202110794-TRAVELLER_PREMIUM-STUDENT-TEXTBOOK-MATHEMATICS-G05-PART1

In the crores place, 5 > 2 > 1. In the ten lakhs place, 8 > 7 > 4 > 3. Thus, 14327818 < 23187542 < 57124721 < 58348975. Therefore, the required ascending order is 14327818, 23187542, 57124721, 58348975. The descending order of numbers is just the reverse of their ascending order. Thus, 58348975 > 57124721 > 23187542 > 14327818. Therefore, the required descending order is 58348975, 57124721, 23187542, 14327818. Example 13: The population of Town A is 36,15,492 and that of Town B is 36,84,947. Which town has more population? Write it in words. Solution: Population of Town A = 36,15,492 Population of Town B = 36,84,947 Comparing the digits in the ten thousands place, we have: 36,84,947 > 36,15,492 Therefore, the population of Town B is more than that of Town A. The population of town B is thirty-six lakh eighty-four thousand nine hundred forty- seven. Form the greatest and the smallest numbers Example 14: Form the greatest and the smallest 8-digit numbers using the digits 1, 2, 5, 7, 3, 8, 4 and 6. Do not repeat any of the digits. Solution: The given digits are 1, 2, 5, 7, 3, 8, 4 and 6. Follow these steps to form the greatest number. Step 1: Arrange the digits in descending order: 8, 7, 6, 5, 4, 3, 2, 1 Step 2: Place the digits in the place value chart from left to right. C TL L T Th Th H TO 8 7 6 5 4 3 21 Numbers 47

So, the greatest 8-digit number that can be formed is 87654321. The steps to form the smallest 8-digit number are as follows: Step 1: Arrange the digits in ascending order: 1, 2, 3, 4, 5, 6, 7, 8 Step 2: Place the digits in the place value chart from left to right. C TL L T Th Th H T O 7 8 123456 So, the smallest 8-digit number that can be formed is 12345678. I Explore (H.O.T.S.) Let us see a few more examples involving large numbers. Example 15: What is the sum of the place values of the digit 7 in the number 79806724? Solution: The place values of 7 in 79806724 are 7 crores (70000000) and 7 hundred (700). Their sum is 70000000 + 700 = 70000700. Example 16: What is the difference between the place value and the face value of the digit 5 in the number 25600017? Solution: The place value of 5 in 25600017 is 5000000 and its face value is 5. Their difference is 5000000 – 5 = 4999995. 3.3 Round off Numbers I Think There is a birthday party at Pooja’s house. 48 children were invited. Her mother ordered 100 bars of chocolate to give 2 bars to each child. Do you think Pooja’s mother ordered enough number of chocolates? I Recall Let us revise comparing 1-digit, 2-digit and 3-digit numbers. Fill in the blanks using > or <: a) 4 ____ 9 b) 42 ____ 52 c) 195 ____ 105 d) 23 ____ 12 e) 100 ____ 200 48

I Remember and Understand Many times, we need not know the exact number. Just to get an idea of the required number, we round off a given number. For example, if we have ` 993, we say that we have about ` 1000. This rounding off may be to the nearest tens, hundreds, thousands, ten thousands and so on. Rounding off a number to the nearest tens • If the digit in the ones place is 0, 1, 2, 3 or 4 (less than 5), we replace it with 0. • If the digit in the ones place is 5, 6, 7, 8 or 9 (greater than or equal to 5), we replace the digit in the ones place with 0. We then add 1 to the digit in the tens place. 4 5 3 6 2 7 8 1 9 0 Example 17: Round off 16768 to the nearest ten. Rounding off numbers to a given place gives Solution: In 16768, the digit in the ones place is 8, its value approximate to which is greater than 5. So, we round off that place. 16768 to 16770. Rounding off a number to the nearest hundreds • If the digit in the tens place is 0, 1, 2, 3 or 4, we replace the digits in the tens and the ones places with zeros (0). • If the digit in the tens place is 5 or greater, we replace the digits in the ones and the tens places with 0. We then increase the digit in the hundreds place by 1. Example 18: Round off the following numbers to the nearest hundred. a) 1745 b) 21750 Solution: a) In 1745, the digit in the tens place is 4 which is less than 5; so, it is rounded off to 1700. b) In 21750, the digit in the tens place is 5. So, it is rounded off to 21800. Numbers 49

Rounding off a number to the nearest thousands • If the digit in the hundreds place is 0, 1, 2, 3 or 4, we replace the digits in the hundreds, tens and ones places with zeros. • If the digit in hundreds place is 5, 6, 7, 8 or 9, we replace the digits in the hundreds, tens and ones places with zeros. We then increase the digit in the thousands place by 1. Example 19: Round off the following numbers to the nearest thousand. a) 24190 b) 54729 Solution: The digits in the hundreds place are: a) 1 < 5. Therefore, 24190 is rounded off to 24000. b) 7 > 5. Therefore, 54729 is rounded off to 55000. ? Train My Brain Round off the following numbers: a) 459 to the nearest hundred. b) 26 to the nearest ten. c) 412 to the nearest hundred. I Apply Let us look at some real-life examples where we use the knowledge of rounding off numbers. Example 20: 27 people were expected to attend a meeting. How many chairs rounded to the nearest ten should be hired? Solution: In 27, the digit in the ones place is greater than 5. So, 27 is rounded off to 30. Therefore, 30 chairs should be hired. Example 21: There are 858 athletes running in a marathon. Each one of them has to be given a bottle of water. How many bottles of water rounded to the nearest hundred would be needed? Solution: In 858, the digit in the tens place is 5. So, 858 is rounded off to 900. Therefore, 900 bottles of water would be needed. 50

Example 22: 7965 students of a school are to be given a flag each to hold. How many flags rounded to the nearest thousand should be brought? Solution: In 7965, the digit in the hundreds place is greater than 5. So, 7965 is rounded off to 8000. Therefore, 8000 flags should be brought. I Explore (H.O.T.S.) Let us practise more of rounding off numbers. Example 23: Round off 67589 to the nearest tens, hundreds, thousands and ten thousands. Solution: 67589 rounded off to the nearest tens is 67590. 67589 rounded off to the nearest hundreds is 67600. 67589 rounded off to the nearest thousands is 68000. 67589 rounded off to the nearest ten thousands is 70000. Example 24: Consider the digits 5, 2, 9 and 6. Form the smallest and the largest 4-digit numbers using the given digits only once. Round off both the numbers to the nearest thousand. Solution: The smallest number that can be formed using the given digits only once is 2569. The largest number that can be formed using the given digits only once is 9652. 2569 rounded off to the nearest thousand is 3000. 9652 rounded off to the nearest thousand is 10000. Maths Munchies We know that each Roman numeral can be repeated only thrice. And, we can place a dash over it to multiply the number by 1000. Following the rules of Roman numerals strictly, we can say that 4999 is the largest number that can be represented in Roman numerals without using the dashes over the letters. Numbers 51

Connect the Dots Social Studies Fun China is the world’s most populous country. It has a population of over 1.35 billion. The population of our country in 2016 was about 1.34 billion. Science Fun The speed of light in a vacuum is 299792458 metres per second. Drill Time 3.1 Large Roman Numerals 1) Write the following in Roman numerals: a) 983 b) 804 c) 1481 d) 294 e) 1000 2) Write the following in Hindu-Arabic numerals: a) CLXX b) LXVII c) DL d) MCML e) LXIX 3) Word problems a) M DCLV parents visited a science fair on Monday and MDCVIII on Tuesday. Find the total number of parents visiting the science fair on Monday and Tuesday. Write the number in Roman numeral. b) In a race, Neha scores LXVI points and Lata scores XXV points. Who wins the race? 3.2 Count Large Numbers Using Indian and International Systems 4) Write the successor and the predecessor of the following numbers. a) 62591 b) 59104 c) 185031 d) 70001 e) 28501 52

5) Separate the periods of the given numbers using commas in both Indian and international systems of numeration. Write their number names. a) 872492853 b) 658392759 c) 124654368 d) 765401954 e) 378954726 6) Compare and put the correct symbol >, < or = in the blanks. a) 4,34,12,456 ______ 4,34,21,456 b) 2,31,98,896 ______ 6,87,98,541 c) 7,97,43,111 ______ 6,12,41,845 d) 1,67,91,941 ______ 1,76,19,149 7) Arrange the numbers in ascending and descending orders. a) 85714781, 57294769, 18372657 b) 17485729, 91845726, 75638462 c) 38593010, 75639205, 75927592 d) 10101010, 11010101, 10010101 8) Expand the following numbers: a) 4794792 b) 74874910 c) 26582948 d) 6572819 e) 75628401 9) Form the greatest and the smallest numbers using the given digits. a) 4, 6, 1, 5, 7, 2, 9 b) 7, 9, 2, 1, 4, 0, 5, 6, 3 c) 8, 2, 4, 1, 6, 5, 3 d) 0, 5, 2, 6, 8, 2 e) 1, 3, 4, 5, 6, 8, 9, 2 10) Word problems a) Factory A spent ` 87561839 in a month and Factory B spent ` 74829104 in the same month. Which factory spent more? b) The names of some countries and their populations are given. Use this information to answer the questions that follow: Afghanistan: 2,91,17,000; Australia: 83,72,930; Canada: 3,42,07,000; Egypt: 7,88,48,000 a) What is the population of Afghanistan? Write the figure in words. b) What is the population of Egypt? Express the figure in words. c) Which country, Australia or Canada, has more population? 3.3 Round off Numbers 11) R ound off the numbers to the nearest tens, hundreds, thousands and ten thousands. a) 75917 b) 57141 c) 87610 d) 36104 e) 17501 Numbers 53

12) Word problems a) A n adult human has 32 teeth. Round off the number to the nearest tens. b) A human body has 206 bones. Round off the number to the nearest hundreds. A Note to Parent Write some large numbers on a board or a piece of paper and make your child write them in the Indian and international systems. 54

4Chapter ONupemrbaetrions I Will Learn About • adding and subtracting large numbers. • multiplying by 2-digit and 3-digit numbers. • dividing by 2-digit numbers. • BODMAS rule. 4.1 Add and Subtract Large Numbers I Think The total population of Pooja’s town is 1234567, out of which 876986 are men. Pooja wanted to know the number of women in the town. Also, 25378 children were born later that year. Pooja now wants to find the total population of the town. Do you know how to find the same? I Recall Recall that we can add or subtract two or more numbers by writing them one below the other. This is called vertical or column addition or subtraction respectively. Let us solve the following to recall addition and subtraction. a) 283 + 115 b) 13652 +12245 c) 9685 – 5443 d) 47645 – 15322 e) 456789 – 23411 55

I Remember and Understand We can add or subtract large numbers by writing In vertical or column addition, them one below the other. Always remember to write the numbers one below place the numbers according to their place values. the other, and start adding the digits from the ones. In Example 1: Solve the following: subtraction, write the smaller number below the bigger a) 403050906 + 444333222 number. b) 963271087 – 365842719 Solution: a) 403050906 + 444333222 TC C TL L T Th Th H T O 1 4 0 3 0 5 09 0 6 + 4 4 4 3 3 32 2 2 8 4 7 3 8 41 2 8 b) 963271087 – 365842719 O TC C TL L T Th Th H T 17 8 15 12 12 6 10 10 7 /7 /9 6/ 3/ 2/ /7 1/ /0 8/ 9 8 – 3 6 5 8 4 27 1 5 9 7 4 2 83 6 We can add three or more numbers in the same manner. Example 2: Solve: 3608926 + 1560863 + 5697528 Solution: C T L L T Th Th H T O 111 1 2 11 6 3 36 0 8 92 8 7 + 15 6 0 86 + 56 9 7 52 108 6 7 31 56

In some problems, we may have both addition and subtraction together. Let us see some examples. Example 3: Simplify the following: a) 39154189 + 46673956 – 58127492 b) 742503 – 346280 + 210028 Solution: a) First add 39154189 and 46673956. Then subtract 58127492 from the sum. C T L L T Th Th H T O 11 111 39 1 5 4189 +46 6 7 3956 85 8 2 8145 C T L L T Th Th H T O 7 15 7 10 14 /8 5/ 8 2 8/ /1 4/ 5 –58 1 2 7492 27 7 0 0653 Therefore, 39154189 + 46673956 – 58127492 = 27700653. b) First subtract 346280 from 742503. Then, add 210028 to the difference. L T Th Th H T O L T Th Th H T O 6 13 12 4 10 1 1 39 6 223 7/ 4/ 2/ 5/ 0/ 3 −3 4 6 2 8 0 +2 1 0 0 2 8 3 9 6 223 6 0 6 251 Therefore, 742503 – 346280 + 210028 = 606251. Number Operations 57

? Train My Brain Solve the following: a) 34870089 + 568834 b) 1123870 + 33498609 c) 99785633 – 7569386 I Apply Let us consider a few real-life examples of addition and subtraction of large numbers. Example 4: Rathan’s father bought two houses, one for ` 11856000 and the other for ` 21248000. How much money did he spend altogether? By how much is the second house more expensive than the first? Solution: Cost of the 1st house = ` 11856000 Cost of the 2nd house = + ` 21248000 Amount Rathan’s father spent altogether = ` 33104000 Cost of the 2nd house = ` 21248000 Cost of the 1st house = – ` 11856000 Their difference = ` 9392000 T herefore, the second house was more expensive than the first house by ` 9392000. Example 5: Rohan, Sohan and Mohan purchased cars worth ` 1712890, ` 1212865 and ` 1616725 respectively. Find the total amount they spent on purchasing the cars. Solution: Amount Rohan spent on purchasing a car = ` 1712890 Amount Sohan spent on purchasing a car = + ` 1212865 Amount Mohan spent on purchasing a car = + ` 1616725 Total amount spent on purchasing the cars = ` 4542480 Therefore, total amount spent by Rohan, Sohan and Mohan is ` 4542480. 58

I Explore (H.O.T.S.) Let us now solve some examples of addition and subtraction by rounding off the numbers. Example 6: Estimate 672406 – 573348 by rounding off the numbers to the nearest hundreds. Solution: Rounding off the numbers to the nearest L T Th Th H T O hundreds, we get 672400 and 573300. Their difference is 672400 – 573300. 6 7 2 400 Therefore, the estimated difference of – 5 7 3 3 0 0 the given numbers is 99100. 0 9 9 100 Example 7: The populations of cities A, B and C are 2871428; 3287654 and 1636741 respectively. Find the total population of the three cities. Round off the total population to the nearest thousands. TL L T Th Th H T O Solution: Population of City A = 2 8 7 1 428 Population of City B = + 3 2 8 7 6 5 4 Population of City C = +1 6 3 6 7 4 1 Total population = 7 7 9 5 823 Total population of the three cities is 7795823. Rounding off to the nearest thousands, the total population of the three cities is 77,96,000. 4.2 Multiply Large Numbers and Divide Numbers by 2-digit Numbers I Think Pooja’s brother saved ` 1224 in twelve days. He saved an equal amount every day. Their mother bought a pair of shirts worth ` 112 from the money he saved. How many days’ savings of Pooja’s brother did her mother spend? Number Operations 59

I Recall We have already learnt how to multiply a 4-digit number by a 1-digit number. Let us recall the basic concepts of multiplication. Properties of Multiplication Identity Property: For any number ‘a’, a × 1 = 1 × a = a. 1 is called the multiplicative identity. For example, 213 × 1 = 1 × 213 = 213. Zero Property: For any number ‘a’, a × 0 = 0 × a = 0. For example, 601 × 0 = 0 × 601 = 0. Commutative Property: If ‘a’ and ‘b’ are any two numbers, then a × b = b × a. For example, 25 × 7 = 175 = 7 × 25. Associative Property: If ‘a’, ‘b’ and ‘c’ are any three numbers, then a × (b × c) = (a × b) × c. For example, 3 × (4 × 5) = (3 × 4) × 5 3 × 20 = 12 × 5 60 = 60 Similarly, we can write two multiplication facts for a division fact. For example, we can write the multiplication facts for 45 ÷ 9 = 5 as 9 × 5 = 45 or 5 × 9 = 45. 45 ÷ 9 = 5 ↓ ↓ ↓ Dividend Divisor Quotient The number that is divided is called the dividend. The number that divides is called the divisor. The number of times the divisor divides the dividend is called the quotient. 60

Factors Factors Multiplicand × Multiplier = Product Multiplicand × Multiplier = Product 9 × 5 = 45 5 × 9 = 45 ↓ ↓ ↓ ↓ ↓ ↓ Divisor Quotient Dividend Divisor Quotient Dividend The part of the dividend that remains without being divided is called the remainder. I Remember and Understand Let us solve a few examples of multiplication of large numbers. Multiplying a 4-digit number by a 2-digit number Example 8: Find the product: 2519 × 34 Solution: T Th Th H T O 12 23 2519 ×34 1 1 0 0 7 6 → 2519 × 4 ones + 7 5 5 7 0 → 2519 × 3 tens 8 5 6 4 6 → 2519 × 34 Therefore, the required product is 85646. Number Operations 61

Multiplying a 4-digit number by a 3-digit number Example 9: Find the product of 3768 and 407. Solution: T L L T Th Th H T O Here we can skip 323 the step ‘3768 × 0’ but, 545 add one more zero in 3768 tens place while multiplying by hundreds digit. ×407 1 2 6 3 7 6 → 3768 × 7 ones + 1 5 0 7 2 0 0 → 3768 × 4 hundreds 1 5 3 3 5 7 6 → 3768 × 407 Example 10: Estimate the product of 58265 and 73. Then multiply and verify your answer. Solution: We can round off the number to find the estimated product. Rounding off the multiplicand 58265 to the nearest thousand, we get 58000. Similarly, rounding off the multiplier 73 to the nearest tens, we get 70. Multiplying the estimated numbers, 58000 and 70, we get 4060000. Now, let us verify the answer by finding the product using vertical method of multiplication. T L L T Th Th H T O 5 143 2 11 5 8265 ×73 1 1 11 1 7 4 7 9 5 → 58265 × 3 ones + 4 0 7 8 5 5 0 → 58265 × 7 tens 4 2 5 3 3 4 5 → 58265 × 73 The actual product is close to the estimated product. Hence, verified. 62

Multiplying a 5-digit number by a 3-digit number Example 11: Find the product of 24367 and 506. Solution: T L L T Th Th H T O 2 133 2 244 2 4367 ×506 1 1 4 6 2 0 2 → 24367 × 6 ones + 1 2 1 8 3 5 0 0 → 24367 × 5 hundreds 1 2 3 2 9 7 0 2 → 24367 × 506 Let us now see a few examples of division of large numbers. Dividing a 4-digit number and a 5-digit number by a 1-digit number Example 12: Divide: a) 2065 ÷ 5 b) 76528 ÷ 4 Solution: a) b) 19132 The remainder must always be 4 13 )4 76528 less than the divisor. )5 2 0 6 5 −2 0 ↓ −4 06 36 − 05 − 36 0 15 − 015 05 000 − 04 12 − 12 08 −8 0 Dividing a 3-digit number and a 4-digit number by a 2-digit number Example 13: Divide: a) 414 ÷ 12 b) 2340 ÷ 15 Number Operations 63

Solution: a) 34 b) 156 ) )12 414 15 2340 −36 ↓ − 15↓ 054 84 − 048 − 75 90 6 − 90 00 Dividing a 5-digit number by a 2-digit number Let us solve some examples to understand dividing a 5-digit number by a 2-digit number. Example 14: Divide: 21809 ÷ 14 Solution: Follow the steps given below to divide 21809 by 14. Steps Solved Solve these Step 1: Write the dividend and divisor as shown. )14 21809 )20 53174 )Divisor Dividend − Step 2: Guess the quotient )14 21809 − by dividing the two leftmost digits by the divisor. 14 × 1 = 14 − 14 × 2 = 28 Find the multiplication fact 14 < 21 < 28 − which has the dividend and So,14 is the number to be the divisor. subtracted from 21. Step 3: Write the factor other Write 1 in the quotient )13 34567 than the dividend and the and 14 below 21, and divisor as the quotient. subtract. Then bring down − the next number in the dividend. − 1 − )14 21809 − -14 78 64

Steps Solved Solve these Step 4: Repeat steps 2 and 3 until all the digits of the 1557 )15 45675 dividend are brought down. )14 21809 − Stop the division when the − remainder < divisor. − 14 − 78 − − 70 80 − 70 109 − 98 11 Step 5: Write the quotient Quotient = 1557 and the remainder. Remainder = 11 Checking for the correctness of division: We can check if our division is correct using a multiplication fact of the division. Step 1: Compare the remainder and divisor. If the remainder is less than the divisor, the division is correct. Step 2: Check if (Quotient × Divisor) + Remainder = Dividend. If it is, the division is correct. Let us now check if our division in example 14 is correct or not. Step 1: Remainder < Divisor Dividend = 21809 Divisor = 14 Quotient = 1557 Remainder = 11 11 < 14 (True) Step 2: (Quotient × Divisor) + 1557 × 14 + 11 = 21809 Remainder = Dividend 21798 + 11 = 21809 21809 = 21809 (True) Since the two checks are true, the division is correct. Note: 1) If remainder > divisor, the division is incorrect. 2) If (Quotient × Divisor) + Remainder is not equal to Dividend, the division is incorrect. Number Operations 65

? Train My Brain Solve the following: a) 2868 ÷ 24 b) 7890 ÷ 12 c) 723 ÷ 15 I Apply We use multiplication and division of large numbers in many real-life situations. Let us see a few examples. Example 15: A farmer has 6350 acres of mango farm. If he needs 58 kg of fertiliser for each acre, how many kilograms of fertiliser does he need in all? Solution: Quantity of fertiliser required for 1 acre of L T Th Th H T O farm = 58 kg 12 Quantity of fertiliser required for 6350 acres 24 of farm 6350 = 6350 × 58 kg ×58 1 = 368300 kg Therefore, the farmer needs 368300 kg of 5 0800 fertiliser in all. +3 1 7500 3 6 8300 Example 16: A machine produces 48660 pens in June. How many pens does it produce Solution: in a day? 1622 Number of days in June = 30 )30 48660 Number of pens produced in the month = 48660 − 30 Number of pens produced in a day = 48660 ÷ 30 186 − 180 Therefore, the machine produces 1622 pens in a day. 66 − 60 60 − 60 00 66

I Explore (H.O.T.S.) Consider these examples. Example 17: Find the missing numbers in the given product. T Th Th H T O 3417 ×63 1 21 + 0 5 20 21 271 Solution: T Th T H T O 3417 ×63 1 0251 +2 0 5 0 2 0 2 1 5271 Example 18: Divide the largest 4-digit number by the largest 2-digit number. Write the Solution: quotient and the remainder. 10 1 The largest 4-digit number is 9999. )99 9 9 9 9 The largest 2-digit number is 99. −99↓ The required division is 9999 ÷ 99. 009 − 000 Quotient = 101; Remainder = 0 99 − 99 00 Number Operations 67

4.3 Bodmas I Think Pooja went to a shop to buy 6 ice creams for ` 40 each. She bargained with the shopkeeper to get the ice creams for ` 5 less. She paid ` 300 to the shopkeeper. How could she calculate quickly the amount she is supposed to get back from the shopkeeper? I Recall We have already learnt about the four operations of numbers. Let us recall them by solving the following: a) 40 + 55 b) 23 – 19 c) 19 × 8 d) 24 ÷ 2 I Remember and Understand We can perform only one mathematical operation on two numbers at any given time. We can either add, subtract, multiply or divide. Sometimes, we have situations in which there are 3 or more numbers and 2 or more operations between them. In such cases, it is difficult to decide which operation is to be carried out first. Consider the following: Solve: 8 ÷ 2 × 2 The two operations in the given problem are division and multiplication. Now this problem can be solved in the following two ways. a) When we first divide and then multiply 8÷2=4 While solving a numerical 4×2=8 problem, read it from left Therefore, 8 ÷ 2 × 2 = 8. to right. However, follow b) When we first multiply and then divide the BODMAS rule to 2×2=4 solve certain parts of the problem first. 8÷4=2 Therefore, 8 ÷ 2 × 2 = 2. 68

From this, we see that we get two different answers depending on the order in which we carry the operations. For the answer to be the same, the same order must be followed by anyone who solves the problem. A simple rule known as the BODMAS rule helps us to decide the order of carrying out the operations. The abbreviation ‘BODMAS’ stands for: ( ) Of ÷ × + – Let us consider a few examples where we can use this rule for easy calculation to get an accurate answer. Example 19: Find the value: 3 × 4 + 5 Solution: The two operations in the given problem are - multiplication and addition. Applying the BODMAS rule, we first multiply the numbers. So, 3 × 4 = 12. Then we add. So, 12 + 5 = 17 Therefore, 3 × 4 + 5 = 17. Example 20: Find the value: 6 × 2 ÷ 2 Solution: Applying the BODMAS rule, we first divide and then multiply. So, 2 ÷ 2 = 1 and 6 × 1 = 6. Therefore, 6 × 2 ÷ 2 = 6. Example 21: Find the value: 18 – (2 + 3) Solution: Applying the BODMAS rule, we first solve the operations in the brackets followed by subtraction. So, 2 + 3 = 5 Number Operations 69

18 – 5 = 13 Therefore, 18 – (2 + 3) = 13. ? Train My Brain Use BODMAS rule to solve the following: a) 4 + 10 × 4 b) 20 + 5 ÷ (4 + 1) c) 4 × 3 – 1 I Apply Let us apply the knowledge of BODMAS to solve the sums involving more than two operations. Example 22: Solve: 4 × 2 + (40 ÷ 4 – 1) Solution: Using BODMAS, we solve it as follows: Solving the brackets: 40 ÷ 4 – 1. Now, there are two operations in the brackets. Applying the rule in the brackets, we first divide and then subtract. So, 40 ÷ 4 = 10; 10 – 1 = 9. Thus, 4 × 2 + (40 ÷ 4 – 1) = 4 × 2 + 9. Again applying BODMAS, perform the multiplication followed by addition. So, 4 × 2 = 8; 8 + 9 = 17. Therefore, 4 × 2 + (40 ÷ 4 – 1) = 17. Example 23: Spot the wrong step and correct it for 8 × 4 × 2 + (36 ÷ 6) – 4. Solution: Wrong way Right way Step 1: Removing the brackets, we get Step 1: Solving the operation in the 8 × 4 × 2 + 36 ÷ 6 – 4 brackets, we get 36 ÷ 6 = 6. Step 2: Solving from the left, we get Step 2: Then we multiply to get 8 × 4 × 2 = 64 8 × 4 × 2 = 64. 64 + 36 = 100 100 ÷ 6 – 4 Thus, 8 × 4 × 2 + (36 ÷ 6) – 4 = 64 + 6 – 4. 100 ÷ 2 = 50 So, 70 – 4 = 66. Thus, 8 × 4 × 2 + (36 ÷ 6) – 4 = 50 Therefore, 8 × 4 × 2 + (36 ÷ 6) – 4 = 66. 70

I Explore (H.O.T.S.) Let us now consider a few examples in which we use BODMAS in real life. Example 24: Reema bought 4 oranges for ` 12 each and 8 apples for ` 10 each. What is the total amount Reema spent? Solution: Amount of oranges = ` 12 × 4 Amount of apples = ` 10 × 8 We can apply BODMAS here to find the total amount as: (12 × 4) + (10 × 8) = 48 + 80 = 128 Thus, Reema spent ` 128 on the fruits. Example 25: After Diwali, Shubhro brought 45 sweets for his friends Piyush, Riya, Vaishnavi, Swati and Pooja. He distributed them equally among his friends. But Piyush being greedy took 2 sweets from Pooja. Using BODMAS, show how many sweets are left with Pooja? Solution: Total number of friends Shubhro has = 5 Number of sweets Shubhro brought for his friends = 45 So, number of sweets everybody gets = 45 ÷ 5 Number of sweets Piyush took from Pooja = 2 Therefore, number of sweets left with Pooja = 45 ÷ 5 – 2 Applying the BODMAS rule, perform the division followed by subtraction. 45 ÷ 5 = 9 and 9 – 2 = 7 Therefore, 7 sweets are left with Pooja. Maths Munchies ‘Incrementation’ and ‘summation’ are two words associated with addition. Incrementation, also known as the successor operation, is the addition of 1 to a number. Summation describes the addition of many numbers, usually more than just two. Decrementation is a word associated with subtraction. Decrementation, also known as the predecessor operation, is the subtraction of 1 from a number. Number Operations 71

Connect the Dots English Fun ‘Subtraction’ is an English word derived from the Latin verb ‘subtrahere’. This is a compound of ‘sub’ meaning ‘from under’ and ‘trahere’ meaning ‘to pull’. Thus, to subtract is to draw from under or to take away. Social Studies Fun The addition of large numbers helps us to calculate the total population of a place. To compute the total population of our country, we add the populations of all the states. Drill Time 4.1 Add and Subtract Large Numbers 1) Add: a) 124523456 + 235454134 b) 2680054 + 1098366 c) 429502940 + 145829401 d) 3456786 + 2576987 e) 45678968 + 76894533 2) Subtract: a) 89372051 – 76419265 b) 5396104 – 2278160 c) 9623175 – 8892431 d) 8235676 – 5629012 e) 987968975 – 156857903 72

3) Word problems a) A shopkeeper sold 10 AC units for ` 67371974 and 10 refrigerator units for ` 67371974. How much money did he earn? b) The area of an airport is 18440000 sq. ft. and another airport is of area 16640000 sq. ft. What is the difference in the areas of the two airports? 4.2 Multiply Large Numbers and Divide Numbers by 2-digit Numbers 4) Multiply: a) 3718 × 20 b) 1287 × 143 c) 87461 × 12 d) 87491 × 123 e) 4221 × 21 5) Divide: a) 3124 ÷ 4 b) 12948 ÷ 2 c) 312 ÷ 12 d) 4326 ÷ 14 e) 58105 ÷ 15 6) Word problems a) A publication house prints 32 books. Each book has 1859 pages. How many pages does it print? b) 9582 t-shirts are dyed in 2 days. How many t-shirts are dyed in a day? 4.3 BODMAS 7) Solve the following: a) 18 ÷ 9 × (2 + 10) b) 4 + (16 – 4) ÷ 3 c) 21 ÷ 3 × 4 + 2 – 4 d) 4 + 10 ÷ 2 – (4 × 2) e) 6 ÷ 3 + (4 × 5) – 4 Number Operations 73

A Note to Parent Provide large numbers such as populations of a few cities in your state and make your child practise addition and subtraction using them. 12,482,523 129,435,625 + 23,217,436 + 230,542,263 3,381,003 326,546,320 + 4,317,872 + 452,332,648 74

5Chapter Playing with Numbers I Will Learn About • factors and multiples of numbers. • prime, composite, twin prime and co-prime numbers. • finding prime factors by factor tree method. • finding H.C.F. and L.C.M. by listing method and common division method. • the relationship between H.C.F. and L.C.M. 5.1 Prime and Composite Numbers I Think Pooja and Rama were playing a card game. They decide to divide the number on the card that they pick up by different numbers. Rama got all the cards which had numbers that could be divided by only 1 and the numbers themselves. Do you know what the numbers that Rama got are called? I Recall Let us revise the concept of prime and composite numbers. Circle the prime numbers and cross out the composite numbers in the given grid. 75

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 I Remember and Understand We know that: 1) the numbers that have only two factors (1 and themselves) are called prime numbers. 2) the numbers that have more than two factors are called composite numbers. 3) the number 1 has only one factor. So, it is neither prime nor composite. Twin prime and co-prime numbers Two prime numbers that differ by 2 are called twin prime numbers. For example, 5 and 7 are twin primes as the difference Multiples of 2 are called between them is 2. even numbers. Numbers Two numbers with 1 as their H.C.F. are called co-prime which are not multiples numbers. of 2 are called odd numbers. For example, the numbers 7 and 13 are co-prime numbers as they have only 1 as their highest common factor. Similarly, the numbers 4 and 5 have 1 as their H.C.F. So, 4 and 5 are co-prime. We have learnt the factor tree method to carry out the prime factorisation of a number. Now, let us learn another method known as the division method. Consider these examples. Example 1: Find the prime factors of 90 using the long division method. 2 90 3 45 Solution: To find the prime factors of 90, follow these steps: 3 15 55 Step 1: Find the least prime number that divides the given number. 1 76

Step 2: Divide the given number by the prime number obtained in step 1. Write the quotient below the number. Step 3: Repeat step 1 (finding the least prime number that divides the quotient) and step 2 (dividing the quotient) until a prime number quotient is obtained. Therefore, the prime factorisation of 90 is 2 × 3 × 3 × 5. Example 2: Carry out the prime factorisation of 84 using the division method. Solution: The prime factorisation of 84 using the division method is as 2 84 shown. 2 42 3 21 84 = 2 × 2 × 3 × 7 7 7 1 Therefore, the prime factorisation of 84 is 2 × 2 × 3 × 7. We have learnt multiplication tables from 1 to 20. Let us recall them quickly. Multiplication Tables 1 2 3 4 5 1×1=1 2×1=2 3×1=3 4×1=4 5×1=5 1×2=2 2×2=4 3×2=6 4×2=8 5 × 2 = 10 1×3=3 2×3=6 3×3=9 4 × 3 = 12 5 × 3 = 15 1×4=4 2×4=8 3 × 4 = 12 4 × 4 = 16 5 × 4 = 20 1×5=5 2 × 5 = 10 3 × 5 = 15 4 × 5 = 20 5 × 5 = 25 1×6=6 2 × 6 = 12 3 × 6 = 18 4 × 6 = 24 5 × 6 = 30 1×7=7 2 × 7 = 14 3 × 7 = 21 4 × 7 = 28 5 × 7 = 35 1×8=8 2 × 8 = 16 3 × 8 = 24 4 × 8 = 32 5 × 8 = 40 1×9=9 2 × 9 = 18 3 × 9 = 27 4 × 9 = 36 5 × 9 = 45 1 × 10 = 10 2 × 10 = 20 3 × 10 = 30 4 × 10 = 40 5 × 10 = 50 6 7 8 9 10 6×1=6 7×1=7 8×1=8 9×1=9 10 × 1 = 10 6 × 2 = 12 7 × 2 = 14 8 × 2 = 16 9 × 2 = 18 10 × 2 = 20 6 × 3 = 18 7 × 3 = 21 8 × 3 = 24 9 × 3 = 27 10 × 3 = 30 6 × 4 = 24 7 × 4 = 28 8 × 4 = 32 9 × 4 = 36 10 × 4 = 40 6 × 5 = 30 7 × 5 = 35 8 × 5 = 40 9 × 5 = 45 10 × 5 = 50 6 × 6 = 36 7 × 6 = 42 8 × 6 = 48 9 × 6 = 54 10 × 6 = 60 6 × 7 = 42 7 × 7 = 49 8 × 7 = 56 9 × 7 = 63 10 × 7 = 70 6 × 8 = 48 7 × 8 = 56 8 × 8 = 64 9 × 8 = 72 10 × 8 = 80 6 × 9 = 54 7 × 9 = 63 8 × 9 = 72 9 × 9 = 81 10 × 9 = 90 6 × 10 = 60 7 × 10 = 70 8 × 10 = 80 9 × 10 = 90 10 × 10 = 100 Playing with Numbers 77

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 Rules of divisibility We can find the numbers that divide a given number without actual division. The rules used to find them are called divisibility rules. They help us to find the factors (divisors) of a given number. Divisor Rule Examples 2 10, 42, 56, 48, 24 3 The ones digit of the given number must be 0, 2, 4, 6 or 8. 4 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 of the given number must be divisible by 4 1400, 3364, 2500, 7204 or both the digits must be zero. 78

Divisor Rule Examples 5 6 The ones digit of the given number must 230, 375, 100, 25 9 be 0 or 5. 10 The number must be divisible by both 2 36, 480, 1200 and 3. 36 (3 + 6 = 9) 11 567 (5 + 6 + 7 = 18; The sum of the digits of the given number 18 = 1 + 8 = 9) must be divisible by 9. 300, 250, 5670 The ones digit of the given number must be 0. Add the digits in the odd places of the For 6050, (6 + 5) – (0 + 0) number. Add the digits in the even = 11 – 0 = 11 places of the number. If the difference of the sums is 0 or divisible by 11, then the For 7260, (7 + 6) – (2 + 0) number is divisible by 11. = 13 – 2 = 11 Let us now apply the divisibility rules to check if a given number is divisible by 2, 3, 4, 5, 6, 9, 10 or 11. Example 3: Which of the numbers 2, 3, 4, 5, 6, 9, 10 and 11 divide 42670? Solution: To check if 2, 3, 4, 5, 6, 9, 10 or 11 divide 42670, apply their divisibility rules. Divisibility by 2: The ones place of 42670 has 0. So, it is divisible by 2. Divisibility by 3: The sum of the digits of 42670 is 4 + 2 + 6 + 7 + 0 =19. 19 is not divisible by 3. So, 42670 is not divisible by 3. Divisibility by 4: T he number formed by the digits in the last two places of 42670 is 70, which is not exactly divisible by 4. So, 42670 is not divisible by 4. Divisibility by 5: The ones place of 42670 has 0. So, it is divisible by 5. Divisibility by 6: 42670 is divisible by 2 but not by 3. So, it is not divisible by 6. Divisibility by 9: T he sum of the digits of 42670 is 4 + 2 + 6 + 7 + 0 = 19, which is not divisible by 9. So, 42670 is not divisible by 9. Divisibility by 10: T he ones place of 42670 has 0. So, it is divisible by 10. Divisibility by 11: Adding the digits in the odd and even places, we get 4 + 6 + 0 = 10 and 2 + 7 = 9. Subtracting 9 from 10, we get 1. So, 42670 is not divisible by 11. Therefore, the numbers that divide 42670 are 2, 5, and 10. Playing with Numbers 79

I Apply Let us apply the knowledge of prime and composite numbers in real life. Example 4: There is a music event in school where the orchestra has to be divided into different groups. 24 musicians played the piano and 30 musicians played the guitar. What is the maximum number of groups that can be made with each group having the same number of musicians? Solution: Number of musicians playing piano = 24 Number of musicians playing guitar = 30 We can find the maximum number of groups by finding the common factors of both the numbers. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24 Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 Common factors of 24 and 30 are 1, 2, 3 and 6. As we need to have the maximum number of groups, the musicians can be divided into 6 groups. Example 5: Vinny visits her aunt every four days and her friend every five days. When will be the next day when she will meet them both on the same day? Solution: Vinny visits her aunt every four days and her friend every five days. To find the next day when she will visit both of them, we find the least common multiple of both the numbers. Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ... Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, ... As 20 is the least common multiple of both the numbers, Vinny will visit both of them on the 20th day. I Explore (H.O.T.S.) Let us now learn to prime factorise large numbers using the division method. Example 6: Prime factorise 1050 using the division method. 80

Solution: 2 1050 3 525 5 175 5 35 77 1 Therefore, prime factorisation of 1050 is 2 × 3 × 5 × 5 × 7. Example 7: Prime factorise 2000 using the division method. Solution: 2 2000 2 1000 2 500 2 250 5.2 5 125 5 25 55 1 Therefore, prime factorisation of 2000 is 2 × 2 × 2 × 2 × 5 × 5 × 5. H.C.F. and L.C.M. I Think Pooja is preparing snacks plates. She has 12 pieces of samosa and 16 rolls. She wants to make all the plates identical without any food left over. How can Pooja do this? I Recall Let us recap the divisibility rules by circling the numbers divisible by the given factors. S. No. Factors Numbers a) 2 b) 4 234 441 949 749 c) 10 782 123 d) 11 200 8911 396 324 222 9401 300 231 121 482 Playing with Numbers 81

I Remember and Understand We have learnt about the H.C.F. and the L.C.M. of numbers. Let us find the H.C.F. and L.C.M. of a few numbers using the listing method, prime factorisation method and common division method. Highest Common Factor (H.C.F.) Let us find the H.C.F. of 16 and 24 by listing, prime factorisation and common division methods. Listing method: In the listing method, we find the factors of the given numbers using the divisibility rules. Factors of 16 = 1, 2, 4, 8, 16 Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24 Common factors of 16 and 24 are 1, 2, 4 and 8. The highest common factor is 8. Therefore, the H.C.F. of 16 and 24 is 8. Prime Factorisation Method Factorisation of 16 Factorisation of 24 2 16 2 24 28 2 12 24 26 22 33 11 Prime factorisation of 16 = 2 × 2 × 2 × 2 2 16, 24 2 8, 12 Prime factorisation of 24 = 2 × 2 × 2 × 3 2 4, 6 2, 3 Therefore, the H.C.F. of 16 and 24 is 2 × 2 × 2 = 8. Common Division Method Common division method is similar to the division method. Step 1: Find the smallest prime number which can divide both numbers. Step 2: Write the prime number in the left and the quotient below the numbers as shown. 82

Step 3: Repeat step 1 until there are no common factors except 1. So, we can find the H.C.F. for these two numbers by If two numbers have no multiplying the prime numbers written in the left. common prime factors, then the H.C.F. is equal to 1. Therefore, the H.C.F. of 16 and 24 is 2 × 2 × 2 = 8. Thus, we can say that, using any of these three methods, listing, prime factorisation and common division method, we get the same H.C.F. Finding the common factors can also be shown as follows: Factors of 16 2 2 2 2 3 Factors of 24 Common factors of 16 and 24 are 2, 2 and 2. Therefore, H.C.F. of 16 and 24 is 2 × 2 × 2 = 8. Least Common Multiple (L.C.M.) Let us find the L.C.M. of 12 and 15 by listing, prime factorisation and common division methods. Listing method Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ... Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, ... Common multiples of 12 and 15 are 60, 120, ... The least common multiple is 60. Therefore, the L.C.M. of 12 and 15 is 60. Playing with Numbers 83

Prime Factorisation Method Factorisation of 15 Factorisation of 12 3 15 55 2 12 1 26 33 1 Prime factorisation of 12 = 3 × 2 × 2 Prime factorisation of 15 = 3 × 5 Therefore, the L.C.M. of 12 and 15 is 3 × 2 × 2 × 5 = 60. Common Division Method To find the L.C.M. of the given numbers, follow these steps: 2 12, 15 2 6, 15 Step 1: Find the least prime number that divides both 3 3, 15 the numbers or at least one of the numbers. 1, 5 Step 2: Write the quotients below the numbers. Step 3: Divide until we get co-prime numbers. Common factors of 12 and 15 are 2, 2, 3 and 5. Therefore, the L.C.M. of 12 and 15 is 2 × 2 × 3 × 5 = 60. ? Train My Brain Solve the following: a) Find the L.C.M. of 20 and 40. b) Find the H.C.F. of 10 and 15. c) Find the L.C.M. of 10 and 30. I Apply The relationship between H.C.F. and L.C.M. We have learnt about the H.C.F. and L.C.M. of different numbers. Let us find their relationship with each other. 84

Example 8: Find the H.C.F. and L.C.M. of 8 and 16. Solution: Factors of 8 = 2 × 2 × 2 Factors of 16 = 2 × 2 × 2 × 2 Common factors = 2, 2, 2 H.C.F. of 8 and 16 = 2 × 2 × 2 = 8 L.C.M. of 8 and 16 = 2 × 2 × 2 × 2 = 16 Product of the H.C.F. and L.C.M. = 8 × 16 = 128 Product of the two numbers = 8 × 16 = 128 Thus, the product of two numbers is equal to the product of their H.C.F. and L.C.M. To understand this better, consider the following example. Example 9: T he H.C.F. of two numbers is 6 and their L.C.M. is 36. If one of the numbers is 12, what is the other number? Solution: Given that L.C.M. of two numbers is 36 and their H.C.F. is 6. One of the numbers is 12. We know that, the product of the H.C.F. and L.C.M. = the product of the two numbers 6 × 36 = 12 × ? 6 × 3 × 12 = 12 × ? 18 × 12 = 12 × ? Therefore, the other number is 18. I Explore (H.O.T.S.) Let us now find the H.C.F. and L.C.M. of three numbers using the common division method. Example 10: Find the L.C.M. of 20, 30 and 40 using the common division method. Solution: Factors of 20, 30 and 40 = 2, 2, 2, 3, 5 2 20, 30, 40 L.C.M. of 20, 30 and 40 = 2 × 2 × 2 × 3 × 5 2 10, 15, 20 Therefore, L.C.M. of 20, 30 and 40 is 120. 2 5, 15, 10 3 5, 15, 5 5 5, 5, 5 1, 1, 1 Playing with Numbers 85

Example 11: Find the H.C.F. of 24, 36 and 48 using the division method. Solution: Factors of 24 Factors of 36 Factors of 48 2 24 2 36 2 48 2 12 2 18 2 24 2 12 26 39 26 33 33 33 1 1 1 Prime factors of 24 = 2, 2, 2, 3 Prime factors of 36 = 2, 2, 3, 3 Prime factors of 48 = 2, 2, 2, 2, 3 Common factors of 24, 36 and 48 are 2, 2 and 3. H.C.F. of 24, 36 and 48 = 2 × 2 × 3 = 12 Maths Munchies 2520 is the smallest number which is exactly divided by 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 having left no remainder at all. Connect the Dots Social Studies Fun 24 hours make a day. In two days there are 48 hours. In three days there are 72 hours. The number of hours in any number of days will always be a multiple of 24. 2, 3, 4, 6, 8, 12 and 24 will always be the factors of that number. How many hours are there in 5 days? Is it a multiple of 24? English Fun In a set of letters, assign each letter with a number. Start the letter ‘A’ with the number 1. Now, decode the word ‘COMPOSITE’ and find which of the letters are prime numbers. 86

Drill Time 5.1 Prime and Composite Numbers 1) Identify which of the following pairs are twin prime or co-prime numbers. a) 11 and 13 b) 13 and 17 c) 11 and 12 d) 29 and 31 e) 7 and 25 2) Find the prime factorisation of the following using the division method. a) 48 b) 78 c) 81 d) 99 e) 86 3) Find whether the following numbers are divisible by 2, 3, 4, 5, 6, 9, 10 and 11. a) 42 b) 67 c) 17 d) 38 e) 29 5.2 H.C.F. and L.C.M. 4) Find the H.C.F. of the following numbers using the listing, prime factorisation and division methods. a) 75, 90 b) 36, 54 c) 12, 15 d) 36, 54 e) 36, 40 5) Find the L.C.M. of the following numbers using the listing, prime factorisation and division methods. a) 20, 25 b) 48, 60 c) 18, 27 d) 18, 30 e) 28, 45 6) Find the missing number. a) H.C.F. = 12; L.C.M. = 36; first number = 12; second number = ? b) H.C.F. = 4; L.C.M. = 96; first number = ?; second number = 32 c) H.C.F. = ?; L.C.M. = 180; first number = 36; second number = 60 d) H.C.F. = 4; L.C.M. = ?; first number = 20; second number = 28 e) H.C.F. = ?; L.C.M. = 36; first number = 9; second number = 12 Playing with Numbers 87

7) Word problem a) Two trains A and B, arrive at a railway platform. Train A arrives every 20 minutes while Train B arrives every 24 minutes. When will both the trains arrive at the same station simultaneously? b) On a picnic, the kids have brought 12 cheese slices and 42 tomato slices. They want to use them to make identical burgers with no slices left. What is the maximum number of burgers they can make? A Note to Parent Let your child pick some pairs of numbers from 1 to 99. Ask him or her to find the H.C.F. and L.C.M. of those pairs. 88

Time6Chapter I Will Learn About • converting larger units of time to smaller units and vice versa. • using operations to solve problems on time intervals. 6.1 Convert Time I Think Pooja’s father spends 120 minutes every week reading the newspaper. Pooja wants to know the number of hours he spends reading the newspaper. How can she find that? I Recall We have learnt about time and its units such as minutes, hours, days and so on. Let us revise them by solving the following. 1) Draw hands on a clock to show: a) 7:33 p.m. b) 4:45 a.m. c) 1:28 p.m. d) 14:50 89

2) Answer these questions. a) How many hours are there in a day? b) How many days are there in a year? c) How many days make a week? d) How many days are there in a leap year? e) How many days does December have? I Remember and Understand We have learnt about the different units such as seconds, minutes, hours, and days to measure time. The larger units of measuring time are weeks, months and years. Let us now learn about conversion of time. Days to hours and hours to days 1 day = 24 hours = 24 h 1 hour = 1 day ÷ 24 Hours to minutes and minutes to seconds 1 hour = 60 minutes = 60 min 1 minute = 60 seconds Seconds to minutes and minutes to hours 1 minute = 1 hour ÷ 60 1 second = 1 minute ÷ 60 1 second = 1 hour ÷ 3600 Consider a few examples of conversion of time. Note: W hile converting a smaller unit of time into a larger unit, the remainder from the division is written in the smaller unit. Example 1: Convert the following into hours: a) 13 days b) 2 days 16 hours Solution: a) 1 day = 24 h Therefore, 13 days = 13 × 24 h = 312 h. 90

b) 1 day = 24 h 2 days 16 h = (2 × 24 h) + 16 h = 48 h + 16 h = 64 h Therefore, 2 days 16 hours = 64 hours. Example 2: Convert the following into minutes: a) 7 hours b) 6 hours 25 minutes Solution: a) 1 hour = 60 minutes Therefore, 7 hours = 7 × 60 min = 420 min. b) 1 hour = 60 minutes 6 hours 25 min = (6 × 60 min) + 25 min = 360 min + 25 min = 385 min Therefore, 6 hours 25 minutes = 385 minutes. Example 3: Convert the following into seconds: a) 5 h b) 28 min c) 3 days d) 6 weeks Solution: a) 1 hour = 60 × 60 s Therefore, 5 h = 5 × 60 × 60 s = 18000 s. b) 1 min = 60 s Therefore, 28 min = 28 × 60 s = 1680 s. c) 1 day = 24 h = 24 × 60 × 60 s Therefore, 3 days = 3 × 24 × 60 × 60 s = 259200 s. d) 1 week = 7 days Therefore, 6 weeks = 6 × 7 × 24 × 60 × 60 s = 3628800 s. Example 4: Convert the following: a) 132 min into hours b) 96 hours into days c) 426 seconds into minutes Solution: a) 1 min = 1 h ÷ 60 Therefore, 132 min = 132 h ÷ 60 = 2 h 12 min. Time 91

b) 1 hour = 1 days ÷ 24 Therefore, 96 hours = 96 h ÷ 24 = 4 days. c) 1 s = 1 min ÷ 60 Therefore, 426 s = 426 min ÷ 60 = 7 min 6 s. ? Train My Brain Answer the following: a) Convert 6600 minutes to hours. b) Convert 9 hours into minutes. c) Convert 8 weeks into days. I Apply Let us solve some real-life examples where conversion of time is used. Example 5: An aeroplane stops for 600 seconds at the Mumbai airport. For how many minutes does it stop? Solution: We know that 1 minute = 60 seconds. So, 1 second = 1 minute ÷ 60. Hence, 600 seconds = 600 minutes ÷ 60 = 10 minutes. Therefore, the aeroplane stops for 10 minutes at the Mumbai airport. Example 6: During a television programme, there were 10 breaks of 48 seconds each. For how many minutes did the breaks last in total? Solution: There are 10 breaks, each of 48 seconds. The total time of the breaks in seconds = 48 seconds × 10 = 480 seconds We know that, 1 minute = 60 seconds. So, 480 seconds = 480 minutes ÷ 60 = 8 minutes. Therefore, the breaks lasted for a total of 8 minutes. 92

Example 7: In January, Seema played for 30 minutes every day. For how much time did she play in that month? Give your answer in seconds. Solution: In January, Seema played for 30 minutes every day. Number of days in January = 31 Number of minutes she played in January = 30 minutes × 31 days = 930 minutes 1 minute = 60 seconds So, 930 minutes = 930 × 60 seconds = 55800 seconds. Therefore, Seema played for 55800 seconds in January. I Explore (H.O.T.S.) Let us learn about the conversion of some more units of time. Consider the following examples. Example 8: Roopa travels for 3 h 30 min each day while her sister travels for 3840 seconds. Who travels for a longer duration? Solution: Time for which Roopa travels = 3 h 30 min Time for which her sister travels = 3840 sec = 3840 min ÷ 60 = 64 min We know that, 60 min is 1 hr. So, 64 min = 1 h 4 min. 3 h 30 min > 1 h 4 min Therefore, Roopa travels for a longer duration. Example 9: Sita takes 5 days 6 hours and 15 minutes to complete her Science project. How much time (in seconds) does she take to complete the project? Solution: Time taken by Sita to complete her science project = 5 days 6 hours and 15 minutes We know that 1 day = 24 hours 1 hour = 60 minutes So, 5 days = 5 × 24 hours = 120 hours. 120 hours + 6 hours = 126 hours = 126 hours × 60 = 7560 minutes Time 93

To find the total time taken in seconds, we know that 1 minute = 60 seconds. So, 7560 minutes + 15 minutes = 7575 minutes = 7575 × 60 = 454500 seconds. Therefore, Sita takes 454500 seconds to complete the project. 6.2 Add and Subtract Time I Think In a week, Pooja spends 3 hours playing football, 160 minutes playing basketball and 1 hour 10 minutes playing tennis. Do you know how much time she spends playing? I Recall We have learnt about the conversion of hours to minutes, minutes to seconds and vice versa. Let us recall them. Complete the given table. Hours Minutes Seconds 2 240 360 13 28800 I Remember and Understand Let us now understand addition and subtraction of time through some examples. While adding time, we add the minutes (smaller units) If the sum of the minutes is 60, first and then the hours (larger units). we convert it to 1 hour and add it to the hours. Sometimes, we may have to regroup the sum of minutes. 94

Let us see an example. Example 10: Add: 1 hour 35 minutes and 2 hours 45 minutes Solution: Steps Solved Solve these Step 1: Write both the numbers one below the other. Hours Minutes Hours Minutes 1 35 Step 2: Add hours and minutes 45 separately, regrouping if needed. +2 Hours Minutes 1 20 1 35 +3 50 +2 45 3 80 Step 3: Check whether the minutes 80 minutes > 60 minutes Hours Minutes in the sum is greater than or equal to 60. If yes, convert it into hours. Step 4: Add the hours obtained in 3 hours 80 minutes 2 30 Step 3 to the hours obtained in Step        +2 20 2. Write the sum under hours. 1 hour 60 + 20 4 hours 20 minutes The sum is 4 hours 20 minutes. While subtracting, we subtract the minutes first (smaller units) and then the hours (larger units). Sometimes, we may have to regroup the hours. Let us see an example. Example 11: Subtract: 2 hours 35 minutes from 3 hours 10 minutes Solution: Solved Solve these Steps Hours Minutes Hours Minutes 3 45 Step 1: Write both the numbers 3 10 20 one below the other, such –1 that the smaller number is –2 35 subtracted from the larger. Time 95

Steps Solved Solve these Step 2: Subtract hours and 10 minutes < 35 minutes. Hours Minutes minutes separately, regroup if So, borrow 1 hour, that needed. is, 60 minutes and add it to the minutes. (10 + 60 = 70) 4 20 –2 40 Step 3: Reduce the hours by Hours Minutes 1 and subtract the minutes as usual. 2 70 Step 4: Subtract the hours and –2 35 write the difference. 35 Hours Minutes Hours Minutes 5 30 2 70 –3 35 –2 35 0 35 The difference is 35 min. Example 12: Subtract 4 h 32 min from 400 min. Hours Minutes Solution: We first convert 400 min to hours and minutes. So, 400 min = 400 ÷ 60 h 6 40 = 6 h 40 min –4 32 28 Subtracting 4 h 32 min from 6 h 40 min Therefore, the difference is 2 h 8 min. ? Train My Brain Solve the following: a) 5 h 40 min + 6 h 25 min b) 11 h 45 min – 10 h 15 min c) 14 h 30 min - 12 h 45 min I Apply Let us solve some examples where addition and subtraction of time are mostly used. 96