51704171_BGM_Traveller G4 Mathematics Textbook FY_Text

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 47

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 48

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 49

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 5 6⁄ 11 − 4 5689 hundreds, subtract and write 6⁄ ⁄ 1⁄ the difference under the 5 8 hundreds place. 7 − 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 50

I Apply Addition and subtraction of 4-digit and 5-digit numbers are useful in our daily life. Here are a few examples. Example 5: Raju had ` 90005 with him. He bought clothes for ` 35289. How much money was left with him? Solution: Amount Raju had = ` 90005 T Th Th H TO Amount Raju spent on buying clothes = ` 35289 Amount left with him = ` 90005 – ` 35289 8⁄ 9⁄ 9⁄ 9⁄ 1⁄5 Therefore, the amount left with Raju is ` 54716. 90 0 05 −35 2 89 7 16 54 Example 6: Preeti drove her car for 26349 km in six weeks and 38614 km in the next eight weeks. How many kilometres in all did she drive in 14 weeks? Solution: Distance Preeti drove in the first six weeks = T Th Th H T O 26349 km 11 2 6349 Distance she drove in the next eight weeks = 38614 km + 3 8614 6 4963 The total distance that Preeti drove = 26349 km + 38614 km Therefore, Preeti drove a total distance of 64963 km in 14 weeks. Example 7: Mohan’s uncle stays 8630 m away from Mohan’s house. Mohan travelled 6212 m of the distance. What is the distance yet to be covered by Mohan to reach his uncle’s house? Solution: Distance between Mohan’s house and his uncle’s house = 8630 m Distance travelled by Mohan = 6212 m Th H T O 2⁄ 10 Remaining distance that Mohan has to travel 86 3 0⁄ −62 1 2 = 8630 m – 6212 m 1 8 24 Therefore, Mohan has to travel 2418 m more to reach his uncle’s house. Addition and Subtraction 51

I Explore (H.O.T.S.) We can frame word problems on addition and subtraction. Example 8: Payal and Suma have 1284 and 5215 stamps respectively. Frame an addition problem. Solution: An addition problem contains words such as - in all, total, altogether and so on. So, the question can be “Payal and Suma have 1284 and 5215 stamps respectively. How many stamps do they have altogether?” Example 9: Frame a word problem based on the given subtraction. 50000 – 49100 = 900 Solution: A subtraction problem contains words such as - difference, left, remaining and so on. So, the question can be “Ramesh had ` 50000. He gave ` 49100 to his brother. Find the amount left with Ramesh”. Maths Munchies Always remember that when we add a number to itself, the sum is double the original number. If we subtract a number from itself, the difference is zero. For example, 2000 + 2000 = 4000 (which is double of 2000) and 2000 – 2000 = 0. Connect the Dots Social Studies Fun In 1557, Robert Recorde shortened “is equal to” to two long, parallel lines. This gave the presently used ‘equal to’ sign. He used this to avoid repeating himself 200 times in his book. English Fun ‘Addition’ is a noun. Write the verb for ‘addition’. Can you say what the adjective form for this word is? Share with your friends. 52

Drill Time 4.1 Add and Subtract 4-digit and 5-digit Numbers 1) Add the following: a) 5624 + 1218 b) 42584 + 23568 c) 4721 + 1311 d) 65312 + 25842 e) 35216 + 42355 2) Subtract the following: a) 5943 – 1256 b) 86531 – 65372 c) 95361 – 46472 d) 11213 – 11206 e) 34536 – 15623 3) Word problems a) Seeta went to purchase a TV from an electronics shop. The price of the TV was ` 25689. She paid the shopkeeper ` 50000. How much money will she receive back? b) Rohan collected 1256 beads for a design. Sohan collected 2563 beads for the same design. How many beads did they collect in all? A Note to Parent Play this fun game with your child. Shuffle a deck of cards. Draw a card randomly from it. Multiply the number on the card by 100. The number obtained is your score. Note it down on a piece of paper. All those playing the game should do the same. Continue the game for a few rounds or till all the cards are drawn from the deck. Add the score obtained by each player. The one with the highest score wins. Addition and Subtraction 53

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. 54

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 55

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 H TO product has two digits. Carry the tens digit 1 635 of the product to the tens place and write ×7 its ones digit under the ones place. 513 ×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. 56

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 57

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. 58

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 59

I Apply Let us solve a few real-life examples involving multiplication of 3-digit numbers. Example 4: There are 18 desks in a classroom. There are 11 such classrooms in the school. There are 13 such schools in the city. How many desks are there in total in all the schools? Th H T O Solution: Number of desks in a classroom = 18 22 Number of classrooms = 11 198 ×13 Number of desks in 11 classrooms = 18 × 11 = 198 11 Number of schools = 13 594 Number of desks in 13 schools = 198 × 13 +1980 2574 Therefore, there are 2574 desks in total in all the schools. Example 5: 354 students went to school from each of the two neighbouring towns. How many students in all went to school? Solution: Number of students from each town = 354 H TO Number of towns = 2 1 Total number of students who went to school = 354 × 2 3 54 ×2 Therefore, 708 students went to the school from both the towns. 7 0 8 Consider the following to multiply a 3-digit number by a 3-digit number. Multiply: 159 × 342 T Th Th H T O 12 23 11 159 ×342 1 1 3 1 8 → Multiply the multiplicand by the ones of the multiplier. 159 × 2 ones. + 6 3 6 0 → Multiply the multiplicand by the tens of the multiplier. 159 × 4 tens. +4 7 7 0 0 → Multiply the multiplicand by the hundreds of the multiplier. 159 × 3 hundreds. 5 4 3 7 8 → Add the products and write the sum. Therefore, 159 × 342 = 54378. 60

I Explore (H.O.T.S.) We can frame word problems based on the multiplication of 2-digit numbers and 3-digit numbers. Example 6: Frame a word problem using 48 × 98. Solution: Using 48 × 98, we can frame a word problem as \"There are 48 chalk pieces in each box. If there are 98 such boxes, find the total number of chalk pieces\". Example 7: Frame a word problem using 792 × 19. Solution: U sing 792 × 19, we can frame a word problem as \"There are 792 candies in a big container. How many candies will 19 such containers have?\" 5.2 Multiply Using Lattice Algorithm I Think Surbhi and her friends visited her aunt's plant nursery. There she saw an arrangement with equal number of pots in rows. She wanted to know the total number of pots using a simple method. Do you know any such method? I Recall We know multiplication of 2-digit numbers by 1-digit numbers using lattice algorithm. Let us recall the concept by solving the following: 2 2× 4 1× 5 0× 4 3 4 Multiplication 61

I Remember and Understand We have learnt to multiply a 2-digit number by a 1-digit number using standard algorithm. Let us now learn to multiply a 2-digit number and a 3-digit number by a 2-digit number using a simpler method than the standard algorithm. This method is called the Lattice algorithm. Example 8: Multiply using lattice algorithm: a) 43 × 52 b) 168 × 48 Solution: To multiply numbers using the lattice algorithm, follow these steps: Steps Solved Solved a) 43 × 52 b) 168 × 48 Step 1: Construct a lattice as shown, such that: (a) Number of rows = Number of digits in the multiplier. (b) Number of columns = Number of digits in the multiplicand. Draw diagonals to divide each box into parts as shown. Step 2: Write the multiplicand along 4 3× 1 6 8× the top of the lattice and the 5 4 multiplier along the right, one digit for each row or column. 2 8 Step 3: Multiply each digit of the 4 3× 1 6 8× multiplicand by the tens digit of the 2015 5 multiplier. Write the products in the 0 2 3 4 cells where the corresponding rows 2 4 4 2 and columns meet. Write the tens digit of the product in the upper 8 half and its ones digit in the lower half as shown. 62

Steps Solved Solved a) 43 × 52 b) 168 × 48 Step 4: Multiply each digit of the multiplicand by the ones digit of 4 3× 16 8× the multiplier. Write the products 2015 5 in the cells as done in the previous 02 3 4 step. 44 2 If the product is a single digit 0 0 2 04 6 8 number, put 0 in the tens place. 8 6 88 4 (2 × 3 = 6) = 06 Step 5: Add the numbers along the 4 3× 16 8 × diagonals from the right to find the 2 3 4 product. Regroup if needed. Write 20 1 02 the sum from left to right. 55 0 44 2 8 10 04 6 28 0 2 6 4 3 28 8 6 8 01 4 Therefore, 6 43 × 52 = 2236. Therefore, 168 × 48 = 8064. 6 3× Solve these × 171 × 3 172 3 4 3 1 Train My Brain 2 Therefore, 63 × 33 = _____. Therefore, 172 × 42 = _____. Therefore, 171 × 31 = _____. ? Train My Brain Multiply using the lattice algorithm method: a) 54 × 3 b) 78 × 21 c) 375 × 27 Multiplication 63

I Apply Let us now see a few real-life examples where we use the lattice algorithm for the 3-digit numbers. 34 5 × 1 Example 9: There are 345 students in each class. 0 0 0 Pooja’s school has 12 such classes. 3 4 5 2 How many students are there in her school? 1 0 0 1 6 8 Solution: Number of students in each class = 345 4 0 Number of such classes in Pooja’s 11 4 0 school = 12 Total number of students = 345 × 12 Therefore, there are 4140 students in Pooja’s school. Example 10: 42 people were sitting in a row of a stadium to enjoy a cricket match. How many people would be there in all if 35 such rows in the stadium were filled? Solution: Number of people sitting in one row = 42 4 2 × 1 0 3 Number of such rows filled = 35 12 6 5 Total number of people in 35 rows 2 40 1 = 42 × 35 0 7 Therefore, there are 1470 people in the 0 stadium. I Explore (H.O.T.S.) We know how to multiply numbers using lattice algorithm. Let us see if we can analyse and solve the following. Example 11: Find the missing numbers in the given 2 3 ?× lattice multiplication. 0 0 1 n2 8 4 23___ × 4___= 9954 8 2 00 1 4? 94 6 954 64

Solution: We see that the box at the top right corner has the number 28. It is the product of 4 and ?. To find the missing number, divide 28 by 4. We get 7. That is, 4 × ? = 28 4 × 7 = 28. So, 7 is the first unknown number. Similarly, the box in the bottom left corner has 04. It is the product of 2 and ?. To find the missing number, divide 4 by 2. We get 2. 5.3 That is, 2 × ? = 04 2 × 2 = 04. So, the second unknown number is 2. Therefore, the required missing numbers are 7 and 2 so that 237 × 42 = 9954. Mental Maths Techniques: Multiplication Let us now understand how to complete multiplication facts mentally. Steps Solved Solve this 14 × 6 12 × 8 Step 1: Check by how much The larger number is 14 The larger number is ____, the larger number is more than and it is 4 more than 10. and it is ______ more than 10. (If the larger of the given 10. numbers is more than 10.) Step 2: Write the larger 10 + 4 = 14 ______ + ______ = _______ number as the sum of 10 and the number obtained in the 10 × 6 = 60 and 10 × ________ = _________ previous step. 4 × 6 = 24 and _______ × ______ = _______ Step 3: Multiply the numbers of the sum separately by the smaller number given, using memorised tables of 1 to 10. Step 4: Add both the products 60 + 24 = 84 ______ + ______ = ______ Therefore, 12 × 8 = ______. from step 3 mentally. This gives Therefore, 14 × 6 = 84. the required product. Multiplication 65

? Train My Brain Find the product in each of the following mentally. a) 18 × 7 b) 13 × 4 c) 11 × 6 d) 24 × 8 e) 76 × 5 Maths Munchies Let us see how to multiply two 2-digit numbers, 13 and 21, quickly. For the multiplicand 13, draw 1 tens as 1 green line and 3 ones as 3 pink lines. For the multiplier 21, draw 2 tens as 2 red lines and 1 ones as 1 yellow line. Count the number of dots at each intersection as shown. Add the numbers in the middle column. Here, 6 + 1 = 7. Write the number starting from the left. So, the product is 273. 1 1 ones of 21 3 2 3 ones of 13 1 tens of 13 2 tens of 21 6 This method is also called as the Chinese-stick multiplication method. Connect the Dots English Fun The word ‘Lattice’ in lattice algorithm is not an English word originally. The word is taken from ‘lattis’, in old French language which itself has been taken from ‘latte’ or ‘lath’, that means the wire mesh used for backing in the old German language. 66

Social Studies Fun The oldest known multiplication table was found written on bamboo strips in China which are said to be 2300 years old. Modern multiplication tables are said to have been written down by the famous Greek mathematician Pythagoras. It is also called the Table of Pythagoras in many other languages. Drill Time 5.1 Multiplication of 2-digit Numbers and 3-digit Numbers 1) Multiply a 3-digit number by a 1-digit number. a) 247 × 7 b) 509 × 1 c) 892 ×9 d) 731 × 1 e) 435 × 4 2) Multiply a 2-digit number by a 2-digit number. a) 17 × 16 b) 39 × 24 c) 99 × 19 d) 43 × 32 e) 76 × 48 3) Multiply a 3-digit number by a 2-digit number. a) 161 × 23 b) 276 × 11 c) 157 × 38 d) 375 × 82 e) 198 × 22 4) Multiply a 3-digit number by a 3-digit number. a) 674 × 124 b) 345 × 238 c) 925 × 759 d) 555 × 725 e) 804 × 682 5) Word problems a) Pranav makes 25 cotton bags in a day. How many bags will he be able to make in the year 2017? (Hint: 2017 is not a leap year.) b) Tanya bought a few gifts of the same kind for Christmas for her 14 friends. If one gift costs ` 27, then how much money in all did she spend for the gifts? Multiplication 67

5.2 Multiply Using Lattice Algorithm 6) Multiply a 2-digit number by a 2-digit number. a) 24 × 32 b) 56 × 15 c) 13 × 39 d) 67 × 51 e) 78 × 42 7) Multiply a 3-digit number by a 2-digit number. a) 158 × 17 b) 451 × 39 c) 651 × 67 d) 721 × 41 e) 329 × 78 8) Word problems a) A movie theatre sold 127 tickets for a movie. The cost of one ticket was ` 85. How much money did the theatre owner earn? b) There are 47 students in Class 3. Answer booklets were given to each student for a Maths exam. If one answer booklet has 15 pages, then how many sheets of paper in all were used for the exam? A Note to Parent Help your child understand and practise the Chinese-stick method of multiplication using straws of different colours. 68

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. 69

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 Dividend = _____ from the dividend and write the Divisor = ______ difference below the product. )5 45 Quotient = ____ Remainder = _____ This difference is called the − 45 remainder. )00 4 36 45 = Dividend − 5 = Divisor 9 = Quotient 0 = Remainder 70

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 Division 71

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 dividend and write it beside the remainder. 1 − )3 57 − −3 Dividend = _____ 2 Divisor = ______ Quotient = ____ Step 6: Find the largest number in the 3 × 8 = 24 Remainder = ___ multiplication table of the divisor that can be subtracted from the 2-digit number in 19 the previous step. )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) 72

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. Dividend = _____ Step 2: Find the largest number in 5 Divisor = ______ the table of the divisor that can be Quotient = ____ subtracted from the 2-digit number )5 265 Remainder = ___ of the dividend. Write the quotient. − 25 )6 366 Write the product of the quotient and 5 × 4 = 20 − the divisor below the dividend. 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. Division 73

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. − Write the product of the quotient and the 1 − divisor under the hundreds place of the dividend. )7 784 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 74

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 Division 75

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 I Apply Division of 2-digit numbers and 3-digit numbers is used in many real-life situations. Let us consider a few examples. Example 5: 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 any house? Solution: Number of students = 634 158 Number of houses = 4 )4 634 Number of students in a house = 634 ÷ 4. − 4↓ 23 Therefore, the number of students in each house = 158 − 20 The remainder in the division is 2. 34 − 32 Therefore, 2 students are not grouped into any house. 02 Example 6: A football game had 99 spectators. If each row in the stadium has only Solution: 9 seats, how many rows did the spectators occupy? 11 Number of spectators = 99 )9 99 Number of seats in each row = 9 − 9↓ Number of rows occupied by the spectators = 99 ÷ 9. 09 Therefore, 11 rows were occupied by the spectators. −9 0 76

I Explore (H.O.T.S.) In all the divisions we have seen so far, we did not have a 0 (zero) in the dividend or the 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 the division of numbers that have a 0 (zero) in the dividend or quotient, through this example. Example 7: Divide: 505 ÷ 5 Solution: Solved Solve this 505 ÷ 5 804 ÷ 4 101 )4 804 )5 505 − − 5↓ − 00 − − 00 05 − 05 00 Maths Munchies Dividing a 2-digit number or a 3-digit number by 10: Observe the pattern in the division of the following examples: a) 562 ÷ 10 = Q 56 and R 2 d) 45 ÷ 10 = Q 4 and R 5 b) 325 ÷ 10 = Q 32 and R 5 e) 72 ÷ 10 = Q 7 and R 2 c) 687 ÷ 10 = Q 68 and R 7 We observe that the ones digit of the dividend is the remainder and the number formed by its remaining digits is the quotient. This helps us to do the divisions quickly. Division 77

Connect the Dots Science Fun There are different kinds of living things other than people on the Earth. All living things are divided into various groups. These divisions help scientists study various aspects about them. Social Studies Fun India is a very large country. It becomes difficult to take care of the citizens of the entire country by a small group of leaders. So, it is divided into 29 states and 7 union territories. Drill Time 6.1 Divide 3-digit Numbers by 1-digit Numbers 1) 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 2) 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 3) 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 4) 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 5) 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 get? b) There are 24 passengers in a bus. 2 passengers can sit on each seat. How many seats in the bus are occupied? 78

A Note to Parent Collect electricity bills of the past three months and show them to your child. Highlight the total amount and the cost per unit of electricity. Explain how the number of units of electricity utilised is calculated. Now help your child to calculate the number of units of electricity that the family has used in the past three months. Division 79

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. 80

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 Playing with Numbers 81

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. 82

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. Playing with Numbers 83

• 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. 84

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. Playing with Numbers 85

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) T he numbers 2, 3, 5 and 7 have only two factors (1 and themselves). Such numbers are called prime numbers. 2) T he 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: 86

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. Playing with Numbers 87

? 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. I Apply Let us apply the knowledge of factors and multiples in day-to-day life. Example 8: 24 students are to be lined up in rows for an assembly. Can they be lined up equally in 7 rows? Solution: 7 is not a factor of 24. In other words, 24 is not a multiple of 7. Therefore, 24 students cannot be lined up equally in 7 rows. Example 9: Suman has 15 toys. She wants to arrange them in shelves such that each shelf has the same number of toys. How many shelves will she need? Solution: Number of toys = 15 Factors of 15 = 1, 3, 5 and 15 We can divide 15 by the factors to find the number of shelves. So, 15 ÷ 1 = 15 15 ÷ 3 = 5 15 ÷ 5 = 3 15 ÷ 15 = 1 Therefore, Suman will need 1, 3, 5 or 15 shelves to organise the toys equally in each shelf. I Explore (H.O.T.S.) Consider these examples. Example 10: Ravi has two ribbons. One ribbon is 4 cm long and the other is 10 cm long. He wants to cut both the ribbons in such a way that each piece is of the same length. What will be the length of each piece? 88

Solution: The length of two ribbons is 4 cm and 10 cm. Factors of 4 = 1, 2, 4 Factors of 10 = 1, 2, 5, 10 Common factors of 4 and 10 are 1 and 2. Therefore, Ravi can cut both the ribbons in pieces of length 1 cm or 2 cm. Example 11: There are 14 students in Class 4A and 21 students in Class 4B. If the students of both the classes need to be arranged in the same number of rows, what is the maximum number of rows possible in each class? Solution: Number of students in Class 4A = 14 Number of students in Class 4B = 21 Factors of 14 = 1, 2, 7, 14 Factors of 21 = 1, 3, 7, 21 Common factors of 14 and 21 are 1 and 7. Therefore, the teacher will need a maximum of 7 rows to arrange these students. 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. Playing with Numbers 89

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. 90

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 Playing with Numbers 91

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 92

I Apply Let us apply the knowledge of H.C.F. and L.C.M. to some real-life situations. Example 20: N ikki organised a party at her house. She went to a grocery store to buy candles and candle stands. The store only had packs that contained 18 candles and 12 stands each. How many packs would she have to buy to ensure that all the candles were placed on a candle stand? Solution: Number of candle stands in a pack = 12 Number of candles in a pack = 18 T o find the number of packs to buy to avoid any additional candle or candle stand, we find the L.C.M. of both the numbers. Prime factors of 12 = 2 × 2 × 3 Prime factors of 18 = 2 × 3 × 3 L.C.M. of 12 and 18 is 2 × 3 × 2 × 3 = 36. To find the number of packs, we divide 36 by 18 and 12. 36 ÷ 12 = 3; 36 ÷ 18 = 2 Therefore, Nikki would need 3 packs of candle stands and 2 packs of candles. Example 21: Rekha baked 20 butterscotch cakes and 24 mango cakes. She wants to pack equal number of cakes in each container. What is the maximum number of containers that Rekha will need? Solution: Number of butterscotch cakes baked = 20 Number of mango cakes baked = 24 To find the total number of containers that can fit the maximum of cakes of each type, we find the H.C.F. of both the numbers. 20 24 2 × 10 2 × 12 2 × 5 2 × 6 2 × 3 Playing with Numbers 93

Prime factorisation of 20 is 2 × 2 × 5. Prime factorisation of 24 is 2 × 2 × 2 × 3. H.C.F. of 20 and 24 = 2 × 2 = 4 Therefore, Rekha will need 4 containers. I Explore (H.O.T.S.) Let us see some more applications of H.C.F. and L.C.M. in real life. Example 22: Rita has a chocolate shop. She gives a free strawberry chocolate to every 14th customer and a free mango chocolate to every 16th customer. Who would be the first customer to receive both the free chocolates? Solution: Every 14th customer receives a free strawberry chocolate. Every 16th customer receives a free mango chocolate. The first customer who receives both the free chocolates is the L.C.M. of 14 and 16. Multiples of 14 = 14, 28, 42, 56, 70, 84, 98, 112, 126, ... Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, ... The L.C.M. of 14 and 16 is 112. T herefore, the 112th customer will receive both a free strawberry chocolate and a free mango chocolate. Example 23: Three bells rings at intervals of 2, 3 and 4 minutes respectively. Find the time when they will ring together. Solution: Intervals at which the bells ring = 2, 3 and 4 minutes To find the time when they will ring together, we find the L.C.M. of the numbers. Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, ... Multiples of 3 = 3, 6, 9, 12, 15, 18, ... Multiples of 4 = 4, 8, 12, 16, ... The L.C.M. of 2, 3 and 4 = 12 Therefore, the bells will ring together after every 12 minutes. 94

Maths Munchies The numbers 60, 72, 84, 90 and 96 are the six 2-digit numbers which have the maximum number of factors. The factors of each one of them are: 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 Connect the Dots Social Studies Fun How do you think traffic controllers control traffic lights correctly? How do they ensure that all the lights do not turn green at the same time? They do it by finding the L.C.M. of the durations of all the traffic lights at a crossing. English Fun State the part of speech of the underlined word in the following sentence: Prime numbers have only two factors. Drill Time c) 32 7.1 Factors and Multiples 1) Find the factors of the following: a) 36 b) 49 d) 40 e) 28 Playing with Numbers 95

2) Find the multiples of the following as given in the brackets: a) 7 (First 8) b) 15 (First 5) c) 100 (First 10) d) 25 (First 4) e) 30 (First 6) 3) Find the highest common factor of the following pairs of numbers: a) 12, 20 b) 15, 27 c) 24, 48 d) 16, 64 e) 30, 45 4) Find the least common multiple of the following pairs of numbers: a) 8, 10 b) 12, 15 c) 16, 20 d) 6, 18 e) 15, 30 7.2 H.C.F. and L.C.M. 5) Prime factorise using the factor tree: a) 18 b) 16 c) 56 d) 48 e) 63 6) Find the following using the prime factorisation method: a) L.C.M. of 32 and 56 b) H.C.F. of 25 and 75 c) H.C.F. of 96 and 108 d) L.C.M. of 45 and 75 A Note to Parent Help your child to collect a few objects such as leaves, pebbles, plastic bottles, flowers and so on from the surroundings. Ask him or her to make a table with their number. Then check for their divisibility by 2, 3, 4, 5, 6, 9 and 10. 96