181010032-Alpine-G4-Textbook-Maths-FY

Solution: Fraction of the cardboard painted by Venu = 3 6 Fraction of the cardboard painted by Raj = 1 3 Area of the cardboard = 144 sq. cm Area of the cardboard painted by Venu = 3 × 144 sq.cm Example 7: 6 Solution: =3 ×144 sq.cm = 432 sq.cm = 72 sq.cm 66 Area of the cardboard painted by Raj = 1 × 144 sq.cm 3 =1× 144 sq.cm = 144 sq.cm = 48 sq.cm 33 Therefore, Venu painted 72 sq.cm of the cardboard and Raj painted 48 sq.cm of the cardboard. Find if 2 of 154 and 4 of 49 are equal to each other or one of them is greater 11 7 than the other. To find if the fractions of the numbers are equal, we first find their valu and compare them. 2 of 154 = 2 × 154 = 2 × 154 = 308 = 28 11 11 11 11 4 of 49 = 4 × 49 = 4 × 49 = 196 = 28 77 77 Therefore, 24 of 49. of 154 = 11 7 Concept 9.2: Conversions of Fractions Think Jasleen knew about fractions in which the numerators were less than their denominators. She wondered if there could be some fractions in which the numerators are greater than their denominators. Do you know of such fractions? Fractions - II 97 Alpine_V3_Maths_G4_TB.indb 97 30-01-2018 16:09:55

Recall In the previous chapter, we have learnt about addition and subtraction of like fractions. Recall that the sum of two like fractions is a like fraction. Let us answer these to recall the concept. a) 2 + 1 = ______ 41 5 5 b) 7 + 7 = ______ c) 1 + 5 = ______ d) 3 + 1 = _______ 11 11 22 e) 1 + 3 = ______ f) 2 + 1 = _______ 88 99 & Remembering and Understanding Consider 1 + 5 = 6 . Here, the sum of two like fractions is a like fraction with its numerator less 888 than its denominator. Such fractions are called proper fractions. Sometimes, it is possible that we get the sum with its numerator greater than the denominator. For example, 7 + 5 = 12 . Here, the sum of two like fractions is a like fraction with its 88 8 numerator greater than its denominator. Such fractions are called improper fractions. Note: In some cases, the sum of the numerators of the like fractions may be equal to the denominator. Then, the fraction is said to be an improper fraction. For example, 3 + 4 = 7 , 3 + 5 = 8 and so on. 7 7 78 8 8 Fractions such as 7 , 8 and so on can also be written as a whole, that is 1. 78 12 8 + 4 . This has a whole  8 We can write 8 as the sum of like fractions as 88  8  and a proper fraction  4  . That is, 12 =1+ 4 = 14 . Such fractions are called mixed fractions.  8  8 8 8 A mixed fraction is also called a mixed number. For example, in the mixed fraction 12 3 , 12 is the whole and 3 is the proper fraction. 88 98 30-01-2018 16:10:00 Alpine_V3_Maths_G4_TB.indb 98

In short, we can say that, Proper fractions – Fractions having the numerators less than the denominators. Improper fractions – Fractions having the numerators greater than the denominators. Mixed fractions – Fractions having whole numbers and proper fractions. Example 8: List out proper fractions, improper fractions and mixed fractions from the following: 13 ,15 7, 11 , 37 ,9 , 65 13, 143 , 75 3 ,1074297 , 72 , 68 2 , 29 , 50 23, 69 , 53 18 9 34 6 14 17 98 4 59 5 32 35 32 30 Solution: From the given fractions, Proper fractions: 13 , 11 , 9 , 29 18 34 14 32 Improper fractions: 37 , 143 , 72 , 69 , 53 6 98 59 32 30 Mixed fractions: 15 7, 6513 , 75 3 , 107 27 , 68 2 , 50 23 9 17 4 49 5 35 We usually write fractions as proper or mixed fractions. So, we need to learn to convert improper fractions to mixed fractions and mixed fractions to improper fractions. Conversion of improper fractions to mixed fractions Let us understand the conversion of improper fractions to mixed fractions by solving a few examples. Example 9: Convert 37 to its mixed fraction form. Solution: 6 To convert improper fractions into mixed fractions, follow these steps. Solved Solve these Steps 37 143 72 69 53 6 98 59 32 30 Step 1: Divide the numerator 6 by the denominator. )6 37 − 36 1 Fractions - II 99 Alpine_V3_Maths_G4_TB.indb 99 30-01-2018 16:10:03

Solved Solve these 37 Steps 6 143 72 69 53 30 98 59 32 Step 2: Write the quotient as The mixed the whole. The remainder is fraction form of the numerator of the proper fraction and the divisor is its 37 is 6 1 . denominator. This gives the 66 required mixed fraction. Conversion of mixed fractions to improper fractions Let us understand the conversion of mixed fractions into improper fractions by solving a few examples. Example 10: Convert 15 7 into its improper fraction. 9 Solution: To convert mixed fractions into improper fractions, follow these steps. Solved Solve these 75 3 Steps 15 7 65 13 4 107 27 9 17 49 Step 1: Multiply the whole by the 15 × 9 = 135 denominator. Step 2: Add the numerator of the proper 135 + 7 = 142 fraction to the product in Step 1. Step 3: Write the sum as the denominator The improper of the proper fraction. fraction form of This given the required improper fraction. 15 7 is 142 9 9. Application Let us now see a few real-life examples involving conversions of fractions. Example 11: Rohan wants to arrange 60 books in his shelf. If only 13 books can be put in a rack, how many racks will be filled by the books? Give your answer as a mixed fraction and as an improper fraction. 100 30-01-2018 16:10:04 Alpine_V3_Maths_G4_TB.indb 100

Solution: Number of books Rohan wants to arrange = 60 Number of books that can be arranged on each rack = 13 Example 12: 8 Number of racks that are filled = 0 ÷ 13 = 4 Solution: 13 Improper fraction equivalent to 4 8 = 60 13 13 On a science fair day, a group of students prepared 12 1 litres of orange 2 juice. Express the number of litres of orange juice as an improper fraction. Number of litres of orange juice made = 12 1 2 Improper fraction equivalent to 12 1 = 12 × 2 +1 = 25 2 2 2 Higher Order Thinking Skills (H.O.T.S.) Conversion of fractions is done when we need to add and subtract fractions. In the previous chapter, we have already learnt addition and subtraction of like (proper) fractions. Let us see a few examples that involve addition and subtraction of improper and mixed fractions. Example 13: Add: a) 42 + 35 b) 50 23 + 16 Solution: 25 25 35 35 a) 42 + 35 25 25 T o add the given like improper fractions, we add their numerators and write the sum on the same denominator. Therefore, 42 + 35 = 42 + 35 = 77 25 25 25 25 W e usually write fractions as proper or mixed fractions. So, we convert the sum into a mixed fraction by dividing the numerator by the denominator. 77 = 3 2 (77 ÷ 25 gives the quotient as 3 and remainder as 2.) 25 25 Therefore, the sum of the given fractions is 3 2 . 25 Alpine_V3_Maths_G4_TB.indb 101 Fractions - II 101 30-01-2018 16:10:05

b) 50 23 + 16 35 35 T o add the given fractions, we have to convert the mixed fraction into improper fraction. So, 50 23 = (50× 35)+ 23 =11777530 + 23 =1773 35 35 35 35 35 T hen add their numerators and write the sum as the numerator. Therefore, 50 23 + 16 = 1773 + 16 = 1773 + 16 = 1789 . 35 35 35 35 35 35 Convert the improper fraction into a mixed fraction. 1789 =51 4 (1789 ÷ 35 gives the quotient as 51 and the remainder as 4.) 35 35 Therefore, the sum of the given fractions is 51 4 . 35 Example 14: Subtract: a) 342 - 135 b) 34 17 - 37 Solution: 25 25 42 42 a) 342 - 135 25 25 To subtract the given improper fractions, we subtract their numerators. We then write the difference as the numerator. Therefore, 342 - 135 = 342−135 = 207 25 25 25 25 A s we usually write fractions as proper or mixed fractions, we convert the difference into a mixed fraction. 207 = 8 7 (207 ÷ 25 the quotient as 8 and the remainder as 7.) 25 25 Therefore, the difference of the given fractions is 8 7 . 25 b) 34 17 - 37 42 42 T o subtract the given fractions, we first convert the mixed fraction into an improper fraction. So, 34 17 = 34× 42 + 17 = 1445 42 42 42 T hen subtract their numerators and write the difference as the numerator. 102 30-01-2018 16:10:07 Alpine_V3_Maths_G4_TB.indb 102

Therefore, 34 17 − 37 = 1445 − 37 = 1445 − 37 = 1408 42 42 42 42 42 42 Again convert the improper fraction into a mixed fraction. 11440485 = 33 22 (1408 ÷ 42 the quotient as 33 and the remainder as 22.) 4422 42 Therefore, the difference of the given fractions is 33 22 . 42 Drill Time Concept 9.1: Fraction of a Number 1) Find the following: a) 1 of 20 b) 3 of 24 c) 3 of 20 d) 4 of 12 e) 2 of 18 2 4 5 6 3 Concept 9.2: Conversions of Fractions 2) Convert the following improper fractions to mixed fractions. a) 35 b) 121 c) 93 d) 100 e) 115 4 12 12 26 20 3) Convert the following mixed fractions to improper fractions. a) 15 6 b) 23 2 c) 40 4 d) 125 9 Traei)n40M35 y Brain 8 3 5 10 4) Word Problems a) A t Sudhir’s birthday party, there are 19 sandwiches to be shared equally among 13 children. What part of the sandwiches will each child get? Give your answer as a mixed fraction. b) I bought 2 1 litres of paint but used only 3 litres. How much paint is left with me? 22 Give your answer as an improper fraction. Alpine_V3_Maths_G4_TB.indb 103 Fractions - II 103 30-01-2018 16:10:10

Chapter Decimals 10 Let Us Learn About • the term ‘decimal’ and its parts. • understanding decimal system. • expanding decimal numbers with place value charts. • converting fractions to decimals and vice versa. Concept 10.1: Conversion Involving Fractions Think Jasleen and her friends participated in the long jump event in their Jasleen – 4.1m Ravi – 2.85m games period. Her sports teacher noted the distance they jumped on a Rajiv – 3.05 m piece of paper as shown here. Amit – 2.50m Jasleen wondered why the numbers had a point between them as in the case of writing money. Do you know what the point means? Recall Recall that in Class 3 we have learnt to measure the lengths, weights and volumes of objects. For example, a pencil is 12.5 cm long. 12. 5 cm 104 30-01-2018 16:10:17 Alpine_V3_Maths_G4_TB.indb 104

A crayon is 5.4 cm long. 5.4 cm The weight of your mathematics textbook is 0.905 kg. A milk packet has 0.250 of milk, and so on. In all these values, we see numbers with a point between them. Have you read price tags on some items when you go shopping? ` 300.75 ` 439.08 They also have numbers with a point between them. Let us learn why a point is used in such numbers. & Remembering and Understanding We know how to write fractions. In this figure, 3 portion is coloured and 7 portion is not coloured. 10 10 3 or 0.3 and the We can write the coloured portion of the figure as 10 portion that is not coloured as 7 or 0.7. 10 Numbers such as 0.3, 0.7, 3.0, 3.1, 4.7, 58.2 and so on are called decimal numbers or simply decimals. Tenths: The figure below is divided into ten equal parts. 1 111 1 1 1 1 1 1 10 10 10 10 10 10 10 10 10 10 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 Each column is of the same size. Then, each of the ten equal parts is 1 . It is read as one-tenth. Fractional form of each equal part is 1 . 10 10 Decimal form of each equal part is 0.1. Decimals 105 Alpine_V3_Maths_G4_TB.indb 105 30-01-2018 16:10:20

We read numbers like 0.1, 0.2, 0.3 … as ‘zero point one’, ‘zero point two’, ‘zero point three’ and so on. Zero is written to indicate the place of the whole number. A decimal number has two parts. 48 . 35 Whole or integral part Decimal part (= or > 0) (< 1) Decimal Point Note: T he numbers in the decimal part are read as separate digits. Recall the place value chart of numbers. 100 × 10 10 × 10 1 × 10 1 Thousands Hundreds Tens Ones 6 2 5 5 3 2 2 6 5 2 We know that in this chart, as we move from right to left, the value of the digit increases 10 1 times. Also, as we move from left to right, the value of a digit becomes times. The place 10 value of the digit becomes one-tenth, read as a tenth. Its value is 0.1 read as ‘zero point one’. 2 is read as ‘two-tenths’, 7 is read as ‘seven–tenths’ and so on. 10 10 We can extend the place value chart to the right as follows: 1 × 1000 1 × 100 1 × 10 1 . 1 10 Thousands Hundreds Tens Ones Decimal Tenths 7 . 2 1 2 4 . 3 30 4 3 . 6 1 5 . 7 The number 3015.7 is read as three thousand and fifteen point seven. Similarly, the other numbers are read as follows: 106 Alpine_V3_Maths_G4_TB.indb 106 30-01-2018 16:10:21

Seven point two; twenty-four point three and one hundred and forty-three point six. The point placed in between the number is called the decimal point. The system of writing numbers using a decimal point is called the decimal system. [Note: ‘Deci’ means 10.] Hundredths: Study this place value chart. Thousands Hundreds Tens Ones Decimal Tenths Hundredths 1 × 10 1 point 1 × 1000 1 × 100 2 1 1 2 8 6 . 10 100 3 . 9 When the number moves right from the tenths place, we get a new place, which is 1 of the tenths place. It is called the ‘hundredths’ place written as 1 and read 10 100 as one-hundredths. Its value is 0.01, read as ‘zero point zero one’. 2 is read as two-hundredths, 5 is read as five-hundredths and so on. 100 100 So, the number in the place value chart is read as ‘two thousand eight hundred and sixty-two point three nine’. Expansion of decimal numbers Using the place value chart, we can expand decimal numbers. Let us see a few examples. Example 1: Expand these decimals. a) 1430.8 b) 359.65 c) 90045.75 d) 654.08 Solution: To expand the given decimal numbers, first write them in the place value chart as shown. S. no Ten Thousands Hundreds Tens Ones Decimal Tenths Hundredths thousands 1 point a) 4 3 0 . 8 b) 9 0 3 5 9 65 c) 0 4 5 . 75 d) 6 5 4 08 . . Alpine_V3_Maths_G4_TB.indb 107 Decimals 107 30-01-2018 16:10:23

Expansions: 1 a) 1430.8 = 1 × 1000 + 4 × 100 + 3 × 10 + 0 × 1 + 8 × 10 b) 359.65 = 3 × 100 + 5 × 10 + 9 × 1 + 6 × 1 + 5 × 1 Example 2: 10 100 c) 90045.75 = 9 × 10000 + 0 × 1000 + 0 × 100 + 4 × 10 + 5 × 1 + 7 × 1 + 5 × 1 10 100 1 1 d) 654.08 = 6 × 100 + 5 × 10 + 4 × 1 + 0 × 10 +8× 100 Solution: Write these as decimals. a) 7 × 1000 + 2 × 100 + 6 × 10 + 3 × 1 + 9 × 1 + 3 × 1 10 100 b) 3 × 10000 + 0 × 1000 + 1 × 100 + 9 × 10 + 6 × 1 + 4 × 1 + 5 × 1 10 100 c) 2 × 1000 + 2 × 100 + 2 × 10 + 2 × 1 + 2 × 1 + 2 × 1 10 100 d) 5 × 100 + 0 × 10 + 0 × 1 + 0 × 1 + 5 × 1 10 100 First write the numbers in the place value chart as shown. S. no Ten Thousands Hundreds Tens Ones Decimal Tenths Hundredths thousands point a) 7 2 63 . 93 b) 3 0 1 96 . 45 c) 2 2 22 . 22 d) 5 00 . 05 Standard forms of the given decimals are: a) 7263.93 b) 30196.45 c) 2222.22 d) 500.05 Conversion of fractions to decimals Fractions can be written as decimals. Consider an example. Example 3: Express these fractions as decimals. Solution: a) 18 2 b) 43 5 c) 26 1 d) 4 9 10 10 10 10 To write the given fractions as decimals, follow these steps. Step 1: Write the integral part as it is. Step 2: Place a point to its right. 108 30-01-2018 16:10:26 Alpine_V3_Maths_G4_TB.indb 108

Step 3: Write the numerator of the proper fraction part. a) 18 2 = 18.2 b) 43 5 = 43.5 10 10 Example 4: c) 26 1 = 26.1 d) 4 9 = 4.9 10 10 Express these fractions as decimals. Solution: a) 25 b) 17 2 c) 43 d) 5 92 100 100 100 100 a) 25 = 25 hundredths = 0.25 100 b) 17 2 = 17 and 2 hundredths = 17.02 100 c) 43 = 43 hundredths = 0.43 100 d) 5 92 = 5 and 92 hundredths = 5.92 100 Shortcut method: Fractions having 10 or 100 as their denominators, can be expressed in their decimal form by following the steps given below. Step 1: Write the numerator. Step 2: Then count the number of zeros in the denominator. Step 3: Place the decimal point after the same number of digits from the right as the number of zeros. For example, the decimal form of 232 =2.32 100 Note: F or the decimal equivalent of a proper fraction, place a 0 as the integral part of the decimal number. Conversion of decimals to fractions To convert a decimal into a fraction, follow these steps. Step 1: Write the number without the decimal. Step 2: Count the number of decimal places (that is, the number of places to the right of the decimal number). Step 3: Write the denominator with 1 followed by as many zeros as the decimal point. Alpine_V3_Maths_G4_TB.indb 109 Decimals 109 30-01-2018 16:10:29

Example 5: Write these decimals as fractions. a) 2.3 b) 13.07 c) 105.43 d) 0.52 Solution: a) 2.3 = 23 b) 1=3.07 1307 10 100 c) Alternate method: 105.43 = 10543 d) 0.52 = 52 100 100 A decimal having an integral part can be written as a mixed fraction. So, 2.3 = 2 and 3 tenths = 2 3 10 13.07 = 13 and 7 hundredths = 13 7 100 105.43 = 105 and 43 hundredths = 105 43 100 Application Let us see a few real-life examples of decimals. Example 6: The amount of money with Sneha and her friends are given in the table. Sneha ` 432.50 Anjali ` 233.20 Rohan ` 515.60 Jay ` 670.80 Write the amounts in words. Solution: To write the decimals in words, the integral part is read as usual. The decimal part is read as digits. Amount In words ` 432.50 four hundred and thirty-two and fifty paise rupe ` 233.20 two hundred and thirty-three and twenty paise rupee ` 515.60 five hundred and fifteen and sixty paise rupe ` 670.80 six hundred and seventy and eighty paise rupee 110 30-01-2018 16:10:31 Alpine_V3_Maths_G4_TB.indb 110

Example 7: The weights of some children in grams are given in the table below: Name Weight in grams Solution: Rahul 23456 Anil 34340 Anjali 28930 Soham 25670 Convert these weights into kilograms. We know that 1 kg = 1000 g. To convert grams to kilograms, we divide it by 1000. So, the weights in kilograms are as follows. Name Weight in grams Weight in kilograms Rahul 23456 23456 1000 = 23.456 Anil 34340 34340 = 34.340 1000 Anjali 28930 28930 = 28.930 Soham 25670 1000 25670 = 25.670 1000 Example 8: Complete this table. S. No Fraction Read as Decimal Read as 0.7 Zero point seven a) 7 7 tenths 10 b) 47 100 c) 3 5 10 d) 0.34 e) 12 and 65 hundredths Alpine_V3_Maths_G4_TB.indb 111 Decimals 111 30-01-2018 16:10:33

Solution: S. No. Fraction Read as Decimal Read as a) 7 tenths 0.7 Zero point seven b) 7 47 hundredths 0.47 Zero point four seven c) 10 3 and 5 tenths 3.5 Three point fiv d) 47 34 hundredths 0.34 Zero point three four e) 100 12 and 65 hundredths Twelve point six fiv 3 150 12.65 34 100 12 65 100 Example 9: Ajay and Vijay represented the coloured part of the figure given as follows: Ajay: 3 Vijay: 0.03 10 Solution: Whose representation is correct? The number of shaded parts as a fraction is 3 or 3 tenths. 10 As a decimal it is 0.3 and not 0.03. So, Ajay’s representation is correct. Higher Order Thinking Skills (H.O.T.S.) Observe the following: 2 tenth=s =2 0.2 10 5 tent=hs =5 0.5 10 8 tent=hs =8 0.8 10 10 tent=hs 1=0 1 10 112 30-01-2018 16:10:36 Alpine_V3_Maths_G4_TB.indb 112

57 hundredths = 57 =0.57 100 hundredths = 100 = 1 100 100 Example 10: Write the decimals that represent the shaded part. a) b) c) d) Solution: a) The fully shaded part represents a whole. So, the decimal that represents the given figure is 1.3 Alpine_V3_Maths_G4_TB.indb 113 Decimals 113 30-01-2018 16:10:38

b) The required decimal is 0.6. c) 10 + 43 = 143 =114.04303 d) 100+100+ 29 =22.2299 10 100 100 100 100 100 100 Example 11: Observe the pattern in these decimals and write the next three numbers in each. a) 0.12, 0.13, 0.14, _________, _________, _________ b) 2.00, 2.10, 2.20, _________, _________, _________ c) 8.5, 9.5, 10.5, _________, _________, _________ d) 23.31, 23.41, 23.51, _________, _________, _________ Solution: a) 0.12, 0.13, 0.14, 0.15, 0.16, 0.17 (increases by 1 hundredths) b) 2.00, 2.10, 2.20, 2.30, 2.40, 2.50 (increases by 1 tenths) c) 8.5, 9.5, 10.5, 11.5, 12.5, 13.5 (increases by ones) d) 23.31, 23.41, 23.51, 23.61, 23.71, 23.81 (increases by 1 tenths) Drill Time Concept 10.1: Conversion involving Fractions 1) Convert the following into fractions: a) 2.56 b) 14.02 c) 105.89 d) 52.60 e) 8.01 e) 834 2) Convert the following into decimals: 100 a) 2 b) 23 23 d) 73 e) 1.87 10 100 c) 10 1000 3) Write the following decimals in words: a) 73.5 b) 413.45 c) 0.73 d) 13.45 4) Word problem The measures of some objects are given in the table. Height of a flag pol 9.50 m Side of a dining table 1.20 m Distance between the two cities 325.75 km Height of a plant 127.80 cm Write these lengths in words. 114 Alpine_V3_Maths_G4_TB.indb 114 30-01-2018 16:10:40

Chapter Money 11 Let Us Learn About • c onverting rupees to paise and vice versa. • problems involving conversion of money. • adding and subtracting money with column method. • m ultiplying and dividing money. Concept 11.1: Conversion of Rupees and Paise Think Jasleen had some play money in the form of notes and coins. While playing, her friend gave her ` 10. Jasleen has to give paise for the amount her friend gave her. How many paise should Jasleen give her friend? Recall We have already learnt to identify currency and coins, conversion of rupees to paise and also that 1 ` = 100 p. Let us answer these to revise the concept of conversion of money. a) ` 62 = __________ paise b) 500 paise = ` __________ c) ` 28 = __________ paise d) 900 paise = ` __________ e) ` 76 = __________ paise f) 200 paise = ` __________ Alpine_V3_Maths_G4_TB.indb 115 115 30-01-2018 16:10:47

& Remembering and Understanding We already know that to change rupees into paise we multiply the rupees by 100. For example, as ` 1 = 100 paise, ` 3 = 3 × 100 paise = 300 paise To convert an amount into paise we multiply the rupees given in the amount by 100 and add the product to the number of paise. To convert paise to rupee just add a decimal point two digits from the right. Let us see a few examples involving conversion between rupee and paise. Example 1: Convert ` 132.28 into paise. Solution: ` 132.28 = ` 132 + 28 p = ` 132 × 100 p + 28 p = 13200 p + 28 p = 13228 p Note: An easy way to convert rupees into paise is to remove the symbol (` and p)and the dot (.) between the rupees and the paise and write the number together. So, ` 132.28 = 13228 p. An amount of more than 100 paise, can be expressed in rupees and paise. To convert paise into rupees and paise, divide the number by 100. Write the quotient as rupees and remainder as paise. Example 2: Convert 24365 paise into rupees and paise. Solution: 24365 p = 24300 p + 65 p = ` 243 + 65 p = ` 243.65 Note: An easy way 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 `. TTh Th H T O So, 24365 = ` 243.65 2 4 36 5 a) C onvert ` 477.95 to paise. Solve these c) Convert 44390 paise into b) Convert ` 892.95 into rupees. paise. _________________________ _________________________ _________________________ 116 30-01-2018 16:10:48 Alpine_V3_Maths_G4_TB.indb 116

Application Now let us solve some examples involving conversion of money. Example 3: Sheeba has ` 223.57. How many paise does she have in all? Solution: Amount with Sheeba = ` 223.57 We know that, ` 1 = 100 paise. ` 223.57 = ` 223 + 57 p = 223 × 100 p + 57 p = (22300 + 57) p = 22357 p Hence, Sheeba has 22357 paise. Example 4: Anish has 2435 p and Beena has ` 23.75. Who has more money? Solution: Amount with Anish = 2435 p Amount with Beena = ` 23.75 To compare the money they have, both the amounts must be in the same units. So, we convert rupees to paise. ` 23.75 = ` 23 × 100 p + 75 p (Since ` 1 = 100 p.) = (2300 + 75) p = 2375 p Clearly, 2435 > 2375. Therefore, Anish has more money. Example 5: Ram has ` 374.50 and Chandu has ` 365.75 in their kiddy banks. Who has less amount and by how much? Solution: Amount with Ram = ` 374.50 Amount with Chandu = ` 365.75 Comparing the rupee part of the amounts, we get 365 < 374. So, ` 365.75 < ` 374.50. Therefore, Chandu has less money. The difference in their amounts = ` 374.50 – ` 365.75 = ` 8.75 Therefore, Chandu has ` 8.75 less than Ram. Alpine_V3_Maths_G4_TB.indb 117 Money 117 30-01-2018 16:10:48

Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of conversion of money. Example 6: Complete the following by writing the number of different coins that can be used to pay ` 10 using different coins. 50 paise coins 1-rupee coins ` 10 2 - rupee coins 2-rupee coins and 1-rupee coins 5-rupee coins Solution: 20 50 paise coins 10 1-rupee coins ` 10 5 2-rupee coins 3 2-rupee coins and 4 1-rupee coins 2 5-rupee coins Example 7: Solution: Write two different ways in which you can pay ` 50. Combination 1: ` 50 = ` 20 + ` 20 + ` 10 Combination 2: ` 50 = ` 10 + ` 10 + ` 10 + ` 10 + ` 10 118 30-01-2018 16:10:49 Alpine_V3_Maths_G4_TB.indb 118

Concept 11.2: Add and Subtract Money with Conversion Think Jasleen went shopping with her elder sister. She bought some groceries for ` 110.50, vegetables for ` 105.50 and stationery for ` 40. They had ` 300. Do you know how much money was left with them after shopping? Recall Recollect that we can add or subtract money just as we add or subtract numbers. 1) To find the total am unt, we write one amount below the other. We see to it that the decimal points are exactly one below the other. We then add the amounts just as we add numbers. 2) To find the differen e in amounts, we write the smaller amount below the bigger one. We see to it that the decimal points are exactly one below the other. We then subtract the smaller amount from the bigger one. Answer the following to revise the concept of addition and subtraction of money. a) ` 22.10 – ` 11.10 = ___________ b) ` 15.30 + ` 31.45 = ___________ c) ` 82.45 – ` 42.30 = __________ d) ` 15.30 – ` 5.20 = __________ e) ` 32 + ` 7.20 = ___________ & Remembering and Understanding To add or subtract a given amount of money, we follow the steps given below. Step 1: Express the given amounts in fig res as decimal numbers. Step 2: Arrange the given amounts in a column. Place the decimal points exactly below one another. Step 3: Add or subtract the amounts as usual. Step 4: In the sum or difference so obtained, put the decimal point exactly below the other decimal points. Alpine_V3_Maths_G4_TB.indb 119 Money 119 30-01-2018 16:10:50

Let us see some examples. Example 8: Add: a) ` 547.38 + ` 130.83 b) ` 239.74 + ` 355.54 Solution: a) `p b) ` p 11 11 74 54 5 4 7.38 2 3 9. 28 + 3 5 5. + 1 3 0.83 ` 5 9 5. ` 6 7 8.21 Example 9: Subtract: a) ` 53354 − ` 24765 b) ` 866.95 − ` 492.58 Solution: a) ` b) ` p 12 12 14 4⁄ 2⁄ 2⁄ 4⁄ 1⁄4 7⁄ 1⁄6 8⁄ 1⁄5 8 6 6.9 5 53354 − 4 9 2.5 8 −24765 ` 28 5 8 9 ` 3 7 4.3 7 Application Let us now see a few real-life situations where addition and subtraction of money are used. Example 10: Anita saved ` 213.60, ` 105.30 and ` 305.45 in three months from her pocket money. How much did she save in all? Solution: Amount saved in the 1st month = ` 213.60 Amount saved in the 2nd month = + ` 105.30 Amount saved in the 3rd month = + ` 305.45 Therefore, the total amount saved in 3 months = ` 624.35 Example 11: Mrs. Gupta had ` 5000 with her. She spent ` 3520.50 for buying different food items. How much money is left with her? Solution: Amount with Mrs. Gupta = ` 5000.00 Amount spent on food items = – ` 3520.50 Therefore, the amount left with Mrs. Gupta = ` 1479.50 120 30-01-2018 16:10:51 Alpine_V3_Maths_G4_TB.indb 120

Higher Order Thinking Skills (H.O.T.S.) Let us see solve a few more real-life examples involving addition and subtraction of money. Example 12: Tanya had ` 525 and her friend Arpan had ` 330. They bought a gift for their brother’s birthday costing ` 495.75. How much amount is left with Tanya and Arpan so that they can continue their shopping? Solution: Amount Tanya had = ` 525 Amount Arpan had = ` 330 Total amount = ` 525 + ` 330 = ` 855 `p Total amount = 855 . 00 The amount spent for the gift = – 495 . 75 359 . 25 Therefore, ` 359.25 is left with Tanya and Arpan. Example 13: The cost 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? Solution: Does he get any change? If yes, how much change does he get? 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. ` 450 < ` 500. So, Suresh gets change from the shopkeeper. The change he gets = ` 500 − ` 450 = ` 50 Concept 11.3: Multiply and Divide Money Think Jasleen knows the cost of one dairy milk chocolate and the cost of five biscuit packets. She could quickly find the cost of 10 dairy milk chocolates and 1 biscuit packet. Can you do such quick calculations? Alpine_V3_Maths_G4_TB.indb 121 Money 121 30-01-2018 16:10:52

Recall Remember that we use multiplication to find cost of many items from the cost of one. Similarly, we use division to find the cost of one item from the cost of many. Multiplying or dividing an amount by a number is similar to the usual multiplication and division of numbers. Answer the following to revise the multiplication and division of numbers. a) 2356 × 10 = __________ b) 72 × 3 = ____________ c) 200 ÷ 4 = ___________ d) 549 ÷ 3 = ___________ e) 621 × 2 = ___________ & Remembering and Understanding Let us understand how to multiply or divide the given amounts of money. When 1 or more items are of the same price, multiply the amount by the number of items to get the total amount. To find out the price of one item, divide the total amount by the number of items Multiplying money To multiply an amount of money by a number, we follow these steps. Step 1: Write the amount in figures with ut the decimal point. Step 2: Multiply it by the given number, as we multiply any two numbers. Step 3: In the product, place the decimal point ( if the amount is a decimal number) after the second digit from the right. Example 14: Multiply: a) ` 14105 by 7 b) ` 312. 97 by 34 c) ` 506. 75 by 125 122 30-01-2018 16:10:52 Alpine_V3_Maths_G4_TB.indb 122

Solution: a) 2 3 b) 2 2 c) 11 1 `14105 13 2 33 2 ×7 ` 312 . 97 ` 506 . 75 `98735 × 34 × 125 1 11 111 1251 . 88 2533 . 75 + 9389 . 10 + 10135 . 00 ` 10640 . 98 + 50675 . 00 ` 63343 . 75 Dividing money To divide an amount by a number, we follow these steps. Step 1: Write the amount as the dividend and the number as the divisor. Step 2: Carry out the division just as we divide any two numbers. Step 3: Place the decimal point in the quotient, immediately after dividing the rupees, that is, digits before the decimal point in the dividend. Example 15: Divide: a) ` 23415 by 7 b) ` 481.65 by 13 c) ` 543.40 by 110 Solution: )a) 3345 )b) 37.05 )c) 4.94 7 23415 13 481.65 110 543.40 − 21↓ − 39↓ − 440 ↓ 24 91 1034 − 21 − 91 − 990 31 06 440 − 28 − 00 − 440 35 65 000 − 35 − 65 00 00 Alpine_V3_Maths_G4_TB.indb 123 Money 123 30-01-2018 16:11:03

Application Let us solve a few real-life examples involving multiplication and division of money. Example 16: A textbook of class 4 costs ` 75.20. What is the ` p cost of 35 such textbooks? Solution: 1 20 Cost of one textbook = ` 75. 20 21 35 75 Cost of 35 such textbooks = ` 75. 20 × 35 . 00 × . 00 Therefore, the cost of 35 textbooks is ` 2632. 1 1 00 76 3 56 + 22 32 `2 6 Example 17: 19 cakes cost ` 332.50. What is the cost of 1 cake? 17.50 Solution: Cost of 19 cakes = ` 332.50 Cost of 1 cake = ` 332.50 ÷ 19 )19 332.50 Therefore, the cost of 1 cake is ` 17. 50. − 19 ↓ 142 − 133 95 − 95 00 Higher Order Thinking Skills (H.O.T.S.) Train My Brain Let us see a few more examples involving multiplication and division of money. Example 18: Multiply the sum of ` 2682 and ` 2296 by 10 . Solution: The sum of ` 2682 and ` 2296 is ` 2682 + ` 2296. `` 1 2682 4978 +2296 × 10 `4 9 7 8 0 4978 Therefore, the sum multiplied by 10 = 4978 × 10 = ` 49780. 124 Alpine_V3_Maths_G4_TB.indb 124 30-01-2018 16:11:07

Example 19: A bag has one bundle of ` 50 notes and one bundle of ` 20 notes. It also has two bundles of ` 10 notes and one bundle of ` 5 notes. What is the total amount of money in the bag? [Note: Each bundle consists of 100 notes.] Solution: Amount in the bundle of ` 50 = 100 × ` 50 (1 bundle) = ` 5000 Amount in the bundle of ` 20 = ` 20 × 100 (1 bundle) = ` 2000 Amount in two bundles of ` 10 = ` 10 × 200 (2 bundles) = ` 2000 Amount in the bundle of ` 5 = ` 5 × 100 (1 bundle) = ` 500 Total money = ` 5000 + ` 2000 + ` 2000 + ` 500 = ` 9500 Therefore, the total amount of money in the bag is ` 9500. Drill Time Concept 11.1: Conversion of Rupees and Paise 1) Convert the following to paise. d) ` 537.58 e) ` 724.80 a) ` 632.18 b) ` 952.74 c) ` 231.48 2) Convert paise to rupees. a) 52865 b) 64287 c) 13495 d) 34567 e) 78654 3) Word problems a) Rehmat has ` 892.64. How many paise does he have in all? b) Andrews has 56700 paise. How much money does he have in all? Express your answer in rupees. Concept 11.2: Add and Subtract Money with Conversion 4) Add: b) ` 3467.45 + ` 2356. 50 c) 25382 p + 65237 p a) ` 875.62 + ` 964.98 e) ` 279.50 + ` 642.90 d) ` 456.23 + ` 123.75 5) Subtract: b) 85732 p – 23784 p c) ` 578.14 – ` 345.89 a) ` 132.75 – ` 112.90 e) ` 784.50 – ` 234.25 d) ` 456.72 – ` 234.34 Alpine_V3_Maths_G4_TB.indb 125 Money 125 30-01-2018 16:11:08

6) Word problems a) Rosy has ` 451.20 and Chetan has ` 495.35 in their piggy banks. Who has more amount and by how much? b) S hane spent ` 213.60, ` 105.30 and ` 305.45 in three months. How much did he spend in all? Concept 11.3: Multiply and Divide Money 7) Multiply: b) 27510 p × 2 c) ` 315.50 × 10 a) ` 152.45 × 5 e) ` 115.50 × 35 d) ` 113.50 × 15 8) Divide: b) 22347 p ÷ 9 c) ` 111.44 ÷ 7 a) ` 126.12 ÷ 3 e) ` 824.40 ÷ 8 d) ` 121.77 ÷ 7 9) Word problems a) A packet of chips costs ` 24.40. How much will 5 such packets cost? b) A football costs ` 159.99. What is the cost of 26 such footballs? 126 30-01-2018 16:11:08 Alpine_V3_Maths_G4_TB.indb 126

Chapter Measurements 12 Let Us Learn About • relation between units of length, weight and capacity. • converting smaller units to larger units. • m ultiplying and dividing length, weight and capacity. Concept 12.1: Multiply and Divide Lengths, Weights and Capacities Think Jasleen had some guests visiting her place. Jasleen’s mother asked her to pour juice from three bottles, each of 1.5 litres, into 15 glasses. What was the total quantity of juice and how much juice was poured in each glass? Recall Let us revise the basic concepts of measurements, their units and the different operations involving measurements. Length: kilometre, centimetre, millimetre Weight: kilogram, gram, milligram Capacity: litre, millilitre Solve the following problems based on addition and subtraction of lengths, weights and capacities. Alpine_V3_Maths_G4_TB.indb 127 127 30-01-2018 16:11:13

a) 560 m 65 cm – 230 m 55 cm = ___________ b) 250 g + 2 kg 500 g = ___________ c) 240 m 22 cm – 220 m 20 cm = ___________ d) 5 ℓ 250 mℓ + 4 ℓ 250 mℓ = ___________ e) 745 km 45 m – 434 km 15 m = ___________ & Remembering and Understanding To convert measures from a larger unit to a smaller unit, we multiply. To convert measures from a smaller unit to a larger unit, we divide. Let us understand the relation between the different units of length, weight and capacity in detail. Relation between units of length, weight and capacity Larger unit – Smaller unit Smaller unit – Larger unit Length 1 m = 1 km 1000 1 km = 1000 m 1 m = 100 cm 1 1 cm = 100 m 1 cm = 10 mm 1 mm = 1 cm 10 1 g = 1000 mg 1 kg = 1000 g Weight 1 mg = 1 g 1 litre = 1000 mℓ Capacity 1000 1 g = 1 kg 1000 1 mℓ = 1 ℓ 1000 1 kilolitre = 1000 litres 1 ℓ = 1 kℓ 1000 128 30-01-2018 16:11:14 Alpine_V3_Maths_G4_TB.indb 128

Conversion of smaller units to larger units Let us understand conversions through a few examples. Example 1: Convert the following: a) 5000 m to km b) 8000 g to kg c) 2000 mℓ to ℓ Solution: Solved Solve these a) C onversion of m into km 9000 m = ________________ km 5000 m = _____________ km 4000 g = ______________ kg 1000 m = 1 km So, 5000 m = 5000 ÷ 1000 m 3000 mℓ = ______________ ℓ = 5 km 5000 m = 5 km b) C onversion of g into kg 8000 g = _____________ kg 1000 g = 1 kg So, 8000 g = 8000 ÷ 1000 g = 8 kg c) Conversion of mℓ into ℓ 2000 mℓ = _____________ ℓ 1000 mℓ = 1 ℓ So, 2000 mℓ = 2000 ÷ 1000 mℓ =2ℓ Multiply and divide length, weight and capacity Interestingly, multiplication and division of lengths, weights and capacities are similar to that of usual numbers. Let us see a few examples. Example 2: Solve: b) 18 km 361 m × 19 c) 7 ℓ 260 mℓ × 37 a) 65 kg 345 g × 28 Alpine_V3_Maths_G4_TB.indb 129 Measurements 129 30-01-2018 16:11:14

Solution: a) 65 kg 345 g × 28 b) 18 km 361 m × 19 c) 7  260 m× 37 km m kg g ℓ mℓ 1 1 34 1 42 345 65 28 73 5 14 × 760 18 361 7 260 1 900 522 660 × 19 × 37 +1 3 0 6 1829 1 1 165 249 50 820 + 183 610 + 217 800 348 859 268 620 Example 3: Solve: a) 15 kg 183 g ÷ 21 b) 3 km 84 m ÷ 12 c) 5 ℓ 882 mℓ ÷ 17 a) 15 kg 183 g ÷ 21 b) 3 km 84 m ÷ 12 c) 5 ℓ 882 mℓ ÷ 17 15 kg 183 g 3 km 84 m 5 ℓ 882 mℓ = 15 × 1000 g + 183 g = 3 × 1000 m + 84 m = 5 × 1000 mℓ + 882 mℓ = 15183 g = 3084 m = 5882 mℓ 346 723 257 )17 5882 )21 15183 )12 3084 − 51 − 147 − 24 048 068 078 − 068 − 042 − 060 0063 0084 0102 − 0102 − 0063 − 0084 0000 0000 0 15 kg183 g ÷ 21 = 723 g 3 km 84 m ÷ 12 = 257 m 5 ℓ 882 mℓ ÷ 17 = 346 mℓ 130 30-01-2018 16:11:16 Alpine_V3_Maths_G4_TB.indb 130

Application Let us solve a few examples based on multiplication and division of length, weight and capacity. Example 4: The distance between two post offices A and B is 58 km 360 m. What is the total distance travelled in four round trips between A and B? Solution: The distance between two post offices A and B is 58 km 360 m Four round trips = 4 times from A to B and 4 times from B to A = 8 times the distance between A and B Therefore, the total distance travelled in four round trips = 58 km 360 m × 8 = 466 km 880 m Example 5: Mrs. Rani has 2 kg of coffee powder. She wants to put it into smaller packets of 25 g each. How many packets will she need? Solution: Weight of coffee powder Mrs. Rani has = 2 kg 1 kg = 1000 g 2 kg = 2 × 1000 g = 2000 g Weight of one small packet = 25 g Therefore, the number of packets she needs = 2000 g ÷ 25 g = 80 Example 6: Rahul has a can of 6112 mℓ juice. If he pours it equally in 16 glasses, what is the quantity of juice in each glass? Solution: Quantity of juice in full can = 6112 mℓ Number of glasses into which the juice is poured = 16 Quantity of juice in each glass = 6112 mℓ ÷ 16 = 382 mℓ Therefore, each glass contains 382 ml of juice. Alpine_V3_Maths_G4_TB.indb 131 Measurements 131 30-01-2018 16:11:16

Higher Order Thinking Skills (H.O.T.S.) Sometimes, we have to use more than one mathematical operation to measure things. Consider these examples. Example 7: 185 kg sugar costing ` 444 is packed in paper bags. Each bag can hold 5 kg of sugar. Find the number of bags needed to pack all the sugar. Also, find the cost of each bag. Solution: Weight of sugar = 185 kg Weight of sugar in the paper bag = 5 kg Number of paper bags needed = 185 kg ÷ 5 kg = 37 Therefore, 37 paper bags of 5 kg sugar each can be made. Cost of 37 bags of sugar = ` 444 Cost of each bag = ` 444 ÷ 37 = ` 12 Therefore, 185 kg sugar can be packed into 37 bags costing ` 12 each. Example 8: A container can hold 13 ℓ 625 mℓ of milk. What is the capacity of 15 such containers? Give your answer in mℓ. Solution: Capacity of one container = 13 ℓ 625 mℓ Capacity of 15 such containers = 13 ℓ 625 mℓ × 15 = 204 ℓ 375 mℓ 1 litre = 1000 mℓ 204 ℓ = 204 × 1000 mℓ = 204000 mℓ 204 ℓ 375 mℓ = 204000 mℓ+ 375 mℓ = 204375 mℓ Therefore, the capacity of 15 cans is 204375 mℓ. Example 9: The distance between two places is 4520 km. Ratan travelled a fourth of the distance by bus paying ` 12 per km. As the bus failed, he hired a car and travelled three-fourths of the distance by paying ` 20 per km. What amount did he spend on travelling? Solution: Total distance = 4520 km 1 of the distance = 1 × 4520 km = 1130 km 4 4 Distance travelled by bus = 1130 km 132 Alpine_V3_Maths_G4_TB.indb 132 30-01-2018 16:11:17

Ratan travelled 1130 km by bus. Cost of ticket per km = ` 12 Cost of ticket for 1130 km = 1130 × ` 12 = ` 13560 Fraction of distance travelled by car = 3 4 Actual distance travelled by car = 3 × 4520 km 4 = 3 × 1130 km = 3390 km Cost of travelling by car per km = ` 20 Cost of travelling 3390 km = 3390 × ` 20 = ` 67800 Total amount spent by Ratan on travelling = ` 13560 + ` 67800 = ` 81360 Drill Time Concept 12.1: Multiply and Divide Lengths, Weights and Capacities 1) Convert: a) 2000 cm to m b) 5000 g to kg c) 5000 m to km d) 8000 mℓ to ℓ 2) Multiply: a) 85 kg 145 g ×10 b) 5 ℓ 225 mℓ × 65 c) 7 m 450 cm × 25 d) 5 km 150 cm × 12 3) Divide: a) 34 kg 450 g by 6 b) 50 ℓ 225 mℓ by 5 c) 17 m 85 cm by 9 d) 42 kg 420 g by 7 Alpine_V3_Maths_G4_TB.indb 133 Measurements 133 30-01-2018 16:11:18

Chapter Data Handling 13 Let Us Learn About • reading and interpreting bar graphs. • d rawing bar graphs based on the given data. Concept 13.1: Bar Graphs Think Jasleen attended a fruit festival conducted for a week in her school. She was asked to give a report on the sale of different fruits per day in the form of a graph. Till then Jasleen only knew how to represent the data as a pictograph. She wanted to find an easier and simpler way of representation. How do you think Jasleen would have given the report? Recall Recall these points: • The information collected for a specific purpose is called data. • The information given as numbers is called numerical data. • The information shown in the form of pictures is called a pictograph. 134 30-01-2018 16:11:23 Alpine_V3_Maths_G4_TB.indb 134

We have already learnt about pictographs. Let us recall them through the following. Let us recall the pictographs through the following example. The favourite sports of Class 4 students are given. Read the pictograph and answer the questions. Key: 1 = 6 students Favourite Sports of Class 4 Students Volleyball Cricket Basketball Kabaddi Football a) The most favourite sports of Class 4 students is _____________. b) The least favourite sports of Class 4 students is _____________. c) The number of students who like to play basketball is___________. d) The number of students who like to play football is _____________. e) The number of students who like to play kabaddi is _____________. Alpine_V3_Maths_G4_TB.indb 135 Data Handling 135 30-01-2018 16:11:30

& Remembering and Understanding While drawing pictographs, we choose a relevant picture to represent the given data. If the data is large, it is tedious and time consuming to draw a pictograph. An easier way of representing data is the bar graph. It uses rectangular bars of the same width. These bars can be a drawn either horizontally or vertically. Bar graphs are drawn on a graph paper. A suitable title is given for the bar graph. Let us understand how to read and interpret bar graphs. Example 1: The marks scored by Kamala in a monthly test are represented using a bar graph as given. Understand the graph and answer the questions that follow. Scale: X-axis: 1 cm = 1 subject; Y-axis: 1 cm = 5 marks Kamala’s Performance in a Monthly Test Marks Scored English Math s Science Social Music Hindi Studies Subjects 136 30-01-2018 16:11:32 Alpine_V3_Maths_G4_TB.indb 136

a) What is the title of the graph? b) In which subject did Kamala perform the best? c) In which subject does Kamala need to improve? d) What are Kamala’s total marks? Solution: a) The title of the graph is “Kamala’s Performance in a Monthly Test”. b) T he height of the bar representing Maths is maximum. It means that, Kamala performed the best in Maths. c) T he height of the bar representing Social Studies is the minimum. Example 2: So, Kamala needs to improve in Social Studies. d) Kamala’s total marks are 35 + 47 + 42 + 28 + 32 + 40 = 224 Information about a primary school is represented in the form of a bar graph as shown. Observe the graph carefully and answer the questions that follow. Scale: X-axis: 1 cm = 1 class; Y-axis: 1 cm = 5 students Strength of Primary School School Class strength Class Alpine_V3_Maths_G4_TB.indb 137 Data Handling 137 30-01-2018 16:11:36

a) What is the total strength of all the 5 classes? b) Which class has the least strength? c) Which class has the greatest strength? d) What is the title of the graph? Solution: a) Total strength is 42 + 36 + 38 + 43 + 45 = 204 b) Class 2 c) Class 5 d) Strength of a Primary School Application We have learnt how to read and interpret bar graphs. Now, let us learn to draw a bar graph. Steps to draw a bar graph: Step 1: Draw one horizontal line and another vertical line, called the axes. They meet at a point called the origin. Step 2: Take a suitable scale such as 1 cm = 5 units. Step 3: On the X-axis, show the items of the data and on the Y-axis show their values. Step 4: Draw bars of equal width on the X-axis. The heights of the rectangles represent the values of the data which are given on the Y- axis. Step 5: Give a relevant title to the bar graph. Let us understand this through an example. Example 3: The following pictograph shows the number of scooters manufactured by a factory in a week. Complete the pictograph. Then draw a bar graph for the same data. Key: 1 = 5 scooters 138 30-01-2018 16:11:36 Alpine_V3_Maths_G4_TB.indb 138

Weekday Scooters manufactured in a week Number of Monday scooters Tuesday Wednesday Thursday Friday Saturday Solution: Total Step 1: Let us follow these steps to draw a bar graph. Step 2: Count the number of pictures in the pictograph. Complete the table by writing the product of the number of pictures and the number of scooters per key. Take a graph paper and draw the X and Y axes meeting each other at one corner as shown. Alpine_V3_Maths_G4_TB.indb 139 Data Handling 139 30-01-2018 16:11:41

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 140 30-01-2018 16:11:45 Alpine_V3_Maths_G4_TB.indb 140

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. Weekly Manufacturing of Scooters Number of scooters manufactured Monday Tuesday Wednesday Thursday Friday Saturday Weekdays Alpine_V3_Maths_G4_TB.indb 141 Data Handling 141 30-01-2018 16:11:46

We can draw the same graph using horizontal bars by interchanging the values on X and Y axes. Weekly Manufacturing of Scooters Weekdays Number of scooters manufactured Example 4: The number of roses sold during a month in Roopa’s shop is given in the table Week Number of roses sold 1st week 148 2nd week 165 3rd week 130 4th week 172 Represent the data in a bar graph. 142 30-01-2018 16:11:49 Alpine_V3_Maths_G4_TB.indb 142

Solution: Scale: X-axis: 1 cm = 1 week;Y-axis: 1 cm = 20 roses Roses sold Weeks Higher Order Thinking Skills (H.O.T.S.) Train My Brain Consider a few real-life examples where we represent data using a bar graph. Example 5: In 2010, the heights of Ramu, Somu, Radha and Swetha were noted as 130 cm, 125 cm, 115 cm and 120 cm respectively. After two years, their heights were again noted as 140 cm, 132 cm, 124 cm and 128 cm respectively. Draw a bar graph to represent the data and answer the questions that follow. a) Who was the tallest among the friends in 2010? b) Who was the shortest among them during 2012? c) How much taller was Ramu than Somu in 2010? d) Whose height has increased the maximum in 2 years? Alpine_V3_Maths_G4_TB.indb 143 Data Handling 143 30-01-2018 16:11:51

e) A rrange the children’s heights in 2010 in ascending order and their heights in Solution: 2012 in descending order. Name Height in 2010 Height in 2012 Ramu 130 cm 140 cm Somu 125 cm 132 cm Radha 115 cm 124 cm Swetha 120 cm 128 cm Scale: On X-axis: 2 cm = 1 student On Y-axis: 1 cm = 20 cm Comparison of Heights Height (in cm) Names of children 144 30-01-2018 16:11:52 Alpine_V3_Maths_G4_TB.indb 144

a) As the bar for Ramu’s height in 2010 is the highest, Ramu is the tallest among the children. b) R adha is the shortest among them in 2012. (Shortest bar in 2012). c) Ramu is 5 cm (130 – 125) taller than Somu. d) Increase in the heights of the children in the two years: Ramu: (140 – 130) cm = 10 cm S omu: (132 – 125) cm = 7 cm Radha: (124 – 115) cm = 9 cm Swetha: (128 – 120) cm = 8 cm 7 cm < 8 cm < 9 cm < 10 cm Therefore, Ramu’s height increased the maximum in 2 years. e) Heights of the children in 2010: 130 cm, 125 cm, 115 cm, 120 cm Ascending order: 115 cm, 120 cm, 125 cm, 130 cm Heights of the children in 2012: 140 cm, 132 cm, 124 cm, 128 cm Descending order: 140 cm, 132 cm, 128 cm, 124 cm Example 6: The weights of four children are noted in 2014 and 2016 as given. Draw a bar graph and answer the questions that follow. Name Weight in 2014 Weight in 2016 Ram 30 kg 34 kg Shyam 34 kg 32 kg Reema 28 kg 31 kg Seema 29 kg 31 kg a) Who weighed the most in 2014 and 2016? b) Whose weight has decreased in 2016 from 2014? c) Name the two children who were of the same weight in 2016. d) Whose weight in 2014 is the same as that of another child in 2016? Alpine_V3_Maths_G4_TB.indb 145 Data Handling 145 30-01-2018 16:11:53

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 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 146 30-01-2018 16:11:54 Alpine_V3_Maths_G4_TB.indb 146