Class 6 Natural and Whole Numbers

ID : in-6-natural-and-whole-numbers [1] Class 6 Natural and Whole Numbers For more such worksheets visit www.edugain.com Answer the questions (1) Find the largest 3-digit number which is exactly divisible by 66. (2) What is the largest number that divides 965 and 380 leaving a remainder of 5? (3) Find the largest number that divides 919 and 313 leaving remainders 7 and 9, respectively. (4) Find the largest number that will divide 153, 69, and 117 leaving the same remainder in each case. (5) Determine the two numbers nearest to 200000 which are exactly divisible by 3, 6, 5 and 7. (6) Two brands of candies are available in the packs of 63 and 35, respectively. Priyanka wants to buy the same number of candies of both the brands. What is the least number of packs of each brand of candies that she will need to buy? (7) The set of consecutive odd numbers starting from 1 till N (i.e. 1,3, 5, 7...N) has a sum of 441. What is the value of N? (8) Two tankers contain 648 liters and 88 liters of petrol, respectively. Find the capacity of the largest measuring container which can measure the petrol of each tanker exactly? (9) Meenakshi has two sheets of paper. One sheet is 456 inches wide and the other sheet is 120 inches wide. She wants to divide the sheets into strips of equal width without wasting any paper such that they are as wide as possible. How wide should she cut the strips? (10) Raj and Balvinder are friends as well as football coaches in the same school. Raj goes to Apeejay school every 4 days while Balvinder goes there every 7 days to deliver coaching classes. If they both delivered the coaching sessions today, how many days, in the next 140 days, will both of them be delivering the coaching sessions on the same day? (11) A lighthouse has two lights - one that flashes every 3 minutes and another that flashes every 1 1 2 minutes. Suppose the lights flash together at noon. What is the first time after 3 PM when they will flash together again? Choose correct answer(s) from the given choices (12) The product of two 2-digit numbers is 1116. If the product of their units digits is 16 and that of their tens digits is 6, then the two numbers are : a. 68, 12 b. 61, 28 c. 21, 86 d. 62, 18 (13) The number 289 can be represented by a 17 x 17 square grid. Out of the following numbers, which number cannot be represented by such a square grid? a. 144 b. 729 c. 422 d. 441

ID : in-6-natural-and-whole-numbers [2] (14) The product of any two numbers is equal to product of their a. HCF and difference of the two numbers b. LCM and the first number c. LCM and HCF d. HCF and the sum of the two numbers Check True/False (15) There is a whole number which does not change the value of any other whole number it is added to. True False 2017 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com

Answers ID : in-6-natural-and-whole-numbers [3] (1) 990 We know that the largest 3-digit number is 999. From the division formula, we know that dividend = quotient divisor + remainer. Thus, if we subtract the remainder from the dividend, the resulting number will be perfectly divisible by the divisor. When 999 is divided by 66, the remainder is 9. Hence, the required number is 999-9 = 990.

(2) 15 ID : in-6-natural-and-whole-numbers [4] We have to find the largest number that divides 965 and 380 leaving a remainder of 5. In other words, we have to find the largest number that divides (965-5) and (380-5) leaving no remainder. Such a number is the Highest Common Factor (H.C.F) of : 960 [i.e. 965-5] and 375 [i.e. 380-5]. Now, let us find the H.C.F of 960 and 375. All the prime factors of 960 : 2 960 2 is a factor of 960 2 480 2 is a factor of 480 2 240 2 is a factor of 240 2 20 2 is a factor of 120 2 60 2 is a factor of 60 2 30 2 is a factor of 30 3 5 3 is a factor of 15 5 5 5 is a factor of 5 Therefore, 960 = 2 2 2 2 2 2 3 5. All the prime factors of 375 : 3 375 3 is a factor of 375 5 25 5 is a factor of 125 5 25 5 is a factor of 25 5 5 5 is a factor of 5 Therefore, 375 = 3 5 5 5. The H.C.F of 960 and 375 = 15. Step 6 Hence, the largest number which divides 965 and 380 leaving a remainder of 5 is 15.

(3) 304 ID : in-6-natural-and-whole-numbers [5] We need to find the largest number that divides 919 leaving a remainder of 7 and divides 313 leaving a remainder of 9. Such a number is the H.C.F of (919-7) and (313-9), i.e., the H.C.F of 912 and 304. Let us now find the H.C.F of 912 and 304. All the prime factors of 912 : 2 912 2 is a factor of 912 2 456 2 is a factor of 456 2 228 2 is a factor of 228 2 14 2 is a factor of 114 3 57 3 is a factor of 57 19 9 19 is a factor of 19 Thus, 912 = 2 2 2 2 3 19. All the prime factors of 304 : 2 304 2 is a factor of 304 2 52 2 is a factor of 152 2 76 2 is a factor of 76 2 38 2 is a factor of 38 19 9 19 is a factor of 19 Thus, 304 = 2 2 2 2 19. Step 6 Thus, the H.C.F of 912 and 304 is = 2 2 2 2 19 = 304. Step 7 Therefore, the largest number which divides 919 and 313 leaving remainders 7 and 9, respectively, is 304. (4) 12

ID : in-6-natural-and-whole-numbers [6] It is given that when 153, 69, and 117 are divided by a number, they leave the same remainder. It means that if we find the difference between these numbers, the remainder will cancel out and the difference of these numbers should be fully divisible by the same number. Let us find the difference between the given numbers : 153-69 = 84 153-117 = 36 69-117 = 48 Now, we have to find the largest number which fully divides 84, 36, and 36. We know that the H.C.F of 84, 36, and 48 is the largest number that will divide 153, 69, and 117, leaving the same remainder in each case. All the prime factors of 84 : 2 84 2 is a factor of 84 2 42 2 is a factor of 42 3 21 3 is a factor of 21 7 7 7 is a factor of 7 84 = 2 2 3 7 All the prime factors of 36 : 2 36 2 is a factor of 36 2 8 2 is a factor of 18 3 9 3 is a factor of 9 3 3 3 is a factor of 3 36 = 2 2 3 3 Step 6 All the prime factors of 48 : 2 48 2 is a factor of 48 2 24 2 is a factor of 24 2 2 2 is a factor of 12 2 6 2 is a factor of 6 3 3 3 is a factor of 3

48 = 2 2 2 2 3 ID : in-6-natural-and-whole-numbers [7] Step 7 The H.C.F of 84, 36, and 48 = 2 2 3 = 12 Step 8 Therefore, 12 is the largest number that will divide 153, 69, and 117, leaving the same remainder in each case. (5) 199920, 200130 Let us first find the smallest number that is exactly divisible by the numbers 3, 6, 5 and 7. This number will be the LCM of the numbers 3, 6, 5 and 7. Let us find the LCM of 3, 6, 5 and 7: 2 7, 6, 5, 3 3 7, 3, 5, 3 5 7, 1, 5, 1 7 7, 1, 1, 1, 1, 1, 1 The LCM is = 2 3 5 7 = 210 Now, the other numbers that are exactly divisible by 3, 6, 5 and 7 will have to be the multiples of their LCM. So, we will have to find the multiples of 210 that are nearest to 200000. On dividing 200000 by 210, we get a remainder of 80. Hence either 200000-80 = 199920 and 200000-80 + 210 = 200130, both will be divisible by 210. Hence, the required numbers are 199920 and 200130.

(6) 5 packs of the first brand, 9 packs of the second brand. ID : in-6-natural-and-whole-numbers [8] Priyanka wants to buy equal number of candies of two brands, but she needs to buy them in the multiples of 63 and 35, respectively. Also, she wants to buy the least number of packs. Thus, the number of chocolates of each brand she needs to buy = L.C.M of 63 and 35. L.C.M of 63 and 35 = 315. Number of packs of the first brand Priyanka needs to buy = Number of candies of the first brand Number of candies in one pack of the first brand = 315 63 = 5 Number of boxes of the second brand Priyanka needs to buy= Number of candies of the second brand Number of candies in one pack of the second brand = 315 35 = 9 Hence, she needs to buy 5 packs of the first brand and 9 packs of the second brand. (7) 41 Let us assume that the count of consecutive odd numbers from 1 till N is n. Then, the value of N = 2n - 1. According to the question, the sum of the consecutive odd numbers from 1 till N = 441. We know that the sum of the first n odd numbers is equal to the square of n. Therefore, n 2 = 441 or n 2 = 21 2 or n = 21 Thus, the value N = 2n - 1 = (2 21) - 1 = 41

(8) 8 liters ID : in-6-natural-and-whole-numbers [9] The container which can measure petrol in both the tanks should be such that its volume in liters should fully divide 648 and 88. Therefore, the capacity of the largest measuring container which can measure the petrol of each tanker exactly is the H.C.F of 648 and 88. Let us find all the prime factors of 648 : 2 648 2 is a factor of 648 2 324 2 is a factor of 324 2 62 2 is a factor of 162 3 81 3 is a factor of 81 3 27 3 is a factor of 27 3 9 3 is a factor of 9 3 3 3 is a factor of 3 648 = 2 2 2 3 3 3 3 Let us now find all the prime factors of 88 : 2 88 2 is a factor of 88 2 44 2 is a factor of 44 2 22 2 is a factor of 22 11 1 11 is a factor of 11 88 = 2 2 2 11 So, the H.C.F of 648 and 88 = 8 Therefore, the largest measuring container which can measure the petrol of each tanker exactly will have a capacity of 8 liters.

(9) 24 inches ID : in-6-natural-and-whole-numbers [10] Let the width of strip be x inches. Since, no cloth should be wasted, 456 inches should be divisible by x inches. Similarly, 120 inches should be divisible by x inches. Also, we should remember that x has to be as large as possible. Therefore, x should be the HCF of 456 and 120. Hence, x = HCF(456, 120) = 24 inches. (10) 5 Given : Raj and Balvinder both delivered their coaching sessions today. Raj delivers coaching sessions every 4 th day. This means he will deliver sessions again after all these days: 4, 8, 12..., and so on (multiples of 4). Balvinder delivers coaching sessions every 7 th day. This means he will deliver sessions again after all these days: 7, 14, 21..., on so on (multiples of 7). In order to find the number of days after which both of them will go together, we need to find a number that is as small as possible and is divisible by both 4 and 7. In other words, we need to calculate the LCM of 4 and 7 which is: 28 days. Now, we know that both of them go to school together every 28 th day. Therefore, the number of days, in the next 140 days, when both of them will deliver sessions on the same day = 140 = 5 days 28 (11) 3:03 PM The first light flashes every 3 minutes and the other light flashes every 1 1 2 minutes. Once the two lights flash together, the amount of time after which the two lights will flash together again is equal to the L.C.M of 3 minutes and 1 1 2 minutes.

ID : in-6-natural-and-whole-numbers [11] Before we calculate the L.C.M of 3 minutes and 1 1 2 minutes, let us convert both the time periods into seconds. Since, 1 minute = 60 seconds, 3 minutes = 3 60 = 180 seconds, and 1 1 2 minutes = 3 2 60 = 90 seconds. Let us now calculate the L.C.M of 180 and 90. All the prime factors of 180: 2 80 2 is a factor of 180 2 90 2 is a factor of 90 3 45 3 is a factor of 45 3 5 3 is a factor of 15 5 5 5 is a factor of 5 Thus, 180 = 2 2 3 3 5. All the prime factors of 90: 2 90 2 is a factor of 90 3 45 3 is a factor of 45 3 5 3 is a factor of 15 5 5 5 is a factor of 5 Thus, 90 = 2 3 3 5. The L.C.M of 180 and 90 = 2 3 3 5 2 = 180. Step 6 180 seconds in minutes = 180 60 minutes = 3 minutes. Step 7 Now, we know that the two lights flash together every 3 minutes. We have been told that the two lights flash together at noon. This means that times when they flash together again are 12:3, 12:6, 12:9, 12:12,... 3:03,... Therefore, the time after 3 PM when the two lights will flash together again = 3:03 PM.

(12) d. 62, 18 ID : in-6-natural-and-whole-numbers [12] Let us assume that the two two-digit numbers are, 10 x 1 + y1 and 10 x 2 + y2, where y1 and y2 are their units digits and, x1 and x2 are their tens digits. Given : y1 y2 = 16 -----(1) x1 x2 = 6 -----(2) (10 x 1 + y1) (10 x 2 + y2) = 1116 -----(3) Now, we notice that there are four variables and only three equations have been specified. We know that four variables cannot be uniquely determined from three equations. Therefore, we will check the given options to find which of them satisfies the above three equations. On checking all the options carefully, we find that the option 62, 18 satisfies all the three equations. 2 8 = 16 6 1 = 6 62 18 = 1116 Therefore, we can say that the two numbers are 62, 18. (13) c. 422 We know that only a number which is a perfect square can be represented on a square grid and the number which is not a perfect square cannot be represented on a square grid. A perfect square is a natural number which is the square of some natural number. In other words, only a natural number whose square root is a natural number is a perfect square. Now, if we look at all the options carefully, we notice that only the number 422 is not a perfect square. All the other numbers are perfect squares (squares of the numbers 17, 21, or 27). Therefore, we can say that the number 422 cannot be represented on a square grid.

(14) c. LCM and HCF ID : in-6-natural-and-whole-numbers [13] Let us consider two random numbers 4 and 6. Their HCF is equal to 2 and LCM is 12. The product of these numbers is 4 6 = 24, and the product of their HCF and LCM is equal to 2 12 = 24. Therefore, we observe that the product of any two numbers is equal to the product of their HCF and LCM. (15) True We know that whole numbers are the numbers 0, 1, 2, 3... If we add the whole number '0' to any other whole number, we get that whole number itself as the sum. For example: 0 + 7 = 7 0 + 1 = 1 Hence, the answer is true.