S o, we have to find the equivalent decimals of the other three decimal numbers such that they have three decimal places. Thus, the required like decimals are: 42.700, 53.280, 261.135, 11.010 b) Unlike decimals: 0.742, 12.06, 8.5, 17.12 Like decimals: 0.742, 12.060, 8.500, 17.120 c) Unlike decimals: 7.23, 2.1, 0.6, 4.382 Like decimals: 7.230, 2.100, 0.600, 4.382 Application Let us see some real-life examples based on decimals. Example 6: If 502 out of 1000 students in a school are girls, then write the decimal equivalent of the fraction of girls in the school. Solution: The total number of students in the school = 1000 Number of girls = 502 The fraction of girls = 502 1000 The required decimal equivalent = 0.502 Example 7: If there is 263 cm of tape on a tape roll, how many metres of tape is on the roll? Give your answer in the decimal form. Solution: The length of the tape on the tape roll = 263 cm Converting into metres = 263 m 100 The required decimal form of the length of the tape = 2.63 m Example 8: The weight of a sugar jar is 5670 g. What is its weight in kilograms? Solution: The weight of a sugar jar = 5670 g Converting into kg = 5670 kg 1000 The required decimal form of the weight of the jar = 5.67 kg Decimals - I 47
Higher Order Thinking Skills (H.O.T.S.) Consider the following examples. Example 9: In a small village of 1000 people, there are 238 women and 450 men. Find the fraction of men, women and children and write them in decimal form. Write the decimals using place value chart. Solution: Total number of people in the village = 1000 Number of women = 238 Fraction of women = 238 1000 Decimal form = 0.238 Number of men = 450 Fraction of men = 450 1000 Decimal form = 0.450 Number of children = 1000 – (238 + 450) = 1000 – 688 = 312 Therefore, fraction of children = 312 1000 Decimal form = 0.312 Women Ones Decimal point Tenths Hundredths Thousandths Men 1 1 1 1 (.) 10 Children 2 100 1000 0 . 4 3 8 0 . 3 5 0 0 . 1 2 48
Concept 11.2: Compare and Order Decimals Think Pooja went to purchase a bag to gift her mother on her birthday. She selected two bags of prices ` 455.80 and ` 455.40. She couldn’t understand which is more expensive of the two. Do you know which is more expensive of the two? Recall We have already learnt about equivalent, like and unlike decimals. To compare two decimals, we should know the concepts of equivalent decimals and like decimals. Let us recall the concepts by answering the following: a) Write four equivalent decimals of 6.1. b) Convert these unlike decimals to like decimals: 32.5, 410.635, 6, 78.7 c) Identify the pair of like decimals: a) 39.12, 56.03 b) 0.14, 0.04 c) 6.75, 83.16 d) 7.101, 6.2 & Remembering and Understanding We know that, 1) A dding any number of 0s to the right side of the decimal point does not change its value. 2) Unlike decimals can be converted to like decimals by adding zeros at the right end. Now, let us learn to compare two decimals through a few examples. Example 10: Which is greater of the given decimals? a) 69.2 and 69.02 b) 77.10 and 77.012 c) 3.5631 and 3.61 Decimals - I 49
Solution: a) 69.2 and 69.02 T o compare two decimals, follow these steps: Step 1: C onvert the given decimal numbers into like decimals. 69.20, 69.02 Step 2: C ompare their integral parts. The decimal with the greater integral part is greater. Here, 69 = 69 Step 3: If the integral parts are the same, then we have to compare the tenths digits. The decimal with the greater digit in the tenths place is greater. If the tenths digits are the same, compare the hundredths digits and so on. 69.20 69.02 6=6 9=9 2>0 Hence, 69.20 > 69.02 Note: Always start comparing from the largest place value in the integral part. b) 77.10 and 77.012 n Step 1: Convert the unlike decimals into like decimals: 77.100, 77.012 Step 2: Compare the integral parts: 77 = 77 Step 3: Compare the tenths digits: 1 > 0 Hence, 77.10 > 77.012. c) 3.5631 and 3.61 Step 1: Convert the unlike decimals into like decimals: 3.5631, 3.6100 Step 2: Compare their integral parts: 3 = 3 Step 3: Compare the tenths digits: 5 < 6 So, 3.5631 < 3.61000. 50
Example 11: Which is smaller between each of the given pairs of decimal numbers? a) 367.80 and 362.801 b) 21.673 and 21.673 c) 11.729 and 11.726 Solution: a) 367.80 and 362.801 Converting the unlike decimals into like decimals: 367.800, 362.801 Comparing their integral parts, 367 > 362 Hence, 362.801 is smaller. b) 21.673 and 21.673 Given decimals are like decimals. Comparing their integral parts: 21 = 21 Compare: Tenths digits: 6 = 6, Hundredths digits: 7 = 7 Thousandths digits: 3 = 3 Hence, 21.673 = 21.673. c) 11.729 and 11.726 Given decimals are like decimals. Comparing their integral parts: 11 = 11 Compare: Tenths digits: 7 = 7 Hundredths digits: 2 = 2 Thousandths digits: 9 > 6 Hence, 11.726 is smaller. Comparing decimal numbers helps us in arranging them in ascending and descending orders. Let us see a few examples. Example 12 : Arrange the following decimal numbers in the ascending order. a) 2.1, 2.01, 3.06, 0.831 b) 15.12, 19.18, 26.7, 1.007 c) 37.502, 36.512, 67.3, 22 Decimals - I 51
Solution: a) 2.1, 2.01, 3.06, 0.831 Like decimals of the given unlike decimals: 2.100, 2.010, 3.060, and 0.831 0.831 < 2.010 < 2.100 < 3.060 Thus, the ascending order is: 0.831, 2.01, 2.1, 3.06 b) 15.12, 19.18, 26.7, 1.007 Like decimals of the given unlike decimals: 15.120,19. 180, 26.700 and 1.007 1.007 < 15.120 < 19.180 < 26.700 Thus, the ascending order is: 1.007, 15.12, 19.18, 26.7 c) 37.502, 36.512, 67.3, 22 Like decimals of the given unlike decimals: 37.502, 36.512, 67.300, and 22.000 22.000 < 36.512 < 37.502 < 67.300 Thus, the ascending order is: 22, 36.512, 37.502, and 67.3 Example 13: Arrange these decimal numbers in the descending order. a) 43.25, 43.2, 43.21, 43.127 b) 63.901, 63.09, 63.009, 6.39 c) 11.2, 11.028, 1.127, 13.02 Solution: a) 43.25, 43.2, 43.21, 43.127 Like decimals of the given decimal numbers: 43.250, 43.200, 43.210, 43.127 43.250 > 43.210 > 43.200 > 43.127 Thus, the descending order is 43.25, 43.21, 43.2, 43.127. b) 63.901, 63.09, 63.009, 6.39 Like decimals of the given decimal numbers: 63.901, 63.090, 63.009, 6.390 63.901 > 63.090 > 63.009 > 6.390 Thus, the descending order is: 63.901, 63.09, 63.009, and 6.39. 52
c) 11.2, 11.028, 1.127, 13.02 Like decimals of the given decimal numbers: 11.200, 11.028,1.127, 13.020 13.020 > 11.200 > 11.028 > 1.127 Thus, the descending order is: 13.02, 11.2, 11.028, and 1.127. Application Let us now see a few real-life examples involving comparison of decimals. Example 14: Ramu saves ` 361.80 while Raghu saves ` 351.90. Who saves more money? Solution: To know who saves more money, we have to find which number is greater between ` 361.80 and ` 351.90. 361.80 351.90 3=3 6>5 361.80 > 351.90 Therefore, Ramu saves more money. Example 15: Leela scores 32.5 marks in English, 48.5 in Mathematics and 32.75 in Science. Arrange her marks in the descending order and find the subject in which she scored the maximum marks. Solution: English Mathematics Science 32.5 48.5 32.75 Like decimals equivalent to the unlike decimals: 32.50, 48.50 and 32.75 Descending order: 48.50, 32.75, 32.50 48.5 is the greatest number. Therefore, Leela scored the maximum marks in Mathematics. Decimals - I 53
Example 16: Ravi bought two watermelons. One of them weighs 12.352 kg and the other weighs 12.365 kg. Which watermelon is heavier? Solution: Weight of one watermelon = 12.352 kg Weight of the second watermelon = 12.365 kg 12.352 12.365 1=1 2=2 3=3 5<6 Hence, 12.352 < 12.365 or in other words, 12.365 > 12.352. Therefore, the 2nd watermelon is heavier. Higher Order Thinking Skills (H.O.T.S.) Let us see another real-life example of comparing and ordering decimals. Example 17: In a swimming competition, there are five competitors. Four of the swimmers have had their turns. Their scores are 9.8 s, 9.75 s, 9.79 s and 9.81 s. What score must the last swimmer get in order to win the competition? Solution: Arrange the given scores in ascending order: 9.75 < 9.79 < 9.80 < 9.81 The lowest decimal is 9.75. Therefore, the last swimmer must get a score less than 9.75 s in order to win the competition. 54
Concept 11.3: Add and Subtract Decimals Think Pooja went to an ice cream parlour to purchase some ice creams. She bought strawberry for ` 25.50, vanilla for ` 15.30 and chocolate for ` 32.20. She gave ` 100 to the shopkeeper. She wanted to calculate the total price before the shopkeeper gives the bill. Since the prices are in decimals, she was unable to calculate. Do you know how to find the total cost of the ice creams that Pooja bought? How much change would she get in return? Recall Addition and subtraction of decimal numbers is similar to that of usual numbers. Let us recall the conversion of unlike decimals to like decimals. Convert the given unlike decimals into like decimals. a) 4.32, 4.031, 4.1, 7.823 b) 0.7, 0.82, 4.513, 0.72 c) 1.82, 7.01, 5.321, 0.8 d) 7.32, 7.310, 7.8, 5.2 & Remembering and Understanding Add and subtract decimal numbers with the thousandths place Addition and subtraction of decimal numbers with the thousandths place is similar to that of decimals with the hundredths place. Before adding or subtracting any decimals, convert the unlike decimals to like decimal. Write the given decimal numbers such that the digits in their same places are exactly one below the other. Note: The decimal points of the numbers must be exactly one below the other. Decimals - I 55
Let us see a few examples. Example 18: a) Find the sum of 173.809 and 23.617. b) Subtract 216.735 from 563.726. b) 12 16 Solution: a) 11 5 2/ 6/ 12 5 6/ 3/ . 7/ 2/ 6 17 3 . 8 0 9 + 23 . 617 –2 1 6 . 7 3 5 197 . 426 346 . 991 Example 19: Solve: a) 294.631 + 306.524 b) 11.904 – 6.207 Solution: a) 1 1 1 b) 11 8 9 14 294 . 631 1/ 1/ . 9/ 0/ 4/ +3 0 6 . 5 2 4 – 6 . 20 7 601 . 155 5 . 69 7 Application Let us see a few real-life examples involving addition and subtraction of decimals. Example 20: Dolly bought 450 g of tomatoes and 150 g of green chillies to make tomato pickle. What is the total weight (in kg) of the vegetables that Dolly bought? Solution: Weight of tomatoes = 450 g Weight of green chillies = 150 g The total weight of vegetables = (450 + 150) g = 600 g 1 We know that 1 kg = 1000 g or 1 g = 1000 kg 600 g = 600 kg = 0.6 kg 1000 Therefore, Dolly bought 0.6 kg of vegetables. 56
Example 21: Vinod purchased a shirt for ` 275.40, a pair of trousers for ` 1462.30 and a pair of shoes for ` 524.95. Find the total money spent by Vinod. Solution: Amount spent by Vinod on a shirt = ` 275.40 Amount spent on a pair of trousers = ` 1462.30 Amount spent on a pair of shoes = ` 524.95 Total amount spent by Vinod = ` 275.40 + ` 1462.30 + ` 524.95 = ` 2262.65 Example 22: By how much should 67.23 be decreased to get 28.59? 16 11 Solution: The required number is the difference of 67.23 5 6/ 1/ 13 and 28.59 = 67.23 – 28.59 6/ 7/ . 2/ 3/ −2 8 . 5 9 Therefore, 67.23 is to be decreased by 38.64 to 38.6 4 get 28.59. Example 23: Mrs. Roopa bought 13.75 litres of milk. She used 9.2 litres of milk for making paneer. Find the quantity of milk left. Solution: Milk bought by Mrs. Roopa = 13.75 litres Milk used to make paneer = 9.2 litres Quantity of milk left = (13.75 – 9.2) litres = 4.55 litres Example 24: Pawan needs 1.40 m of cloth for a shirt and 2.45 m of cloth for a pair of trousers. His father gave a piece of cloth which is 0.65 m less than that needed. What was the length of the cloth given by Pawan’s father? Solution: Cloth needed for a shirt = 1.40 m Cloth needed for a pair of trousers = 2.45 m Total cloth needed = 1.40 m + 2.45 m = 3.85 m The length by which the cloth was short by = 0.65 m The length of the cloth given by Pawan’s father = 3.85 m – 0.65 m = 3.20 m Therefore, 3.20 m of cloth was given by Pawan’s father. Decimals - I 57
Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples involving addition and subtraction of decimals. Example 25: Subtract the sum of 6.223 and 37.512 from the sum of 42.106 and 5.07. Solution: We should find (42.106 + 5.07) – (6.223 + 37.512) Step 1: Add 42.106 and 5.07 Step 2: Add 6.223 and 37.512 Step 3: Subtract the sum in step 2 from 1 the sum in step 1. 1 6 11 4 /7 . /1 7 6 4 2.1 0 6 0 6.2 2 3 + 5 . 0 7 0 +3 7 . 5 1 2 −4 3 . 7 3 5 4 7.1 7 6 4 3.7 3 5 0 3.4 4 1 Example 26: Pooja saw a doll in the showcase of a mall. The cost of the doll was ` 85.65. She wanted to buy it, but she fell short of ` 5.75. How much money did Pooja have? Solution: Cost of the doll = ` 85.65 Amount she fell short of = ` 5.75 Money that Pooja had = ` 85.65 – ` 5.75 = ` 79.90 Therefore, Pooja had ` 79.90 Drill Time Concept 11.1: Like and Unlike Decimals 1) Write the following numbers in the decimal place value chart. a) 42.874 b) 315.097 c) 2795.741 d) 127.243 2) Write the expanded forms of the given decimals and then write them in words. a) 3578.048 b) 450.981 c) 32.62 d) 432.789 58
3) Convert the given unlike decimals into like decimals. a) 52.7, 25.28, 321.265, 101.51 b) 42.52, 4.7, 32.472, 48.8 c) 1.7, 32.4, 328.732, 1.82 d) 7.42, 1.821, 7.01, 432.2 4) Word problems a) 4 75 out of 800 parents who attended a parent-teacher meet were women. Write the decimal equivalent of the fraction of men who attended the meet. b) The distance between two places is 3564 m. What is the distance in kilometres? Concept 11.2: Compare and Order Decimals 5) Which is greater of the given pairs of decimal numbers? a) 75.9 and 75.09 b) 32.87 and 32.087 c) 3.52 and 3.5221 d) 372.12 and 347.127 6) Which is smaller of the given decimals? a) 52.6 and 52.609 b) 87.48 and 87.489 c) 585.23 and 585.235 d) 538.72 and 583.72 7) Arrange these decimal numbers in the ascending and descending orders. a) 7.9, 7.91, 8.5, 8.2 b) 26.25, 26.85, 32.96, 12.951 c) 63.824, 63.756, 63.714, 63.823 d) 253.98, 253.74, 253.945, 253.732 8) Word problems a) R iya scores 35.5 marks in English, 58.50 in Mathematics and 58.75 in Science. Arrange her marks in the descending order and find the subject in which she scored the maximum marks. b) Ravi bought two pumpkins. One of them weighs 8.556 kg and the other weighs 8.305 kg. Which pumpkin is heavier? Concept 11.3: Add and Subtract Decimals 9) Add: a) 528.364 and 974.623 b) 523.97 and 49.25 c) 23.547 and 14.974 d) 242.57 and 132.60 Decimals - I 59
10) Subtract: a) 954.367 – 412.650 b) 234.45 – 142.52 c) 74.812 – 35.634 d) 732.532 – 522.147 11) Word problems a) By how much should 41.65 be increased to get 98.53? b) S aritha bought 56.6 litres of water. She used 9.2 litres of water for washing her uniform. Find the quantity of water left. 60
Chapter Decimals - II 12 Let Us Learn About • multiplying and dividing decimals by 1-digit and 2-digit numbers. • m ultiplying decimals by 10, 100 and 1000. • multiplying and dividing a decimal number by another decimal number. • the relationship between percentages, decimals and fractions. Concept 12.1: Multiply and Divide Decimals Think Pooja bought six different types of toys for ` 236.95 each. She calculated the total cost and paid the amount to the shopkeeper. Pooja then went to a sweet shop where 410.750 kg of a sweet was prepared. She wanted to know the number of 250 g packs that can be made from it. Do you know how to find the total cost of the toys? Can you calculate how many packs of sweets can be made? Recall Multiplication and division of decimal numbers are similar to that of usual numbers. Let us recall multiplication and division of numbers by answering the following. Solve: a) 267 × 14 b) 3218 × 34 c) 7424 × 14 d) 576 ÷ 12 e) 265 ÷ 5 f) 384 ÷ 4 61
& Remembering and Understanding Multiplication of decimals is similar to multiplication of numbers. When two decimal numbers are multiplied, a) count the total number of digits after decimal point in both the numbers. Say it is ‘n’. b) multiply the two decimal numbers as usual and place the decimal point in the product after ‘n’ digits from the right. Multiply decimals by 1-digit and 2-digit numbers Let us understand the multiplication of decimals through a few examples. Example 1: Solve: a) 25.146 × 23 b) 276.32 × 6 Solution: a) 25.146 × 23 T Th Th H TO 1 1 1 46 11 23 25 1 38 20 × 58 1 + 75 4 502 9 5 7 8.3 Therefore, 25.146 × 23 = 578.358 b) 276.32 × 6 Step 1: To multiply the given numbers, follow the steps outlined here. Multiply the numbers without considering the decimal point. T Th Th H T O 4 3 11 2 7 6 32 ×6 1 6 5 7 92 62
Step 2: Count the number of decimal places in the given number. The number of decimal places in 276.32 is two. Step 3: Count from the right, the number of digits in the product as the number of decimal places in the given number. Then place the decimal point. Therefore, 276.32 × 6 = 1657.92. Multiply decimals by 10,100 and 1000 Example 2: Solve: a) 3.4567 × 10 b) 3.4567 × 100 c) 3.4567 × 1000 Solution: To multiply a decimal number by 10, 100 and 1000, follow the steps. Step 1: Write the decimal number as it is. Step 2: Shift the decimal point to the right by as many digits as the number of zeros in the multiplier. Therefore, a) 3 .4567 × 10 = 34.567 (The decimal point is shifted to the right by one digit as the multiplier is 10.) b) 3.4567 × 100 = 345.67 (The decimal point is shifted to the right by two digits as the multiplier is 100.) c) 3 .4567 × 1000 = 3456.7 (The decimal point is shifted to the right by three digits as the multiplier is 1000.) Multiply a decimal number by another decimal number Multiplication of a decimal number by another decimal number is similar to multiplication of a decimal number by a number. Let us understand this through an example. Example 3: Solve: 7.12 × 3.7 Solution: Step1: Multiply the numbers without considering the decimal point. 1 7 12 × 37 11 4 9 84 +2 1 3 6 0 26 3 44 Decimals - II 63
Step 2: Count the number of decimal places in both the multiplicand and the multiplier and add them. The number of decimal places in 7.12 is two The number of decimal places in 3.7 is three Total number of decimal places = 2 + 1 = 3 Step 3: Count as many digits in the product from the right as the total number of decimal places. Then place the decimal point. Therefore, 7.12 × 3.7 = 26.344. Divide decimal numbers by 1-digit and 2-digit numbers Division of decimal numbers is similar to the division of usual numbers. Let us understand this through a few examples. Example 4: Divide: a) 147.9 ÷ 3 b) 64.2 ÷ 6 Solution: Step 1: Follow the steps to divide. Divide the decimal number (dividend) by the 1-digit number (divisor) as usual. Step 2: Place the decimal point in the quotient exactly above the decimal point in Example 5: t he dividend. a) 49.3 b) 10.7 )3 147.9 )6 64.2 −12 −6 27 042 − 27 − 42 09 00 − 09 b) 56.96 ÷ 32 00 Divide: a) 20.475 ÷ 25 Solution : a) 0.819 b) 1.78 )25 20.475 )32 56.96 − 200 − 32 47 249 − 25 − 224 225 256 − 225 − 256 000 000 64
Divide decimals by 10,100 and 1000 Example 6: Solve: a) 3.4567 ÷ 10 b) 3.4567 ÷ 100 c) 3.4567 ÷ 1000 Solution: To divide a decimal number by 10, 100 and 1000, follow these steps: Step 1: Write the decimal number as it is. Step 2: Shift the decimal point to the left by as many digits as the number of zeros in the divisor. Therefore, a) 3 .4567 ÷ 10 = 0.34567 (The decimal point is shifted to the left by one digit as the divisor is 10.) b) 3 .4567 ÷ 100 = 0.034567 (The decimal point is shifted to the left by two digits as the divisor is 100.) c) 3 .4567 ÷ 1000 = 0.0034567 (The decimal point is shifted to the left by three digits as the divisor is 1000.) Use decimals to continue division of numbers resulting in remainders Recall that sometimes we get remainders in the division of numbers. We can use the decimal point to divide the remainder up to the desired number of decimal places. Let us understand this through a few examples. Example 7: Solve: 54487 ÷ 46 Solution: To divide the given numbers, follow the steps given here. Step 1: Divide as usual, till you get a remainder. 1184 )46 54487 − 46 84 − 46 388 − 368 207 − 184 23 Step 2: Place a point to the right of the quotient. Add a zero to the right of the remainder and continue the division. Decimals - II 65
1184.5 )46 54487 − 46 84 − 46 388 − 368 207 − 184 230 − 230 000 In this case, the division is stopped after one decimal place as the remainder is zero. In some cases, the division continues for more than three decimal places. But usually, we divide up to three decimal places. We then round off the quotient to two decimal places. Example 8: Divide the following up to two decimal places. a) 91158 ÷ 28 b) 78323 ÷ 15 Solution: a) 3255.642 b) 5221.533 )28 91158 )15 78323 − 84 − 75 71 33 − 56 − 30 155 32 − 140 − 30 158 23 − 140 − 15 180 80 − 168 − 75 120 50 − 112 − 45 80 50 − 56 − 45 4 5 Therefore, 91158 ÷ 28 = 3255.64 and 78323 ÷ 15 = 5221.53 after rounding off to two decimal places. 66
Divide a decimal number by another decimal number Let us understand the division of a decimal number by another through an example. Example 9: Solve: 3.0525 ÷ 5.5 Solution: To divide a decimal number by another, follow these steps. Step 1: Convert the decimals into fractions. Step 2: 3.0525 = 30525 and 5.5 = 55 Step 3: 10000 10 Step 4: Find the reciprocal of the divisor. 55 10 Reciprocal of is . 10 55 Multiply dividend by the reciprocal of divisor. 30525 10 30525 555 × = = 10000 55 55´1000 1000 Convert the fraction to a decimal number. 555 = 0.555 1000 Therefore, 3.0525 ÷ 5.5 = 0.555 Application Let us see a few real-life examples where we use multiplication and division of decimal numbers. Example 10: Sania bought 10 dozen bananas, each dozen costing ` 120.50. What amount does Sania have to pay altogether? Solution: Cost of 1 dozen bananas = ` 120.50 Cost of 10 dozen bananas = ` 120.50 × 10 = ` 1205.00 Therefore, Sania has to pay ` 1205.00 altogether. Example 11: The weight of a bag of wheat is 19.85 kg. Find the weight of 14 such bags. Solution: The weight of one wheat bag = 19.85 kg Weight of 14 such bags = 19.85 kg × 14 = 277.9 kg Therefore, 14 bags weigh 277.9 kg. Decimals - II 67
Example 12: Mr. Arun purchases 3 kg of potatoes for ` 8.70. What is the cost of 1 kg potatoes? Solution: Cost of 3 kg potatoes = ` 8.70 Cost of 1 kg potatoes = ` 8.70 ÷ 3 = ` 2.90 Therefore, 1 kg potatoes costs ` 2.90. Example 13: Rahul bought 8 bags for ` 56.24. Find the cost of one bag. Solution: Cost of 8 bags = ` 56.24 Cost of one bag = ` 56.24 ÷ 8 = ` 7.03 Therefore, the cost of a bag is ` 7.03. Higher Order Thinking Skills (H.O.T.S.) Let us solve a few more examples of multiplication and division of decimal numbers. Example 14: Find the missing numbers. a) ____ × 100 = 467.2 b) 53.052 × ____ = 530.52 c) ____ × 10 = 3764 d) 901.5 × ____ = 90150 Solution: a) 4.672 b) 10 c) 376.4 d) 100 Example 15: Roopa bought 6 bags for ` 246.12 and Pooja bought 8 bags for ` 348.16. Who paid less for each bag and by how much? Solution: Cost of 6 bags that Roopa bought = ` 246.12 Cost of one bag = ` 246.12 ÷ 6 = ` 41.02 Cost of 8 bags that Pooja bought = ` 348.16 Cost of one bag = ` 348.16 ÷ 8 = ` 43.52 As ` 41.02 < ` 43.52, Roopa paid less for the bags. The amount that Roopa paid less for each bag =` 43.52 – ` 41.02 = ` 2.50 68
Concept 12.2: Percentages Think Pooja and her mother went to a shopping mall. There she saw a banner as shown here and asked her mother about the sign written beside 50. Do you know what sign it is? Recall We have already learnt how to convert a decimal into a fraction. Let us recall the concept by writing the decimals given below as fractions. a) 0.76 b) 0.34 c) 0.57 d) 0.45 e) 0.92 & Remembering and Understanding Observe the following fractions. Train My Brain a) 4 b) 72 c) 82 d) 14 100 100 100 100 All the fractions have the denominator 100. Such a fraction with denominator 100 is expressed as a percentage. Convert fraction into percentage To convert a fraction into a percentage, multiply the fraction by 100%. Consider these examples. Example 16: Convert the following fractions into percentages. a) 76 b) 34 c) 43 d) 5 e) 3 100 100 50 10 5 Decimals - II 69
Solution: S.No. Fraction Conversion Percent Read as a) 76 76 100 × 100% = 76% 76 % Seventy-six percent 100 b) 34 34 34 % Thirty-four percent 100 100 × 100% = 34% 43 43 86 % Eighty-six percent c) 50 × 100% 50 = 43 × 2% = 86% d) 5 5 ×100% 50 % Fifty percent 10 10 = 5 × 10% = 50% e) 3 3 60 % Sixty percent 5 5 × 100% = 3 × 20% = 60% Convert percentage into fraction To convert a percentage into a fraction, divide by 100. Consider these examples. Example 17: Convert the following percentages into fractions. a) 73% b) 1% c) 6.5% d) 10% e) 18.6% Solution: S.No. Percent Fraction a) 73% 73 100 b) 1% 1 100 c) 6.5% 6.5 = 65 d) 10% 100 1000 10 100 e) 18.6% 18.6 = 186 100 1000 70
Convert percentage into decimal To convert a percentage into a decimal, write the number and place a decimal point after two digits from its right. Consider these examples. Example 18: Convert the following percentages into decimals. Solution: a) 200% b) 13% c) 150% d) 25% e) 300% S.No. Percent Decimal a) 200% 2.0 b) 13% 0.13 c) 150% 1.5 d) 25% 0.25 e) 300% 3.0 Application Let us now see a few real-life examples in which conversion of decimals and percentages are used. Example 19: There are 40 mangoes and 60 apples in a basket. What are the percentages of mangoes and apples in the basket? Solution: Total number of fruits = 40 + 60 =100 Fraction of mangoes = 40 100 Example 20: Percentage of mangoes = 40 × 100 % = 40% 100 Fraction of apples = 60 100 Percentage of mangoes = 60 × 100 % = 60 % 100 Madhu secured 9 in Mathematics and 46 in Science. In which subject did 10 50 Madhu perform better? Solution: Mathematics is out of 10 marks and Science is out of 50 marks. So, we cannot compare both directly. We have to find the percentage of his scores in the two subjects to compare them. Decimals - II 71
9 as a percentage = 9 × 100 % = 90% 10 10 46 as a percentage = 46 × 100% = 92% 50 50 92% > 90% Therefore, Madhu performed better in Science. Higher Order Thinking Skills (H.O.T.S.) Let us solve a few more examples of percentages. Example 21: Out of a class of 35 students, 28 students know swimming. What percent of students do not know swimming? S olution: Total number of students = 35 Number of students who know swimming = 28 Number of students who do not know swimming = 35 – 28 = 7 Fraction of students who do not know swimming = 7 35 Percent of students who do not know swimming = 7 × 100 % = 1 × 100 % 35 5 = 20% Example 22: Sanjay spent 35% of his income on rent, 25% on household expenses, 30% on savings and remaining on petrol. How much did he spend on petrol? Solution: Let Sanjay’s income be ` 100. Percent of income spent on rent = 35% Amount spent on rent = ` 35 Percent of income spent on household expenses = 25% Amount spent on household expenses = ` 25 Percent of income spent on savings = 30% Amount saved = ` 30 Remaining amount spent on petrol = ` [100 – (35 + 25 + 30)] = ` 100 – ` 90 = ` 10 Therefore, Sanjay spent 10% of his income on petrol. 72
Drill Time Concept 12.1: Multiply and Divide Decimals 1) Multiply the following: a) 2.498 × 10 b) 32.64 × 53 c) 5.645 × 1000 d) 682.93 × 2.8 e)1.742 × 3.81 2) Divide the following: a) 6.78 ÷ 10 b) 88.8 ÷ 20 c) 124.8 ÷ 16 d) 3.0525 ÷ 5.5 e) 832.5 ÷ 2.5 3) Word problems a) T he daily wages of a labourer is ` 120.85. Find the daily wages of 26 labourers. b) T he consumption of petrol by a truck per 2 weeks is 101.25 litres. How many litres of petrol does it consume per day? Concept 12.2: Percentages 4) Convert these fractions into percentages. d) 83 e) 32 a) 21 b) 67 c) 20 150 100 50 100 45 5) Convert these percentages into fractions. d) 53% e) 24% a) 0.45% b) 7.6% c) 43% Percent 6) Complete the table. 26% S.No Decimal Fraction a) 1.5 b) 8 c) 0.65 10 d) e) 18 100 Decimals - II 73
7) Word problems a) T here are 30 blue ribbons and 50 red ribbons in a box. What are the percentages of blue ribbons and red ribbons in the box? b) Of a group of 25 children in a locality,18 know singing. What percent of children do not know singing? 74
Chapter Measurements 13 Let Us Learn About • p erimeter of a rectangle and a square. • area of a rectangle, a square and a triangle. • volume of a cube and a cuboid. Concept 13.1: Perimeter, Area and Volume Think Pooja uses 17.5 cm of wool to make a friendship bracelet. How can Pooja find out how much wool she will need to make 3 such friendship bracelets? Recall Recall that figures or shapes such as triangle, square and rectangle are called flat shapes. These are also known as 2D shapes. square rectangle triangle circle 75
In the previous class, we have also learnt about solids such as: H G Edge Corner H G C Face Face D C D E F E F B A B Corner Edge A cuboid cube Corner Flat Face Curved Curved Curved face face face Flat Face cylinder cone sphere A 2D shape has edges and corners. A 3D shape has curved or flat faces, corners and edges. Let us revise the concept by naming the corners and sides of the given 2D shapes. a) b) c) R D CH G P QA BE F & Remembering and Understanding We have learnt about the simple closed figures made of straight line segments. Let us now learn about their perimeters and areas. Perimeter The sum of the lengths of a closed figure is called its perimeter. It is the distance all the way around a two-dimensional shape. It is calculated by adding the length of all sides together and is denoted by ‘P’. 76
Perimeter of a rectangle: Consider the given rectangle ABCD. The two opposite sides AB & CD are called lengths D C They are represented by ‘’. The two opposite sides BC & DA are called breadths b B They are represented by ‘b’. A Perimeter of a rectangle = Sum of all its sides = AB + BC + CD + DA = AB + CD + BC + DA = 2 length () + 2 breadth (b) = 2 (length + breadth) S Therefore, P = 2 ( + b) Perimeter of a square: Consider the given square ABCD. DC Perimeter of a square = Sum of all its sides MeasuremS ents S = AB + BC + CD + DA = side + side + side + side AB = 4 × side S =4s Since perimeter is the total length of all the sides of a closed figure, its units are the units of length such as cm, m or km. Example 1: Find the perimeter of the given rectangles. a) 2 cm b) 3 cm c) 4 cm 5 cm 8 cm d) 4 cm 3 cm 13 cm Measurements 77
Solution: a) P erimeter of a rectangle = 2 ( + b) units b) Perimeter of a rectangle = 2 ( + b) units Given length = 4 cm and breadth = 2 cm Given length = 5 cm and breadth = 3 cm P = 2 (4 cm + 2 cm) = 2 × 6 cm = 12 cm P = 2 (5 cm + 3 cm) = 2 × 8 cm = 16 cm T herefore, perimeter of the given T herefore, perimeter of the given rectangle is 12 cm. rectangle is 16 cm. c) P erimeter of a rectangle = 2 ( + b) units d) Perimeter of a rectangle = 2 ( + b)units Given length = 3 cm and breadth = 8 cm G iven length = 13 cm and breadth = 4 cm P = 2 (3 cm + 8 cm) = 2 × 11 cm = 22 cm P = 2 (13 cm + 4 cm) = 2 × 17 cm = 34 cm T herefore, perimeter of the given T herefore, perimeter of the given rectangle is 22 cm. rectangle is 34 cm. Example 2: Find the perimeter of the given squares. a) b) 2 cm 9 cm Solution: a) Perimeter of a square = 4 × s units b) Perimeter of a square = 4 × s units Given side = 9 cm Given side = 2 cm P = 4 × 9 cm P = 4 × 2 cm = 36 cm = 8 cm Therefore, perimeter of the given square is Therefore, perimeter of the given square is 36 cm. 8 cm. Area Area specifies the region in the number of squares of unit side. So, its units are square units, written in short as sq. units. The lengths of the sides of a 2D figure are usually in cm and m. So, the units of its area are sq. cm or cm2 and sq. m or m2 respectively. Area of an object is the amount of surface or region covered by it. It is denoted by ‘A’. 78
Area of a rectangle: Observe the given rectangle. 1 cm 1 cm The region covered by the rectangle = The total region covered by 15 squares of size 1 cm × 1 cm = 15 × 1 cm × 1 cm = 15 cm2 or 15 sq. cm The rectangle can be considered to have a length of 5 cm and breadth of 3 cm. Thus, area of a rectangle = 15 sq. cm = 5 cm × 3 cm = length × breadth = × b sq. units Area of a square: Observe the given square. The surface region covered by the square = The total region covered by 9 squares of size 1cm × 1 cm. = 9 × 1 cm × 1 cm 1 cm = 9 cm2 or 9 sq. cm 1 cm = 3 cm × 3 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 = s2 sq. units 1 cm Area of a triangle: Observe the triangle formed from a rectangle in 1 cm these figures. The diagonal of a rectangle or square divides it into two equal triangles. T hus, the surface or region covered by the triangle formed = 1 of the total region covered by the rectangle. 2 Thus, the area of a triangle = 1 the area of the rectangle 1 cm 2 1 cm =1 × b 2 Observe the triangle formed from a square. Measurements 79
T he area of the triangle thus formed = 1 of the area of the square Area in square units 2 1) Rectangle : × b =21 s2 2) Square : s × s 3) Triangle : 1 b × h Since, we cannot use two different formulae for the area of a triangle; we give one common formula as 2 Area of a triangle = 1 × base (length or side) × height (breadth or side) 2 =1 × bh sq. units 2 Example 3: Find the area of the given figures: 9.5 cm a) b) c) 3.6 cm 12 cm 11 cm 9 cm d) 7.8 cm n 8.5 cm g) f) 7.8 cm e) 3.5 cm 13 cm 10 cm 3.5 cm 15 cm 9.8 cm Solution: a) Area of a rectangle = × b b) Area of a rectangle = × b = 11 cm × 3.6 cm = 12 cm × 9.5 cm = 39.6 sq. cm = 114 sq. cm c) Area of a triangle = 1 × b × h d) Area of a square = s2 2 = 7.8 cm × 7.8 cm =1 × 8.5 cm × 9 cm 2 = 60.84 sq. cm = 38.25 sq. cm 80
e) Area of a square = s2 f) Area of a rectangle = × b = 3.5 cm × 3.5 cm = 15 cm × 13 cm = 12.25 sq. cm = 195 sq. cm g) Area of a triangle = 1 × b ×h= 1 × 9.8 cm × 10 cm = 49 sq. cm 2 2 Volume The matter contained in a solid object is called its volume. In other words, the space occupied by a solid or a 3D shape is called its volume. Since, the solid has three dimensions, its volume is given by their product. The volume of a solid is the number of unit cubes that fit exactly in a solid. Unit cube: A cube of edge 1 cm is called a unit cube. Let us now find the volumes of a cube and cuboid. Volume of a cube: Top Layer Observe the given cube. The number of unit cubes contained in it is Bottom Layer 4 (bottom layer) + 4 (top layer) = 8. So, the volume of the given cube is 8 cubic units, written in short as 8 cu. units. Volume of the cube = 8 cubic units =8 × 1 cubic units = {(2 × 2 × 2) × (1 × 1 × 1)} cubic units = (2 × 1) × (2 × 1) × (2 × 1) cubic units = side × side × side cu. units Layer 4 Layer 3 =s3 cu. units Layer 2 Layer 1 Volume of a cuboid: Observe the given cuboid. The number of unit cubes contained in it is 4 per layer. The number of layers is 4. Measurements 81
So, the total number of unit cubes in the given cuboid is 4 × 4 = 16. Thus, the volume of the cuboid is 16 cubic units, written in short as 16 cu. units. Volume of a cuboid = 16 cu. units. =2 × 2 × 4 cu. units = length × breadth (width) × height cu. units =bh cu. units. The edges of cubes and cuboids are measured in centimetres or metres. So, their volume is obtained in cubic centimetres or cubic metres written as cu. cm or cu. m. Note: 1) In solids, breadth is usually termed as width. 2) Height is sometimes taken as depth. Example 4: Find the volume of the given figures where = 1 cubic cm. a) b) Solution: a) Volume of a cube = s × s × s = 5 cm × 5 cm × 5 cm = 125 cu. cm b) Volume of a cube = s × s × s = 3 cm × 3 cm × 3 cm = 27 cu. cm 82
Application Let us see how we can apply the concepts of perimeter, area and volume in real life. Example 5: Find the perimeters of the given figures. a) 3.3 cm b) 8.9 cm 7.2 cm 3.8 cm4.5 cm2 cm 2 cm 3.4 cm 1.8 cm 4.3 cm 1.8 cm 7.8 cm 4.9 cm Solution: a) Perimeter of the given figure = Sum of all its sides = 7.8 cm + 3.4 cm + 4.5 cm + 3.8 cm + 3.3 cm + 7.2 cm = 30 cm b) Perimeter of the given figure = Sum of all its sides = (4.9 + 4.3 + 2 + 1.8 + 8.9 + 1.8 + 2 + 4.3) cm = 30 cm Example 6: Find out the volume of the following solids. The volume of each cube is 1 cu. cm. a) b) Solution: a) Number of cubes at the bottom = 7 Number of cubes at the top = 1 Total number of cubes = 7 + 1 = 8 Volume of 1 cube = 1 cu. cm Therefore, volume of the given solid = 8 × 1 cu. cm = 8 cu. cm Measurements 83
b) Number of cubes at the bottom = 5 Number of cubes at the top = 2 Total number of cubes = 5 + 2 = 7 Volume of 1 cube = 1 cu. cm Therefore, volume of the given solid = 7 × 1 cu. cm = 7 cu. cm Example 7: Adil made a park on the two sides of his house. 2m park 2m The length of his house is 15 m and breadth is 10 m. Adil’s house 10 m The breadth of the park is 2 m. Find the area of the park. Solution: Adil’s park can be divided into two different 15 m 2m rectangles. 17 m 1 We know the breadth of the park = 2 m 2 park Therefore, length of rectangle 1 = 15 m + 2 m 10 m = 17 m Adil’s house Length of rectangle 2 = 10 m 15 m 2m Area of rectangle 1 = × b = 17 m × 2 m = 34 sq. m. Area of rectangle 2 = × b = 10 m × 2 m = 20 sq. m. Therefore, the total area of the park = 34 sq. m. + 20 sq. m. = 54 sq. m. Higher Order Thinking Skills (H.O.T.S.) Let us see some more examples of perimeter, area and volume. Example 8: Find the length of the side and area of a square whose perimeter is 68 m. Solution: Perimeter of the square = 4 s = 68 m Side of the square = 68 m ÷ 4 = 17 m Area of the square = s2 = 17 m × 17 m = 289 sq. m 84
Example 9: A swimming pool is in the shape of a cuboid. Its length is 45 m, breadth is 18 m and volume is 4860 cu. m. Find the depth of the swimming pool. Solution: Volume of the cuboid = × b× h Length of the swimming pool = 45 m Breadth of the swimming pool = 18 m Depth of the swimming pool = ? Volume of the cuboidal swimming pool = 4860 cu. m × b× depth (height) = 4860 cu. m h = 4860 cu. m ÷ ( × b) 4860 m3 = 45 ×18 m2 = 6 m Therefore, the depth of swimming pool is 6 m. Drill Time Concept 13.1: Perimeter, Area and Volume 1) Find the perimeter and area of the following: a) Square with side 8 cm b) Rectangle with = 12 cm and b = 2 cm c) A B d) P 3 cm 10 cm 8 cm R D C 6 cm 14 cm Q 2) Find the volume of the following: a) A cube with side 9 cm b) A cube with side 7 cm c) A cuboid with = 2 cm, b = 3 cm, h = 4 cm d) A cuboid with = 3 cm, b = 2 cm, h = 5 cm e) A cube with side 5 cm Measurements 85
3) Word problems a) If two squares of side 30 cm each are joined to form a rectangle, what is the area of the rectangle? b) A box has a square base of side 5 cm. What is its volume if the height of the box is 8 cm? c) W hich has a greater volume - a cuboid of length 12 cm, breadth 8 cm and height 5 cm or a cube of edge 7 cm? 86
Chapter Data Handling 14 Let Us Learn About • the term ‘circle graph’. • interpreting and constructing circle graphs. Concept 14.1: Circle Graphs Think Other gases Pooja was reading a Science lesson from her sister’s book. She saw a chart displaying the composition Oxygen of gases in the air as shown. She did not understand Nitrogen what it was and how to read it. She asked her sister about it. What do you understand about the chart? Recall Let us recall the concepts of fractions and decimals that we have already learnt. Complete the table. One has been done for you. 87
Percent Fraction (in lowest terms) Decimal 15% 0.15 15 = 3 12% 100 20 28% 3 4 35% & Remembering and Understanding In Class 4, we have learnt in detail about bar graphs. Now, let us learn about one more type of representation of data, that is, circle graph. A circle divided into parts to show the fraction of each category of data is called a circle graph or circle chart. They are also called pie charts. Let us understand this through some examples. Example 1: Students of class V spent a week at a summer camp. At the end of the week, the students were asked about their favourite part of the camp. The given circle graph shows their responses. Observe the graph and answer the questions that follow. Favourite Activity of Students Art and Craft Class Dancing Riding a Horse Playing Cricket 88
a) Which activity did the students enjoy the most? b) Between the two classes, Dancing and Art & Craft which one was more preferred more by the students? c) What fraction of the students chose horse riding as their favourite activity? d) Which activity did the students enjoy the least? e) W hat fraction of the students chose cricket as their favourite activity? Solution: a) The students enjoyed horse riding activity the most. b) Dancing class was preferred more by the students. c) Half of the students chose horse riding as their favourite activity. d) The students enjoyed Art & Craft activity the least. e) One fourth of the students chose cricket as their favourite activity. Example 2: Time spent by Arun on homework is given as a circle graph. Observe the graph carefully and answer the questions that follow. Time Spent by Arun on Homework Spelling - 12% Reading - 8% Social Studies - 15% Maths - 30% English - 15% Science a) On which subject did Arun spend the maximum time? b) On which subject did Arun spend the least time? c) On which two subjects did Arun spend the same amount of time? Solution: d) What percent of time was spent on Science? e) What is the total time spent on reading and spelling? a) Arun spent the maximum time on Maths. b) Arun spent the least time on Reading. c) Arun spent the same amount of time on English and Social Studies. Data Handling 89
d) Percentage of time spent on Science = [100 – (15 + 15 + 30 + 12 + 8)] % = [100 – 80]% = 20% e) The total time spent on reading and spelling is 12% + 8% = 20% Application We have learnt to interpret the circle graphs. Now, let us learn to draw the circle graph. • If information is given in the form of a paragraph, we would first draw a table and then construct the circle graph. To draw a circle graph, we need to determine the following: • what is the ‘whole’ • how many different parts or groups are there in the data • what percentage is each part or data group of the whole Steps for constructing circle graphs 1) Determine the parts and the whole: Determine if there is a “whole” for the data. Then determine what different parts or data groups of the whole are. 2) Calculate percentages: For data that is not already given as a percentage, convert the amounts for each part or data group size into a percentage of the whole. 3) Draw the graph: Draw a circle of any convenient radius. Divide the circle into corresponding equal parts for each data group. Try to make the part sizes look as close to the percentage of the circle as the percentage of the data group. 4) Title and label the graph: Label the parts with the data group name and percentage. Then add a title to the graph. This is same as the title of the table. Example 3: In a class of 48 students, 6 students come to school by car, 18 students come by bus, 12 students come by bicycle and 12 students come by auto. Draw a circle graph according to the data. Solution: Follow these steps to draw a circle graph. Step 1: Represent the data in tabular form. 90
Class Means of transport Number of students Car 6 Bus 18 Bicycle 12 Auto 12 Step 2: Determine the fraction or percentage of the given data groups. Step 3: Class Means of Number of Fraction Percentage transport students 6 61 1 Car 48 = 8 8 × 100% = 12.5% 18 3 × 100% = 37.5% Bus 18 = 3 8 12 48 8 2 × 100% = 25% Bicycle 8 12 = 2 48 8 Auto 12 12 = 2 2 48 8 8 × 100% = 25% Draw a circle and divide it into 8 equal parts. Step 4: From step 2 we can conclude that 1 part out of 8 parts can be coloured for car. Similarly, 3, 2 and 2 parts out of 8 can be coloured for bus, bicycle and auto respectively. Data Handling 91
Means of Transport of Students Auto Car 25% 12.5% Bicycle Bus 25% 37.5% Example 4: Mani’s monthly earnings is ` 9500. He spends 40% of it on food, 10% on rent, Solution: 10% on paying bills (electricity, mobile, water bill, etc.). He spends 20% on his children’s education and 10% on petrol and saves 10%. Represent the data in the form of a circle graph. 1) Represent the data in tabular form. Mani’s monthly expenditure Food 40% 4 10 Rent 1 10% 10 1 Bills 10% 10 Education 20% 2 10 Petrol 10% 1 10 Savings 10% 1 10 2) D raw a circle of any convenient radius. Divide it into required parts to represent the data groups. 92
Petrol Mani’s Monthly Expenditure 10% Savings 10% Education 20% Food 40% Bills 10% Rent 10% Higher Order Thinking Skills (H.O.T.S.) Let us draw a few more circle graphs for the data given. Example 5: The marks obtained in a test by the students of Class V are given. Prepare the tally marks, draw a circle graph for the same and answer the questions that follow. 34, 32, 32, 30, 30, 30, 32, 34, 32, 30, 34, 30, 32, 34, 30, 30, 43, 43, 34, 32, 32, 30, 34, 34, 32, 30, 30, 43, 34, 30, 34, 32, 30, 43, 30, 43, 30, 34, 30, 32 a) Which mark is obtained by most of the students? b) Which mark is obtained by the least number of students? c) How many students obtained the highest mark? d) How many students obtained the least mark? e) What percent of students obtained 30 marks? f) How many students are there in the class? Solution: Marks of Class V students Marks Tally marks Number of Students Fraction 30 15 = 3 15 40 8 32 10 10 = 2 40 8 Data Handling 93
Marks of Class V students Marks Tally marks Number of Students Fraction 34 43 10 10 = 2 40 8 5 5 =1 Performance of Class V Students 40 8 10 students (scored 32 marks) 15 students (scored 30 marks) 10 students (scored 34 marks) 5 students (scored 43 marks) a) 30 marks b) 43 marks c ) 5 students d) 15 students Example 6: e) 15 × 100% = 37.5% f) 40 students 40 The different aspirations of the students of a class are as given: Doctor, Engineer, Scientist, Doctor, Businessman, Engineer, Doctor, Engineer, Engineer, CA, Businessman, Businessman, Doctor, Engineer, Scientist, Doctor, Engineer, Engineer, Scientist, Doctor, Engineer, Scientist, Businessman, Engineer, Scientist, Engineer, Doctor, Engineer, Scientist, CA, Engineer, Doctor. Draw a circle graph from the tally marks table and answer these questions: a) Which profession is aspired by the maximum number of students? b) W hich profession is aspired by the least number of students? c) How many students want to become businessmen? d) How many students want to become a scientist? e) How many students are there in the class altogether? 94
Solution: Aspirations of students of a class Profession Tally marks Number of Fraction of students Doctor students Engineer 8 8 = 4 = 25% 32 16 12 12 = 6 = 37.50% 32 16 Scientist 6 6 = 3 = 18.75% Businessman 32 16 4 42 32 = 16 = 12.50% CA 2 2 = 1 = 6.25% 32 16 We divide a circle into 16 parts and represent these as fractions of the total number of students. a) Engineer b) CA c) 4 students d) 6 students e) 32 students are there in the class. Aspiration of Students of a Class 6.25% Businessman CA 12.50% Scientist Doctor 18.75% 25% Engineer 37.50 Data Handling 95
Drill Time Concept 14.1: Circle Graphs Solve the following: a) Draw the circle graph based on the following data. Pets owned by people Percentage of people Dog 30% Cat 15% Fish 20% 10% Rabbit 25% Parrot b) Raghu spends his day as given in the table here. At school 8 hours Playing 2 hours Doing homework 2 hours Reading 1 hour Routine work 3 hours Sleeping 8 hours Draw a circle graph and colour the groups with different colours. (Hint: Divide the circle into 24 equal parts as 24 hours = 1 day) c) The data shows the students with their favourite sports. Draw a circle graph based on the data. Favourite sport Percentage of students Football 25% Cricket 50% 25% Basketball 96
Search
