9.2 Like, Unlike and Equivalent Fractions 5) Write four equivalent fractions for each of the following. a) 1 b) 4 c) 3 d) 4 2 7 10 11 6) Shade the regions to represent equivalent fractions. a) ª 1 and 2 º «¬ 5 10 ¼» b) ª1 and 2º ¬« 2 4 »¼ 7) Word problems 1 6 a) Zoya has 24 notebooks. 1 of them are unruled and of them are four-ruled. 2 How many books are (a) unruled and (b) four-ruled? b) A circular disc is divided into 12 equal parts. Venu shaded 1 of the disc pink 4 1 and 3 of the disc green. How many parts of the disc are shaded? How many parts are not shaded? 8) Identify like and unlike fractions from the following. a) 2 , 2 , 1 , 5 , 2 , 7 , 6 , 2 b) 7 , 4 4 , 2 , 4 2 3, 6 83286 889 9 5, 9 9 7, 4, 4 9 c) 6 ,154 ,5 ,147 ,187 ,174 ,197 ,124 d) 3 , 4 , 1 , 3 , 1 ,141 14 17 5 5 5 7 9 9) Arrange the following fractions in ascending order. a) 131, 111, 171, 4 b) 3 ,123 ,9 ,153 c) 1 , 3 , 4 , 2 d) 1 ,184 ,7 ,194 11 13 13 7 7 7 7 14 14 Fractions 47
10) Arrange the following fractions in descending order. a) 1 , 8 , 7 , 4 b) 3 ,167 ,10 ,187 c) 271, 291, 221,1231 d) 1 , 7 , 8 , 3 9 9 9 9 17 17 20 20 20 20 11) Add: a) 2 + 5 b) 3+ 16 c) 9 + 4 d) 8 +4 e) 1 + 2 7 7 11 11 5 5 17 17 13 13 12) Subtract: a) 15 − 7 b) 9 − 5 c) 11 − 3 d) 7 − 4 e) 13 − 12 66 8 8 40 40 45 45 30 30 13) Word problems a) Colour each figure to represent the given fraction and compare them. 57 88 b) Akansha ate 1 of a cake in the morning and 2 of it in the evening. What 5 5 part of the cake is remaining? A Note to Parent Fractions are present all around us. The easiest way to make a child relate to fractions is through food items. Cut fruits such as apples and oranges in different equal parts and use them to help your child understand fractions. 48
Money10Chapter I Will Learn About • conversion between Rupees and Paise. 10.1 Conversion between Rupees and Paise I Think Surbhi had some play money in the form of notes and coins. While playing, her friend gave her ` 10. Surbhi has to give paise for the amount her friend gave her. How many paise should Surbhi give her friend? I Recall Let us recall what we have learnt about money in the previous class. Observe the rates (per kg) of vegetables in the box given. Ramesh buys a few of them in different quantities. Write the rates and the amount to complete the bill for the vegetables he bought. 49
Pumpkin Potato Cabbage ` 99 ` 15 ` 100 Green peas ` 120 Onion Chillies Carrot ` 30 ` 68 ` 60 Tomato Brinjal Broccoli Amount ` 24.50 ` 60 ` 100 `p S.No Vegetables Bill Rate per item 1 Pumpkin Quantity 2 Tomato (in kg) 3 Carrot 4 Broccoli 1 5 Brinjal 1 2 2 2 Total I Remember and Understand We already know that ` 1 = 100 paise. To convert rupees into paise, we multiply the rupees by 100. For example, ` 3 = 3 × 100 paise = 300 paise. To convert an amount in ‘rupees’ and ‘paise’ into ‘paise’, we multiply the rupees by 100 and add the product to the number of paise. Let us see a few examples involving the conversion between rupees and paise. Example 1: Convert ` 132.28 into paise. Money can be added or Solution: ` 132.28 = ` 132 + 28 p subtracted easily using the = 132 × 100 p + 28 p column method. The rupees = 13200 p + 28 p and paise should be written = 13228 p with the dot (.) exactly one below the other. 50
Note: An easy method to convert rupees into paise is to remove the rupee symbol of rupees and the dot (.) and write the rupees and paise together. So, ` 132.28 = 13228 p. An amount of more than 100 paise, can be expressed in rupee and paise. The last two digits in the number is paise and the rest of the digits represent rupees. Example 2: Convert 24365 paise into rupees and paise. TTh Th H T O Solution: 24365 p = 24365 ÷ 100 = ` 243 + 65 p = ` 243.65 2 4 365 Note: An easy method to convert ‘paise’ into ‘rupees’ and ‘paise’ is to just put a dot (.) after two digits (ones and tens places) from the right and express it as `. So, 24365 p = ` 243.65. We add or subtract money just as we add or subtract numbers. Let us see a few examples where we add or subtract money with conversion. Example 3: Add 54738 paise and ` 130.83. ` p 1 Solution: T o add 54738 paise and ` 130.83, we convert 1 38 83 paise into rupees. 5 4 7. 21 54738 paise = ` 547.38 + 1 3 0. ` 6 7 8. By adding using the column method, we get ` 678.21 as the sum. Example 4: Subtract ` 247.65 from 53354 paise. `p Solution: To subtract ` 247.65 from 53354 paise, 12 12 14 we convert paise into rupees. 422 4 14 53354 paise = ` 533.54 5 3 3.5 4 By subtracting using the column method, − 2 4 7.6 5 we get ` 285.89 as the difference. ` 2 8 5.8 9 ? Train My Brain Solve the following: a) Convert 67923 paise into rupees. b) Convert ` 890.03 into paise. c) Add 22341 paise and ` 367.91 Money 51
I Apply Now let us solve some real-life examples involving money. Example 5: Sheeba spent ` 50 to buy 5 bars of chocolate. What is the cost of each bar of chocolate? Solution: Amount Sheeba spent on buying chocolates = ` 50 Number of chocolate bars Sheeba bought = 5 To find the cost of one chocolate bar, we divide ` 50 by 5. We know that the quotient when 50 is divided by 5 is 10. Therefore, each chocolate bar costs ` 10. Example 6: Ramu wants to buy 7 kg of apples costing ` 99.50 per kg. If he has ` 650, does he have enough amount? ` p Cost of 1 kg apples = ` 99.50 63 Solution: Quantity of apples Ramu wanted to buy = 7 kg 9 9 . 5 0 Total cost needed to buy 7 kg of apples = × 7 ` 99.50 × 7 kg ` 696.50 The total cost of 7 kg of apples is ` 696.50. But, Ramu has only ` 650 with him. As, ` 696.50 > ` 650, Ramu does not have enough money with him to buy 7 kg of apples. I Explore (H.O.T.S.) Let us see some more real-life examples involving addition and subtraction of money. Example 7: Tanya had ` 525 and her sister Tanvi had ` 330. They bought a gift costing ` 495.75 for their brother on his birthday. How much amount is left with them? Solution: Amount Tanya had = ` 525 Amount Tanvi had = ` 330 Total amount they had = ` 525 + ` 330 = ` 855 Total amount = `p The amount spent for gift = Amount left with them = 855 . 00 – 495 . 75 359 . 25 Therefore, Tanya and Tanvi have ` 359.25 left with them. 52
Example 8: The costs of three items are ` 125, ` 150 and ` 175. Suresh has only notes of ` 100. If he buys the three items, how many notes must he give the shopkeeper? Does he get any change back? If yes, how much change does he get? Solution: Total cost of the three items = ` 125 + ` 150 + ` 175 = ` 450 The denomination of money Suresh has = ` 100 The nearest hundred, greater than the cost of the three items is ` 500. So, the number of notes that Suresh has to give the shopkeeper is 5. As, ` 450 < ` 500, Suresh gets change from the shopkeeper. The change he gets = ` 500 − ` 450 = ` 50 Maths Munchies To convert rupees to paise, move the decimal point two places to the right. Connect the Dots English Fun Apart from Hindi and English, which language appears on the front side of a currency note? Fifteen other languages appear on the reverse side of an Indian rupee note. List the names of the other languages. Social Studies Fun The earliest metal coins were used in China. Try to find out different coins with their values and their shapes. Money 53
Drill Time 10.1 Conversion between Rupees and Paise 1) Convert the following to paise. a) ` 632.18 b) ` 952.74 c) ` 231.48 d) ` 537.58 e) ` 724.80 2) Convert paise to rupees. a) 52865 b) 64287 c) 13495 d) 34567 e) 78654 3) Add: b) ` 3467.45 + ` 2356.50 a) ` 875.62 + 96498 p d) ` 456.23 + 27950 p c) 25382 p + ` 652.37 e) ` 123.75 + 642.90 p 4) Subtract: b) 85732 p – ` 237.84 a) ` 132.75 – 11290 p d) ` 456.72 – 23434 p c) ` 578.14 – ` 345.89 e) ` 784.50 – ` 234.25 5) Word problems a) 5 packets of chips cost ` 20. How much will one such packet cost? b) A football costs ` 159.99. What is the cost of 26 such footballs? A Note to Parent Show your child currency notes of different denominations such as ` 10, ` 20, ` 100 and so on. Also, show them some shopping bills to make them understand how addition and subtraction of money are useful in our day-to-day life. 54
Measurement 60 Chapter 5575 5011 Measurement70 65Cm 45 40 35 30 25 20 15 10 I Will Learn About 5 0 20 • conversion of units of length, weight and volume. • estimation and verification of length, weight and volume. • area and perimeter of simple shapes. 11.1 Conversion of Units of Length I Think Surbhi went with her mother to a shop to buy cloth for her frock. Her mother asked the shopkeeper to give two metres of the cloth. How do you think the shopkeeper should measure two metres of the cloth? I Recall We know that people sometimes measure lengths of objects using their hands or feet. But the size of the body parts differ from one person to another. So, the length of the same object also differs when measured by different people. Suppose a boy and a man measure the same object. We see that the measures of the object are different. So, measures such as hand span, cubit, leg span and so on are called non-standard units. 55
To express measurements correctly, standard units were developed. The measurement of objects remains the same anywhere in the world, when these standard units are used. Measures of Length: Centimetre: It is a unit of length used to measure the lengths of a pencil, the sides of a book and so on. We write centimetres in short as cm. Metre: It is the standard unit of length. It is used to measure the lengths of a piece of cloth, a wall and so on. We write metres in short as m. Kilometre: It is a unit of length larger than a metre. It is used to measure the distance between two places, length of a river and so on. We write it as km. 5 cm 3m 2 km I Remember and Understand We can convert one unit of measurement into another using the relation between them, i.e., 1 m = 100 cm To convert measures from a larger unit to a smaller unit, we 1 km = 1000 m multiply. Let us understand the conversion through a few To convert measures from a examples. smaller unit to a larger unit, we divide. Example 1: Convert: a) 5 m 7 cm into cm b) 6 km 4 m into m c) 815 cm into m and cm d) 4805 m into km and m 56
Solution: Solve these Solved 2 m 9 cm = _______________ cm a) To convert metres into centimetres, multiply by 100. 3 km 4 m = 5 m 7 cm = ____________ cm _______________ m 1 m = 100 cm 270 cm = ________ m ________ cm So, 5 m = 5 × 100 cm 6045 m = = 500 cm ________ km________ m 5 m 7 cm = (500 + 7) cm = 507 cm Therefore, 5 m 7 cm is 507 cm. b) To convert kilometres into metres, multiply by 1000. 6 km 4 m = ___________ m 1 km = 1000 m So, 6 km = 6 × 1000 m = 6000 m 6 km 4 m = (6000 + 4) m = 6004 m Therefore, 6 km 4 m = 6004 m. c) To convert centimetres into metres, divide by 100. 815 cm = _________ m ________ cm 815 cm = (800 + 15) cm 100 cm = 1 m So, 800 cm = 800 ÷ 100 m =8m Therefore, 815 cm = 8 m 15 cm. d) To convert metres into kilometres, divide by 1000. 4805 m = _________ km ________ m 4805 cm = (4000 + 805) km, 1000 m = 1 km So, 4000 m = 4000 ÷ 1000 km = 4 km Therefore, 4805 cm = 4 km 805 m We can add or subtract lengths just as we add or subtract numbers. Remember to write the units after the sum or difference. Note: Suppose the length of an object is expressed in km and m and there are only 2 digits in the metre. Place a 0 at the hundreds place and make it a 3-digit number. We do this because 1000 m = 1 km. Measurement 57
Addition of lengths Example 2: Add: a) 25 m 16 cm and 32 m 30 cm b) 34 km 450 m and 125 km 235 m Solution: Follow these steps to add the given lengths: Steps Solved Solved Solve these Step 1: Write the m cm km m m cm numbers in columns 25 16 34 450 19 27 as shown. + 32 30 + 125 235 + 40 20 Step 2: Add the m cm km m km m 34 450 12 150 numbers under the 25 16 + 125 235 + 14 340 685 smaller unit and write + 32 30 the sum. 46 Step 3: Add the m cm km m km m numbers under the 25 16 34 450 10 100 larger unit and write + 32 30 + 125 235 + 100 100 the sum. 57 46 159 685 Subtraction of lengths Example 3: Subtract: a) 125 m 20 cm from 232 m 30 cm b) 234 km 15 m from 425 km 355 m Solution: Follow these steps to subtract the given lengths: Steps Solved Solved Solve these Step 1: Write the m cm km m km m numbers in columns 232 30 425 355 14 350 as shown. − 125 20 − 234 015 − 12 150 Step 2: Subtract the m cm km m m cm numbers under the 232 30 425 355 26 42 smaller unit and write − 125 20 − 234 015 − 13 21 the difference. 10 340 58
Steps Solved Solved Solve these Step 3: Subtract the m cm km m m cm numbers under the 2 12 3 12 larger unit and write 23 2 30 425 355 59 26 the difference. − 12 5 20 − 234 015 − 39 14 10 7 10 191 340 \\ \\ \\ \\ ? Train My Brain b) Convert 8 km into m. d) 42 m 30 cm – 30 m 20 cm Solve the following: a) Convert 5 m 7 cm into cm. c) 10 km 20 m + 20 km 10 m I Apply Let us solve some real-life examples where we use the addition and subtraction of lengths. Example 4: Reema rode her bicycle for 9 km 6 m. Prajakta rode her bicycle for 8 km 24 m. Calculate the distance travelled by them in metres. Solution: The distance travelled by Reema on her bicycle = 9 km 6 m The distance travelled by Prajakta on her bicycle = 8 km 24 m Distance travelled altogether = 9 km 6 m + 8 km 24 m km m 9 006 = 17 km 30 m + 8 024 17 030 We know that 1 km = 1000 m So, 17 km = 17 × 1000 m = 17000 m 17 km 30 m = (17000 + 30) m = 17030 m Therefore, Reema and Prajakta rode a distance of 17030 metres altogether. Example 5: Sunny bought a rope of length 20 m 12 cm. Bunny bought another rope of length 12 m 20 cm. What is the total length of the rope that they bought? Measurement 59
Solution: The length of the rope bought by Sunny = 20 m 12 cm m cm 20 12 The length of the rope bought by Bunny = 12 m 20 cm 12 20 The total length of the ropes = 20 m 12 cm + 12 m 20 cm + 32 32 Therefore, the total length of the rope bought by both of them is 32 m 32 cm. Example 6: Raj’s house was at a distance of 36 km 119 m from his uncle’s house. He travelled by car for 14 km 116 m from his uncle’s house. How much more distance should Raj cover to reach his home? Solution: Distance between Raj’s house and his uncle’s house = 36 km 119 m Distance travelled by Raj from his uncle’s house = 14 km 116 m Distance left to be covered by Raj to reach his house km m = 36 km 119 m – 14 km 116 m 36 119 Therefore, the distance yet to be covered by Raj to − 1 4 1 1 6 22 003 reach his home is 22 km 3 m. I Explore (H.O.T.S.) Let us now see some more examples where we use the concept of standard units of lengths. Example 7: Isha’s mother told her that she will get one chocolate for every 100 metres she runs. She ran 400 m on the first day. At the end of the second day, her mother gave her 16 chocolates. Calculate the total distance covered by Isha on the second day. Express your answer in km and m. Solution: Number of chocolates that Isha gets for every 100m she runs = 1 The number of chocolates Isha got from her mother = 16 The total distance ran by Isha on the two days = 16 × 100 m = 1600 m Distance Isha ran on the first day = 400 m Distance Isha ran on the second day = 1600 m – 400 m = 1200 m = 1000 m + 200 m = 1 km 200 m Therefore, Isha ran 1 km 200 m on the second day. Example 8: The figure given below is a map. It shows the different ways to reach the places shown. 60
Swati’s house 3 km 7 km 500 m Market 4 km 350 m 3 km 250 m 3 km 800 m Playground 2 km 250m School 6 km 700 m Rahul’s house Look at the map and answer these questions. a) Whose house is closer to the playground and by what distance? b) Which is the shortest route to playground from Rahul’s house? c) What is the shortest distance from Swati’s house to Rahul’s house? d) What is closer to Rahul’s house: School or market? Solution: From the map, we see that a) S wati’s house is closer to the playground than Rahul’s house. The difference between the distances = 6 km 700 m – 4 km 350 m = 2 km 350 m T herefore, Swati’s house is closer to the playground than Rahul’s house by 2 km 350 m. b) T he distance between Rahul’s house and the playground by the direct road = 6 km 700 m. T he distance between Rahul’s house and the playground through the school = 3 km 250 m + 2 km 250 m = 5 km 500 m T herefore, the shortest route to the playground from Rahul’s house is through the school. c) Here, we consider any three possibilities. T he distance from Swati’s house to Rahul’s house through the market = 7 km 500 m + 3 km 800 m = 11 km 300 m T he distance from Swati’s house to Rahul’s house through the school Measurement 61
= 3 km + 3 km 250 m = 6 km 250 m T he distance from Swati’s house to Rahul’s house through the playground = 4 km 350 m + 6 km 700 m = 11 km 50 m Therefore, the shortest distance from Swati’s house to Rahul’s house is 6 km 250 m. d) The distance between Rahul’s house and school = 3 km 250 m The distance between Rahul’s house and market = 3 km 800 m Therefore, school is closer to Rahul’s house. 11.2 Standard Units of Mass and Volume I Think Surbhi went to the market to buy 1 litre of oil. The shopkeeper gave her two bottles of 500 millilitres each. Did Surbhi get the correct quantity of oil? I Recall The weight of an object is the measure of its heaviness. Different objects have different weights. We use standard units to measure the weights of objects around us. The standard unit of weight is kilogram. We write kilogram in short as ‘kg’. Another unit of weight is gram. We write gram in short as ‘g’. The smallest unit of weight is milligram. We write milligram in short as ‘mg’. Milligram (mg) is the unit used for weighing medicines, tablets and so on. Gram (g) is used for weighing objects such as pens, books, spices and so on. 62
Kilogram (kg) is used for weighing heavier objects such as rice, wheat, flour and so on. The quantity of liquid (water, oil, milk and so on) that a container can hold is called its capacity or volume. Similar to weight, the standard units of capacity are millilitres, litres and kilolitres. The standard unit of capacity or volume is litre, denoted by ‘ℓ’. The unit smaller than a litre that is used for measuring capacity is called millilitre. We write it in short as ‘mℓ’. We use kilolitres to measure liquids of quantity greater than litres. It is written as ‘kℓ’. I Remember and Understand Sometimes, to measure the weight of an object, we need Relation between the smaller unit instead of the larger unit. For this, we need to units of weight and convert the units for appropriate measurement. Let us see how volume we can convert the units of mass. 1 g = 1000 mg Conversion of weights 1 kg = 1000 g We can convert one unit of measurement into another using the relation between them. 1 ℓ = 1000 mℓ 1 kℓ = 1000 mℓ Let us understand the conversion through a few examples. Example 9: Convert the following: a) 4 kg to g b) 3 kg 150 g to g Solution: Solved Solve these a) To convert kilograms to grams,multiply by 1000. 6 kg to grams 4 kg to grams 1 kg = 1000 g So, 4 kg = 4 × 1000 g = 4000 g. Measurement 63
Solved Solve these b) T o convert kilograms and grams to grams, convert 4 kg 20 g to grams kilograms to grams and add it to the grams. 3 kg 150 g to grams 1 kg = 1000 g So, 3 kg = 3 × 1000 g = 3000 g. 3 kg 150 g = 3000 g + 150 g = 3150 g Conversion of volume We can also convert one unit of measurement into another using the relation between them. Let us understand the conversion through a few examples. Conversion of larger units to smaller units To convert litres into millilitres, multiply by 1000. Example 10: Convert the following: a) 3 ℓ to mℓ b) 2ℓ 269 mℓ to mℓ Solution: Solved Solve these a) To convert litres into millilitres, multiply by 1000. 7 ℓ to millilitres 3 ℓ to millilitres Train My Brain 1 ℓ = 1000 mℓ 3 ℓ = 3 × 1000 mℓ = 3000 mℓ b) To convert litres and millilitres into millilitres, convert 3 ℓ 750 mℓ to millilitres litres to millilitres and add it to the millilitres. 2ℓ 269 mℓ to millilitres 1 ℓ = 1000 mℓ So, 2 ℓ = 2 × 1000 mℓ = 2000 mℓ. 2 ℓ 269 mℓ = 2000 mℓ + 269 mℓ = 2269 mℓ 64
? Train My Brain Solve the following: a) Convert 5 kg into g. b) Convert 10 kg 250 g into g. c) Convert 8 ℓ into mℓ. d) Convert 34 ℓ 420 mℓ into mℓ. I Apply We add or subtract weights or volume just as we add or subtract numbers. Remember to write the unit after the sum or difference. Note: Suppose the weight of an object is expressed in g and mg and there are only 2 digits in the milligram. Place a 0 in the hundreds place and make it a 3-digit number. We do this because 1000 mg = 1 g. Addition of weights Example 11: Add: a) 15 g 150 mg and 23 g 285 mg b) 17 kg 706 g and 108 kg 189 g Solution: Follow these steps to add the given weights: Steps Solved Solved Solve these Step 1: Write the g mg kg g kg g numbers in the 15 150 17 706 11 230 columns as shown. + 23 285 + 108 189 + 8 760 Step 2: Add the g mg kg g g mg numbers under the 1 1 smaller unit and 26 190 write the sum. 15 150 17 706 + 23 260 + 23 285 + 108 189 Step 3: Add the g mg numbers under 435 895 the larger unit and 33 333 write the sum. g mg kg g + 22 333 1 11 17 706 15 150 + 108 189 + 23 285 125 895 38 435 Measurement 65
Subtraction of weights Example 12: Subtract: a) 153 g 100 mg from 262 g 300 mg b) 234 kg 150 g from 355 kg 305 g Solution: Follow these steps to subtract the given weights: Steps Solved Solved Solve these Step 1: Write the g mg kg g kg g numbers in columns 262 300 355 305 505 600 as shown. − 153 100 − 234 150 − 200 400 Step 2: Subtract the g mg kg g g mg numbers under the 2 10 smaller unit and 262 300 15 260 write the difference. − 153 100 3 5 5 \\3 \\0 5 − 15 260 − 234 150 200 155 Step 3: Subtract the g mg kg g g mg numbers under the 5 12 2 10 larger unit and write 2 \\6 \\2 3 0 0 23 555 the difference. − 153 100 3 5 5 \\3 \\0 5 − 16 454 109 200 − 234 150 121 155 Addition of volumes Example 13: Add: 13 ℓ 450 mℓ and 32 ℓ 300 mℓ Solution: Follow these steps to add the given volumes: Steps Solved Solve these ℓ mℓ Step 1: Write the ℓ mℓ 21 200 numbers in columns as 13 450 shown. + 32 300 + 11 303 Step 2: Add the ℓ mℓ ℓ mℓ numbers under the 13 450 24 129 smaller unit and write + 32 300 + 31 110 the sum. 750 66
Steps Solved Solve these Step 3: Add the ℓ mℓ ℓ mℓ numbers under the 13 450 52 000 larger unit and write + 32 300 + 41 000 the sum. 45 750 Subtraction of volumes Example 14: Subtract: 351 ℓ 200 mℓ from 864 ℓ 350 mℓ Solution: Follow these steps to subtract the given volumes: Steps Solved Solve these Step 1: Write the numbers ℓ mℓ ℓ mℓ in columns as shown. 864 350 316 186 − 351 200 − 116 205 Step 2: Subtract the ℓ mℓ ℓ mℓ numbers under the 864 350 119 209 smaller unit and write the − 351 200 − 11 101 difference. 150 Step 3: Subtract the ℓ mℓ ℓ mℓ numbers under the 864 350 291 112 larger unit and write the − 351 200 − 180 100 difference. 513 150 I Explore (H.O.T.S.) Let us now see the use of the standard units of weight and volume in real-life situations. Example 15: Kiran bought 13 kg flour in the first month and 10 kg 240 g in the second month. Venu bought 11 kg 750 g in the first month and 12 kg 400 g in the second month. Who bought more quantity of flour in two months and by how much? Solution: Quantity of flour bought by Kiran = 13 kg + 10 kg 240 g = 23 kg 240 g Measurement 67
= 23 × 1000 g + 240 g (As 1 kg = 1000 g) = 23240 g Quantity of flour bought by Venu = 11 kg 750 g + 12 kg 400 g = 24 kg 150 g = 24 × 1000 g + 150 g (As 1 kg = 1000 g) = 24150 g = 24150 g > 23240 g Therefore, Venu bought more quantity of flour than Kiran. The difference in their weights is (24150 – 23240) g = 910 g. Example 16: Suresh bought a few apples, grapes and a watermelon. The total weight of the fruits in his bag is 3 kg 750 g. The weight of the apples is 1 kg 100 g and that of the grapes is 1 kg 150 g. What is the weight of the watermelon? Solution: Suresh had 3 kinds of fruits, apples, grapes and a watermelon in his bag. Weight of apples = 1 kg 100 g kg g Weight of grapes = 1 kg 150 g 1 100 Total weight of apples and grapes +1 150 = 1 kg 100 g + 1 kg 150 g 2 250 Therefore, the weight of apples and grapes together is 2 kg 250 g. Weight of watermelon = Total weight of the fruits − Total weight of apples and grapes kg g The total weight of all the fruits in the bag = 3 kg 750 g 3 750 Weight of apples and grapes together = 2 kg 250 g − 2 250 Weight of the watermelon = 3 kg 750 g – 2 kg 250 g 1 500 Therefore, the weight of the watermelon is 1 kg 500 g. Example 17: Chandu, the milkman, has only 5 ℓ and 3 ℓ measures. How will he sell 4 ℓ of milk to Gita? (Hint: Find the difference between 5 ℓ and 3 ℓ.) Solution: Chandu first pours milk in 5 ℓ measure. He then transfers some of it into the 3 ℓ measure till its full. Then the quantity of milk left in the 5 ℓ measure is 2 ℓ. This 2 ℓ milk can be transferred into Gita’s vessel. He repeats the same process once more. Thus, in this way, he sells 4 ℓ of milk to Gita. 68
Example 18: A container has some juice. A glass has a capacity of 200 mℓ. How many glasses of juice must be poured to have 2 ℓ of juice? Solution: Capacity of the glass = 200 mℓ Quantity of juice needed = 2 ℓ = 2 × 1000 mℓ = 2000 mℓ 2000 = 200 × 10 Therefore, 10 glasses of juice must be poured to make 2 ℓ. 11.3 Area and Perimeter of Simple Shapes I Think Surbhi has to paste a chart paper on a cardboard and then decorate its borders with a colourful tape. She knows the quantity of paper and the length of the tape that she needs. How do you think she could find that? I Recall Recall that figures or shapes such as triangle, square and rectangle are called flat shapes. These are also known as 2D shapes. All these shapes have different features. Write down the features of the shapes given. Shape Features AB D C X Y WZ Measurement 69
Shape Features T SU •O I Remember and Understand We have learnt about the simple closed figures The perimeter is the length made of straight line segments. of the outline of a shape. To find the perimeter of a Let us now learn about their perimeters. rectangle or square, we add the lengths of all its The sum of the lengths of the sides of a closed figure four sides. is called its perimeter. It is denoted by ‘P’. Example 19: Find the perimeter of each of the given shapes. a) J 4 cm K b) D 3 cm E c) X 2 cm 2 cm 3 cm 3 cm 5 cm M 4 cm Solution: 4 cm L G 3 cm F a) Perimeter of rectangle = JK + KL + LM + MJ Z 3 cm y = 4 cm + 2 cm + 4 cm + 2 cm = 12 cm b) Perimeter of square = DE + EF + FG + GD = 3 cm + 3 cm + 3 cm + 3 cm = 12 cm c) Perimeter of triangle = XY + YZ + ZX = 5 cm + 3 cm + 4 cm = 12 cm Let us now learn a common method to find the perimeters of a few 2D shapes. Perimeter of a rectangle Consider the given rectangle ABCD. 70
The two opposite sides AB & CD are called lengths. They are represented by ‘ℓ’ units. The two opposite sides BC & DA are called breadths. They are represented by ‘b’ units. Perimeter of a rectangle = Sum of all its sides ℓ AB = AB + BC + CD + DA = length (ℓ) + breadth (b) b b + length (ℓ) + breadth (b) D C Therefore, perimeter (P) = 2 × ℓ + 2 × b units. ℓ Example 20: Find the perimeter of the given rectangle. M 5 cm N 3 cm 3 cm P 5 cm O Perimeter of rectangle = 2 × ℓ + 2 × b Solution: = 2 × 5 cm + 2 × 3 cm = 10 cm + 6 cm = 16 cm Therefore, the perimeter of the rectangle MNOP is 16 cm. Perimeter of a square Consider the given square ABCD. s B A s Perimeter of a square = Sum of all its sides s = AB + BC + CD + DA = 4 × side units Ds C Therefore, perimeter (P) of a square is 4 × s units. Example 21: Find the perimeter of the given square. Q 2 cm R 2 cm 2 cm T 2 cm S Solution: Perimeter of a square = 4 × s = 4 × 2 cm = 8 cm Therefore, the perimeter of the square QRST is 8 cm. Measurement 71
? Train My Brain Find the perimeter of the following: V a) S 4 cm T b) N 3 cm O c) 3 cm 4 cm 4 cm 4 cm 4 cm 2 cm 2 cm V 4 cm U Q P X 3 cm W I Apply Let us now see how to find the area of a given shape. Area specifies the region of the total number of squares of unit side it covers. So, the units of area is square units, written in short as sq. units. The lengths of the sides of a 2D figure are usually in cm or m. So, the units of its area are sq. cm or cm2 or sq. m or m2 respectively. Area of an object is the amount of surface or region covered by it. It is denoted by ‘A’. Area of a rectangle: Observe the given rectangle. The area of 1 square = 1 cm × 1 cm = 1 sq. cm The region covered by the rectangle = the total region covered by 15 squares of size 1 sq. cm each = 15 × 1 sq. cm 1 cm = 15 cm2 or 15 sq. cm 1 cm The rectangle can be considered to have a length of 5 cm and breadth of 3 cm. Thus, area of a rectangle = length × breadth = × b sq. units = 5 cm × 3 cm = 15 sq. cm Area of a square: Observe the given square. The area of 1 square = 1 cm × 1 cm = 1 sq. cm The surface region covered by the square = The total region covered by 9 squares of size 1 sq. cm each = 9 × 1 sq. cm 72
= 9 cm2 or 9 sq. cm 1 cm = 3 cm × 3 cm 1 cm The square can be considered to have a side of 3 cm. Thus, area of a square = 9 sq. cm = 3 cm × 3 cm = side × side = s × s sq. units Example 22: Find the areas of the given figures: a) b) 9 cm 4 cm 11 cm 9 cm Solution: a) Area of a rectangle = × b b) Area of a square = s × s = 11 cm × 4 cm = 9 cm × 9 cm = 44 sq. cm = 81 sq. cm The areas of the other shapes can be found out by adding the number of unit squares they cover. Example 23: Find the areas of each of given figures if each square is of area 1 sq. cm. a) b) c) Solution: a) This figure has 6 full and 6 half squares (3 squares). Therefore, its area = (6 + 3) sq. cm = 9 sq. cm. b) This figure has 6 full and 4 half squares (2 squares). Therefore, its area = (6 + 2) sq. cm = 8 sq. cm. c) This figure has 14 full and 2 half squares (1 square). Therefore, its area = (14 + 1) sq. cm = 15 sq. cm. Measurement 73
I Explore (H.O.T.S.) Let us see some more examples of perimeter and area. Example 24: Find the length of a rectangle with breadth 6 cm and perimeter 32 cm. Solution: Breadth of a rectangle = 6 cm Its perimeter = 32 cm Perimeter of a rectangle (P) = 2 × length + 2 × breadth 32 = 2 × length + 2 × 6 32 = 2 × length + 12 32 = 20 + 12 = (2 × 10) + 12 Therefore, the length of the rectangle is 10 cm. Example 25: Find the length of the side and area of a square whose perimeter is 68 m. Solution: Perimeter of the square = 4 × side = 68 m Side of the square = 4 × 17 m = 68 m Side of the square = 17 m Area of the square = s × s = 17 m × 17 m = 289 sq. m. Therefore the side of the square is 17 m and its area measures 289 sq. m. Maths Munchies The blood in our body also has a unit of measurement called ‘pint’ or ‘unit’. An adult body contains 8 to 10 pints of blood. 1 pint is equal to 473 mℓ. Therefore, our body has 3784 mℓ to 4730 mℓ of blood. Connect the Dots Social Studies Fun India measures about 3,200 kilometres from north to south. The length from west to east is about 2,900 kilometres. 74
Science Fun Dwarf Willow is one of the smallest woody plants in the world. It grows to only 1 to 6 cm in height. It has round and shiny green leaves, 1 to 2 cm in length and breadth. Drill Time 11.1 Conversion of Units of Length 1) Word problem Roopa’s house and the places close to it are shown on the map. 2 km 2 km Hospital Playground 4 km 2 km 1 km 250 m Market Post Office 5 km 500 m Roopa’s house School 2 km 450 m 4 km 6 km 3 km 300 m Airport Study the map and answer these questions. a) T he shortest route from Roopa’s house to the market is through __________ and is __________ km. b) The shortest route from Roopa’s house to the airport is _________ km. c) Which is the shortest route from the post office to the market? Measurement 75
d) Roopa went to the post office from her school through the shortest route. What is the distance she travelled? 2) Convert into centimetres. a) 3 m b) 9 m c) 2 m 45 cm d) 5 m 20 cm e) 8 m 36 cm 3) Convert into metres. a) 4 km b) 15 km c) 5 km 555 m d) 6 km 112 m e) 1 km 100 m 4) Solve the following: a) 24 m 13 cm + 13 m 45 cm b) 31 m + 18 m 59 cm c) 10 km 100 m + 20 km 200 m d) 88 km 100 m − 10 km 800 m e) 26 m 14 cm – 20 m 10 cm 11.2 Standard Units of Mass and Volume 5) Convert into grams. a) 14 kg b) 29 kg c) 14 kg 300 g d) 75 kg 226 g e) 10 kg 112 g 6) Solve the following: a) 28 kg 421 g + 30 kg 232 g b) 42 kg 876 g + 31 kg 111 g c) 44 kg 444 g – 22 kg 222 g d) 43 g 230 mg – 11 g 100 mg 7) Word problem Maya bought a few vegetables with the given quantities: Brinjal – 2 kg 250 g; Onion – 1 kg 750 g; Potato – 1 kg 250 g Find the total weight of vegetables in her bag. 8) Convert into millilitres. a) 13 ℓ b) 28 ℓ c) 13 ℓ 400 mℓ d) 64 ℓ 206 mℓ e) 14 ℓ 142 mℓ 9) Solve the following: a) 28 ℓ 421 mℓ + 40 ℓ 262 mℓ b) 41 ℓ 836 mℓ + 41 ℓ 113 mℓ c) 30 ℓ 320 mℓ + 20 ℓ 300 mℓ d) 33 ℓ 530 mℓ – 11 ℓ 300 mℓ e) 66 ℓ 666 mℓ – 44 ℓ 444 mℓ 76
10) Word problem Aarthi has a jug with some buttermilk. She uses glasses which can hold 150 mℓ. How many glasses must she fill so that she has 3 ℓ of buttermilk? 11.3 Area and Perimeter of Simple Shapes 11) Find the perimeter of the given figures. a) 2 cm b) 3 cm 4 cm 3 cm c) d) 8 cm8 cm4 cm 3 cm 13 cm 12) Find the area of the given figures. c) a) b) 13 cm 8 cm 4 cm 4 cm 8 cm 15 cm A Note to Parent Ask your child to weigh different things present at home such as a pencil, a flower vase or some utensils. This will help him or her to understand the usage of mg, g and kg for different objects. Measurement 77
CHAPTER55 13 12Chapter Data Handling 50 45 40 35 30 I Will Learn About 25 20 15 10 5 0 • reading bar graphs. • drawing bar graphs based on the given data. 12.1 Bar Graphs I Think Surbhi 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. Surbhi knew how to represent the data as a pictograph. She knew that it would take a lot of effort to draw a pictograph. She wanted to represent it in an easier and simpler way. How do you think Surbhi would have given the report? I Recall We know that: • the information collected for a specific purpose is called data. • the information given as numbers is called the numerical data. • the information shown in the form of pictures is called a pictograph. 78
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 game of Class 4 students Volleyball Cricket Basketball Kabaddi Football a) The most favourite game of Class 4 students is _____________ . b) The least favourite game 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 _____________ . I Remember and Understand While drawing pictographs, we choose a relevant picture to represent the given data. If the data is large, it is difficult and time consuming to draw a pictograph. Data Handling 79
An easier way of representing data is the bar graph. In a bar graph, we draw rectangular bars of the same width. The bars in a bar graph can be drawn either We draw bar graphs on a graph paper and give them horizontally or vertically. suitable titles. First, we decide the scale of the graph. The markings on Bar graphs are useful in the axes of the graphs is the scale. We can adjust the comparing data. scale to draw the bar graph within the given range. 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 English Maths Science Social Music Hindi Studies Subjects a) What is the title of the graph? b) In which subject did Kamala perform the best? 80
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. So, Kamala needs to improve in Social Studies. d) Kamala’s total marks are 35 + 47 + 42 + 28 + 32 + 40 = 224 Example 2: 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 l oohcS Class strength Class a) What is the total strength of all the 5 classes? b) Which class has the least strength? Data Handling 81
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 ? Train My Brain Fill in the blanks. a) A bar graph is used to represent ___________________. b) _________________ bars are used in a bar graph. c) Bar graphs are drawn on __________. I Apply We have learnt how to read and interpret bar graphs. Now, let us learn to draw a bar graph using a graph paper. Steps to draw a bar graph on a graph paper: Step 1: Draw a horizontal line and 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. On the Y-axis show the values of the items. 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. 82
Key: 1 = 5 scooters Number of Weekday Scooters manufactured in a week scooters Monday 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 83
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 84
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 Step 5: Monday Tuesday Wednesday Thursday Friday Saturday Weekdays 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 Weekdays Data Handling 85
We can draw the same graph using horizontal bars by interchanging the values on X and Y axes. Scale: X-axis: 1 cm = 5 scooters; Y-axis: 1 cm = 1 day 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. 86
Solution: Scale: X-axis: 1 cm = 1 week; Y-axis: 1 cm = 20 roses Sale of Roses Roses sold Weeks I Explore (H.O.T.S.) Consider a few real-life examples where we represent daTtraauisningMaybaBr rgaraipnh. 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? e) Arrange the children’s heights in 2010 in ascending order and their heights in 2012 in descending order. Data Handling 87
Solution: 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 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. 88
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? e) W rite the weights of the children in 2014 in descending order and their Solution: weights in 2016 in ascending order. Scale: On X-axis: 2 cm = 1 student; Y-axis: 1 cm = 5 kg Data Handling 89
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 Maths Munchies We can show the bar graphs as histograms when the data keeps changing with respect to the time. The bar graphs are usually separated while the histograms do not have space between the bars and are adjacent to each other. Sometimes, you can have bar graphs without any space. However, a histogram cannot have space between the bars. 90
Connect the Dots Social Studies Fun A Climate graph is usually shown as a bar graph. It is a combination of temperature and rainfall. The months are shown on the X-axis while the amount of rainfall and temperatures are shown on the Y-axis. English Fun Make a list of your favourite authors. Count the number of books that you know of each author. Use this data to draw a bar graph. (Some names of authors for reference: J K Rowling, Ruskin Bond, C S Lewis, Charles Dickens, R K Narayan and so on.) Drill Time 12.1 Bar Graphs 1) The score of students in an essay writing competition are given in the table. Draw a bar graph. Subject Piyush Suman Vaishnavi Pooja Marks scored 65 72 82 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 Mathematics 60 85 Science 88 80 Social Studies 54 65 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? Data Handling 91
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. 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? A Note to Parent From newspapers or magazines, find out the bar graphs and explain what they are about. Explain the terms mentioned in the bar graph first, to give the background and then form basic questions from the same. You may choose articles of common interest like cars, bikes, movies, travel, hobbies and so on. 92
Search
