by classklapTM MATHEMATICS 2 TEXTBOOK - PART ALPINE SERIES Enhanced Edition 5 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________ Alpine_Maths_G1_TB_ToC.indd 1 12/12/2018 2:46:42 PM
Preface IMAX Program partners with schools, supporting them with learning materials and processes that are all crafted to work together as an interconnected system to drive learning. IMAX Program presents the latest version of this series – updated and revised after considering the perceptive feedback and comments shared by our experienced reviewers and users. This series endeavours to be faithful to the spirit of the prescribed board curriculum. 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 larger aim of the curriculum 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 Mathematics textbooks and workbooks offer the following features: S tructured as per Bloom’s taxonomy to help organise the learning process according to the different levels involved S tudent engagement through simple, age-appropriate language S upported 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 W ord 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 IMAX 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 Alpine_Maths_G1_TB_ToC.indd 2 12/14/2018 12:10:03 PM
Textbook Features Let Us Learn About Think Contains the list of learning objectives Introduces the concept and to be covered in the chapter arouses curiosity among students Recall Discusses the prerequisite knowledge for the concept from the previous academic year/chapter/ concept/term Remembering and Understanding Explains the elements in detail that form the Application basis of the concept Ensures that students are engaged in learning throughout Connects the concept to real-life situations by enabling students to apply what has been learnt through the practice questions Higher Order Thinking Skills (H.O.T.S.) Encourages students to extend the concept learnt to advanced scenarios Drill Time Additional practice questions at the end of every chapter
Contents 5Class 8 Money 8.1 Unitary Method in Money............................................................................................ 1 9 Fractions - I 9.1 Equivalence of Fractions.............................................................................................. 7 9.2 Fraction in its Lowest Terms........................................................................................ 11 9.3 Compare Unlike Fractions......................................................................................... 15 9.4 Add and Subtract Unlike Fractions........................................................................... 20 10 Fractions - II 10.1 Add and Subtract Mixed Fractions........................................................................... 26 10.2 Multiply Fractions........................................................................................................ 31 10.3 Reciprocals of Fractions............................................................................................. 35 11 Decimals - I 11.1 Like and Unlike Decimals........................................................................................... 42 11.2 Compare and Order Decimals................................................................................. 49 11.3 Add and Subtract Decimals...................................................................................... 55 12 Decimals - II 12.1 Multiply and Divide Decimals.................................................................................... 61 12.2 Percentages................................................................................................................ 69 13 Measurements 13.1 Perimeter, Area and Volume..................................................................................... 75 14 Data Handling 14.1 Circle Graphs.............................................................................................................. 87
Chapter Money 8 Let Us Learn About • u nitary method in money. • problems based on unitary method in money. • c hart of exchange rates. Concept 8.1: Unitary Method in Money 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? 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
& Remembering and Understanding We use multiplication to find the value of many units from the value of a single unit. We use division to find the value of a single unit from the value of many units. We can also find the value of different number of units by following these two steps. Step 1: Find the value of a single unit from the value of the given 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. The method of finding the value of one unit and then finding the value of many units is called the Unitary Method. 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. 2
Example 3: The cost of 25 notebooks is ` 525. How many notebooks can be bought for ` 1575? Solution: Cost of 25 notebooks = ` 525 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. Application 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. Money 3
Example 5: A set of eight plastic sharpeners costs ` 48.80. A set of six steel sharpeners costs Solution: ` 48.60. Which set of sharpeners is expensive? `6.10 `8.10 Cost of 8 plastic sharpeners = ` 48.80 Cost of 1 plastic sharpener = ` 48.80 ÷ 8 )8 `48.80 )6 `48.60 Cost of 6 steel sharpeners = ` 48.60 − 48 − 48 8 6 −8 −6 Cost of 1 steel sharpener = ` 48.60 ÷ 6 0 0 −0 −0 Since ` 8.10 > ` 6.10, the set of steel sharpeners is expensive. 0 0 Example 6: The cost of 17 kg of guavas is ` 552.50. 3 2.50 )What is the cost of 10 kg of guavas? 17 552.50 Solution: Cost of 17 kg of guavas = ` 552.50 - 51 42 Cost of 1 kg of guavas = ` 552.50 ÷ 17 - 34 = ` 32.50 85 - 85 Cost of 10 kg of guavas = ` 32.50 × 10 00 = ` 325 - 00 Therefore, 10 kg of guavas cost ` 325. 00 Higher Order Thinking Skills (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 4
Let us see a few examples on the exchange of money using this chart. 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. 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) T he cost of five textbooks is ` 525. How many textbooks can be bought for ` 945? Money 5
d) Rahul 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? 6
Chapter Fractions - I 9 Let Us Learn About • e quivalent 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 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? 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 & Remembering and Understanding Fractions that denote the same part of a whole are called equivalent fractions. Let us now understand the process used to check the equivalence of the given fractions. Cross-multiplication: To check if two fractions are equivalent, we cross multiply them. In cross-multiplication, we multiply the numerator of the first fraction by the denominator of the second. Then, we multiply the denominator of the first by the numerator of 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 5 15 36 Solution: 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 Solution: 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. Application 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? Fractions - I 9
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 Example 4: As the cross products are equal, the given fractions are equivalent. Solution: Therefore, Kiran spent the same amount of time for both the activities. 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? 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. Higher Order Thinking Skills (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. Solution: a) 15 = [ ] b) [ ] = 3 25 5 49 7 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 = [ ] 10
Therefore, the missing number in the brackets is 3. b) [ ] × 7 = 49 × 3 [ ] × 7 = 7 × 7 × 3 [ ] × 7 = 7 × 21 [ ] = 21 Therefore, the missing number in the brackets is 21. Example 6: If the given fractions are equivalent, find the missing denominators in the brackets. a) 14 7 18 = 9 28 = [ ] b) [ ] 27 Solution: 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. Concept 9.2: Fraction in its Lowest Terms 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? Fractions - I 11
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 12
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 [Using division method] 16 = 16 ÷ 8 = 2 32 32 ÷ 8 4 Clearly, 1 < 2. So, 1 < 2 . 44 Therefore, Vijay ate more. Fractions - I 13
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 6.) 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.) 14
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? Fractions - I 15
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 16
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 . 7 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 Fractions - I 17
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? 18
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: Fractions - I 19
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? 20
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. Example 21: Solve: a) 3 + 1 b) 7 + 2 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 Fractions - I 21
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 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 n 9 Total number of parts of the second strip = 7 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 63 63 50 63 + 63 - 50 126 - 50 76 = + – = == 63 63 63 63 63 63 22
Example 24: Manasa ate a quarter of a chocolate bar and her sister ate two-thirds of it. Solution: How much chocolate did they eat in all? How much chocolate is remaining? 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 Fractions - I 23
Part of the population that was women and children = 1 + 1 = 3 + 2 = 5 4 6 12 12 12 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 = 51+53 185 = = 1 – 1 + 1 – = 1 – 15 -8 = 15 - 8 = 7 3 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 24
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) S udheer 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? Fractions - I 25
Chapter Fractions - II 10 Let Us Learn About • the terms ‘mixed’, ‘proper’ and ‘improper’ fractions. • a dding and subtracting mixed fractions. • multiplying 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. 26
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 + 32 Add: 2 5 7 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 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 Fractions - II 27
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 32 -23 = 23 13 subtract the improper fractions. 7 5 7 -5 [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. 28
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 – 2 = 7 5 35 Application 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 5 biscuits and Arun ate 8 4 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 = 20 = 6 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 Fractions - II 29
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 21 Total distance she travelled = 34 3 km + 16 km 4 104 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 than 2 ? 39 65 Solution: 1 – 39 2 = 247 – 197 = 1235 −1182 The required number = 41 6 56 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 4 7 4= 28 = 767 = 27 11 28 28 Therefore, 22 3 is less than 50 1 by 27 11 . 4 7 28 30
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. • Fraction 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 14 23 21 f) 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. Fractions - II 31
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 32
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 11 54 6 5 6×5 6 6 6 45 60 5 Fractions - II 33
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 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. 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 34
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 Fractions - II 35
& 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) 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) 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. 36
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 ÷ 5 means to find the number of one-fifths in 10. 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 3 11 1 Fractions - II 37
Step 4: Reduce the product to its lowest terms. 45 = 45 1 Therefore, 15 ÷ 1 = 45. Step 1: 3 Step 2: b) 75 ÷ 3 Step 3: 5 75 Write the number as a fraction as 75 = 1 3 5. Find the reciprocal of the divisor. The reciprocal of 5 is 3 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. 1 × 3 The H.C.F. of 75 and 3 is 3. Cancelling 3 and 75 by 3, we get 25 75 × 5 = 25 × 5 = 125. 1 31 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: 3 ÷ 67 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 The reciprocal of 67 is 1 . the divisor. 1 67 Step 3: Multiply the dividend 1 11 1 by the reciprocal of the ÷ 67 = × = divisor. 3 3 67 3 × 67 38
Steps Solved Solve this 1 3 Step 4: Reduce the product ÷ 54 to its lowest terms. ÷ 67 5 3 11 = 3 × 67 201 Therefore, 1 ÷ 67 = 1 . 3 201 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 The reciprocal of 1 is 21 . the divisor. 21 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? Fractions - II 39
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 kg × 1 = 3 kg = 3 50 = 3 × 50 g = 150 g 5 4 20 × 1000 g 20 1 Therefore, each bowl contains 150 g of sugar. 16 8 Example 21: There is 25 litres of orange juice in a bottle. 25 litres of it is poured in each 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 × reciprocal of 8 = 16 × 25 = 2 25 25 25 1 8 1 Therefore, 2 glasses are filled. 40
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 3 + 12 3 2) Solve: b) 10 1 – 5 3 27 1 – 2 1 c) 7 2 – 4 1 d) 12 3 – 11 2 a) 4 3 7 84 89 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 2 23 11 6) Divide: d) 7 by 49 a) 16 by 1 b) 14 by 2 c) 1 by 3 4 7 42 Fractions - II 41
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. • converting unlike decimals to like decimals. • a dding 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. 42
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: 1 m = 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 Decimals - I 43
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? 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 44
Solution: Thousands Hundreds Tens Ones Decimal Tenths Hundredths Thousandths point 1 1 1 10 1000 100 10 1 (.) 4 100 1000 5 7 a) 13 . 7 b) 4 50 . 2 8 c) 2 1 53 . 0 6 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 = F ive 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) 4 6.937 = 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 unlike decimals. Decimals - I 45
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 46
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