷ 67 = 1 × 1 = 1 by the reciprocal of the 3 3 67 3 × 67 divisor. 11 Step 4: Reduce the product = to its lowest terms. 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: Steps Solved Solve this 1÷ 1 3 210 3 21 25 ÷ 75 Step 1: Find the reciprocal The reciprocal of 1 is 21 . of the divisor. 21 1 Step 2: Multiply the dividend 1 1 1 21 by the reciprocal of the 3 ÷ 21 = 3 × 1 divisor. 1 7 Step 3: Reduce the product × 21 = 7 into its lowest terms. 31 Therefore, 1 ÷ 1 = 7. 3 21 40

10.3 I Explore (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? 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. She 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 13 kg = 3 50 =5 kg × 4 = 20 × 1000 g = 3 × 50 g = 150 g 20 1 Therefore, each bowl contains 150 g of sugar. 16 8 Example 21: There is litres of orange juice in a bottle. 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 =1265 × reciprocal of 8 16 × 25 = 2 25 = 25 1 8 1 Therefore, 2 glasses are filled. Fractions - II 41

Maths Munchies Another method to find the sum or difference of mixed fractions. 213 We can add mixed fraction by adding the integral parts and then the proper fractions a) 2 3 + 3 2 Step 1: 57 Add the integral parts of the mixed fractions. 2 + 3 = 5 Step 2: Add the proper fraction parts by using the L.C.M. 3 2 7 × 3 + 5 × 2 21+10 31 5 +7 = = 35 = 35 [L.C.M. of 5 and 7 is 35.] 35 Step 3: If we get an improper fraction here, convert it to a mixed fraction. Step 4: Add the sums obtained in steps 1 and 2 to get the required sum. 3 2 31 Therefore, 2 5 +37 =5 . 35 Connect the Dots English Fun 5 The English alphabet has 26 letters. Out of these, 26 are vowels. Show the rest of the consonants in the form of a fraction. Science Fun In our atmosphere, 78 part is made up of Nitrogen, 21 100 100 by oxygen and 1 part is composed of other gases. 100 42

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 c) 7 2 – 4 1 2) Solve: 27 84 1 – 2 1 d) 12 3 – 11 2 a) 4 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. d) 7 × 45 15 49 a) 22 × 26 b) 4 × 16 c) 3 × 51 13 44 12 24 17 21 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 1 47 c) 42 by 3 A Note to Parent While having meals, engage your child with the concept of fractions as you serve food items such as chapatti and so on. Remember to emphasise the fact that fractions deal with equal parts. Fractions - II 43

Chapter Decimals - I 11 I Will Learn About • c onverting fractions to decimals and vice versa. • decimal place value chart and expanding the decimal numbers. • equivalent, like and unlike decimals. • converting unlike decimals to like decimals. • adding and subtracting decimals. Concept 11.1: Like and Unlike Decimals I 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? 11.1 I Recall In class 4, we have learnt decimals and fractions with their conversions. Let us recall them. 44

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. 11.1 I Remember and Understand 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. Consider the following example. The place value of a number increases by ten when we Example 1: Convert 1396 m into km. move from right to left and Solution: decreases by ten when we 1 m = 1 km move from left to right. 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 100. It is called the hundredths place. Decimals - I 45

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? Solution: c) which digit is in the hundredths place? d) which digit is in the thousandths place? e) which digit is in the ones place? Example 3: 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. Write the following numbers in the decimal place value chart. a) 13.457 b) 450.72 c) 2153.068 46

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