202110792-TRAVELLER_PREMIUM-STUDENT-TEXTBOOK-MATHEMATICS-G04-latest

MATHEMATICS TEXTBOOK – Grade 4 Name: _________________________ Section: ________Roll No: _______ School: ________________________

Contents Part 1 4 Addition and Subtraction 4.1  Add and Subtract 4-digit and 5-digit Numbers������������������������������������������������������ 43 5 Multiplication 5.1 Multiplication of 2-digit Numbers and 3-digit Numbers ��������������������������������������� 52 6 Division 6.1 D ivide 3-digit Numbers by 1-digit Numbers���������������������������������������������������������� 66

4Chapter SAdudbittrioanctainond I Will Learn About • adding and subtracting 4-digit and 5-digit numbers. • applying two operations addition and subtraction in solving real-life situations. • framing word problems based on mathematical statements. 4.1 Add and Subtract 4-digit and 5-digit Numbers I Think Surbhi tried to log into her father’s computer. Unfortunately, it was locked and she did not know the password. She saw the hint as given below: 6738 − 2345 = _______ Can you solve it? I Recall We have learnt the addition and subtraction of 4-digit numbers without regrouping. 43

Let us recall the steps followed. Step 1: Arrange the numbers one below the other according to their places. For subtraction, ensure that the smaller number is placed below the larger number. Step 2: Start adding or subtracting the digits from the ones place. Step 3: Write the answer. Let us use these steps to solve the following: a) Th H T O b) Th H T O c) Th H T O 4216 1335 5985 +1251 +1234 +2003 d) Th H T O e) Th H T O f) Th H T O 7453 4472 6134 −1322 −1322 −1021 I Remember and Understand Let us see the examples of addition and subtraction of While subtracting numbers 4-digit and 5-digit numbers with regrouping. by the vertical method, always write the smaller Add 4-digit numbers number below the bigger number. We regroup the sum when it is equal to or more than 10. Example 1: Add 1456 and 1546. Solution: Arrange the numbers one below the other. Add and regroup if necessary. Solved Step 1: Add the ones. Step 2: Add the tens. Th H T O Th H T O 1 11 1456 1456 +1 5 4 6 +1 5 4 6 2 02 44

Solved Step 3: Add the hundreds. Step 4: Add the thousands. Th H T O Th H T O 111 111 1456 1456 +1 5 4 6 +1 5 4 6 002 3002 Th H T O Solve these Th H T O Th H T O 1758 4592 +5662 2678 +1456 +1332 Subtract 4-digit numbers In the subtraction of 4-digit numbers, we can regroup the digits in thousands, hundreds and tens places. Let us see an example. Example 2: Subtract: 4868 from 7437 Solution: Write the smaller number below the larger number. Steps Solved Solve these Step 1: Subtract the ones. But, 7 – 8 is not possible as 7 < 8. So, Th H T O Th H T O regroup the tens digit, 3. 2 17 3 tens = 2 tens + 1 ten. Borrow 1 ten to 1654 the ones place and add it (1 ten = 10 7 4 \\3 \\7 –1246 ones) to the ones place. Reduce the –4868 tens digit by 1. Then subtract the ones and write the answer. 9 So, 17 – 8 = 9. Addition and Subtraction 45

Steps Solved Solve these Step 2: Subtract the tens. But, 2 – 6 is Th H T O not possible as 2 < 6. So, regroup the Th H T O hundreds digit, 4. 12 5674 4 hundreds = 3 hundreds + 1 hundred. –2382 Borrow 1 hundred to the tens place and 3 \\2 17 add it (1 hundred = 10 tens) to the tens 7 4\\ \\3 7\\ Th H T O place. Reduce the hundreds digit by –4868 1. Then subtract the tens and write the 7468 answer. 69 –4837 So, 12 – 6 = 6. Th H T O Th H T O Step 3: Subtract the hundreds. But, 13 12 3 – 8 is not possible as 3 < 8. So, regroup 9276 the thousands digit, 7. 7 thousands = 6 \\3 2\\ 17 –5147 6 thousands + 1 thousand. Borrow 1 \\7 \\4 3\\ \\7 thousand to the hundreds place and –4868 add it (1 thousand = 10 hundreds) to the hundreds place. Reduce the thousands 569 digit by 1. Then subtract the hundreds and write the answer. Th H T O 13 12 So, 13 – 8 = 5. Step 4: Subtract the thousands and write 6 \\3 \\2 17 the answer. \\7 4\\ \\3 \\7 –4868 6–4=2 2569 Therefore, 7437 - 4868 = 2569. Add 5-digit numbers Example 3: Add: 48415 and 20098 Solution: Arrange the numbers in columns, one below the other. Steps Solved Solve these T Th Th H T O Step 1: Add the ones. Write T Th Th H T O the sum under the ones. 1 5 7383 Regroup if needed. + 3 1347 4 8415 + 2 0098 3 46

Steps Solved Solve these Step 2: Add the tens and T Th Th H T O also the carry forward (if T Th Th H T O any) from the previous step. 11 2 5347 Write the sum under the tens. 4 8415 Regroup if needed. + 2 0098 + 6 2567 Step 3: Add the hundreds 13 T Th Th H T O and also the carry forward 1 7298 (if any) from the previous T Th Th H T O step. Write the sum under the 11 + 2 6543 hundreds place. Regroup if needed. 4 8415 + 2 0098 Step 4: Add the thousands and also the carry forward 513 (if any) from the previous step. Write the sum under the T Th Th H T O thousands place. Regroup if 11 needed. 4 8415 + 2 0098 8513 Step 5: Add the ten T Th Th H T O T Th Th H T O thousands and also the carry 11 forward (if any) from the 3 4765 previous step. Write the sum 4 8 4 15 + 2 1178 under the ten thousands + 2 0 0 98 place. 6 8 5 13 Therefore, 48415 + 20098 = 68513. Subtract 5-digit numbers Example 4: Subtract: 56718 – 16754 Solution: Arrange the numbers in columns, one below the other. Steps Solved Solve these T Th Th H T O T Th Th H T O Step 1: Subtract the ones and write the difference 5 6718 9 7 0 54 under the ones place. − 1 6754 − 2 3 5 67 4 Addition and Subtraction 47

Steps Solved Solve these T Th Th H T O Step 2: Subtract the tens. T Th Th H T O That is, 1 − 5, which is not 7 5400 possible. 6⁄ 1⁄1 − 3 2689 Regroup the hundreds to 5 6718 T Th Th H T O tens, subtract and write the − 1 6754 difference under the tens 5 4635 place. 64 − 1 2789 Step 3: Subtract the hundreds. That is, 6 − 7, T Th Th H T O T Th Th H T O which is not possible. 16 8 9576 Regroup the thousands to 6⁄⁄ − 4 5689 hundreds, subtract and write 5 5 11 8 the difference under the 7 hundreds place. 6⁄ 1⁄ − 1 6754 964 Step 4: Subtract the T Th Th H T O thousands. That is, 5 − 6, which is not possible. 1⁄5 1⁄6 4⁄ 5⁄ 6⁄ 1⁄1 Regroup the ten thousands to thousands, subtract and 5 6 7 18 write the difference under − 1 6 7 54 the thousands place. Step 5: Subtract the ten 9 9 64 thousands, and write the difference under the ten T Th Th H T O thousands place. 4⁄ 15⁄⁄5 16⁄⁄6 1⁄1 Therefore, 56718 – 16754 = 39964. 5 6 7 18 − 1 6 7 54 3 9 9 64 ? Train My Brain Solve the following: a) 3456 + 2709 b) 42361 + 18194 c) 97972 – 10402 48

5Chapter Multiplication I Will Learn About • multiplication of 2-digit and 3-digit numbers. • multiplication using lattice algo- rithm. • mental multiplication. 5.1 Multiplication of 2-digit Numbers and 3-digit Numbers I Think Surbhi went to a stadium to watch a football match with her parents. She observed that the seats are arranged in many rows and columns. All the seats were occupied. She wanted to guess the total number of people who watched the match that day. How will she be able to do that? I Recall We have learnt to multiply 2-digit numbers by 1-digit numbers. 52

Let us solve the following to revise the concept of multiplication. a) T O b) H T O c) H T O d) H T O e) H T O 39 56 89 75 90 ×2 ×3 ×4 ×5 ×4 I Remember and Understand We have learnt the multiplication tables from 2 to 10. Let us now learn the multiplication tables from 11 to 20. Multiplication Tables 11 12 13 14 15 11 × 1 = 11 12 × 1 = 12 13 × 1 = 13 14 × 1 = 14 15 × 1 = 15 11 × 2 = 22 12 × 2 = 24 13 × 2 = 26 14 × 2 = 28 15 × 2 = 30 11 × 3 = 33 12 × 3 = 36 13 × 3 = 39 14 × 3 = 42 15 × 3 = 45 11 × 4 = 44 12 × 4 = 48 13 × 4 = 52 14 × 4 = 56 15 × 4 = 60 11 × 5 = 55 12 × 5 = 60 13 × 5 = 65 14 × 5 = 70 15 × 5 = 75 11 × 6 = 66 12 × 6 = 72 13 × 6 = 78 14 × 6 = 84 15 × 6 = 90 11 × 7 = 77 12 × 7 = 84 13 × 7 = 91 14 × 7 = 98 15 × 7 = 105 11 × 8 = 88 12 × 8 = 96 13 × 8 = 104 14 × 8 = 112 15 × 8 = 120 11 × 9 = 99 12 × 9 = 108 13 × 9 = 117 14 × 9 = 126 15 × 9 = 135 11 × 10 = 110 12 × 10 = 120 13 × 10 = 130 14 × 10 = 140 15 × 10 = 150 16 17 18 19 20 16 × 1 = 16 17 × 1 = 17 18 × 1 = 18 19 × 1 = 19 20 × 1 = 20 16 × 2 = 32 17 × 2 = 34 18 × 2 = 36 19 × 2 = 38 20 × 2 = 40 16 × 3 = 48 17 × 3 = 51 18 × 3 = 54 19 × 3 = 57 20 × 3 = 60 16 × 4 = 64 17 × 4 = 68 18 × 4 = 72 19 × 4 = 76 20 × 4 = 80 16 × 5 = 80 17 × 5 = 85 18 × 5 = 90 19 × 5 = 95 20 × 5 = 100 16 × 6 = 96 17 × 6 = 102 18 × 6 = 108 19 × 6 = 114 20 × 6 = 120 16 × 7 = 112 17 × 7 = 119 18 × 7 = 126 19 × 7 = 133 20 × 7 = 140 16 × 8 = 128 17 × 8 = 136 18 × 8 = 144 19 × 8 = 152 20 × 8 = 160 16 × 9 = 144 17 × 9 = 153 18 × 9 = 162 19 × 9 = 171 20 × 9 = 180 16 × 10 = 160 17 × 10 = 170 18 × 10 = 180 19 × 10 = 190 20 × 10 = 200 Multiplication 53

Let us now learn to multiply: Standard algorithm is the 1) 3-digit numbers by 1-digit numbers. method of multiplication 2) 2-digit numbers by 2-digit numbers. in which the product is 3) 3-digit numbers by 2-digit numbers. regrouped as ones and tens. Multiply a 3-digit number by a 1-digit number When a 3-digit number is multiplied by a 1-digit number, we may get a 2-digit product in any or all of the places. We regroup these products and carry over the tens digit of the product to the next place. Let us understand this better through the following example. Example 1: Multiply: 513 × 5 Solution: Follow these steps to multiply the given numbers. Steps Solved Solve these H TO Step 1: Multiply the ones. Regroup if the HT O product has two digits. Carry the tens digit 1 635 of the product to the tens place and write 3 ×7 its ones digit under the ones place. 51 5 × 5   Step 2: Multiply the tens. Add the carry H TO H TO over (if any) to the product. Regroup if the 1 product has two digits. Carry the tens digit 444 of the product to the hundreds place and 513 ×8 write its ones digit under the tens place. ×5 65 Step 3: Multiply the hundreds. Add the Th H T O H TO carry over (if any) to the product. Regroup 1 if the product has two digits. Carry the 342 tens digit of the product to the thousands 513 ×5 place. Write its ones digit under the ×5 hundreds place and the tens digit under the thousands place. 2565 Multiply a 2-digit number by a 2-digit number Let us multiply 2-digit numbers by 2-digit numbers through a step-by-step procedure. Consider the following example. 54

Example 2: Multiply: 24 × 34 Solution: Follow these steps to multiply the given numbers. Steps Solved Solve these Step 1: Multiply the ones by the ones digit H TO of the multiplier. 4 × 4 = 16 TO 41 1 ×22 Write 6 in the ones place of the product. 24 Write 1 in the tens place as carry over. ×34 H TO 52 Step 2: Multiply the tens digit by the ones 6 digit of the multiplier. 2 × 4 = 8 ×23 H TO Add the carry over from the previous step 1 H TO to the product. So, 8 + 1 = 9. Write 9 in the 24 63 tens place of the product. ×34 ×23 Step 3: Write 0 in the ones place under 96 the product obtained from the previous steps. Now multiply the ones digit by the H TO tens digit of the multiplier. 1 3 × 4 = 12 1 Write 2 in the tens place, below 9 and 24 carry the tens digit,1, to the tens place. ×34 Step 4: Multiply the tens by the tens digit 96 of the multiplier. 20 3×2=6 H TO 1 Add the carry over from the previous 1 step. So, 6 + 1 = 7. Write 7 in the hundreds 24 place. ×34 Step 5: Add the products and write the 96 sum, which is the required product. 720 Th H T O 1 1 24 ×34 196 +720 816 Multiplication 55

Multiply a 3-digit number by a 2-digit number Multiplication of 3-digit numbers by 2-digit numbers is similar to multiplication of two 2-digit numbers. It may sometimes involve regrouping too. Let us understand this concept through a step-by-step procedure. Consider the following example. Example 3: Multiply: 243 × 34 Solution: Arrange the numbers in columns, as shown. Steps Solved Solve these Step 1: Multiply the ones by the ones digit H TO of the multiplier. 3 × 4 = 12 H TO 1 453 Write 2 in the ones place of the product. ×13 Carry the 1 to the tens place. 243 ×34 Step 2: Multiply the tens digit by the ones digit of the multiplier. 4 × 4 = 16 2 Add the carry over from the previous H TO step. So, 16 + 1 = 17. Write 7 in the tens place of the product and 1 in the 11 hundreds place as carry over. 243 ×34 72 Step 3: Multiply the hundreds by the ones H TO H TO digit of the multiplier. 2 × 4 = 8 11 263 Add the carry over from the previous 243 ×23 step. So, 8 + 1 = 9. Write 9 in the hundreds ×34 place of the product. 972 Step 4: Write 0 in the ones place below HTO the product obtained from the previous 11 steps. 243 ×34 Multiply the ones by the tens digit of the 972 multiplier. Write the product under the tens place. 90 3×3=9 Write 9 below 7 in the tens place of the previous product. 56

Steps Solved Solve these H TO Step 5: Multiply the tens by the tens digit H TO 141 of the multiplier. ×22 1 4 × 3 = 12 H TO 11 352 Write 2 in hundreds place of the 243 ×23 product and 1 in hundreds place of the ×34 multiplicand as carry over. 972 290 Step 6: Multiply the hundreds by the tens Th H T O digit of the multiplier. 2×3=6 1 Add the carry over from the previous 11 step. So, 6 + 1 = 7. Write 7 in the thousands 243 place of the product. ×34 972 7290 Step 7: Add the products and write the Th H TO sum. Regroup if necessary. The sum is the 1 required product. 11 243 ×34 11 972 +7 2 9 0 8 2 6 2 ? Train My Brain Solve the following: a) 222 × 8 b) 92 × 32 c) 632 × 22 Multiplication 57

Division6Chapter I Will Learn About • dividing 2-digit and 3-digit numbers by 1-digit numbers. • solving real-life examples based on division. 6.1 Divide 3-digit Numbers by 1-digit Numbers I Think Surbhi and seven of her friends want to share 350 papers equally among themselves. Do you think she can divide the papers without some of them being left? I Recall In grade 3, we have learnt about equal grouping and repeated subtraction. We have also learnt about the relation between multiplication and division. Equal grouping: Having equal number of objects in each group into which a collection of objects is divided is called equal grouping. Repeated subtraction: Subtracting the same numbers over and over from a given number is called repeated subtraction. 66

Let us recall these concepts by answering the following questions. a) 15 – 5 = 10; _____ – 5 = 5; 5 – _____ = 0 b) 9 – _____ = 6; 6 – 3 = _____; ______ – ______ = 0 c) 7 × 5 = 35 is the multiplication fact of 35 ÷ 7 = _____ . d) 40 ÷ 8 = 5 is the division fact of 5 × _____ = 40. e) 28 ÷ 4 = __________ I Remember and Understand Dividing a number by 1 gives the We can make equal groups and divide vertically. Multiplication quotient same as tables can be used to divide numbers vertically. Let us see an the dividend. example. Divide a 2-digit number by a 1-digit number (1-digit quotient) Example 1: 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 the 5)45 )8 56 Dividend = _____ divisor as shown: Divisor = ______ 45 = 5 × 9 − Quotient = ____ )Divisor Dividend 9 Remainder = _____ Step 2: Find the multiplication fact )5 45 in the table of the divisor which has the dividend as the product. − 45 Step 3: Write the factor other than the divisor as the quotient. Write the product in the multiplication fact below the dividend. Step 4: Subtract the product 9 )4 36 Dividend = _____ from the dividend and write the Divisor = ______ difference below the product. )5 45 − Quotient = ____ Remainder = _____ This difference is called the − 45 remainder. 00 45 = Dividend 5 = Divisor 9 = Quotient 0 = Remainder Division 67

Note: If the remainder is zero, the divisor is said to divide the dividend exactly. Checking for the correctness of division The multiplication fact of the division is used to check its correctness. Step 1: Compare the remainder and the divisor. The remainder must always be less than the divisor. Step 2: Check if (Quotient × Divisor) + Remainder = Dividend. If this is true, the division is correct. Let us now check if the division in example 1 is correct or not. Step 1: Remainder < Divisor 0 < 5 (True) Step 2: Quotient × Divisor 9 × 5 = 45 Step 3: (Quotient × Divisor) + Remainder 45 + 0 = 45 Step 4: (Quotient × Divisor) + Remainder = Dividend 45 = 45 (True) Therefore, the division is correct. Note: a) The division is incomplete if Remainder > Divisor (OR) Remainder = Divisor. b) The division is incorrect if (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 an example. Example 2: Divide: 57 ÷ 3 Solution: Follow these steps to divide a 2-digit number by a 1-digit number. Steps Solved Solve these 5>3 Step 1: Check if the tens digit of the )5 60 dividend is greater than the divisor. 1 − Step 2: Divide the tens and write the )3 57 quotient. − −3 Write the product of quotient and divisor, Dividend = _____ below the tens digit of the dividend. 1 Divisor = ______ Quotient = ____ Step 3: Subtract and write the difference )3 57 Remainder = ___ (= Remainder). −3 2 68

Steps Solved Solve these Step 4: Check if difference < divisor is true. 2 < 3 (True) )3 42 Step 5: Bring down the ones digit of the 19 dividend and write it beside the remainder. − )3 57 − Step 6: Find the largest number in the − 3↓ multiplication table of the divisor that can 27 Dividend = _____ be subtracted from the 2-digit number in Divisor = ______ the previous step. 3 × 8 = 24 Quotient = ____ Remainder = ___ 19 )3 57 − 3↓ 27 3 × 9 = 27 3 × 10 = 30 24 < 27 < 30. So, 27 is the required number. Step 7: Write the factor of the required 19 )3 39 number, other than the divisor, as the quotient. Write the product of the divisor )3 57 − and the quotient below the 2-digit number. Subtract and write the difference. − 3↓ − 27 Step 8: If remainder < divisor is true, stop the Dividend = _____ division. − 27 Divisor = ______ If the remainder is equal to or greater than 00 Quotient = ____ the divisor, the division is incorrect. So, Remainder = ___ check the division and correct it to get a 0 < 3 (True) remainder less than the divisor. Step 9: Write the quotient and the Quotient = 19 remainder. Remainder = 0 Step 10: Check if (Divisor × Quotient) + 3 × 19 = 57 Remainder = Dividend is true. 57 + 0 = 57 (If not, then the division is incorrect.) 57 = 57 (True) Division 69

Divide a 3-digit number by a 1-digit number (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 an example. Example 3: Divide: 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 5)265 )4 244 of the dividend is greater than the divisor. If it is not, consider its tens 2 is not greater than 5. − digit also. So, consider 26. − Step 2: Find the largest number in 5 the table of the divisor that can be Dividend = _____ subtracted from the 2-digit number )5 265 Divisor = ______ of the dividend. Write the quotient. Quotient = ____ Write the product of the quotient and − 25 Remainder = ___ the divisor below the dividend. 5 × 4 = 20 )6 366 5 × 5 = 25 5 × 6 = 30 − 25 < 26 − Step 3: Subtract and write the 5 remainder. )5 265 − 25 1 Step 4: Check if remainder < divisor 1 < 5 (True) is true. (If not, then the division is incomplete.) 5 Step 5: Bring down the ones digit )5 265 of the dividend. Write it beside the remainder. − 25↓ 15 Step 6: Find the largest number in the 5 Dividend = _____ multiplication table of the divisor that Divisor = ______ can be subtracted from the 2-digit )5 265 Quotient = ____ number in the previous step. Remainder = ___ − 25↓ 15 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 15 is the required number. 70

Steps Solved Solve these 53 Step 7: Write the factor other than )9 378 the divisor, as the quotient. Write )5 265 the product of the divisor and the − quotient below the 2-digit number. − 25↓ Then, subtract them. 15 − Step 8: Check if remainder < divisor is − 15 Dividend = _____ true. Stop the division. (If not, then the 00 Divisor = ______ division is incomplete.) Quotient = ____ Step 9: Write the quotient and the 0 < 5 (True) Remainder = ___ remainder. Step 10: Check if (Divisor × Quotient) Quotient = 53 + Remainder = Dividend is true. (If Remainder = 0 not, then the division is incorrect.) 5 × 53 = 265 265 + 0 = 265 265 = 265 (True) 3-digit quotients Example 4: 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 the dividend is greater than or equal to the divisor. )7 784 )8 984 Step 2: Divide the hundreds and write the 7=7 − quotient in the hundreds place. 1 − Write the product of the quotient and the − divisor under the hundreds place of the )7 784 dividend. Step 3: Subtract and write the remainder. −7 1 )7 784 −7 0 Step 4: Check if remainder < divisor is true. 0 < 7 (True) Dividend = _____ Divisor = ______ Step 5: Bring down the next digit of the 1 Quotient = ____ dividend. Check if it is greater than or equal Remainder = ___ to the divisor. )7 7 84 −7↓ 08 8>7 Division 71

Steps Solved Solve these Step 6: Find the largest number in the 11 )5 965 multiplication table of the divisor that can be subtracted from the 2-digit number in the )7 784 − previous step. Write the factor other than the divisor as the − 7↓ − quotient. 08 Write the product of the quotient and the − divisor below it. −7 Dividend = _____ 7×1=7<8 Divisor = ______ The required Quotient = ____ number is 7. Remainder = ___ Step 7: Subtract and write the remainder. 11 )2 246 Bring down the next digit (ones digit) of the dividend. )7 784 − Check if the dividend is greater than or equal to the divisor. If so, then continue with the − 7↓ − division. 08 − Step 8: Find the largest number in the −7 multiplication table of the divisor that can 14 Dividend = _____ be subtracted from the 2-digit number in the Divisor = ______ previous step. 14 > 7 Quotient = ____ Write the factor other than the divisor as the Remainder = ___ quotient. 112 Write the product of the quotient and the divisor below it. )7 784 − 7↓ 08 −7 14 − 14 Step 9: Subtract and write the remainder. 7 × 2 = 14 Check if it is less than the divisor. Stop the The required division. number is 14. 112 )7 784 − 7↓ 08 −7 14 − 14 00 72

Steps Solved Solve these Step 10: Write the quotient and the Quotient = 112 remainder. Remainder = 0 Step 11: Check if (Divisor × Quotient) + 7 × 112 = 784 Remainder = Dividend is true. (If not, then the division is incorrect.) 784 + 0 = 784 784 = 784 (True) ? Train My Brain Solve the following: a) 12 ÷ 4 b) 648 ÷ 8 c) 744 ÷ 4 Division 73

Contents Part 2 7 Playing with Numbers 7.1 Factors and Multiples������������������������������������������������������������������������������1 7.2 H .C.F. and L.C.M.�����������������������������������������������������������������������������������10 9 Fractions 9.1 Introduction to Fractions����������������������������������������������������������������������28 9.2 L ike, Unlike and Equivalent Fractions��������������������������������������������������38

7Chapter Playing with Numbers I Will Learn About • finding factors and multiples of numbers. • prime and composite numbers. • finding prime factors of numbers. • calculating the H.C.F. and L.C.M. of numbers. 7.1 Factors and Multiples I Think Manoj and Surbhi were given a set of 200 books each to arrange in racks. Manoj plans to arrange the books in 10 racks while Surbhi plans to organise them in 20 racks. Are both the arrangements possible? Will there be any book left to organise? I Recall We have learnt multiplication and division in the previous grade. In the multiplication of two numbers, • the number that is multiplied is called multiplicand. • the number that multiplies the multiplicand is called the multiplier. • the answer obtained after multiplying is called the product. 1

Multiplication Fact ↓ ↓ ↓ Multiplicand Multiplier Product Similarly, in the division of two numbers, • the number that is divided is called the dividend. • the number that divides is called the divisor. • the answer obtained after division is called the quotient. Division Fact ↓ ↓ ↓ Dividend Divisor Quotient I Remember and Understand Factors of a number Raj arranged a bunch of 12 leaves in different ways as given below. 1 row of 12 leaves 2 rows of 6 leaves 2

3 rows of 4 leaves We can represent the arrangement of the leaves as: 1 row × 12 leaves = 12 leaves 2 rows × 6 leaves = 12 leaves 3 rows × 4 leaves = 12 leaves Similarly, 4 rows of 3 leaves = 12 leaves 6 rows of 2 leaves = 12 leaves 12 rows of 1 leaf = 12 leaves The numbers that divide a given number exactly (that is, without leaving any remainders) are called the factors of that number. In other words, the numbers which are multiplied to give a product are called the factors of the product. For example, in the given example, 1, 2, 3, 4, 6 and 12 are the factors of 12. Let us now find the factors of some numbers. Example 1: Find the factors of: a) 16 b) 18 Solution: a) T o find the factors of a given number, express it as a product of two numbers as shown: 16 = 1 × 16 =2 × 8 =4 × 4 Then write each factor only once. Therefore, the factors of 16 are 1, 2, 4, 8 and 16. b) 18 = 1 × 18 =2 × 9 =3 × 6 Therefore, the factors of 18 are 1, 2, 3, 6, 9 and 18. Playing with Numbers 3

Example 2: Find the common factors of 10 and 15. Solution: First write the factors of 10 and 15. 10 = 1 × 10 and 10 = 2 × 5 So, the factors of 10 are 1, 2, 5 and 10. 15 = 1 × 15 and 15 = 3 × 5 So, the factors of 15 are 1, 3, 5 and 15. Therefore, the common factors of 10 and 15 are 1 and 5. We can find the factors of a number by multiplication or by division. Example 3: Find the factors of 30. Solution: Factors of 30 Using multiplication 1 × 30 = 30 2 × 15 = 30 3 × 10 = 30 5 × 6 = 30 Therefore, the factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Using division 30 ÷ 1 = 30 30 ÷ 2 = 15 30 ÷ 3 = 10 30 ÷ 5 = 6 The different quotients and divisors of the given number are its factors. Therefore, the factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Facts on Factors • 1 is the smallest factor of a number. • 1 is a factor of every number. • Every number is a factor of itself. • The greatest factor of a number is the number itself. • Factors of a number are less than or equal to the number itself. 4

• Every number (other than 1) has at least two factors: 1 and the number itself. • The factors of a number are finite. Multiples of a number In the multiplication table of 2, the products obtained The products obtained are 2, 4, 6, 8, 10 and so on. These are called the multiples when a number is of 2. Similarly, multiplied by 1, 2, 3, 4, 5…. a) 3, 6, 9, 12, 15, 18, … are the multiples of 3. are called the multiples b) 5, 10, 15, 20, 25, 30, … are the multiples of 5. of that number. Let us now find the multiples of some numbers. Example 4: Find the first six multiples of: a) 4 b) 6 c) 7 Solution: The first six multiples of a number are the products when the number is multiplied by 1, 2, 3, 4, 5 and 6. a) 4 × 1 = 4, 4 × 2 = 8, 4 × 3 = 12, 4 × 4 = 16, 4 × 5 = 20, 4 × 6 = 24. Therefore, the first six multiples of 4 are 4, 8, 12, 16, 20 and 24. Now, complete these: b) 6 × 1 = 6, 6 × ______ = ______, 6 × ______ = ______ , 6 × ______ = ______ , ... Therefore, the first six multiples of 6 are ____ , ____ , ____ , ____ , ____ and ____. c) 7 × 1 = 7, 7 × ______ = ______, 7 × ______ = ______ , 7 × ______ = ______ , ... Therefore, the first six multiples of 7 are ____, ____ , ____ , ____ , ____ and ____. Example 5: Find the first three common multiples of 10 and 15. Solution: Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90,100,…. Multiples of 15 are 15, 30, 45, 60, 75, 90, 105,…. Therefore, the first three common multiples of 10 and 15 are 30, 60 and 90. Facts on Multiples • Every number is a multiple of itself. • Every number is a multiple of 1. • The smallest multiple of a number is the number itself. • The multiples of a number are greater than or equal to the number itself. • The multiples of a given natural number is unlimited. • The greatest multiple of a natural number cannot be determined. Playing with Numbers 5

Rules for divisibility We can find the numbers that can divide a given number exactly with the help of the rules for divisibility. Divisor Rule Examples 2 The ones digit of the given number must 10, 42, 56, 48, 24 be 0, 2, 4, 6 or 8. 3 The sum of the digits of the given number 36 (3 + 6 = 9) must be divisible by 3. 48 (4 + 8 = 12) The number formed by the last two digits 4 of the given number must be divisible by 1400, 3364, 2500, 7204 4 or both the digits must be zero. 5 The ones digit of the given number must 230, 375, 100, 25 be 0 or 5. 6 The number must be divisible by both 2 36, 480, 1200 and 3. 9 The sum of the digits of the given number 36 (3 + 6 = 9) must be divisible by 9. 144 (1 + 4 + 4 = 9) 10 The ones digit of the given number must 300, 250, 5670 be 0. Let us now apply the divisibility rules to check if a given number is divisible by 2, 3, 4, 5, 6, 9 or 10. Example 6: Which of the numbers 2, 3, 4, 5, 6, 9 and 10 exactly divide 4260? Solution: To check if 2, 3, 4, 5, 6, 9 or 10, divide 4260, apply their divisibility rules. Divisibility by 2: The ones place of 4260 has 0. So, it is divisible by 2. Divisibility by 3: The sum of the digits of 4260 is 4 + 2 + 6 + 0 = 12. 12 is divisible by 3. So, 4260 is divisible by 3. Divisibility by 4: T he number formed by the digits in the last two places of 4260 is 60, which is exactly divisible by 4. So, 4260 is divisible by 4. Divisibility by 5: The ones place of 4260 has 0. So, it is divisible by 5. Divisibility by 6: 4260 is divisible by 2 and 3. So, it is divisible by 6. Divisibility by 9: The sum of the digits of 4260 is 4 + 2 + 6 + 0 = 12, which is not divisible by 9. So, 4260 is not divisible by 9. Divisibility by 10: The ones place of 4260 has 0. So, it is divisible by 10. T herefore, the numbers that exactly divide 4260 are 2, 3, 4, 5, 6 and 10. 6

Example 7: Complete this table by putting a tick for the numbers which are factors of the given number. Put a cross for the rest of the numbers. Number 2 3 Divisible by 9 10 456 23 464 390 Solution: Apply the divisibility rules to check if the given numbers are divisible by the given factors. Tick the box if the number is a factor and cross the box if it is not. Number 2 3 Divisible by 9 10 456 23        464        390        Prime and composite numbers Observe the following table which gives the factors of numbers from 1 to 10. Number Factors Number of Number Factors Number of factors factors 1 1 1 6 1, 2, 3, 6 4 2 1, 2 2 7 1, 7 2 3 1, 3 8 4 1, 2, 4 2 9 1, 2, 4, 8 4 5 1, 5 10 1, 3, 9 3 3 2 1, 2, 5, 10 4 From the given table, we find that - 1) The numbers 2, 3, 5 and 7 have only two factors (1 and themselves). Such numbers are called prime numbers. 2) The numbers 4, 6, 8, 9 and 10 have three or more factors (more than two factors). Such numbers are called composite numbers. 3) The number 1 has only one factor. So, it is neither prime nor composite. Sieve of Eratosthenes Eratosthenes was a Greek mathematician. He invented a method to find prime numbers between any two given numbers. This is called the sieve of Eratosthenes. Steps to find prime numbers between 1 and 100 using the sieve of Eratosthenes: Playing with Numbers 7

Step 1: Write a grid of numbers 1 to 100. Step 2: Cross out 1 as it is neither prime nor composite. Step 3: Circle 2 as it is the first prime number. Then cross out all the multiples of 2. Step 4: Circle 3 as it is the next prime number. Then cross out all the multiples of 3. Step 5: Circle 5 as it is the next prime number. Then cross out all the multiples of 5. Step 6: Circle 7 as it is the next prime number. Then cross out all the multiples of 7. Continue this process till all the numbers between 1 and 100 are either circled or crossed out. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 The circled numbers are prime numbers and the crossed out numbers are composite numbers. There are 25 prime numbers between 1 and 100. These are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Note: 1) All prime numbers (except 2) are odd. 2) 2 is the only even prime number. 8

? Train My Brain Answer the following: a) Find the first five multiples of 7 and 12. b) Find all the factors of 26 and 28. c) Find the common factors of 15 and 30. Playing with Numbers 9

7.2 H.C.F. and L.C.M. I Think Surbhi starts jumping from 3 and continues her steps in multiples of 3 as 3, 6, 9,... Suraj jumps in steps of 2. At which step-number will both of them meet? I Recall Let us revise the concept of factors by finding the common factors of the following pairs of numbers: a) 12, 9 b) 15, 10 c) 16, 12 d) 24, 16 e) 15, 21 I Remember and Understand Prime numbers have only 1 and themselves as their factors. Composite numbers have more than two factors. So, we can express composite numbers as the products of prime numbers or composite numbers. 10

For example, 5 = 1 × 5; 9 = 1 × 9 and 3 × 3; 20 = 1× 20, 2 × 10 and 4 × 5. We can express all composite numbers as the products of their prime factors. Prime Factorisation of Numbers To prime factorise a number, we use factor trees. Let us see a few examples to understand this better. Example 12: Carry out the prime factorisation of 36 by the factor tree method. Solution: To carry out the prime factorisation of a given number, draw a factor tree Step 1: as shown. Step 2: Express the given number as a product of 36 Step 3: two factors. One of these factors is the least 2 × 18 number (other than 1) that can divide it. The second factor may be prime or composite. If the second factor is a composite number, 2 ×2 ×9 express it as a product of two factors. One of these factors is the least number (other than 2 × 2 × 3 × 3 1) that can divide it. The second factor may be prime or composite. Repeat the process till the factors cannot be split further. In other words, repeat the process till the factors do not have any common factor other than 1. Step 4: Then write the given number as the product of all the prime numbers. Therefore, the prime factorisation of 36 is 2 × 2 × 3 × 3. Note: A factor tree must be drawn using a prime number as one of the factors of the number at each step. Example 13: Carry out the prime factorisation of 54 by the factor tree method. Solution: The prime factorisation of 54 using a factor tree is as shown below: 54 2 × 27 2×3×9 2×3×3×3 Therefore, the prime prime factorisation of 54 is 2 × 3 × 3 × 3. Playing with Numbers 11

Highest Common Factor (H.C.F.) and Least Common Multiple (L.C.M.) Finding factors and multiples helps us to find the Highest Common Factor (H.C.F.) and the Least Common Multiple (L.C.M.) of the given numbers. H.C.F.: The highest common factor of two or more numbers is the greatest number that divides the numbers exactly (without leaving a remainder). L.C.M.: The least common multiple of two or more numbers is the smallest number that can be divided by the numbers exactly (without leaving a remainder). Example 14: Find the highest common factor of 12 and 18. Solution: 12 = 1 × 12, 12 = 2 × 6 and 12 = 3 × 4 So, the factors of 12 are 1, 2, 3, 4, 6 and 12. 18 = 1 × 18, 18 = 2 × 9 and 18 = 3 × 6 So, the factors of 18 are 1, 2, 3, 6, 9 and 18. The common factors of 12 and 18 are 1, 2, 3 and 6. Therefore, the highest common factor of 12 and 18 is 6. Example 15: Find the least common multiple of 12 and 18. Solution: The multiples of 12 are 12, 24, 36, 48, 60, 72, …. The multiples of 18 are 18, 36, 54, 72, …. The common multiples of 12 and 18 are 36, 72, ….. Therefore, the least common multiple of 12 and 18 is 36. Let us now complete these tables of H.C.F. and L.C.M. of the given numbers. Example 16: Complete the H.C.F. table given. Some H.C.F. values are given for you. Numbers 10 12 18 30 2 2 3 12 6 15 15 Solution: Numbers 2 3 10 12 18 30 12 15 2222 1333 2 12 6 6 5 3 3 15 12

Example 17: Complete the L.C.M. table below. Some L.C.M. values are given for you. Solution: Numbers 10 12 18 30 2 18 3 12 12 15 30 Numbers 10 12 18 30 2 10 12 18 30 3 30 12 18 30 12 60 12 36 60 15 30 60 90 30 Finding H.C.F. and L.C.M. using the prime factorisation method Let us now find the H.C.F. of two numbers using the prime factorisation method. Consider these examples. Example 18: Find the H.C.F. of 48 and 54 by the prime • The product of two factorisation method. numbers is equal to the product of H.C.F. and Solution: The prime factorisation of 48 is L.C.M. 2 × 2 × 2 × 2 × 3. • The L.C.M. of the given The prime factorisation of 54 is numbers is always a 2 × 3 × 3 × 3. multiple of their H.C.F. Therefore, the H. C. F of 48 and 54 is 2 × 3 = 6. Example 19: Find the L.C.M. of 18 and 24 by the prime factorisation method. Solution: The prime factorisation of 18 is 2 × 3 × 3. The prime factorisation of 24 is 2 × 2 × 2 × 3. Therefore, the L.C.M. of 18 and 24 is 2 × 3 × 2 × 2 × 3 = 72. ? Train My Brain Find the prime factors of: a) 28 b) 20 c) 14 Playing with Numbers 13

CHAPTER Fractions9 9Chapter Fractions 1 4 I Will Learn About • fractions as a part of a whole and its representation. • fractions of a collection. • like, unlike, unit and equivalent fractions. • addition and subtraction of like fractions. 9.1 Introduction to Fractions I Think Surbhi and her three friends, Joseph, Salma and Rehan, went on a picnic. Rehan brought only one apple with him. He wanted to share it equally with everyone. What part of the apple does each of them get? I Recall Look at the rectangle given in the next page. We can divide the whole rectangle into many equal parts as shown. 28

1 part: 2 equal parts: 3 equal parts: 4 equal parts: 5 equal parts: and so on. I Remember and Understand Let us understand the concept of parts of a whole. Suppose we want to share an apple with our friends. First, we count their number. Then, we cut the apple into as many equal pieces as the number of persons. Thus, each person gets an equal part of the apple after division. Fraction as a part of a whole A complete or full object is called a whole. Observe the following parts of a chocolate bar: Whole 2 equal parts 3 equal parts 4 equal parts Fractions 29

We can divide a whole into equal parts as shown. 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 equal part 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 whole 2 is divided into. Note: 1 and 1 make a whole. 2 2 Thirds In figure (a), observe that the three parts are not equal. We can also divide a whole into three equal parts as shown in figures (b) and (c). 11 1 33 3 Three unequal parts Three equal parts Fig. (a) Fig. (b) Fig (c) The shaded grey part in figure (c) is one out of three equal parts. So, each equal part is called a third or 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). We write one-third as 1 and two-thirds as 2 . 3 3 Note: 1 , 1 and 1 or 1 and 2 make a whole. 3 3 3 3 3 30

Fourths Similarly, fold a square piece of paper into four equal parts. Each of them is called a fourth or a quarter. In figure (d), the four parts are not equal. In figure (e), each equal part is called a fourth or a quarter and is written as 1 . 4 1 4 1 4 1 4 1 4 Four parts Four equal parts Fig (d) Fig (e) Two out of four equal parts are called two-fourths and three out of four equal parts are called three-fourths, written as 2 and 3 respectively. 44 Note: Each of 1 and 3 1 1 1 and 1 or 1 1 and 2 make a whole. 4 4; 4, 4, 4 4 4, 4 4 The total number of equal parts a whole is divided into is called the denominator. The number of such equal parts taken is called the numerator. A fraction is a part of a whole. Numbers of the form Numerator are For example, 1 , 1, 1 , 2 and so on are fractions. 2 3 4 3 Denominator Let us now see a few examples. called fractions. Example 1: Identify the numerator and denominator in each of the following fractions: a) 1 b) 1 c) 1 2 3 4 Fractions 31

Solution: S. No Fractions Numerator Denominator a) 1 1 2 2 1 1 3 b) 3 1 1 4 c) 4 Example 2: Identify the fraction for the shaded parts in the figures given. a) b) Solution: Steps a) Solved b) Solve this Total number of equal parts Step 1: Count the number of Total number of = _______ equal parts the figure is divided equal parts = 8 into (Denominator). Number of parts shaded = Step 2: Count the number of Number of parts ______ shaded parts (Numerator). shaded = 5 Step 3: Write the fractions. Fraction = Fraction = 5 Numerator 8 Denominator Fraction of a collection 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. 32

Each equal part has 5 pens, as 10 ÷ 2 = 5. So, 1 of 10 pens is 5 pens. 2 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 bananas is 4 bananas. 3 1 3 1 3 1 3 Finding a fourth (or a quarter) One-fourth is 1 out of 4 equal parts. In this figure, there are 8 books. To find a fourth, we divide the number of books into 4 equal parts. Fractions 33

1 1 1 1 4 44 4 Each equal part has 2 books, as 8 ÷ 4 = 2. So, 1 of 8 books is 2 books. 4 Let us see a few examples to find the fraction of a collection. Example 3: Write the fraction of the coloured shapes in each of the following. Shapes Fractions a) b) c) Solution: The fractions of the coloured parts of the given shapes are – a) b) Shapes Fractions c) 2 6 3 6 5 8 Example 4: Colour the given shapes to represent the given fractions. Shapes Fractions a) 1 5 b) 2 7 c) 3 4 34

Solution: We can colour the shapes to represent the fractions as – Shapes Fractions a) 1 5 b) 2 7 c) 3 4 ? Train My Brain What fraction of the collection are: a) Chocolate cupcakes b) Strawberry cupcakes c) Blueberry cupcakes Fractions 35

9.2 Like, Unlike and Equivalent Fractions I Think Surbhi cuts 3 cakes into 18 equal pieces. Zeenal cuts a cake into 6 equal pieces. Did both of them cut the cakes into equal number of pieces? I Recall Let us recall the concept of fractions by finding the fraction of the parts not shaded in these figures. a) = ____________ b) = ____________ c) = ___________ d) = ____________ 38

I Remember and Understand In the previous concept, we have learnt about fractions. Now let us learn the different types of fractions. Like fractions: Fractions such as 1 , 2 and 3 , that have the same denominator are 88 8 called like fractions. Unlike fractions: Fractions such as 1 , 2 and 3 that have different denominators are 84 7 called unlike fractions. To understand like and unlike fractions, consider the following example. Example 11: Identify the like and unlike fractions from the following. 3 , 3 , 1 , 5 , 6 , 1 , 4 Fractions with numerator Solution: 7 5 7 7 7 4 11 ‘1’ are called unit 3 15 6 fractions. For example, 7 , 7 , 7 and 7 have the same denominator. 1 , 1 , 1 and so on. 234 Therefore, they are like fractions. 3 , 1 and 4 along with 3 or 1 or 5 or 6 5 4 11 7 7 7 7 have different denominators. Therefore, they are unlike fractions. Add and subtract like fractions While adding or subtracting like fractions, add or subtract only their numerators. Write the sum or difference with the same denominator. Let us understand this through some examples. Example 12: Solve: a) 3 + 1 b) 4 + 5 c) 8 – 4 d) 33 – 25 8 8 13 13 9 9 37 37 Solution: 3 1 3 +1 4 a) 8 + 8 = 8 = 8 4 5 4+5 9 b) 13 + 13 = 13 = 13 84 4 c) 9 – 9 = 9 d) 33 – 25 = 33 − 25 = 8 37 37 37 37 Fractions 39

Equivalent fractions Fractions that denote the same part of a whole are called equivalent fractions. Let us now understand what equivalent fractions are. Suppose a bar of chocolate is cut as shown. Ram eats 1 of the chocolate. 5 Then the piece of chocolate that he eats is: 2 Raj eats 10 of the chocolate. Then the piece of chocolate that he eats is: We see that both the pieces of chocolate are of the same size. So, we say that the fractions 1 and 2 are equivalent. We write them as 1 = 2 . 5 10 5 10 Example 13: Shade the regions to show equivalent fractions. a) 12 [ 3 and 6 ] b) 12 [ 4 and 8 ] Solution: 1 2 a) 3 6 40

b) 1 2 4 8 Example 14: Find the figures that represent equivalent fractions. Also, mention the fractions. a) b) c) d) Solution: The fraction represented by trheeprsehsaednetsd24p.aTrht eofshfigaudreedap) aisrt21o.f figure d) The shaded part of figure b) represents 1 . 2 So, the shaded parts of figures a), b) and d) represent equivalent fractions. Methods to find equivalent fractions There are two methods to find equivalent fractions. Let us learn them through a few examples. Example 15: Find two fractions equivalent to the given fractions. 2 12 a) 11 b) 16 Solution: To find fractions equivalent to the given fractions, we either multiply or divide both the numerator and the denominator by the same number. a) 2 11 W e see that 2 and 11 do not have any common factors. So, we 2 cannot divide them to get an equivalent fraction of 11. T herefore, we multiply both the numerator and the denominator by the same number, say 5. 2 2×5 = 10 11= 11× 5 55 Therefore, 10 is a fraction equivalent to 121. 55 Fractions 41

S imilarly, we multiply both the numerator and the denominator by the same number, say 2. 2×2 = 4 11×2 22 Therefore, 4 is a fraction equivalent to 121. 22 b) 12 16 W e see that 12 and 16 have common factors 2 and 4. So, dividing both the numerator and the denominator by 2 and 4, we get fractions equivalent to the given fraction 12 . 16 12 ÷ 2 = 6; 12 ÷ 4 = 3 16 ÷ 2 8 16 ÷ 4 4 Therefore, 6 and 3 are the fractions equivalent to 12 . 8 4 16 ? Train My Brain Identify the like fractions from each of the following. a) 1 , 3 , 62 , 64 b) 151,161, 2 ,181, 67 ,171, 71 c) 1 , 5 , 2 , 3 , 18 , 3, 3, 6 4 6 7 3 25 25 5 25 4 7 25 42