Alpine-G5-Textbook-Maths-FY

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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 151 Decimals - II 147 03-Feb-18 2:37:03 PM

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 the dividend. a) 49.3 b) 10.7 )3 147.9 )6 64.2 −12 −6 27 042 − 27 − 42 09 00 − 09 Example 5: 00 b) 56.96 ÷ 32 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 − 256 000 000 148 181010044-Alpine-G5-Textbook-Maths-FY.pdf 152 03-Feb-18 2:37:03 PM

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. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 153 Decimals - II 149 03-Feb-18 2:37:03 PM

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. 150 181010044-Alpine-G5-Textbook-Maths-FY.pdf 154 03-Feb-18 2:37:03 PM

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. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 155 Decimals - II 151 03-Feb-18 2:37:03 PM

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 = ` 2.50 152 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 156

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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 157 Decimals - II 153 03-Feb-18 2:37:04 PM

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% c) 43 43 86 % Eighty-six percent 50 50 × 100% = 43 × 2% = 86% d) 5 5 ×100% 50 % Fifty percent 10 10 = 5 × 10% = 50% 3 3 60 % Sixty percent e) 5 × 100% 5 = 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 154 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 158

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 155 181010044-Alpine-G5-Textbook-Maths-FY.pdf 159 03-Feb-18 2:37:04 PM

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? Solution: 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. 156 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 160

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 fortnight 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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 161 Decimals - II 157 03-Feb-18 2:37:04 PM

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? 158 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 162

Chapter Measurements 13 Let Us Learn About • p erimeter of a rectangle and a square. • area of a rectangle, a square and a triangle. • v olume 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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 163 159 03-Feb-18 2:37:04 PM

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’. 160 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 164

Perimeter of a rectangle: Consider the given rectangle ABCD. The two opposite sides AB & CD are called lengths D C B They are represented by ‘’. The two opposite sides BC & DA are called breadths 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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 165 Measurements 161 03-Feb-18 2:37:04 PM

Solution: a) Perimeter 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) P erimeter 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’. 162 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 166

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. Thus, 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. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 167 Measurements 163 03-Feb-18 2:37:04 PM

The 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 164 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 168

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 22 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. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 169 Measurements 165 03-Feb-18 2:37:04 PM

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 166 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 170

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 167 181010044-Alpine-G5-Textbook-Maths-FY.pdf 171 03-Feb-18 2:37:04 PM

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 168 03-Feb-18 2:37:04 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 172

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 6 cm 10 cm Q 8 cm R D 14 cm C 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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 173 Measurements 169 03-Feb-18 2:37:05 PM

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) Which has a greater volume - a cuboid of length 12 cm, breadth 8 cm and height 5 cm or a cube of edge 7 cm? 170 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 174

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. 181010044-Alpine-G5-Textbook-Maths-FY.pdf 175 171 03-Feb-18 2:37:05 PM

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. Art and Craft Class Dancing Riding a Horse Playing Cricket 172 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 176

a) W hich activity did the students enjoy the most? b) B etween 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) What 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 173 181010044-Alpine-G5-Textbook-Maths-FY.pdf 177 03-Feb-18 2:37:05 PM

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. 174 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 178

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 175 181010044-Alpine-G5-Textbook-Maths-FY.pdf 179 03-Feb-18 2:37:05 PM

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 4 40% 10 Rent 1 10% 10 Bills 10% 1 10 Education 20% 2 10 Petrol 10% 1 10 Savings 10% 1 10 2) Draw a circle of any convenient radius. Divide it into required parts to represent the data groups. 176 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 180

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 = 2 10 40 8 181010044-Alpine-G5-Textbook-Maths-FY.pdf 181 Data Handling 177 03-Feb-18 2:37:05 PM

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 15 students 10 students 5 students a) 30 marks b) 43 marks c) 5 students d) 15 students e) 15 × 100% = 37.5% f) 40 students 40 Example 6: 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) Which 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? 178 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 182

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 181010044-Alpine-G5-Textbook-Maths-FY.pdf 183 Data Handling 179 03-Feb-18 2:37:05 PM

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 180 03-Feb-18 2:37:05 PM 181010044-Alpine-G5-Textbook-Maths-FY.pdf 184