181910055_Maple G5_Textbook Integrated_Term2

& Remembering and Understanding Let us now understand the addition and subtraction of time through some examples. While adding time, we add the minutes (smaller units) first and then the hours (larger units). Sometimes, we may have to regroup the sum of the minutes. If the sum of minutes is 60, we convert it to 1 hour and add it to the hours. Let us see an example. Example 10: Add: 1 hour 35 minutes and 2 hours 45 minutes Solution: Steps Solved Solve these Step 1: Write both the numbers one Hours Minutes below the other. 1 35 +2 45 Step 2: Add hours and minutes Hours Minutes Hours Minutes separately, regrouping if needed. 1 35 1 20 +2 45 +3 50 3 80 Step 3: Check whether the minutes 80 minutes > 60 minutes Hours Minutes in the sum is greater than or equal to 60. If yes, then convert it into 2 30 hours. Step 4: Add the hours obtained 3 hours 80 minutes +2 20 in step 3 to the hours obtained in        step 2. 1 60+ 20 4 hours 20 minutes The sum is 4 hours 20 minutes. While subtracting, we subtract the minutes first (smaller units) and then the hours (larger units). Sometimes, we may have to regroup the hours. Let us see an example. Time 7 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 51 2/17/2018 4:21:59 PM

Example 11: Subtract: 2 hours 35 minutes from 3 hours 10 minutes Solution: Steps Solved Solve these Step 1: Write both the numbers one below the other, such that Hours Minutes Hours Minutes the smaller number is subtracted 3 45 from the larger one. 3 10 1 20 –2 35 – Step 2: Subtract hours and 10 minutes < 35 minutes. So, minutes separately, regrouping if borrow 1 hour, that is, 60 needed. minutes and add it to the minutes. (10 + 60 = 70) Hours Minutes Step 3: Reduce the hours by Hours Minutes – 4 20 1 and subtract the minutes as – 2 40 usual. 2 70 –2 35 35 Step 4: Subtract the hours and Hours Minutes Hours Minutes write the difference. 5 30 2 70 3 35 –2 35 0 35 The difference is 35 min. Example 12: Subtract 4 h 42 min from 380 min. Solution: We first convert 380 min to hours and minutes. Hours Minutes 1 5 80 380 min = (300 × 60 h) + 80 min = 5 h 80 min –4 42 1 38 Therefore, the difference is 1 h 38 min. 8 2/17/2018 4:21:59 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 52

Application Now let us solve a few examples where the addition and subtraction of time are mostly used. Example 13: A courier boy delivered letters for 2 hours 35 minutes and parcels for 3 hours 28 minutes in a day. For how long was he on the job? Solution: Time spent in delivering letters = 2 h 35 min Hours Minutes Time spent in delivering parcels = 3 h 28 min Total time spent on the job = 2 35 2 h 35 min + 3 h 28 min = 5 h 63 min +3 28 5 63 63 > 60 63 min = 1 h 3 min Therefore, the total time spent on job = (5 h + 1 h) + 3 min = 6 h 3 min. Example 14: On Saturday, Rima’s drawing class lasted for 2 hours 20 minutes, while on Sunday, it lasted for 1 hour 40 minutes. How much longer was the drawing class on Saturday? Solution: To find how much longer the drawing class on Saturday was, we must subtract 1 hour 40 minutes from 2 hours 20 minutes. Hours Minutes 2 h 20 min can be written as 1 h 80 min by regrouping. 1 80 So, on Saturday, Rima’s drawing class lasted 40 minutes – 1 40 longer. 0 40 Higher Order Thinking Skills (H.O.T.S.) Now let us solve a few more examples involving addition and subtraction of time. Example 15: Mr. Roy spends 1 hour 30 minutes in his garden every day. Mr. Pavan does the same for 50 minutes. How much more time does Mr. Roy spend than Mr. Pavan in his garden? Give your answer in seconds. Solution: Time spent by Mr. Roy in his garden = 1 hour 30 minutes Time spent by Mr. Pavan in his garden = 50 minutes To find the required, subtract 50 minutes from 1 hour 30 minutes. Time 9 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 53 2/17/2018 4:21:59 PM

Now, we need to find the answer in seconds. Hours Minutes 1 minute = 60 seconds 1 30 40 minutes = 40 × 60 seconds = 2400 seconds Therefore, Mr. Roy spends 2400 seconds more in his –0 50 0 40 garden. Example 16: Sohan started preparing for his exam from 16th July. The exams were scheduled to begin 25 days later. On which date were the exams scheduled to begin? Solution: Start date of exam preparation = 16th July Preparation day for exams includes 16th July. So, subtract 15 days from 31 days of July. Number of days of preparation in July = 31 – 15 = 16 Days of preparation left in the month of August = 25 – 16 = 9 Therefore, the date when the exam begins is 10th August. Drill Time Concept 7.1: Convert Time 1) Convert into days. a) 4 years 5 weeks b) 3 years 10 days c) 2 years 15 days d) 4 years 20 days e) 1 year 3 weeks 2) Convert the given time to hours. a) 240 minutes b) 360 minutes and 3600 seconds c) 180 minutes d) 300 minutes and 3600 seconds 3) Word problems a) A bus takes 1 hour and 25 minutes to reach a bus stand. It stops 5 times for 45 seconds at each stop to pick up passengers. For how many minutes did the bus stop? b) Amit reached his house from school in 110 minutes. Find the time taken by Amit to reach his house in hours and minutes. 10 2/17/2018 4:21:59 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 54

Concept 7.2: Add and Subtract Time 4) Add: a) 2 hours 40 minutes and 1 hour 33 minutes b) 3 hours 26 minutes and 2 hours 22 minutes c) 4 hours 31 minutes and 1 hour 28 minutes 5) Subtract: a) 1 hour 30 minutes from 3 hours 75 minutes b) 2 hours 20 minutes from 5 hours 60 minutes c) 1 hour 40 minutes from 6 hours 49 minutes 6) Word problems a) S ohail takes 2 hours 30 minutes to complete his homework and Aditya does the same homework in 145 minutes. Who takes less time to complete homework? b) Preeti spends 70 minutes on the playground and Andy spends 1 hour 900 seconds on the playground. How much more time Andy does spend than Preeti on the playground? Time 11 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 55 2/17/2018 4:21:59 PM

Chapter Money 8 Let Us Learn About • unitary method in money. • problems based on unitary method in money. • chart 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 12 2/17/2018 4:21:59 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 56

& 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. Money 13 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 57 2/17/2018 4:21:59 PM

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. 14 2/17/2018 4:21:59 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 58

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. Solution: What is the cost of 10 kg of guavas? 32.50 Cost of 17 kg of guavas = ` 552.50 )17 552.50 - 51 Cost of 1 kg of guavas = ` 552.50 ÷ 17 42 = ` 32.50 - 34 Cost of 10 kg of guavas = ` 32.50 × 10 85 = ` 325 - 85 Therefore, 10 kg of guavas cost ` 325. 00 - 00 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 Money 15 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 59 2/17/2018 4:21:59 PM

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? 16 2/17/2018 4:21:59 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 60

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? Money 17 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 61 2/17/2018 4:22:00 PM

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. • a dding 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 = . 18 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 62

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 Fractions - I 19 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 63 2/17/2018 4:22:00 PM

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? 20 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 64

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 = [ ] Therefore, the missing number in the brackets is 3. Fractions - I 21 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 65 2/17/2018 4:22:00 PM

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? 22 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 66

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 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 Fractions - I 23 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 67 2/17/2018 4:22:00 PM

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 be26co÷m13e=s 2 . 65 65 ÷13 5 Application Let us solve a few real-life examples that involve reducing fractions to their lowest terms. Example 9: Jai ate 4 of a watermelon and Vijay ate 16 of another watermelon of the 16 32 same size. Did they eat the same quantity of watermelon? If not, who ate Solution: more? Fraction of watermelon Jai ate = 4 16 Fraction of watermelon Vijay ate = 16 32 To compare the fractions, we must reduce them to their lowest terms so that we get like fractions. 4 4÷4 1 [H.C.F. of 4 and 16 is 4.] 16 = 16 ÷ 4 = 4 16 = 16 ÷ 8 = 2 [Using division method] 32 32 ÷ 8 4 Clearly, 1 < 2. So, 1 < 2 . 44 Therefore, Vijay ate more. 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 24 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 68 2/17/2018 4:22:00 PM

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 5.) Example 11: Clearly, 4 is greater than 3 . Solution: 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.) 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.) Fractions - I 25 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 69 2/17/2018 4:22:00 PM

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? 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 26 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 70

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 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 . Step 2: Compare their 11 11× 7 77 numerators and find which is greater or lesser. 33 > 28 So, 33 > 28 . 77 77 Thus, 3 > 4 . 5 11 Fractions - I 27 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 71 2/17/2018 4:22:00 PM

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 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? 28 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 72

Solution: Fraction of the apple Esha ate in the morning = 1 4 Fraction of the apple she ate in the evening = 2 Step 1: 3 To find when she ate a larger part we must compare the two fractions. 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? 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 29 9 7 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 73 Fractions - I

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: 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 30 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 74

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? 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 c) 22 + 7 15 10 13 39 100 10 Solution: a) 3 + 1 = 6 + 3 15 10 30 30 Fractions - I 31 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 75 2/17/2018 4:22:00 PM

[L.C.M. of 15 and 10 is 30.] 6+3 9 3 = 30 = 30 = 10 [H.C.F. of 9 and 30 is 3.] 7 2 21 2 21+ 2 23 b) 13 + 39 = 39 + 39 = 39 = 39 [L.C.M. of 13 and 39 is 39.] c) 22 + 7 = 22 + 70 22 + 70 92 23 == = 100 10 100 100 100 100 25 [ The L.C.M. of 100 and 10 is 100 and the H.C.F. of 92 and 100 is 4.] Example 22: Solve: a) 8 – 4 b) 17 – 5 c) 14 – 17 9 11 30 24 25 50 Solution: a) 8 – 4 88 36 [L.C.M. of 9 and 11 is 99.] = – 9 11 99 99 88 - 36 52 == 99 99 b) 17 – 5 = 68 – 25 [L.C.M. of 24 and 30 is 120.] [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 9 32 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 76 2/17/2018 4:22:00 PM

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 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 n 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 Fractions - I 33 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 77 2/17/2018 4:22:00 PM

Part of the population of the town that was women = 1 4 Part of the population of the town that was children = 1 6 Part of the population that was men and women = 5 + 1 = 5 + 2 = 7 8 4 8 8 8 Part of the population that was men and children = 5 + 1 = 15 + 4 = 19 8 6 24 24 24 Part of the population that was women and children = 1 + 1 = 3 + 2 = 5 4 6 12 12 12 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 34 2/17/2018 4:22:00 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 78

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 Concept 9.4: Add and Subtract Unlike Fractions 5) Solve: a) 3 + 5 b) 4 + 3 c) 4 + 1 d) 19 + 5 e) 2 + 6 4 13 14 12 15 10 100 10 16 30 6) Solve: a) 4 – 3 b) 14 – 3 c) 13 – 14 d) 3 – 4 e) 15 – 16 9 11 30 24 30 60 15 30 20 40 7) Word problems a) U sha played the keyboard for 7 of an hour and did her homework for 5 of 30 12 an hour. Did she spend the same amount of time for both the activities? b) William ate 3 of a chocolate bar and Wasim ate 1 of the chocolate. Did they 16 4 eat the same part of the chocolate? Who ate less? c) M ani and Roja were painting a rectangular cardboard each. Mani painted 15 of the cardboard in an hour and Roja painted 18 of the cardboard in the 25 30 same time. Who is more efficient? d) Sudheer saves ` 360 per month from his salary of ` 3600. Hari saves ` 200 per month from his salary of ` 2400. What fraction of their salary did each of them save? e) P avani used 450 cm of satin ribbon from a bundle of satin ribbon of length 3000 cm. What fraction of the satin ribbon is used? Fractions - I 35 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 79 2/17/2018 4:22:01 PM

Chapter Fractions - II 10 Let Us Learn About • the terms ‘mixed’, ‘proper’ and ‘improper’ fractions. • adding and subtracting mixed fractions. • m ultiplying and dividing fractions by fractions. • finding the reciprocals of fractions. Concept 10.1: Add and Subtract Mixed Fractions Think Pooja has learnt addition and subtraction of unlike fractions. She has also learnt the conversion of improper fractions to mixed fractions and vice-versa. She was curious to know if she could add and subtract improper fractions and mixed fractions too. How do you think Pooja can add or subtract mixed fractions? Recall We have learnt about the types of fractions. Let us recall them here. 1) A fraction whose numerator is greater than the denominator is called an improper fraction. 2) A fraction whose denominator is greater than the numerator is called a proper fraction. 3) The combination of a whole number and a fraction is called a mixed fraction. 36 2/17/2018 4:22:01 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 80

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 d) 8 10 9 – 23 5 – 4 3 – 11 e) 2 15 f) 6 5 & Remembering and Understanding A mixed fraction can be converted into an improper fraction by multiplying the whole number part by the fraction’s denominator and then adding the product to the numerator. Then we write the result on top of the denominator. The addition and subtraction of mixed fractions are similar to that of unlike fractions. Let us understand the same through the following examples. Example 1: 3 + 3 2 Add: 2 5 7 Solved Solve this Steps 23 + 32 12 1 + 15 1 57 43 Step 1: Convert all the mixed 2 3 = 2 × 5 + 3 = 13 ; fractions into improper fractions. 55 5 3 2 = 3 ×7 + 2 = 23 77 7 Step 2: Find the L.C.M. and add the 2 3 + 3 2 = 13 + 23 improper fractions. 5 75 7 [L.C.M. of 5 and 7 is 35.] = 7 ×13 + 5 × 23 35 = 91+115 = 206 35 35 Fractions - II 37 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 81 2/17/2018 4:22:01 PM

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. 38 2/17/2018 4:22:01 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 82

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. Therefore, 3 2 – 2 3 = 24 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 20 123 3 [L.C.M. of 1 and 20 is 20.] = 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 39 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 83 2/17/2018 4:22:01 PM

Solution: Total distance to be covered by Veena = 65 3 km 5 Distance covered by her in the first 2 hours = 34 2 km 3 Distance covered by her in the next hour = 16 1 km 2 14 Total distance she travelled = 34 3 km + 16 4 km 104 km + 65 km = 416 +195 km = 611 km = 50 11 km 3 4 12 12 12 Distance yet to be covered = 65 3 km – 50 11 km 5 12 = 328 km – 611 km 5 12 = 3936 − 3055 km [L.C.M. of 5 and 12 is 60.] 60 = 881 km = 14 41 km 60 60 Therefore, the distance Veena has to cover is 14 41 km. 60 Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of addition and subtraction of mixed fractions. Example 5: By how much is 41 1 greater 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 40 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 84 2/17/2018 4:22:01 PM

Concept 10.2: Multiply Fractions Think Pooja and each of her 15 friends had a bar of chocolate. Each of them ate 5 of the 12 chocolate. How much of the chocolate did they eat in all? How do you think Pooja can find this? Recall Recall that when we find the fraction of a number, we multiply the number by the fraction. After multiplication, we simplify the product to its lowest terms. Similarly, we can multiply a fraction by another fraction too. • 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 41 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 85 2/17/2018 4:22:01 PM

Example 7: Find the following: a) 23 of 90 b) 15 of 128 45 32 Solution: a) 23 of 90 = 23 × 90 = 23 × 90 = 2070 = 46 45 45 45 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 Therefore, product of the fractions = Product of numerators 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. 42 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 86 2/17/2018 4:22:01 PM

Example 8: Solve: 23 × 15 Solution: 45 46 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) 1 2 × 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 Application Let us see some real-life examples where we can use multiplication of fractions. Fractions - II 43 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 87 2/17/2018 4:22:01 PM

Example 10: Tina had 1 kg of flour. She used 1 of it for a recipe. How many grams of 6 10 flour did she use? Solution: Quantity of flour Tina had = 1 kg 6 Part of the flour used by her for a recipe = 1 10 Quantity of flour used by Tina = 1 of 1 kg = 1 × 1 kg = 1×1 kg 10 6 10 6 10 ×6 = 1 kg = 1 × 1000 g = 16.67 g 60 60 Example 11: Mohan saves one-fourth of his monthly salary of ` 5500. Arjun saves two-fifths of his monthly salary of ` 4500. Who saves more and by how much? Solution: Fraction of salary saved by Mohan = 1 of ` 5500. 4 1 = 4/ × 1375 = ` 1375 5500 1 2 × ` 4500 = 2 × ` 900 = ` 1800 Fraction of salary saved by Arjun = 5 Since ` 1800 is more than ` 1375, Arjun saves more. The difference in savings = ` 1800 – ` 1375 = ` 425 Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of multiplication of fractions. Example 12: Swetha cut a big watermelon into two equal parts. Jaya cut a part into 16 equal pieces and ate 4 of them. Vijay cut a part into 32 equal pieces and ate 16 of them. Who ate more watermelon? Solution: Each equal part of the watermelon = 1 2 Fraction of watermelon Jaya ate = 4 of 1 = 4 × 1 = 1 × 1 = 1 16 2 16 2 4 2 8 Fraction of watermelon Vijay ate = 16 of 1 = 16 × 1 = 1 × 1 = 1 32 2 32 2 2 2 4 Comparing the fractions, we see that 1 < 1 or 1 > 1 . 8 4 48 Therefore, Vijay ate more. 44 2/17/2018 4:22:01 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 88

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 & 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. Fractions - II 45 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 89 2/17/2018 4:22:01 PM

To find the reciprocal of a fraction, we interchange its numerator and denominator. • The reciprocal of a number is a fraction. For example, the reciprocal of 20 is 1 . 20 • The reciprocal of a unit fraction is a number, For example, the reciprocal of 1 is 7. 7 • The reciprocal of a proper fraction is an improper fraction. It can be left as it is or converted into a mixed fraction, For example, the reciprocal of 3 is 7 or 2 1 . 7 3 3 • The reciprocal of an improper fraction is a proper fraction, For example, the reciprocal of 9 is 5 . 59 • The reciprocal of a mixed fraction is a proper fraction, For example, the reciprocal of 2 3 is 8 . 8 19 Note: 1) The reciprocal of 1 is 1. 2) The reciprocal of 0 does not exist as division by zero is not defined. 3) N umbers such as 4, 6, 9 and so on are converted into improper fractions by writing them as 4 , 6 , 9 before finding their reciprocals. 111 4) F ractions 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. 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. 46 2/17/2018 4:22:01 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 90

The multiplicative inverses of the given fractions are: a) 1 b) 9 c) no multiplicative inverse 5 68 d) 1 e) 3 100 Note: 0 has no reciprocal or multiplicative inverse because we cannot multiply any number by it to get 1. Zero multiplied by any number is zero. Therefore, 0 is the only number that does not have a multiplicative inverse. Application Divide a number by a fraction The division of a number by another means to find the number of divisors present in the 1 dividend. For example, 8 ÷ 4 means to find the number of fours in 8. Similarly, 10 ÷ means to 5 find the number of one-fifths in 10. Let us understand the division by fractions through some examples. 1 b) 75 ÷ 3 Example 16: Divide: a) 15 ÷ 5 3 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. Step 4: 15 ÷ 1 = 15 × 3 = 45 3 11 1 Step 1: Reduce the product to its lowest terms. 45 = 45 Step 2: 1 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. Fractions - II 47 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 91 2/17/2018 4:22:01 PM

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 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: 3 ÷ 67 Solution: To divide the given, follow these steps: Steps Solved Solve this 1 3 Step 1: Write the number as a ÷ 54 fraction. ÷ 67 5 3 67 67 = 1 Step 2: Find the reciprocal of the 67 1 divisor. The reciprocal of 1 is 67 . Step 3: Multiply the dividend by 1 11 1 the reciprocal of the divisor. ÷ 67 = × = 3 3 67 3 × 67 Step 4: Reduce the product to its lowest terms. 11 = 3 × 67 201 11 Therefore, 3 ÷ 67 = 201 . Divide a fraction by another fraction Division of a fraction by another fraction is similar to the division of a number by a fraction. Let us understand this through some examples. Example 18: Solve: 1 ÷ 1 3 21 48 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 92 2/17/2018 4:22:01 PM

Solution: To solve the given sums, follow these steps: Solved Solve this 3 210 Steps 1 ÷ 1 25 ÷ 75 3 21 Step 1: Find the reciprocal of 1 21 the divisor. The reciprocal of 21 is 1 . Step 2: Multiply the dividend by 1 1 1 21 3 ÷ 21 = 3 × 1 the reciprocal of the divisor. Step 3: Reduce the product 1 7 into its lowest terms. × 21 = 7 31 11 Therefore, 3 ÷ 21 = 7. Higher Order Thinking Skills (H.O.T.S.) Let us see some real-life examples using division of fractions. Example 19: Sakshi had 7 apples. She cut them into quarters. How many pieces did she get? Solution: To find the number of pieces that Sakshi will get, we must find the number of quarters in 7. That is, we must divide the total number of apples by the size of each piece of apple. Number of quarter pieces = 7 ÷ 1 = 7 × reciprocal of 1 = 7 × 4 = 28 44 Therefore, Sakshi got 28 pieces of apple. Example 20: Nani had 3 of a kilogram of sugar. He poured it equally into 4 bowls. How 5 many grams of sugar is in each bowl? Solution: Total quantity of sugar = 3 kg 5 Number of bowls = 4 Quantity of sugar in each bowl = 3 kg ÷ 4 = 3 kg × reciprocal of 4 55 = 3 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. Fractions - II 49 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 93 2/17/2018 4:22:01 PM

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 16 8 16 × 25 = 2 = 25 × reciprocal of 25 = 25 1 81 Therefore, 2 glasses are filled. 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 + 12 2 75 85 75 33 2) Solve: c) 7 2 – 4 1 d) 12 3 – 11 2 84 89 a) 4 1 – 2 1 b) 10 1 – 5 3 37 27 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) by a) 16 by 1 b) 14 by 2 c) 1 by 3 7 49 4 7 42 50 2/17/2018 4:22:01 PM NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 94

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EVS-I (SCIENCE) TERM 2 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 97 2/17/2018 4:22:02 PM

Contents 5Class 6 Plants and Environment����������������������������������������������������������������������������������������������� 1 7 Food for Animals����������������������������������������������������������������������������������������������������������� 7 8 Food Production��������������������������������������������������������������������������������������������������������� 11 9 Forests as Shelter�������������������������������������������������������������������������������������������������������� 15 Inside the Lab – B�������������������������������������������������������������������������������������������������������������� 20 Activity B1: Seed Germination Activity B2: Food Web NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 98 2/17/2018 4:22:02 PM

Lesson Plants and Environment 6 Let Us Learn About r habitats and habits of plants. u adaptations of plants. a protection of plant environments. h sacred groves. Think Seema planted a lotus plant in her garden and watered it. But it drooped down and dried up within a few days. What can be the reason for this? Remembering Plants can be found almost all over the Earth. They grow on land as well as in water. These places are called their habitats. Let us learn more about the habitats of plants. TERRESTRIAL PLANTS The plants that grow on land are known as terrestrial plants. They grow in different areas like mountains, plains, deserts, swampy areas, coastal areas and so on. NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 99 1 2/17/2018 4:22:02 PM

plants on mountains plants in plains plants in deserts plants in swampy areas plants in coastal areas AQUATIC PLANTS Plants that grow in water are called aquatic plants. They are of three kinds – floating, fixed and underwater plants. Floating plants: These plants are found floating freely on water. They are water lettuce not attached to the bottom of the water body. For example, water lettuce, lotus water hyacinth and so on. Fixed plants: These plants have roots that are fixed to the soil at the bottom of the water body. Their leaves and flowers float on the surface of the water to get oxygen from the air and sunlight. They have broad and wax coated leaves. Waxy coating prevents the leaves from rotting due to water. For example, lotus, water lily and so on. Underwater plants: These plants grow completely under the water. They take in carbon dioxide from the water. For example, sea grass, tape grass and so on. Now, let us learn about some habits of plants. Plants also differ according to their sea grass food habits. Plants that make food on their own: Most green plants make their own food. They absorb water and nutrients from the soil with the help of roots. Leaves produce food by combining carbon dioxide and water using energy from sunlight. Plants which depend on other plants: Some plants like cuscuta and sandalwood tree absorb water and nutrients from the roots of other plants. Such plants that depend on other plants for their food are called parasitic plants. cuscuta plant 2 NR_BGM_181910055_Maple G5_Textbook Integrated_Term2_text.pdf 100 2/17/2018 4:22:02 PM