1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text_Reduced

Division8CHAPTER I Will Learn Concepts 8.1: Div1i.s1io: nVaerstiEcqeusaalnGdroDuiapginognaalsndofRTewlao-tDeimDievinssioionntaol SMhualptipeslication 8.2: Divide 2-digit and 3-digit Numbers by 1-digit Numbers JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___101 / 1840

Concept 8.1: Division as Equal Grouping and Relate Division to Multiplication Think Neena wants to distribute 8 sticks equally among 4 of her friends. How can she distribute? To answer this question, we must know equal grouping. Recall In the previous chapter, we have learnt multiplication. Multiplication is finding the total number of objects that have been grouped equally. Let us use this to distribute objects equally in groups. Consider 12 bars of chocolate. The different ways in which they can be distributed are as follows. Distributing in 1 group: 1 × 12 = 12 Distributing in 2 groups: 2 × 6 = 12 Distributing in 3 groups: 3 × 4 = 12 Distributing in 4 groups: 4 × 3 = 12 96 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___102 / 1804

Distributing in 6 groups: 6 × 2 = 12 Distributing in 12 groups: 12 × 1 = 12 Distributing a given number of objects into equal groups is called division. We can understand division better by using equal sharing and equal grouping. & Remembering & Understanding Equal sharing means having equal number of objects or things in a group. We use division to find the number of things in a group and the number of groups. Suppose 9 balloons are to be shared equally by 3 girls. Let us use repeated subtraction to distribute the balloons. 1st round: 1 balloon is taken by each girl. 9 – 3 = 6. So, 6 balloons remain. 2nd round: 1 more balloon is taken by each girl. Now, each of them has 2 balloons. 6 – 3 = 3. So, 3 balloons remain. 3rd round: 1 more balloon is taken by each girl. Now, each of them has 3 balloons. 3 – 3 = 0. So, 0 balloons remain. Each girl gets 3 balloons. We can write it as 9 divided by 3 equals 3. Division 97 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___103 / 1804

The symbol for ‘is divided by’ is ÷. 9 divided by 3 equals 3 is written as 9÷ 3 = 3 [Total number] ÷ [Number in each group] = [Number of groups] Dividend ÷ Divisor = Quotient In a division, the number that is divided is called the dividend. The number that divides is called the divisor. The answer in division is called the quotient. The number (part of the dividend) that remains is called the remainder. 9 ÷ 3 = 3 is called a division fact. In this, 9 is the dividend, 3 is the divisor and 3 is the quotient. Note: Representing the dividend, divisor and quotient using the symbols ÷ and = is called a division fact. We use multiplication tables to find the quotient in a division. We find the factor which when multiplied by the divisor gives the dividend. Let us understand this through a few examples. Example 1: 18 pens are to be shared equally by 3 children. How many pens does each of them get? Solution: Total number of pens = 18 Number of children = 3 Example 2: Number of pens each child gets = 18 ÷ 3 = 6 (since 6 × 3 = 18) Solution: Therefore, each child gets 6 pens. 10 flowers are put in some vases. If each vase has 2 flowers, how many vases are used? Number of flowers = 10 Number of flowers in each vase = 2 98 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___104 / 1840

Number of vases used = 10 ÷ 2 = 5 (since 2 × 5 = 10) Therefore, 5 vases are used to put 10 flowers. We get two division facts from a multiplication fact. The divisor and the quotient are the factors of the dividend. Observe the following: Dividend ÷ Divisor = Quotient Multiplicand × Multiplier = Product 18 ÷ 6 =3 6 × 3 = 18 ↓ ↓ ↓ ↓ ↓ ↓ Divisor Quotient Dividend Product Factor Factor (Multiplicand) (Multiplier) From the multiplication fact 6 × 3 = 18, we can write two division facts: a) 18 ÷ 3 = 6 and b) 18 ÷ 6 = 3 Multiplication and division are reverse operations. Let us now understand this through an activity. We can show a multiplication fact on the number line. For example, 3 × 5 = 15 means 5 times 3 is 15. To show 5 times 3 on the number line, we take steps of 3 for 5 times. We go forward from 0 to 15. Similarly, we can show the division fact 15 ÷ 3 = 5 on the number line . To show 15 divided by 3 on the number line, we take steps of 3 for 5 times. We go backward from 15 to 0 as shown. Division 99 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___105 / 1840

Application Equal sharing and equal grouping are used in some real-life situations. Consider the following situations. Example 3: 25 buttons are to be stitched on 5 shirts. If each shirt has the same number of buttons, how many buttons are there on each shirt? Solution: Total number of buttons = 25 Number of shirts = 5 The division fact for 25 buttons distributed to 5 shirts = 25 ÷ 5 = 5. Therefore, each shirt has 5 buttons on it. Example 4: 24 marbles are to be divided in 4 friends. How many marbles will each Solution: friend get? Total number of marbles = 24 Number of friends = 4 Number of marbles each friend will get = 24 ÷ 4 = 6 Therefore, each friend will get 6 marbles. Higher Order Thinking Skills (H.O.T.S.) Division is used in many situations in our day-to-day lives. Let us see some examples. Example 5: Aman spends 14 hours a week for tennis practice. He spends 21 Solution: hours a week for doing homework and 48 hours a week at school. How much time does he spend in a day for these activities? (Hint: 1 week = 7 days. The school works for 6 days a week.) Time spent for tennis practice per day = 14 hours ÷ 7 = 2 hours Time spent for doing homework per day = 21 hours ÷ 7 = 3 hours Time spent at school per day = 48 hours ÷ 6 = 8 hours (School works for 6 days a week) Thus, the total time spent by Aman in a day for all the activities = (2 + 3 + 8) hours = 13 hours (except Sunday) 100 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___106 / 1840

Example 6: Deepa shares 15 lollipops among her 5 friends. Instead, if she shares among only 3 of them, how many more lollipops does each of them get? Solution: Number of lollipops = 15 If Deepa shares the lollipops among her five friends, the number of lollipops each of them would get = 15 ÷ 5 = 3 If Deepa shares the lollipops among only three of them, the number of lollipops each of them gets = 15 ÷ 3 = 5 Difference in the number of lollipops = 5 – 3 = 2 Therefore, her friends would get 2 more lollipops. Concept 8.2: Divide 2-digit and 3-digit Numbers by 1-digit Numbers Think Neena has 732 stickers. She wants to distribute them equally among her three friends. How will she distribute? To answer this, we must learn how to divide a 3-digit number by a 1-digit number. Recall In the previous section, we have learnt that division is related to multiplication. For every division fact, we can write two multiplication facts. For example, the two multiplication facts of 35 ÷ 7 = 5 are: a) 7 × 5 = 35 and b) 5 × 7 = 35. Let us answer these to recall the concept of division. a) The number which divides a given number is called _________________. b) The answer we get when we divide a number by another is called _________________. Division 101 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___107 / 1840

c) The division facts for the multiplication fact 2 × 4 = 8 are ________________ and __________________. d) 14 ÷ 2 = __________ e) 10 ÷ 5 = __________ & Remembering & Understanding We can make equal shares or groups and divide with the help of vertical arrangement. Let us see some examples. 1) Dividing a 2-digit number by a 1-digit number (1-digit quotient) Example 7: Divide: 45 ÷ 5 Solution: Follow these steps to divide a 2-digit number by a 1-digit number. Steps Solved Solve these Step 1: Write the dividend and 5)45 Dividend = _____ )divisor as shown: Divisor Dividend Divisor = ______ Quotient = ____ Step 2: Find the multiplication fact 45 = 5 × 9 8) 56 Remainder = _____ which has the dividend and divisor. - Step 3: Write the other factor as 9 quotient. Write the product of the factors below the dividend. 5) 45 − 45 Step 4: Subtract the product 9 4) 36 Dividend = _____ from the dividend and write the Divisor = ______ difference below the product. 5) 45 - Quotient = ____ This difference is called the Remainder = _____ remainder. − 45 00 45 = Dividend 5 = Divisor 9 = Quotient 0 = Remainder Note: If the remainder is zero, the divisor is said to divide the dividend exactly. 102 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___108 / 1840

Checking for correctness of division: The multiplication fact of the division is used to check its correctness. Step 1: Compare the remainder and divisor. The remainder must always be less than the divisor. Step 2: Check if (Quotient × Divisor) + Remainder = Dividend Let us now check if our division in example 7 is correct or not. Step 1: Remainder < Divisor 0 < 5 (True) Step 2: Quotient × Divisor 9×5 Step 3: (Quotient × Divisor) + Remainder = 45 + 0 = 45 = Dividend Note: The division is incorrect if a) Remainder > or = divisor b) (Quotient × Divisor) + Remainder ≠ Dividend 2-digit quotients In the examples we have seen so far, the quotients are 1-digit numbers. In some divisions, the quotients may be 2-digit numbers. Let us see some examples. Example 8: Divide: 57 ÷ 3 Solution: Follow these steps to divide a 2-digit number by a 1-digit number. Steps Solved Solve these Step 1: Check if the tens digit of the 5>3 5) 60 dividend is greater than the divisor. 1 Step 2: Divide the tens and write the − quotient. 3) 57 Write the product of quotient and divisor, − below the tens digit of the dividend. −3 Division 103 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___109 / 1840

Steps Solved Solve these Step 3: Subtract and write the difference. 1 Dividend = _____ Step 4: Check if difference < divisor is true. Divisor = ______ Step 5: Bring down the ones digit of the 3) 57 Quotient = ____ dividend and write it beside the remainder. Remainder = ___ −3 2 2 < 3 (True) 1 3) 57 − 3↓ 27 Step 6: Find the largest number in the 3 × 8 = 24 19 3) 42 multiplication table of the divisor that can 3 × 9 = 27 be subtracted from the 2-digit number in 3× 10 = 30 3) 57 − the previous step. 24 < 27 < 30. So, 27 is − 3↓ 27 the required number. Step 7: Write the factor of required 19 − number, other than the divisor, as quotient. Write the product of divisor and quotient 3) 57 below the 2-digit number. Subtract and write the difference. −T23ra↓7 in My Brain − 27 Dividend = _____ 00 Step 8: Check if remainder < divisor is true. 0 < 3 (True) Divisor = ______ Stop the division. Quotient = ____ (If this is false, the division is incorrect.) Quotient = 19 Remainder = ___ Remainder = 0 Step 9: Write the quotient and remainder. 3 × 19 + 0 = 57 57 + 0 = 57 Step 10: Check if (Divisor × Quotient) + 57 = 57 (True) Remainder = Dividend is True. (If this is false, the division is incorrect.) 104 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___110 / 1804

2) Divide 3-digit numbers by 1-digit numbers (2-digit Quotients) Dividing a 3-digit number by a 1-digit number is similar to dividing a 2-digit number by a 1-digit number. Let us understand this through a few examples. Example 9: Divide: a) 265 ÷ 5 Solution: Follow these steps to divide a 3-digit number by a 1-digit number. Steps Solved Solve these Step 1: Check if the hundreds digit 4) 244 of the dividend is greater than the 5)265 − divisor. If it is not, consider the tens digit too. 2 is not greater than 5. So, consider 26. Step 2: Find the largest number that 5 − can be subtracted from the 2-digit number of the dividend. Write the 5) 265 Dividend = _____ quotient. Divisor = ______ Write the product of the quotient − 25 Quotient = ____ and divisor below the dividend. Remainder = ___ 5 × 4 = 20 5 × 5 = 25 9) 378 5 × 6 = 30 − 25 < 26 − Step 3: Subtract and write the 5 difference. 5) 265 − 25 1 Step 4: Check if difference < divisor 1 < 5 (True) is true. (If it is false, the division is incorrect.) 5 Step 5: Bring down the ones digit 5) 265 of the dividend. Write it beside the remainder. − 25↓ 15 Division 105 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___111 / 1804

Steps Solved Solve these Dividend = _____ Step 6: Find the largest number in 5 Divisor = ______ the multiplication table of the divisor Quotient = ____ that can be subtracted from the 5) 265 Remainder = ___ 2-digit number in the previous step. − 25↓ 5) 245 15 − 5 × 2 = 10 5 × 3 = 15 − 5 × 4 = 20 15 is the required Dividend = _____ number. Divisor = ______ Quotient = ____ Step 7: Write the factor of required 53 Remainder = ___ number, other than the divisor, as quotient. Write the product of divisor 5) 265 and quotient below the 2-digit number. Then, subtract them. − 25↓ 15 − 15 00 Step 8: Check if remainder < divisor is 0 < 5 (True) true. Stop the division. (If this is false, the division is incorrect.) Step 9: Write the quotient and Quotient = 53 remainder. Remainder = 0 Step 10: Check if (Divisor × 5 ×53 + 0 = 265 Quotient) + Remainder = Dividend 265 + 0 = 265 is true. (If this is false, the division is 265 = 265 (True) incorrect.) 3-digit quotients Example 10: Divide: 784 by 7 Solution: Follow these steps to divide a 3-digit number by a 1-digit number. Steps Solved Solve these Step 1: Check if the hundreds digit of 7)784 the dividend is greater than or equal to the divisor. 7=7 106 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___112 / 1804

Steps Solved Solve these 1 Step 2: Divide the hundreds and 8) 984 write the quotient in the hundreds 7) 784 place. − Write the product of the quotient −7 − and the divisor under the hundreds − place of the divided. 1 Dividend = _____ Step 3: Subtract and write the 7) 784 Divisor = ______ difference. Quotient = ____ −7 Remainder = ___ 0 5) 965 Step 4: Check if difference < divisor 0 < 7 (True) is true. − Step 5: Bring down the next digit of 1 − the dividend. Check if it is greater − than or equal to the divisor. 7) 7 84 − 7↓ 08 8>7 Step 6: Find the largest number in the 11 multiplication table of the divisor that can be subtracted from the 2-digit 7) 784 number in the previous step. Write the factor other than the divisor − 7↓ as quotient. 08 −7 Write the product of the quotient and the divisor below it. 7×1=7<8 The required number is 7. Step 7: Subtract and write the 11 difference. Bring down the next digit (ones digit) of the dividend. 7) 784 Check if the dividend is greater than or equal to the divisor. − 7↓ 08 −7 14 14 > 7 Division 107 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___113 / 1840

Steps Solved Solve these Dividend = _____ Step 8: Find the largest number in 112 Divisor = ______ the multiplication table of the divisor Quotient = ____ that can be subtracted from the 7) 784 Remainder = ___ 2-digit number in the previous step. Write the factor other than the − 7↓ divisor as the quotient. 08 Write the product of the quotient and the divisor below it. −7 14 − 14 Step 9: Subtract and write the 7 × 2 = 14 2) 246 difference. The required number Check if it is less than the divisor. − Stop the division. is 14. − 112 − 7) 784 Dividend = _____ − 7↓ Divisor = ______ 08 Quotient = ____ Remainder = ___ −7 14 − 14 00 Step 10: Write the quotient and the Quotient = 112 remainder. Remainder = 0 Step 11: Check if (Divisor × Quotient) + Remainder = Dividend is true. (If it 7 × 112 + 0 = 784 784 + 0 = 784 is false, the division is incorrect.) 784 = 784 (True) Application Division of 2-digit numbers and 3-digit numbers is used in many real-life situations. Let us consider a few examples. Example 11: A school has 634 students, who are equally grouped into 4 houses. How many students are there in a house? Are there any students who are not grouped into a house? 108 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___114 / 1840

Solution: Number of students = 634 158 Number of houses = 4 4) 634 Number of students in a house = 634 ÷ 4 − 4↓ Number of students in each house = 158 23 The remainder in the division is 2. − 20 34 Therefore, 2 students are not grouped into any house. − 32 Example 12: A football game had 99 spectators. If each row has only 02 9 seats, how many rows would the spectators occupy? 11 Solution: Number of spectators = 99 9) 99 Number of seats in each row = 9 − 9↓ 09 Number of rows occupied by the spectators = 99 ÷ 9 = 11 −9 Therefore, 11 rows were occupied by the spectators. 0 Higher Order Thinking Skills (H.O.T.S.) In all the divisions we have seen so far, we did not have a 0 (zero) in dividend or quotient. When a dividend has a zero, we place a 0 in the quotient in the corresponding place. Then, get the next digit of the dividend down and continue the division. Let us now understand division of numbers that have a 0 (zero) in dividend or quotients, through these examples. Example 13: Divide: 505 ÷ 5 Solution: Follow these steps for division of numbers having 0 in dividend. Solved Solve this 101 4) 804 5) 505 − − 5↓ − 00 − − 00 05 − 05 00 Division 109 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___115 / 1804

Drill Time Concept 8.1: Division as Equal Grouping and Relate Division to Multiplication 1) Divide the number in equal groups a) 16 in 4 equal groups b) 18 in 9 equal groups c) 20 in 5 equal groups d) 32 in 8 equal groups e) 10 in 2 equal groups 2) Word Problems a) 26 students are to be divided in 2 groups. How many students will be there in each group? b) 14 pencils must be distributed among 7 children. How many pencils will each student receive? Concept 8.2: Divide 2-digit and 3-digit Numbers by 1-digit Numbers 3) Divide 2-digit numbers by 1-digit numbers (1-digit quotient) a) 12 ÷ 2 b) 24 ÷ 6 c) 36 ÷ 6 d) 40 ÷ 8 e) 10 ÷ 5 4) Divide 2-digit numbers by 1-digit numbers (2-digit quotient) a) 12 ÷ 1 b) 99 ÷ 3 c) 48 ÷ 2 d) 65 ÷ 5 e) 52 ÷ 4 5) Divide 3-digit numbers by 1-digit numbers (2-digit quotient) a) 123 ÷ 3 b) 102 ÷ 2 c) 497 ÷ 7 d) 111 ÷ 3 e) 256 ÷ 4 6) Divide 3-digit numbers by 1-digit numbers (3-digit quotient) a) 456 ÷ 2 b) 112 ÷ 1 c) 306 ÷ 3 d) 448 ÷ 4 e) 555 ÷ 5 7) Word Problems a) There are 260 chocolates in a jar that have to be distributed equally among 4 students. How many chocolates will each student receive? b) There are 24 people in a bus. Each row in the bus can seat 2 people. How many rows in the bus are occupied? 110 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___116 / 1840

Fraction9CHAPTER 1 4 I Will Learn CCoonncceeppttss 9.1: Fra1c.1t:ioVneartsicaePs aarntdofDaiaWgohnoalels of Two-Dimensional Shapes 9.2: Fraction of a Collection JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___117 / 1840

Concept 9.1: Fraction as a Part of a Whole Think Neena and her three friends share an apple equally. What part of the apple does each of them get? To answer this question, we must learn to find fraction of a whole. Recall Look at the rectangle shown below. We can divide the whole rectangle into many equal parts. Consider the following: 2 equal parts: 3 equal parts: 4 equal parts: 5 equal parts: and so on. Let us understand the concept of parts of a whole through an activity. & Remembering & Understanding Suppose we want to share an apple with our friends. First, we count our friends with whom we want to share the apple. Then, we cut it into as many equal pieces as the number of persons. Thus, each person gets an equal part of the apple after division. 112 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___118 / 1840

Parts of a whole A complete or full object is called a whole. Observe the following parts of a chocolate: Whole 2 equal parts 3 equal parts 4 equal parts We can divide a whole into equal parts as shown above. Each such division has a different name. To understand this better, let us do an activity. Activity: Halves Take a square piece of paper. Fold it into two equal parts as shown. Each of the equal parts is called a ‘half’. ‘Half’ means 1 out of 2 equal parts. Putting these 2 equal parts together makes the complete piece of paper. We write ‘1 out of 2 equal parts’ as 1 . 2 In 1 , 1 is the number of parts taken and 2 is the total number of equal parts the 2 whole is divided into. Note: 1 and 1 make a whole. 2 2 Fractions 113 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___119 / 1840

Thirds In figure (a), observe that the three parts are not equal. We can also divide a whole into three equal parts. Fold a rectangular piece of paper as shown in figures (b) and (c). 1 11 3 33 Three parts Three equal parts Fig (c) Fig (a) Fig (b) Each equal part is called a third or one-third. The shaded part in figure (c) is one out of three equal parts. So, we call it a one-third. Two out of three equal parts of figure (c) are not shaded. We call it two-thirds (short form of 2 one-thirds). 1 2 3 3 We write one-third as and two-thirds as . Note: 1 , 1 and 1 or 1 and 2 makes a whole. 3 3 3 3 3 Fourths Similarly, fold a square piece of paper into four equal parts. Each of them is called a fourth or a quarter. In figure (a), the four parts are not equal. In figure (b), each equal part is called a fourth or a quarter and is written as 1 . 4 Two out of four equal parts are called two-fourths and three out of four equal parts 2 3 are called three-fourths, written as 4 and 4 respectively. Note: Each of 1 and 4 3 ; 1 , 1 , 1 and 1 and 1 , 1 and 2 makes a whole. 4 4 4 4 4 4 4 4 114 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___120 / 1804

The total number of equal parts a 1 4 whole is divided into is called the denominator. 1 4 The number of such equal parts taken is called the numerator. 1 4 Representing the parts of a whole as 1 Numerator is called a fraction. 4 Denominator Four equal parts Thus, a fraction is a part of a whole. Four parts Fig (b) Fig (a) For example, 1 , 1 , 1 , 2 and so on 2 3 4 3 are fractions. Let us now see a few examples. Example 1: Identify the numerator and denominator in each of the following fractions: a) 1 b) 1 1 2 3 c) 4 Solution: S. No Fractions Numerator Denominator a) 12 b) 1 13 2 1 3 1 1 4 c) 4 Example 2: Identify the fraction for the shaded parts in the figures below. a) b) Fractions 115 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___121 / 1840

Solution: Steps Solved Solve this a) b) Step 1: Count the number of Total number of Total number of equal parts = _______ equal parts, the figure is divided equal parts = 8 Number of parts shaded = into (Denominator). ______ Step 2: Count the number of Number of parts shaded parts (Numerator). shaded = 5 Step 3: Write the fractions. 5 Fraction = Numerator Fraction = 8 Denominator Example 3: The circular disc shown in the figure is divided into equal parts. What fraction of the disc is painted yellow? Write the fraction of the disc that is painted white. Solution: Total number of equal parts of the disc is 16. Number of parts painted yellow is 3. The fraction of the disc that is painted yellow = Number of parts painted yellow = 3 Total number of equal parts 16 The fraction of the disc that is painted white = Number of parts painted white = 6 Total number of equal parts 16 Example 4: Find the fraction of parts not shaded in the following figures. a) b) c) Solution: a) Total number of equal parts = 2 116 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___122 / 1804

Number of parts not shaded = 1 Fraction = Number of parts not shaded = 1 Total number of equal parts 2 Complete the table given. Steps b) c) Total number of equal parts Number of parts not shaded Fraction = Number of parts not shaded Total number of equal parts Application We have learnt to identify the fraction of a whole using the shaded parts. We can learn to shade a figure to represent a given fraction. Let us see some examples. Example 5: Shade a square to represent these fractions: Solution: 1 2 31 a) 4 b) 3 c) 5 d) 2 We can represent the fractions as: Steps Solved Solve these 1 1 2 Step 1: Identify the 4 23 denominator and 35 Denominator the numerator. = Denominator Denominator Denominator Numerator = Step 2: Draw the =4 = = required shape. Numerator Numerator = Numerator = Divide it into =1 as many equal parts as the denominator. Fractions 117 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___123 / 1804

Steps Solved Solve these 1 1 2 4 23 35 Step 3: Shade the number of equal parts as the numerator. This shaded part represents the given fraction. Example 6: Colour the shapes to represent the given fractions. Fractions Shapes 1 4 Solution: 2 5 Shapes 1 2 We can represent the fractions as: Fractions 1 4 2 5 1 2 118 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___124 / 1804

Higher Order Thinking Skills (H.O.T.S.) Let us see some examples of real-life situations involving fractions. Example 7: A square shaped garden has coconut trees in a quarter of its land. It has mango trees in two quarters and neem trees in another quarter. Draw a figure of the garden and represent its parts. Solution: Fraction of the garden covered by So, the square garden is as 1 shown in the figure. coconut trees = Quarter = 4 Fraction of the garden covered by Coconut 1 mango trees = 2 Quarters = 2 trees Mango Fraction of the garden covered by trees 1 Neem neem trees = Quarter = 4 trees Example 8: Answer the following questions: a) How many one-sixths are there in a whole? b) How many one-fifths are there in a whole? c) How many halves make a whole? Solution: a) There are 6 one-sixths in a whole. b) There are 5 one-fifths in a whole. c) 2 halves make a whole. 11 66 11 11 55 22 1 1 11 6 6 55 1 1 1 6 6 5 Fractions 119 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___125 / 1840

Concept 9.2: Fraction of a Collection Think Neena has a bunch of roses. Some of them are red, some white and some yellow. Neena wants to find the fraction of roses of each colour. How can she find that? To answer that, we must know the fraction of a collection. Recall We know that a complete or a full object is called a whole. We also know that we can divide a whole into equal number of parts. Let us answer these to revise the concept. Divide these into equal number of groups as given in the brackets. Draw circles around them. a) Trai n M[y2Bgrroauipns] b) [3 groups] c) [2 groups] d) [5 groups] 120 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___126 / 1840

& Remembering & Understanding Finding a half We can find different fractions of a collection. Suppose there are 10 pens in a box. To find a half of them, we divide them into two equal parts. Each equal part is a half. 1 Each equal part has 5 pens, as 10 ÷ 2 = 5. So, 2 of 10 is 5. Finding a third One-third is 1 out of 3 equal parts. In the given figure, there are12 bananas. To find a third, we divide them into three equal parts. Each equal part is a third. Each equal part has 4 bananas, as 12 ÷ 3 = 4. So, 1 of 12 is 4. 3 Finding a fourth (or a quarter) One-fourth is 1 out of 4 equal parts. In the figure, there are 8 books. Fractions 121 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___127 / 1840

To find a fourth, divide the number of books into 4 equal parts. 11 1 1 44 4 4 1 Each equal part has 2 books, as 8 ÷ 4 = 2. So, 4 of 8 is 2. Let us see a few examples to find the fraction of a collection. Example 9: Find the fraction of the coloured parts of the shapes. Shapes Fractions Solution: The fractions of the coloured parts of the shapes are – Shapes Fractions 2 6 3 6 5 8 122 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___128 / 1804

Example 10: Colour the shapes according to the given fractions. Fractions Shapes 1 5 2 7 3 4 Solution: We can colour the shapes according to the fractions as – Shapes Fractions 1 5 2 7 3 4 Application We can apply the knowledge of fractions in many real-life situations. Let us see a few examples. Example 11: Raju brought a bag of 30 marbles. Out of these, 10 marbles are blue. How many marbles are blue? Solution: Total number of marbles = 30 Fraction of blue marbles = 1 of 30 = 30 ÷ 3 = 10 3 Total number of blue marbles = 10 Fractions 123 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___129 / 1840

Example 12: A basket has 64 flowers. Half of them are roses, a quarter of them are marigolds and a quarter of them are lotus. How many each of roses, marigolds and lotus are there in the basket? Solution: Total number of flowers = 64 Half of the flowers are roses. 1 The number of roses = 2 of 64 = 64 ÷ 2 = 32 A quarter of the flowers are marigolds. 1 The number of marigolds = 4 of 64 = 64 ÷ 4 = 16 A quarter of the flowers are lotus. 1 The number of lotus = 4 of 64 = 64 ÷ 4 = 16 Therefore, there are 32 roses, 16 marigolds and 16 lotus in the basket. Example 13: A set of 48 pens have 13 blue, 15 red and 11 black ink pens. The remaining are green ink pens. What fraction of the pens is green? Solution: Total number of pens = 48 Total number of blue, red and black ink pens = 13 + 15 + 11 = 39 Number of green ink pens = 48 – 39 = 9 9 48 Fraction of green ink pens = = Number of green ink pens = Total number of pens Example 14: A bunch of 15 balloons has 2 green balloons, 3 blue balloons and 10 red balloons. Write the fraction of balloons of each colour. Solution: Total number of balloons= 15 Number of green balloons = 2 2 So, fraction of green balloons = 15 124 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___130 / 1804

Number of blue balloons = 3 So, fraction of blue balloons = 3 15 Number of red balloons = 10 So, fraction of red balloons = 10 15 Higher Order Thinking Skills (H.O.T.S.) In some real-life situations, we need to find a fraction of some goods such as fruits, vegetables, milk, oil and so on. Let us now see some such examples. Example 15: One kilogram of apples costs ` 16 and one kilogram of papaya costs 1 ` 20. If Rita buys 1 kg of apples and 4 kg of papaya, how much did 2 she spend? Solution: Cost of 1 kg apples = ` 16 Cost of 1 kg apples = 1 of ` 16 = ` 16 ÷ 2=`8 2 2 (To find a half, we divide by 2) Cost of 1 kg papaya = ` 20 Cost of 1 kg papaya = 1 of ` 20 = ` 20 ÷ 4 = ` 5 4 4 (To find a fourth, we divide by 4) Therefore, the money spent by Rita = ` 8 + ` 5 = ` 13 Example 16: Sujay completed 2 of his Maths homework. If he had to solve 25 5 problems, how many did he complete? Solution: Fraction of homework completed = 2 5 Total number of problems to be solved = 25 2 Number of problems Sujay solved = 5 of 25 = (25 ÷ 5) × 2 = 5 × 2 = 10 Therefore, Sujay has solved 10 problems. Fractions 125 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___131 / 1804

Drill Time Concept 9.1: Fractions as a Part of a Whole 1) Find the numerator and the denominator in each of these fractions. 2 b) 1 2 4 5 a) 5 7 c) 3 d) 9 e) 7 2) Identify the fractions of the shaded parts. a) b) c) d) e) Concept 9.2: Fraction of a Collection c) 3) Find fraction of coloured parts. a) b) d) e) 4) Find 1 and 1 of the following collection. 2 4 5) Word Problems a) A circular disc is divided into 12 equal parts. Venu shaded 1 of the disc pink 4 1 and 3 of the disc green. How many parts of the disc are shaded? How many parts are not shaded? b) Rajesh has 24 notebooks. 1 of them are unruled and 1 of them are four- 2 6 ruled. How many books are (a) unruled and (b) four-ruled? 126 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___132 / 1804

Money10CHAPTER I Will Learn CCoonncceepptts 10.1: C1o.n1:veVret rRtuicpeeseanindtoDPiaagisoenals of Two-Dimensional Shapes 10.2: Add and Subtract Money With Conversion 10.3: Multiply and Divide Money 10.4: Make Rate Charts and Bills JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___133 / 1804

Concept 10.1: Convert Rupee into Paise Think Neena has ` 38 in her piggy bank. She wants to know how many paise she has. Do you know? To answer this question, we must know how to convert rupees to paise. Recall We have learnt to identify different coins and currency notes. We have also learnt that 100 paise make a rupee. learn more about money. 1 rupee = 100 paise 100 p = 1 rupee Let us revise the concept about money. a) The value of the given coin is: [ ] (A) ` 1 (B) ` 2 (C) ` 5 (D) ` 10 b) The ` 500 note among the following is: [ ] (A) (B) (C) (D) 128 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___134 / 1804

c) The combination that has the greatest value is: [ ] (A) (B) (C) (D) & Remembering & Understanding Let us understand the conversion of rupees to paise through an activity. Activity: The students must take their play money (having all play notes and coins). As the teacher writes the rupees on the board, each student picks the exact number of paise in it. There can be many combinations for the same amount of rupees. For example, 1 rupee is 100 paise. So, the students may take two 50 paise coins. Let us understand the conversion through some examples. Example 1: Convert the given rupees into paise: a) ` 2 b) ` 5 c) ` 9 Solution: We know that 1 rupee = 100 paise a) ` 2 = 2 × 100 paise = 200 paise Money 129 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___135 / 1840

b) ` 5 = 5 × 100 paise = 500 paise c) ` 9 = 9 × 100 paise = 900 paise Similarly, we can convert paise into rupees. Converting paise into rupees is the reverse process of conversion of rupees into paise. Example 2: Convert 360 paise to rupees. Solution: We can convert paise to rupees as: Steps Solved Solve this 380 paise Step 1: Write the given paise 360 paise = 300 paise + as hundreds of paise. 60 paise Step 2: Rearrange 300 paise 300 paise = (3 × 100) as a product of 100 paise. paise + 60 paise Step 3: Write in rupees. ` 3 + 60 paise = ` 3.60 Application Let us see some real-life examples involving conversion of rupees into paise and that of paise into rupees. Example 3: Anil has ` 10 with him. How many paise does he have? Solution: 1 rupee = 100 paise So, 10 rupees = 10 × 100 paise = 1000 paise Anil has 1000 paise with him. Example 4: Raj has 670 paise. How many rupees does he have? Solution: Amount with Raj = 670 paise 130 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___136 / 1804

= 600 paise + 70 paise = (6 × 100) paise + 70 paise = ` 6 + 70 paise = ` 6.70 Therefore, Raj has ` 6.70. Higher Order Thinking Skills (H.O.T.S.) Observe these examples where conversion of rupees to paise and that of paise to rupees are mostly useful. Example 5: Vani has ` 4.30, Gita has ` 5.50 and Ravi has 470 paise. Who has the least amount of money? Solution: Money Vani has = ` 4.30 Money Gita has = ` 5.50 Money Ravi has = 470 paise To compare money, all the money must be in the same unit. So, let us first convert the amounts in rupees to paise. ` 4.30 = (4 × 100) + 30 = 430 paise ` 5.50 = (5 × 100) + 50 = 550 paise Now, arranging the money in ascending order, we get 430 < 470 < 550. Therefore, Vani has the least money. Example 6: Ram has ` 1.10, Shyam has ` 1.40 and Rishi has ` 2.00. Arrange the amount in ascending order. Who has the most money? Solution: Amount Ram has = ` 1.10 Amount Shyam has = ` 1.40 Amount Rishi has = ` 2.00 To compare the money, all of them must be in the same unit. So, let us convert the amounts in rupees to paise. ` 1.10 = (1 × 100) + 10 = 110 paise Money 131 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___137 / 1804

` 1.40 = (1 × 100) + 40 = 140 paise ` 2.00 = (2 × 100) = 200 paise Arranging the money in descending order we get, 200 > 140 > 110. Therefore, Rishi has more money than Ram and Shyam. Concept 10.2: Add and Subtract Money With Conversion Think Neena’s father bought a toy car for ` 56 and a toy bus for ` 43. How much did he spend altogether? How much change does he get if he gives ` 100 to the shopkeeper? To answer this question, we must know how to add and subtract money. Recall Recall that two or more numbers are added by writing them one below the other. This method of addition is called the column method. We know that rupees and paise are separated using a dot or a point. In the column method, we write money in such a way that the dots or points are placed exactly one below the other. The rupees are under rupees and the paise are under the paise. Let us recall a few concepts about money through these questions. a) 50 paise + 50 paise = ________________ b) ` 20 + ` 5 + 50 paise = ______________ c) ` 50 – ` 10 = _______________ d) ` 20 + ` 10 = _______________ e) ` 50 – ` 20 = _______________ 132 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___138 / 1804

& Remembering & Understanding Addition and subtraction of money is similar to addition and subtraction of numbers. In column method, we write numbers one below the other and add or subtract as needed. Let us understand this through some examples. Example 7: Add: ` 14.65 and ` 23.80 Solution: We can add two amounts as: Steps Solved Solve these Step 1: Write the given numbers `p `p with the points exactly one below 1 4. 6 5 the other, as shown. + 2 3. 8 0 4 1. 5 0 + 4 5. 7 5 Step 2: First add the paise. Regroup `p `p the sum if needed. Write the sum 1 3 8. 4 5 under paise. Place the dot just 1 4. 6 5 + 3 5. 6 0 below the dot. + 2 3. 8 0 `p Step 3: Add the rupees. Add the . 45 2 3. 6 5 carry forward (if any) from the + 1 4. 5 2 previous step. Write the sum under `p rupees. 1 1 4. 6 5 Step 4: Write the sum of the given + 2 3. 8 0 amounts. 3 8. 4 5 ` 14.65 + ` 23. 80 = ` 38.45 Example 8: Write in columns and subtract: ` 56.50 from ` 73.50 Solution: We can subtract the amounts as: Money 133 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___139 / 1840

Steps Solved Solve these Step 1: Write the given numbers with the dots exactly one below the `p `p other, as shown. 7 3. 5 0 8 0. 7 5 − 5 6. 5 0 − 4 1. 5 0 Step 2: First subtract the paise. Regroup if needed. Write the `p `p difference under paise. Place the 7 3. 5 0 6 0. 7 5 dot just below the dot. − 5 6. 5 0 − 3 2. 5 0 Step 3: Subtract the rupees. Write the difference under rupees. 00 Step 4: Write the difference of the `p given amounts. 6 13 7 3. 5 0 − 5 6. 5 0 1 7. 0 0 ` 73. 50 - ` 56. 50 = ` 17.00 Application Look at some real-life examples where we apply addition and subtraction of money. Example 9: Arun has ` 45.50 with him. He gave ` 23.50 to Amar. How much money is Arun left with? Solution: Amount Arun has = ` 45.50 `p Amount Arun gave to Amar = ` 23.50 4 5. 5 0 Difference in the amounts − 2 3. 5 0 2 2. 0 0 = ` 45.50 – ` 23.50 = ` 22 Therefore, the amount left with Arun = ` 22 Example 10: Ramu has ` 12.75 with him. His friend has ` 28.50 with him. What is the amount both of them have? Solution: Amount Ramu has = ` 12.75 134 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___140 / 1840

Amount Ramu’s friend has = ` 28.50 `p To find the total amount we have to add both the 11 amounts. 1 2. 7 5 Total amount with Ramu and his friend is ` 41.25. + 2 8. 5 0 4 1. 2 5 Higher Order Thinking Skills (H.O.T.S.) In some situations, we may need to add and subtract amounts together. In such cases, we need to identify which operation is to be carried out first. Let us see a few examples. Example 11: Surya went to a water park with his parents. Roller coaster: ` 35 The ticket for each ride is as given. Surya went on two rides. He gave ` 60 and got Break dance: ` 32 a change of ` 5. Which two rides did he go on? Water ride: ` 20 Solution: Surya gave ` 60. The change he got is ` 5. The money spent for two rides = ` 60 – ` 5 = ` 55 So, we must add and check which two tickets cost ` 55. ` 35 + ` 32 = ` 67 which is not ` 55. ` 32 + ` 20 = ` 52 which is not ` 55. ` 35 + ` 20 = ` 55 Therefore, the two rides that Surya went on are roller coaster and water ride. Example 12: Add ` 20 and ` 10.50. Subtract the sum from ` 40. `p 2 0. 0 0 Solution: First add ` 20 and ` 10.50. ` 20 + ` 10.50 = ` 30.50 + 1 0. 5 0 Now, let us find the difference between 3 0. 5 0 ` 30.50 and ` 40. Therefore, ` 40 – ` 30.50 = ` 9.50 ......... `p 4 0. 0 0 − 3 0. 5 0 0 9. 5 0 Money 135 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___141 / 1840

Concept 10.3: Multiply and Divide Money Think Neena's father gave her ` 150 on three occasions. Neena wants to share the total amount equally with her brother. How should she find the total amount? How much will Neena and her brother get? To answer tRhies qcuaeslltion, we must learn how to multiply and divide money. While multiplying, we begin from ones place and move to the tens and hundreds places. Sometimes, we may need to regroup the products. We begin division from the largest place and move to the ones place of the number. Let us answer these to revise the concept. a) 32 × 4 = _____ b) 11 × 6 = _____ c) 20 ÷ 2 = _____ b) 48 ÷ 3 = _____ e) 10 × 6 = _____ f) 24 ÷ 8 = _____ & Remembering & Understanding Train My Brain Multiplication and division of money is similar to that of numbers. To multiply money, first multiply the numbers under paise, and place the point. Then multiply the number under rupees. To divide money, we divide the numbers under rupees and place the point in the quotient. Then, divide the number under paise. Now, let us understand multiplying and dividing money through a few examples. Example 13: Multiply ` 72.50 by 8. Solution: Follow the steps to find the total amount. `. p 24 Step 1: Multiply the paise. Regroup if needed. 7 2. 50 Place the point in the product, just below the × 8 point in the given number. 580 . 00 Step 2: Multiply the rupees. Regroup if needed. 136 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___142 / 1840

Example 14: Divide: ` 35.70 by 7 5.10 Solution: Follow the steps to divide the amount. Step 1: Divide the rupees. )7 35.70 Step 2: Place point. Step 3: Divide the paise. − 35 07 −7 00 − 00 00 Application We apply multiplication and division of money in many real- life situations. Let us see some examples. Example 15: A dozen bananas cost ` 48. a) What is the cost of three dozen bananas? b) What is the cost of one banana? Solution: One dozen = 12 a) Cost of one dozen bananas = ` 48 2 Cost of three dozen bananas = ` 48 × 3 = ` 144 48 b) Cost of one dozen (12) bananas = ` 48 ×3 Cost of one banana = ` 48 ÷ 12 = ` 4 144 (Recall that 10 × 4 = 40. Then 11 × 4 = 44 and 12 × 4 = 48) Example 16: Rahul went to buy chocolates. If a chocolate costs ` 20.60, how much Solution: money would 4 such chocolates cost? `p Cost of one chocolate = ` 20.60 2 Cost of 4 chocolates = ` 20.60 × 4 = ` 82.40 2 0. 6 0 ×4 8 2. 4 0 Money 137 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___143 / 1840

Higher Order Thinking Skills (H.O.T.S.) In some situations, we have to carry out more than one operation on money. Consider the following examples. Example 17: Nidhi buys 4 bunches of flowers each costing ` 54. She buys 6 candy bars for her brothers at the cost of ` 5.50 each. If she has ` 7.50 left with her, how much did she have in the beginning? Solution: Cost of a bunch of flowers = ` 54 Cost of 4 bunches = ` 54 × 4 = ` 216 Cost of each candy bar = ` 5.50 Cost of 6 candy bars = ` 5.50 × 6 = ` 33 Total cost of the things she bought = ` 216 + ` 33 = ` 249 Amount she is left with = ` 7.50 Therefore, amount she had in the beginning = ` 249 + ` 7.50 = ` 256.50 Example 18: Bhanu bought some items for ` 362. She has some ` 100 notes. How many notes should she give the shop keeper? Solution: Cost of the items ` 362 = ` 300 + ` 62 = (` 100 x 3) + ` 62 As ` 362 has to be paid, Bhanu has to give 3 + 1 = 4 notes of ` 100. Concept 10.4: Make Rate Charts and Bills Think Neena went to a mall with her parents. She buys a a pair of jeans, 2 shirts, a story book and a ball. How much should she pay? She was given a bill for what she has bought. Can you prepare a bill similar to the one given to her? To answer this question, we must know how to prepare rate charts and bills. 138 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___144 / 1840

Recall Recall that we make lists of items when we go shopping. The lists can be of provisions, stationery and items like vegetables or fruits. We can compare the list of items and the items we received. We can compare their rates and add them to get the total amount to be paid. Let us answer these to revise addition and multiplication of money. a) ` 12 × 2 = __________ b) ` 20 × 3 = __________ c) ` 25 × 4 = __________ d) ` 12 + ` 20 = __________ e) ` 30 + ` 40 = __________ f) ` 21 + ` 10 = & Remembering & Understanding 1) Making bills To make a bill of items, we write the rate of the object and the quantity in the bill. We write the product of the rate and the quantity. We add the products to find the total bill amount. Addition of amounts is similar to the addition of numbers of two or more digits. Let us understand how to make bills through a few examples. Example 19: The rates of some items in a stationery shop are given in the box. Money 139 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___145 / 1804

Sunil buys a few items as given in the list. Make a bill for the items he bought. Item Quantity Pencil box 1 Erasers 2 Sharpeners 4 Pens 3 Notebooks 4 Solution: Follow the steps to find the bill. Step 1: Write the items and their quantity in the bill. Step 2: Then write the cost per item. Step 3: Find the total cost of each item by multiplying the number of items by its rate. Step 4: Find the total bill by adding the amount for all the items bought. S.No Item Bill Rate per Amount item `p 1 Pencil box Quantity `24.50 24 50 2 Erasers `5.00 3 1 `3.00 10 00 4 Sharpeners 2 `4.00 12 00 5 Pens 4 `10.00 12 00 3 Total 40 00 Notebooks 4 98 50 2) Making rate charts When rates of all items are written on the item, it is difficult to see and compare them. So, we need to make a rate chart. In this chart, we write the rate of each item. Example 20: Anil and his friends are playing with play money. Anil runs a supermarket. Some items in his supermarket are given below with their rates. 140 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___146 / 1804

` 275/- ` 280/- ` 50/- ` 34/- ` 102/- ` 149.50/- ` 24.50/- ` 140/- per kg ` 45/- per kg ` 130/- kg ` 50/- ` 40/- per dozen He makes a rate chart to display the price of each item. How will the rate chart look? Solution: 1. Draw a table. 2. Write each item and its rate in columns. Item Rate ` Item Rate ` Boost 275.00 1kg rice 45.00 Kitkat 50.00 1kg dal 140.00 Tea powder 102.50 1 biscuit packet 24.50 Horlicks 280.00 paneer 50.00 Soap 34.00 1 kg apples 130.00 Honey 149.50 1 dozen bananas 40.00 Money 141 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___147 / 1804

Application Let us learn how to make rate charts and bills and use them in our daily life with an activity. Go to a vegetable store. Suppose you see the rate chart of all vegetables with their rates per kg. Buy some vegetables and make the bill. Vegetables Rate per kg Bill in ` Cabbage 24.00 Item Rate per kg ` Paise Brinjal 30.00 Okra 40.00 1 kg brinjal 30.00 30 00 Potato 40.00 Tomato 20.00 2 kg potato 40.00 80 00 Onion 22.00 2 kg tomato 20.00 40 00 1 kg onion 22.00 22 00 Total 172 00 The bill for the items you bought would be as shown. Write the rates carefully, by considering the quantity. Find the total bill by adding total cost of each vegetable. Here, the total bill is ` 172.00. Example 21: Ashish went to Amar Bakery with his friends. The rate chart of the items available there is as given. Item Rate Item Rate in ` in ` Cheese Cheese Burger 55.00 Pizza 125.00 Chicken 60.00 Chicken 150.00 Burger Pizza Veg 50.00 Burger 142 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___148 / 1804

Item Rate Item Rate in ` in ` Veg Pizza 110.00 Veg Roll 40.00 Hot Dog 45.00 Chicken 50.00 Roll Solution: What can they buy in this bakery, if they have ` 250? (Write 3 different options and make a bill for one of the options.) To write three different options for Ashish and his friends to choose, see that the sum of the rates does not exceed ` 250. The three options could be: a) 2 Chicken Burgers and 1 Cheese Pizza b) 2 Cheese Pizzas c) 2 Veg Rolls, 1 Veg Pizza, 1 Chicken Burger Let us now make a bill for 3rd option. Find the cost and write the total. Amar Bakery Bill Item Rate per item ` p 80 00 2 Veg. rolls ` 40.00 110 00 60 00 1 Veg. pizza ` 110.00 250 00 1 Chicken burger ` 60.00 Total Higher Order Thinking Skills (H.O.T.S.) Seeing the rate chart in a shop, we calculate mentally the amount for the items we buy and give money to the shopkeeper. Let us now see an example. Money 143 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___149 / 1804

Example 22: Sneha went to an ice cream 1000 ml tub of ice-cream Rate in ` shop and saw the rate Butter Scotch 150.00 chart given. Sneha took 2 Vanilla 120.00 Butter Scotch, 2 Mango, 1 Strawberry 130.00 Chocolate and 1 Vanilla Mango 140.00 ice cream tubs. What is the Chocolate 160.00 total bill ? Make the bill. If she gave ` 1000, how much did she get as change? Solution: Write the items, number of each item and their rates. Multiply them to find cost of each flavour of ice-cream. Find the total by adding all the amounts. Ice cream shop Item Quantity Rate per tub ` paise Butter Scotch 2 ` 150.00 300 00 Mango 2 ` 140.00 280 00 1 ` 160.00 160 00 Chocolate 1 ` 120.00 120 00 Vanilla Total 860 00 Amount Sneha gave = ` 1000 Total bill amount = ` 860 The amount she received as change = ` 1000 – ` 860 = ` 140 Drill Time Concept 10.1: Convert Rupee into Paise 1) Convert rupees to paise a) ` 34 b) ` 12 c) ` 80 d) ` 29 e) ` 10 2) Convert paise to rupees a) 320 paise b) 140 paise c) 450 paise d) 298 paise e) 100 paise 144 JSNR_BGM_1010020-Alpine-G3-FoundationMax-Maths-FY_REVISED_Text.pdf___150 / 1840