Grade 6 LCM and HCF. Answer t he quest ions. Choose correct answer(s) f rom given choice. For more such worksheets visit

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1 ID : gb-6-lcm-and-hcf [1] Grade 6 LCM and HCF For more such worksheets visit Answer t he quest ions (1) Ingravio-Tage is a comet that orbits around the sun once in 56 years, and Sunerva-Primo is a comet that orbits around the sun once in 42 years. The last time they were seen together in the sky was in In which year will they be seen together in the sky next? (2) Find the highest common f actor (HCF) of f ollowing: A) 189, 84 B) 30, 33 C) 192, 276 D) 242, 187 E) 34, 38 F) 200, 224 (3) Determine the two numbers nearest to 6000 which are exactly divisible by 7, 8, 9, 3, 5, 6. (4) Finlay and Abigail are f riends and cricket coach too. Finlay goes to Modern school every 3 days and Abigail goes to Modern every 7 days to deliver coaching classes. If they both delivered the coaching sessions today, how many days in the next 126 days they both would take the session on the same day? (5) Which is the largest number that divides 530 and 302 leaving remainder 2 and 2 respectively? (6) What is the largest number that divides 313 and 145 leaving remainder 5? (7) 95 cola cans and 145 f ruit juice cans need to be stacked in a school canteen. If each stack is to be of the same height and is to contain cans of the same type, what would be the greatest number of cans each stack can have? Choose correct answer(s) f rom given choice (8) The product of any two numbers is equal to product of their a. LCM and the f irst number b. LCM and HCF c. HCF and the sum of the two numbers d. HCF and either of the two numbers (9) The lowest natural number which when divided by 10, 40, 6, 10 leaves the remainder of 4 in each case is a. 124 b. 123 c. 129 d. none of these

2 ID : gb-6-lcm-and-hcf [2] Fill in the blanks (10) T wo tankers contain 150 litres and 140 litres of petrol respectively. T he largest measuring container which can measure the petrol of either tanker exactly will have a capacity of litres. (11) The length, breadth and height of a room are 11 m 40 cm, 5 m 28 cm and 7 m 8 cm respectively. The length of the longest rod which can measure the dimensions of the room exactly is cm (12) A rectangular courtyard with length 5 m 49 cm and breadth 3 m 60 cm is to be paved with square stones of the same size. The least number of such stones required are. (13) The least common multiple (LCM) of 18 and 72 is. (14) Two brands of chocolates are available in packs of 72 and 24 respectively. If Abigail wants to buy equal number of chocolates of each brands, then the least number of boxes of each chocolate brand that she needs to buy are and respectively. (15) Kian saves 34.8 every day. The minimum number of days in which he will be able to save an exact amount of Rupees is Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : gb-6-lcm-and-hcf [3] (1) 2167 We have been told that the comet Ingravio-Tage orbits around the sun once in 56 years, and the comet Sunerva-Primo orbits around the sun once in 42 years. Once the two comets are seen together, the number of years af ter which they will be seen together in the sky is equal to the LCM of 56 and 42. Let us now calculate the LCM of 56 and 42. All prime f actors of 56: is a factor of is a factor of is a factor of is a factor of 7 Thus, 56 = All prime f actors of 42: is a factor of is a factor of is a factor of 7 Thus, 42 = Thus, the LCM of 56 and 42 is = = 168. Step 6 The year when they were least seen together in the sky = Theref ore, the year in which they will be seen together in the sky next = = (2) A) 21 B) 3 C) 12 D) 11 E) 2

4 F) 8 ID : gb-6-lcm-and-hcf [4] (3) 5040, 7560 (4) 6 Finlay goes to Modern every 3 rd days and Abigail goes to Modern every 7 th days. So like this they will both will take session on every 21 days. In order to f ind the number of days af ter which both of them goes together, we need to f ind a number which is as small as possible and divisible by both 3 and 7. Theref ore we need to f ind LCM of 3 and 7 which is 21 days. Now since both of them goes to school every 21 st days. Theref ore number of session in next 126 days they will take = 126 / 21 = 6

5 (5) 12 ID : gb-6-lcm-and-hcf [5] We need to f ind the largest number that divides 530 leaving remainder 2 and divides 302 leaving remainder 2. Such number is the H.C.F of (530-2) and (302-2), that is, the HCF of 528 and 300. Let us now f ind the H.C.F of 528 and 300. All prime f actors of 528: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 11 Thus, 528 = All prime f actors of 300: is a factor of is a factor of is a factor of is a factor of is a factor of 5 Thus, 300 = Step 6 The H.C.F of 528 and 300 is = = 12. Step 7 Theref ore, the largest number which divides 530 and 302 leaving remainders 2 and 2 respectively is 12.

6 (6) 28 ID : gb-6-lcm-and-hcf [6] We have to f ind the largest number that divides 313 and 145 leaving remainder 5. In other words, we have to f ind the largest number that divides (313-5) and (145-5) leaving no remainder. Such number is the Highest Common Factor (HCF) of : 308 [i.e., 313-5] and 140 [i.e., 145-5]. Let us now f ind the HCF of 308 and 140. All prime f actors of 308: is a factor of is a factor of is a factor of is a factor of 11 Theref ore, 308 = All prime f actors of 140: is a factor of is a factor of is a factor of is a factor of 7 Theref ore, 140 = The HCF of 308 and 140 is = = 28. Step 6 Thus, the largest number which divides 313 and 145 leaving remainder 5 is 28.

7 (7) 5 ID : gb-6-lcm-and-hcf [7] Since each stack needs to be of the same height and is to contain cans of the same type, the number of cans in each stack should be a f actor of both 95 and 145. Since the number of cans in each stack should additionally be greatest possible, it should be the Highest Common Factor(HCF) of 95 and 145. Lets f ind the HCF of 95 and 145. All prime f actors of 95: is a factor of is a factor of 19 Thus, 95 = 5 19 All prime f actors of 145: is a factor of is a factor of 29 Thus, 145 = 5 29 The HCF of 95 and 145 is = 5 = 5 Step 6 The greatest number of cans each stack can have 5. (8) b. LCM and HCF

8 (9) a. 124 ID : gb-6-lcm-and-hcf [8] The lowest number which is divisible by 10, 40, 6, 10, is the Least Common Multiple (LCM) of 10, 40, 6, 10 The LCM of 10, 40, 6, 10 = 120. Now we need to f ind the number which leaves the remainder of 4 when divided by these numbers. This number should be 4 more than the LCM. Theref ore, the required number = LCM (10, 40, 6, 10) + 4 = = 124

9 (10) 10 ID : gb-6-lcm-and-hcf [9] The container which can measure petrol of both tanks, should be such that its volume in litres should f ully divide 150 and 140. T heref ore, capacity of the largest measuring container which can measure the petrol of either tanker exactly is the HCF of 150 and 140. Let us f ind all prime f actors of 150: is a factor of is a factor of is a factor of is a factor of = Let us now f ind all prime f actors of 140: is a factor of is a factor of is a factor of is a factor of = The HCF of 150 and 140 is = 2 5 = 10 T heref ore, the largest measuring container which can measure the petrol of either tanker exactly will have a capacity of 10 liters.

10 (11) 12 ID : gb-6-lcm-and-hcf [10] According to the question, the length, breadth and height of the room are 11 m 40 cm, 5 m 28 cm and 7 m 8 cm respectively. Now you have to convert each dimension into same unit. Since you know that 1m = 100 cm, the length, breadth and height of the room in centimeters is 1140 cm, 528 cm and 708 cm respectively. The length of the rod which can be used to exactly measure all three dimensions of a room should be a f actor of all three dimensions, that is, a f actor common to all three dimensions. Since we have been asked to f ind the length of longest such rod, the length should be equal to HCF of all three dimensions of the room. Let us now f ind the HCF of 1140, 528 and 708. All prime f actors of 1140: is a factor of is a factor of is a factor of is a factor of is a factor of 19 Theref ore, 1140 = All prime f actors of 528: is a factor of is a factor of is a factor of is a factor of is a factor of is a factor of 11 Theref ore, 528 = Step 6 All prime f actors of 708: is a factor of 708

11 is a factor of 354 ID : gb-6-lcm-and-hcf [11] is a factor of is a factor of 59 Theref ore, 708 = Step 7 The HCF of 1140, 528 and 708 is = = 12. Step 8 Hence, the length of the longest rod which can measure the dimensions of the room exactly is 12 cm.

12 (12) 2440 ID : gb-6-lcm-and-hcf [12] According to the question, the length and breadth of the courtyard is 5 m 49 cm and 3 m 60 cm respectively. Let us f irst convert each dimension into centimeters. Since we know that 1m = 100 cm, the length and breadth of the courtyard in centimeters is 549 cm and 360 cm respectively. The pavement has to be done with square stones. To perf ectly f it the courtyard, the size of the square stones should be a f actor of both the length and breadth of the courtyard. Please ref er to this f igure to see why it is so. Since we need to minimise the number of stones needed, the size of the stones should be the largest possible. This means the size of the stones should be the highest common f actor (HCF) of 549 cm and 360 cm, the length and breadth of the courtyard in centimeters. The HCF of 549 and 360 is 9. Thus, the needed size of square stones = 9 cm The number of stones needed = Area of the couryard Area of a single stone = 9 9 = 2440 The least number of such stones required are (13) 72

13 (14) 1 3 ID : gb-6-lcm-and-hcf [13] Abigail wants to buy equal number of chocolates of two brands, but she needs to buy them in multiples of 72 and 24 respectively. Also, she wants to buy the least number of boxes. Thus, the number of chocolates of each brand she will need to buy = L.C.M of 72 and 24. L.C.M of 72 and 24 = 72. Number of boxes of the f irst brand Abigail will need to buy = Number of chocolates of the f irst brand he will need to buy Number of chocolates in one box of f irst brand = = 1 Number of boxes of the second brand Abigail will need to buy= Number of chocolates of the second brand he will need to buy Number of chocolates in one box of second brand = = 3 Thus, she will need to buy 1 box of f irst brand and 3 boxes of second brand.

14 (15) 5 ID : gb-6-lcm-and-hcf [14] We need to f ind the number of days af ter which he will have exact amount (no decimal) If we observer 34.8, it has two parts, 34 and 0.8 The f irst part 34 multiplied by any number of days will always result in f ull amount (no decimal), so we need not worry about this part Now we need to f ind the least number of days, which when multiplied to 0.8, results in a f ull amount. We know that this amount will be a multiple of 1 and will also be a multiple of 0.8. So smallest number which is divisible by both 1 and 0.8 will be HCF of 1 and 0.8. Since we are not taught calculating HCF of non-natural numbers, we can f irst calculate HCF of 100 and 80, and then divide the result by 100. HCF of 100 and 80 = 400 HCF of 1 and 0.8 = 400/100 = 4 Theref ore number of days required to save 4, by saving 0.8 daily, = 4 / 0.8 = 5 days