Grade 6 LCM and HCF. Answer the questions. Choose correct answer(s) from the given choices. For more such worksheets visit

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1 ID : eu-6-lcm-and-hcf [1] Grade 6 LCM and HCF For more such worksheets visit Answer the questions (1) Find the greatest number that divides 1283, 402 and 767 leaving remainders 9, 10, and 11, respectively. (2) 4 bells ring at intervals of 54, 324, 90, 63 seconds, respectively. If the bells ring together at 2 o'clock, when will they ring together again? How many times will they ring together in 3402 minutes? (3) Find the largest number that divides 357 and 120 leaving remainders 2 and 5, respectively. (4) A rectangular courtyard with length 3 m 42 cm and breadth 2 m 4 cm is to be paved with square stones of the same size. Find the least number of such stones required. (5) Two brands of muffins are available in the packs of 48 and 24, respectively. Lera wants to buy the same number of muffins of both the brands. What is the least number of packs of each brand of muffins that she will need to buy? (6) Shell-Waters is a comet that orbits around the sun once in 70 years, and Ingravio-Tage is a comet that orbits around the sun once in 56 years. The last time they were seen together in the sky was in In which year will they be next seen together in the sky? (7) What is the largest number that divides 1028 and 115 leaving a remainder of 5? (8) Find the LCM (least common multiple) of the following numbers 36, 84, 6, 21, 4 (9) Two tankers contain 477 liters and 153 liters of petrol, respectively. Find the capacity of the largest measuring container which can measure the petrol of each tanker exactly? (10) Determine the two numbers nearest to which are exactly divisible by 8, 9, 5, 6, 2, 3 and 4. (11) Find the highest common factor (HCF) of following: A) 100, 96 B) 198, 330 C) 300, 110 D) 64, 112 E) 270, 110 F) 36, 54 (12) Determine the two numbers nearest to which are exactly divisible by 7, 5, 8, 4, 3, 2 and 6. Choose correct answer(s) from the given choices (13) If a number is divisible by 25 and 10, it will be necessarily divisible by : a. 48 b. 52 c. 55 d. 50

2 ID : eu-6-lcm-and-hcf [2] Fill in the blanks (14) The least common multiple (LCM) of 48 and 54 is. (15) The least common multiple (LCM) of 14 and 56 is Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : eu-6-lcm-and-hcf [3] (1) 14 We know that the greatest number that divides 1283, 402, and 767 leaving remainders 9, 10, and 11, respectively is the same number that divides the numbers (1283-9), (402-10), and (767-11) leaving no remainder. In other words, we need to find the H.C.F of the following three numbers: 1274 [Simplify ], 392 [Simplify ], 756 [Simplify ]. Now, let us find the H.C.F using the standard method. Let's find the prime factors of 1274 using the standard method: is a factor of is a factor of is a factor of is a factor of 13 Thus, 1274 = Let's find the prime factors of 392 using the standard method: is a factor of is a factor of is a factor of is a factor of is a factor of 7 Thus, 392 = Let's find the prime factors of 756 using the standard method: is a factor of is a factor of is a factor of is a factor of 63

4 is a factor of 21 ID : eu-6-lcm-and-hcf [4] is a factor of 7 Thus, 756 = Step 6 So, the H.C.F of 1274, 392, and 756 = 14 Step 7 Therefore, the greatest number which divides 1283, 402, and 767 leaving remainders 9, 10, and 11, respectively is 14.

5 (2) 5:09:00, 18 ID : eu-6-lcm-and-hcf [5] The 4 bells ring after intervals of 54, 324, 90, 63 seconds, respectively. If the first bell rings just now, it will ring again after : 54 1 = 54 seconds 54 2 = 108 seconds 54 3 = 162 seconds... and so on. This means, the number of seconds after which the 1 st bell rings will be a multiple of 54. Similarly, the number of seconds after which the 2 nd, 3 rd and 4 th bells ring will be a multiple of 324, 90 and 63, respectively. This means, the number of seconds after which all the 4 bells ring together must be a common multiple of all the 4: 54, 324, 90, 63. We need to find the very first time after 2 o'clock, when the 4 bells will ring together. This will happen after a certain number of seconds past 2 O'clock, which is the least common multiple (L.C.M) of 54, 324, 90, 63. The L.C.M of 54, 324, 90, 63 is seconds seconds in minutes = = 189 minutes. 60 Therefore, all the 4 bells will ring together at 189 minutes past 2 o'clock, i.e., at 5:09:00. Step 6 We just found out that the 4 bells ring together every 189 minutes. Therefore, the number of times they will ring together in 3402 minutes = 3402 = 18 times. 189

6 (3) 5 ID : eu-6-lcm-and-hcf [6] We need to find the largest number that divides 357 leaving a remainder of 2 and divides 120 leaving a remainder of 5. Such a number is the H.C.F of (357-2) and (120-5), i.e., the H.C.F of 355 and 115. Let us now find the H.C.F of 355 and 115. All the prime factors of 355 : is a factor of is a factor of 71 Thus, 355 = All the prime factors of 115 : is a factor of is a factor of 23 Thus, 115 = Step 6 Thus, the H.C.F of 355 and 115 is = 5 = 5. Step 7 Therefore, the largest number which divides 357 and 120 leaving remainders 2 and 5, respectively, is 5.

7 (4) 1938 ID : eu-6-lcm-and-hcf [7] According to the question, the length and breadth of the courtyard is 3 m 42 cm and 2 m 4 cm respectively. Let us first convert each dimension into centimeters. Since we know that 1m = 100 cm, the length and breadth of the courtyard in centimeters is 342 cm and 204 cm respectively. The pavement has to be done with square stones. To perfectly fit the courtyard, the size of the square stones should be a factor of both the length and breadth of the courtyard. Please refer to this figure 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 factor (HCF) of 342 cm and 204 cm, the length and breadth of the courtyard in centimeters. The HCF of 342 and 204 is 6. Thus, the needed size of square stones = 6 cm The number of stones needed = Area of the couryard Area of a single stone = 6 6 = 1938 The least number of such stones required are 1938.

8 (5) 1 pack of the first brand, 2 packs of the second brand. ID : eu-6-lcm-and-hcf [8] Lera wants to buy equal number of muffins of two brands, but she needs to buy them in the multiples of 48 and 24, 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 48 and 24. L.C.M of 48 and 24 = 48. Number of packs of the first brand Lera needs to buy = Number of muffins of the first brand Number of muffins in one pack of the first brand = = 1 Number of boxes of the second brand Lera needs to buy= Number of muffins of the second brand Number of muffins in one pack of the second brand = = 2 Hence, she needs to buy 1 pack of the first brand and 2 packs of the second brand.

9 (6) 2275 ID : eu-6-lcm-and-hcf [9] We have been told that the comet Shell-Waters orbits around the sun once in 70 years, and the comet Ingravio-Tage orbits around the sun once in 56 years. Once the two comets are seen together, the number of years after which they will be seen together in the sky is equal to the L.C.M of 70 and 56. Let us now calculate the L.C.M of 70 and 56. All the prime factors of 70 : is a factor of is a factor of is a factor of 7 Thus, 70 = All the prime factors of 56 : is a factor of is a factor of is a factor of is a factor of 7 Thus, 56 = Thus, the L.C.M of 70 and 56 is = = 280. Step 6 The year when they were last seen together in the sky : Therefore, the year in which they will be next seen together in the sky : , or 2275.

10 (7) 11 ID : eu-6-lcm-and-hcf [10] We have to find the largest number that divides 1028 and 115 leaving a remainder of 5. In other words, we have to find the largest number that divides (1028-5) and (115-5) leaving no remainder. Such a number is the Highest Common Factor (H.C.F) of : 1023 [i.e ] and 110 [i.e ]. Now, let us find the H.C.F of 1023 and 110. All the prime factors of 1023 : is a factor of is a factor of is a factor of 31 Therefore, 1023 = All the prime factors of 110 : is a factor of is a factor of is a factor of 11 Therefore, 110 = The H.C.F of 1023 and 110 = 11. Step 6 Hence, the largest number which divides 1028 and 115 leaving a remainder of 5 is 11.

11 (8) 252 ID : eu-6-lcm-and-hcf [11] Let us find the LCM of 36, 84, 6, 21, 4: 2 84, 36, 21, 6, , 18, 21, 3, , 9, 21, 3, 1 3 7, 3, 7, 1, 1 7 7, 1, 7, 1, 1, 1, 1, 1, 1 The LCM is = = 252 (9) 9 liters The container which can measure petrol in both the tanks should be such that its volume in liters should fully divide 477 and 153. Therefore, the capacity of the largest measuring container which can measure the petrol of each tanker exactly is the H.C.F of 477 and 153. Let us find all the prime factors of 477 : is a factor of is a factor of is a factor of = Let us now find all the prime factors of 153 : is a factor of is a factor of is a factor of = So, the H.C.F of 477 and 153 = 9 Therefore, the largest measuring container which can measure the petrol of each tanker exactly will have a capacity of 9 liters.

12 (10) , ID : eu-6-lcm-and-hcf [12] Let us first find the smallest number that is exactly divisible by the numbers 8, 9, 5, 6, 2, 3 and 4. Such number will be the LCM of the numbers 8, 9, 5, 6, 2, 3 and 4. Let us find the LCM of 8, 9, 5, 6, 2, 3 and 4: 2 9, 8, 6, 5, 4, 3, 2 2 9, 4, 3, 5, 2, 3, 1 2 9, 2, 3, 5, 1, 3, 1 3 9, 1, 3, 5, 1, 3, 1 3 3, 1, 1, 5, 1, 1, 1 5, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1 The LCM is = = 360 Now, the other numbers that are exactly divisible by 8, 9, 5, 6, 2, 3 and 4 will have to be the multiples of their LCM. So, we will have to find the multiples of 360 that are nearest to On dividing by 360, we get a remainder of 80. Hence = and = , both will be divisible by 360. Hence, the required numbers are and

13 (11) A) 4 ID : eu-6-lcm-and-hcf [13] Let us find the HCF of 100 and 96: Factors of 100 = Factors of 96 = The HCF of 100 and 96 is 2 2 = 4. B) 66 Let us find the HCF of 198 and 330: Factors of 198 = Factors of 330 = The HCF of 198 and 330 is = 66.

14 C) 10 ID : eu-6-lcm-and-hcf [14] Let us find the HCF of 300 and 110: Factors of 300 = Factors of 110 = The HCF of 300 and 110 is 2 5 = 10. D) 16 Let us find the HCF of 64 and 112: Factors of 64 = Factors of 112 = The HCF of 64 and 112 is = 16.

15 E) 10 ID : eu-6-lcm-and-hcf [15] Let us find the HCF of 270 and 110: Factors of 270 = Factors of 110 = The HCF of 270 and 110 is 2 5 = 10. F) 18 Let us find the HCF of 36 and 54: Factors of 36 = Factors of 54 = The HCF of 36 and 54 is = 18.

16 (12) 79800, ID : eu-6-lcm-and-hcf [16] Let us first find the smallest number that is exactly divisible by the numbers 7, 5, 8, 4, 3, 2 and 6. This number will be the LCM of the numbers 7, 5, 8, 4, 3, 2 and 6. Let us find the LCM of 7, 5, 8, 4, 3, 2 and 6: 2 8, 7, 6, 5, 4, 3, 2 2 4, 7, 3, 5, 2, 3, 1 2 2, 7, 3, 5, 1, 3, 1 3, 7, 3, 5, 1, 3, 1 5, 7, 1, 5, 1, 1, 1 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 The LCM is = = 840 Now, the other numbers that are exactly divisible by 7, 5, 8, 4, 3, 2 and 6 will have to be the multiples of their LCM. So, we will have to find the multiples of 840 that are nearest to On dividing by 840, we get a remainder of 200. Hence either = and = 80640, both will be divisible by 840. Hence, the required numbers are and (13) d. 50 If a number is divisible by two different numbers, it is necessarily divisible by their L.C.M. L.C.M of 25 and 10 = 50. Thus, if a number is divisible by 25 and 10, it will be necessarily divisible by 50.

17 ID : eu-6-lcm-and-hcf [17] (14) 432 Let us find the LCM of 48 and , , , , , 3 3 9, 1 3 3, 1, 1 The LCM is = = 432. (15) 56 Let us find the LCM of 14 and 56: 2 56, , 7 2 4, 7 7 7, 7, 1 The LCM is = = 56