Solution: a) 2 3 b) 2 2 c) 11 1 `14105 13 2 33 2 ×7 ` 312 . 97 ` 506 . 75 `98735 × 34 × 125 1 11 111 1251 . 88 2533 . 75 + 9389 . 10 + 10135 . 00 ` 10640 . 98 + 50675 . 00 ` 63343 . 75 Dividing money To divide an amount by a number, we follow these steps. Step 1: Write the amount as the dividend and the number as the divisor. Step 2: Carry out the division just as we divide any two numbers. Step 3: Place the decimal point in the quotient, immediately after dividing the rupees, that is, digits before the decimal point in the dividend. Example 15: Divide: a) ` 23415 by 7 b) ` 481.65 by 13 c) ` 543.40 by 110 Solution: )a) 3345 ) )b) 37.05 c) 4.94 7 23415 13 481.65 110 543.40 − 21↓ − 39↓ − 440↓ 24 91 1034 − 21 − 91 − 990 31 06 440 − 28 − 00 − 440 35 65 000 − 35 − 65 00 00 Money 47
Application Let us solve a few real-life examples involving multiplication and division of money. Example 16: A textbook of class 4 costs ` 75.20. What is the ` p cost of 35 such textbooks? Solution: 1 20 Cost of one textbook = ` 75. 20 21 35 75 Cost of 35 such textbooks = ` 75. 20 × 35 . 00 × . 00 Therefore, the cost of 35 textbooks is ` 2632. 1 1 00 76 3 56 + 22 32 `2 6 Example 17: 19 cakes cost ` 332.50. What is the cost of 1 cake? 17.50 Solution: Cost of 19 cakes = ` 332.50 Cost of 1 cake = ` 332.50 ÷ 19 )19 332.50 Therefore, the cost of 1 cake is ` 17. 50. − 19 ↓ 142 − 133 95 − 95 00 Higher Order Thinking Skills (H.O.T.S.) Train My Brain Let us see a few more examples involving multiplication and division of money. Example 18: Multiply the sum of ` 2682 and ` 2296 by 10. Solution: The sum of ` 2682 and ` 2296 is ` 2682 + ` 2296. `` 1 2682 4978 +2296 × 10 `4 9 7 8 0 4978 Therefore, the sum multiplied by 10 = 4978 × 10 = ` 49780. 48
Example 19: A bag has one bundle of ` 50 notes and one bundle of ` 20 notes. It also has two bundles of ` 10 notes and one bundle of ` 5 notes. What is the total amount of money in the bag? [Note: Each bundle consists of 100 notes.] Solution: Amount in the bundle of ` 50 = 100 × ` 50 (1 bundle) = ` 5000 Amount in the bundle of ` 20 = ` 20 × 100 (1 bundle) = ` 2000 Amount in two bundles of ` 10 = ` 10 × 200 (2 bundles) = ` 2000 Amount in the bundle of ` 5 = ` 5 × 100 (1 bundle) = ` 500 Total money = ` 5000 + ` 2000 + ` 2000 + ` 500 = ` 9500 Therefore, the total amount of money in the bag is ` 9500. Drill Time Concept 11.1: Conversion of Rupees and Paise 1) Convert the following to paise. d) ` 537.58 e) ` 724.80 a) ` 632.18 b) ` 952.74 c) ` 231.48 2) Convert paise to rupees. a) 52865 b) 64287 c) 13495 d) 34567 e) 78654 3) Word problems a) Rehmat has ` 892.64. How many paise does he have in all? b) Andrews has 56700 paise. How much money does he have in all? Express your answer in rupees. Concept 11.2: Add and Subtract Money with Conversion 4) Add: b) ` 3467.45 + ` 2356. 50 c) 25382 p + 65237 p a) ` 875.62 + ` 964.98 e) ` 279.50 + ` 642.90 d) ` 456.23 + ` 123.75 5) Subtract: b) 85732 p – 23784 p c) ` 578.14 – ` 345.89 a) ` 132.75 – ` 112.90 e) ` 784.50 – ` 234.25 d) ` 456.72 – ` 234.34 Money 49
6) Word problems a) Rosy has ` 451.20 and Chetan has ` 495.35 in their piggy banks. Who has more amount and by how much? b) Shane spent ` 213.60, ` 105.30 and ` 305.45 in three months. How much did he spend in all? Concept 11.3: Multiply and Divide Money 7) Multiply: b) 27510 p × 2 c) ` 315.50 × 10 a) ` 152.45 × 5 e) ` 115.50 × 35 d) ` 113.50 × 15 8) Divide: b) 22347 p ÷ 9 c) ` 111.44 ÷ 7 a) ` 126.12 ÷ 3 e) ` 824.40 ÷ 8 d) ` 121.77 ÷ 7 9) Word problems a) A packet of chips costs ` 24.40. How much will 5 such packets cost? b) A football costs ` 159.99. What is the cost of 26 such footballs? 50
Chapter Measurements 12 Let Us Learn About • relation between units of length, weight and capacity. • converting smaller units to larger units. • multiplying and dividing length, weight and capacity. Concept 12.1: Multiply and Divide Lengths, Weights and Capacities Think Jasleen had some guests visiting her place. Jasleen’s mother asked her to pour juice from three bottles, each of 1.5 litres, into 15 glasses. What was the total quantity of juice and how much juice was poured in each glass? Recall Let us revise the basic concepts of measurements, their units and the different operations involving measurements. Length: kilometre, centimetre, millimetre Weight: kilogram, gram, milligram Capacity: litre, millilitre Solve the following problems based on addition and subtraction of lengths, weights and capacities. 51
a) 560 m 65 cm – 230 m 55 cm = ___________ b) 250 g + 2 kg 500 g = ___________ c) 240 m 22 cm – 220 m 20 cm = ___________ d) 5 ℓ 250 mℓ + 4 ℓ 250 mℓ = ___________ e) 745 km 45 m – 434 km 15 m = ___________ & Remembering and Understanding To convert measures from a larger unit to a smaller unit, we multiply. To convert measures from a smaller unit to a larger unit, we divide. Let us understand the relation between the different units of length, weight and capacity in detail. Relation between units of length, weight and capacity Larger unit – Smaller unit Smaller unit – Larger unit Length 1m= 1 km 1000 1 km = 1000 m 1 m = 100 cm 1 1 cm = 100 m 1 cm = 10 mm 1 mm = 1 cm 10 1 g = 1000 mg 1 kg = 1000 g Weight 1 mg = 1 g 1 litre = 1000 mℓ Capacity 1000 1g= 1 kg 1000 1 mℓ = 1 ℓ 1000 1 kilolitre = 1000 litres 1ℓ= 1 kℓ 1000 52
Conversion of smaller units to larger units Let us understand conversions through a few examples. Example 1: Convert the following: a) 5000 m to km b) 8000 g to kg c) 2000 mℓ to ℓ Solution: Solved Solve these a) Conversion of m into km 9000 m = ________________ km 5000 m = _____________ km 4000 g = ______________ kg 1000 m = 1 km So, 5000 m = 5000 ÷ 1000 m 3000 mℓ = ______________ ℓ = 5 km 5000 m = 5 km b) Conversion of g into kg 8000 g = _____________ kg 1000 g = 1 kg So, 8000 g = 8000 ÷ 1000 g = 8 kg c) Conversion of mℓ into ℓ 2000 mℓ = _____________ ℓ 1000 mℓ = 1 ℓ So, 2000 mℓ = 2000 ÷ 1000 mℓ =2ℓ Multiply and divide length, weight and capacity Interestingly, multiplication and division of lengths, weights and capacities are similar to that of usual numbers. Let us see a few examples. Example 2: Solve: b) 18 km 361 m × 19 c) 7 ℓ 260 mℓ × 37 a) 65 kg 345 g × 28 Measurements 53
Solution: a) 65 kg 345 g × 28 b) 18 km 361 m × 19 c) 7 260 m× 37 km m kg g ℓ mℓ 1 1 34 1 42 345 65 28 73 5 14 × 760 18 361 7 260 1 900 522 660 × 19 × 37 +1 3 0 6 1829 1 1 165 249 50 820 + 183 610 + 217 800 348 859 268 620 Example 3: Solve: a) 15 kg 183 g ÷ 21 b) 3 km 84 m ÷ 12 c) 5 ℓ 882 mℓ ÷ 17 a) 15 kg 183 g ÷ 21 b) 3 km 84 m ÷ 12 c) 5 ℓ 882 mℓ ÷ 17 15 kg 183 g 3 km 84 m 5 ℓ 882 mℓ = 15 × 1000 g + 183 g = 3 × 1000 m + 84 m = 5 × 1000 mℓ + 882 mℓ = 15183 g = 3084 m = 5882 mℓ 346 723 257 )17 5882 )21 15183 )12 3084 − 51 − 14 7 − 24 048 068 078 − 068 − 042 − 060 0063 0084 0102 − 0102 − 0063 − 0084 0000 0000 0 15 kg183 g ÷ 21 = 723 g 3 km 84 m ÷ 12 = 257 m 5 ℓ 882 mℓ ÷ 17 = 346 mℓ 54
Application Let us solve a few examples based on multiplication and division of length, weight and capacity. Example 4: The distance between two post offices A and B is 58 km 360 m. What is the total distance travelled in four round trips between A and B? Solution: The distance between two post offices A and B is 58 km 360 m. Four round trips = 4 times from A to B and 4 times from B to A = 8 times the distance between A and B Therefore, the total distance travelled in four round trips = 58 km 360 m × 8 = 466 km 880 m Example 5: Mrs. Rani has 2 kg of coffee powder. She wants to put it into smaller packets of 25 g each. How many packets will she need? Solution: Weight of coffee powder Mrs. Rani has = 2 kg 1 kg = 1000 g 2 kg = 2 × 1000 g = 2000 g Weight of one small packet = 25 g Therefore, the number of packets she needs = 2000 g ÷ 25 g = 80 Example 6: Rahul has a can of 6112 mℓ juice. If he pours it equally in 16 glasses, what is the quantity of juice in each glass? Solution: Quantity of juice in full can = 6112 mℓ Number of glasses into which the juice is poured = 16 Quantity of juice in each glass = 6112 mℓ ÷ 16 = 382 mℓ Therefore, each glass contains 382 ml of juice. Measurements 55
Higher Order Thinking Skills (H.O.T.S.) Sometimes, we have to use more than one mathematical operation to measure things. Consider these examples. Example 7: 185 kg sugar costing ` 444 is packed in paper bags. Each bag can hold 5 kg of sugar. Find the number of bags needed to pack all the sugar. Also, find the cost of each bag. Solution: Weight of sugar = 185 kg Weight of sugar in the paper bag = 5 kg Number of paper bags needed = 185 kg ÷ 5 kg = 37 Therefore, 37 paper bags of 5 kg sugar each can be made. Cost of 37 bags of sugar = ` 444 Cost of each bag = ` 444 ÷ 37 = ` 12 Therefore, 185 kg sugar can be packed into 37 bags costing ` 12 each. Example 8: A container can hold 13 ℓ 625 mℓ of milk. What is the capacity of 15 such containers? Give your answer in mℓ. Solution: Capacity of one container = 13 ℓ 625 mℓ Capacity of 15 such containers = 13 ℓ 625 mℓ × 15 = 204 ℓ 375 mℓ 1 litre = 1000 mℓ 204 ℓ = 204 × 1000 mℓ = 204000 mℓ 204 ℓ 375 mℓ = 204000 mℓ+ 375 mℓ = 204375 mℓ Therefore, the capacity of 15 cans is 204375 mℓ. Example 9: The distance between two places is 4520 km. Ratan travelled a fourth of the distance by bus paying ` 12 per km. As the bus failed, he hired a car and travelled three-fourths of the distance by paying ` 20 per km. What amount did he spend on travelling? Solution: Total distance = 4520 km 1 of the distance = 1 × 4520 km = 1130 km 4 4 Distance travelled by bus = 1130 km 56
Ratan travelled 1130 km by bus. Cost of ticket per km = ` 12 Cost of ticket for 1130 km = 1130 × ` 12 = ` 13560 Fraction of distance travelled by car = 3 4 Actual distance travelled by car = 3 × 4520 km 4 = 3 × 1130 km = 3390 km Cost of travelling by car per km = ` 20 Cost of travelling 3390 km = 3390 × ` 20 = ` 67800 Total amount spent by Ratan on travelling = ` 13560 + ` 67800 = ` 81360 Drill Time Concept 12.1: Multiply and Divide Lengths, Weights and Capacities 1) Convert: a) 2000 cm to m b) 5000 g to kg c) 5000 m to km d) 8000 mℓ to ℓ 2) Multiply: a) 85 kg 145 g ×10 b) 5 ℓ 225 mℓ × 65 c) 11 m 50 cm × 25 d) 5 km 150 cm × 12 3) Divide: a) 34 kg 450 g by 6 b) 50 ℓ 225 mℓ by 5 c) 17 m 85 cm by 9 d) 42 kg 420 g by 7 Measurements 57
Chapter Data Handling 13 Let Us Learn About • reading and interpreting bar graphs. • d rawing bar graphs based on the given data. Concept 13.1: Bar Graphs Think Jasleen attended a fruit festival conducted for a week in her school. She was asked to give a report on the sale of different fruits per day in the form of a graph. Till then Jasleen only knew how to represent the data as a pictograph. She wanted to find an easier and simpler way of representation. How do you think Jasleen would have given the report? Recall Recall these points: • The information collected for a specific purpose is called data. • The information given as numbers is called numerical data. • The information shown in the form of pictures is called a pictograph. 58
We have already learnt about pictographs. Let us recall them through the following. Let us recall the pictographs through the following example. The favourite sports of Class 4 students are given. Read the pictograph and answer the questions. Key: 1 = 6 students Favourite Sports of Class 4 Students Volleyball Cricket Basketball Kabaddi Football a) The most favourite sports of Class 4 students is _____________. b) The least favourite sports of Class 4 students is _____________. c) The number of students who like to play basketball is___________. d) The number of students who like to play football is _____________. e) The number of students who like to play kabaddi is _____________. Data Handling 59
& Remembering and Understanding While drawing pictographs, we choose a relevant picture to represent the given data. If the data is large, it is tedious and time consuming to draw a pictograph. An easier way of representing data is the bar graph. It uses rectangular bars of the same width. These bars can be a drawn either horizontally or vertically. Bar graphs are drawn on a graph paper. A suitable title is given for the bar graph. Let us understand how to read and interpret bar graphs. Example 1: The marks scored by Kamala in a monthly test are represented using a bar graph as given. Understand the graph and answer the questions that follow. Scale: X-axis: 1 cm = 1 subject; Y-axis: 1 cm = 5 marks Kamala’s Performance in a Monthly Test Marks Scored X - axis English Maths Science Social Music Hindi Studies Subjects 60
a) What is the title of the graph? b) In which subject did Kamala perform the best? c) In which subject does Kamala need to improve? d) What are Kamala’s total marks? Solution: a) The title of the graph is “Kamala’s Performance in a Monthly Test”. b) T he height of the bar representing Maths is maximum. It means that, Kamala performed the best in Maths. c) The height of the bar representing Social Studies is the minimum. Example 2: So, Kamala needs to improve in Social Studies. d) Kamala’s total marks are 35 + 47 + 42 + 28 + 32 + 40 = 224 Information about a primary school is represented in the form of a bar graph as shown. Observe the graph carefully and answer the questions that follow. Scale: X-axis: 1 cm = 1 class; Y-axis: 1 cm = 5 students Strength of Primary School Class strength Class Data Handling 61
a) What is the total strength of all the 5 classes? b) Which class has the least strength? c) Which class has the greatest strength? d) What is the title of the graph? Solution: a) Total strength is 42 + 36 + 38 + 43 + 45 = 204 b) Class 2 c) Class 5 d) Strength of a Primary School Application We have learnt how to read and interpret bar graphs. Now, let us learn to draw a bar graph. Steps to draw a bar graph: Step 1: Draw one horizontal line and another vertical line, called the axes. They meet at a point called the origin. Step 2: Take a suitable scale such as 1 cm = 5 units. Step 3: On the X-axis, show the items of the data and on the Y-axis show their values. Step 4: Draw bars of equal width on the X-axis. The heights of the rectangles represent the values of the data which are given on the Y- axis. Step 5: Give a relevant title to the bar graph. Let us understand this through an example. Example 3: The following pictograph shows the number of scooters manufactured by a factory in a week. Complete the pictograph. Then draw a bar graph for the same data. Key: 1 = 5 scooters 62
Weekday Scooters manufactured in a week Number of Monday scooters Tuesday Wednesday Thursday Friday Saturday Solution: Total Step 1: Let us follow these steps to draw a bar graph. Step 2: Count the number of pictures in the pictograph. Complete the table by writing the product of the number of pictures and the number of scooters per key. Take a graph paper and draw the X and Y axes meeting each other at one corner as shown. Data Handling 63
Step 3: Choose a suitable scale. Since the maximum number of scooters is 30 and the minimum is 10, we can take the scale as 1 cm = 5 scooters. Mark weekdays on the X-axis as 1 cm = 1 weekday. Mark the number of scooters manufactured on the Y-axis from 0 to 35. Number of scooters manufactured Mon Tues Wed Thurs Fri Sat Weekdays 64
Step 4: On the X-axis, mark 30, 15, 20, 25, 20 and 10 against the Y-axis as shown. We can plot these points two points apart. Number of scooters manufactured Mon Tue Wed Thurs Fri Sat Weekdays Step 5: Draw vertical rectangular bars from these points for each weekday on the X-axis. Give a suitable title to the graph. Scale: X-axis: 1 cm = 1 day; Y-axis: 1 cm = 5 scooters Weekly Manufacturing of Scooters Number of scooters manufactured Monday Tuesday Wednesday Thursday Friday Saturday 65 Weekdays Data Handling
We can draw the same graph using horizontal bars by interchanging the values on X and Y axes. Weekly Manufacturing of Scooters Weekdays Number of scooters manufactured Example 4: The number of roses sold during a month in Roopa’s shop is given in the table Week Number of roses sold 1st week 148 2nd week 165 3rd week 130 4th week 172 Represent the data in a bar graph. 66
Solution: Scale: X-axis: 1 cm = 1 week;Y-axis: 1 cm = 20 roses Roses sold Weeks Higher Order Thinking Skills (H.O.T.S.) Train My Brain Consider a few real-life examples where we represent data using a bar graph. Example 5: In 2010, the heights of Ramu, Somu, Radha and Swetha were noted as 130 cm, 125 cm, 115 cm and 120 cm respectively. After two years, their heights were again noted as 140 cm, 132 cm, 124 cm and 128 cm respectively. Draw a bar graph to represent the data and answer the questions that follow. a) Who was the tallest among the friends in 2010? b) Who was the shortest among them during 2012? c) How much taller was Ramu than Somu in 2010? d) Whose height has increased the maximum in 2 years? Data Handling 67
e) A rrange the children’s heights in 2010 in ascending order and their heights in Solution: 2012 in descending order. Name Height in 2010 Height in 2012 Ramu 130 cm 140 cm Somu 125 cm 132 cm Radha 115 cm 124 cm Swetha 120 cm 128 cm Scale: On X-axis: 2 cm = 1 student On Y-axis: 1 cm = 20 cm Comparison of Heights Height (in cm) Names of children 68
a) A s the bar for Ramu’s height in 2010 is the highest, Ramu is the tallest among the children. b) R adha is the shortest among them in 2012. (Shortest bar in 2012). c) Ramu is 5 cm (130 – 125) taller than Somu. d) Increase in the heights of the children in the two years: Ramu: (140 – 130) cm = 10 cm S omu: (132 – 125) cm = 7 cm Radha: (124 – 115) cm = 9 cm Swetha: (128 – 120) cm = 8 cm 7 cm < 8 cm < 9 cm < 10 cm Therefore, Ramu’s height increased the maximum in 2 years. e) Heights of the children in 2010: 130 cm, 125 cm, 115 cm, 120 cm Ascending order: 115 cm, 120 cm, 125 cm, 130 cm Heights of the children in 2012: 140 cm, 132 cm, 124 cm, 128 cm Descending order: 140 cm, 132 cm, 128 cm, 124 cm Example 6: The weights of four children are noted in 2014 and 2016 as given. Draw a bar graph and answer the questions that follow. Name Weight in 2014 Weight in 2016 Ram 30 kg 34 kg Shyam 34 kg 32 kg Reema 28 kg 31 kg Seema 29 kg 31 kg a) Who weighed the most in 2014 and 2016? b) Whose weight has decreased in 2016 from 2014? c) Name the two children who were of the same weight in 2016. d) Whose weight in 2014 is the same as that of another child in 2016? Data Handling 69
e) W rite the weights of the children in 2014 in descending order and their Solution: weights in 2016 in ascending order. Scale: O n X-axis: 2 cm = 1 student; Y-axis: 1 cm = 5 kg Comparison of Weights Comparison of Weights Weights (in kg) Ram Shyam Reema Seema Names of children a) Shyam was the heaviest in 2014 and Ram was the heaviest in 2016. b) Shyam’s weight decreased in 2016 from 2014. c) Reema and Seema are of the same weight in 2016. d) Shyam’s weight in 2014 is equal to Ram’s weight in 2016. e) Weights in 2014: 30 kg, 34 kg, 28 kg, 29 kg Descending order: 34 kg, 30 kg, 29 kg, 28 kg Weights in 2016: 34 kg, 32 kg, 31 kg and 31 kg Ascending order: 31 kg, 31 kg, 32 kg, 34 kg 70
Drill Time Concept 13.1: Bar Graphs 1) The score of students in an essay writing competition are given in the table. Draw a bar graph. Subject Marks scored Piyush 65 Suman 72 Vaishnavi 82 Pooja 93 2) The table shows the marks secured by Rajeev in Test 1 and Test 2. Subject Marks in Test 1 Marks in Test 2 Hindi 65 68 English 78 80 60 85 Mathematics 88 80 Science 54 65 Social Studies Compare his performance in the two tests by drawing a bar graph and answer the questions that follow. a) Find Rajeev’s total marks in Test 1 and Test 2 separately. b) In which of the two tests did he perform well with respect to Mathematics? c) In which subject(s) has he improved from Test 1 to Test 2? d) In which of the two tests has Rajeev got less marks? 3) The approximate monthly attendance of Grade 4 is shown in the pictograph given. Draw a bar graph and answer the questions that follow. Data Handling 71
Month Attendance June July August September October November Key: 1 = 10 students a) In which month is the attendance maximum? b) In which month is the attendance minimum? c) In which months is the attendance less than 45? 72
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