Recent questions tagged lcm-&-hcf

Description : The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A.10 B.14 C.23 D.30 E.None of these

Last Answer : Answer – C (23) Explanation – L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23

Description : The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A.1 B.2 C.3 D.5 E.None of these

Last Answer : Answer – B (2) Explanation – Let the numbers 13a and 13b. Then, 13a x 13b = 2028 ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4).

Description : The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Last Answer : Answer – C (111) Explanation – Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

Description : The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A.534 B.486 C.544 D.548 E.None of these

Last Answer : Answer – D (548) Explanation – Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

Description : The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A.30 B.22 C.40 D.60 E.None of these

Last Answer : Answer – C (40) Explanation – Let the numbers be 2x and 3x. Then, their L.C.M. = 6x. So, 6x = 48 or x = 8. The numbers are 16 and 24. Hence, required sum = (16 + 24) = 40.

Description : A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point ... A.15 minutes 15 seconds B.42 minutes 30 seconds C.42 minutes D.46 minutes 12 seconds E.None of these

Last Answer : Answer – D (46 minutes 12 seconds) Explanation – L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 see i.e., 46 min. 12 sec

Description : The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A.269 B.275 C.308 D.310 E.None of these

Last Answer : Answer – C (308) Explanation – Other number =[11 x 7700]/275 = 308

Description : The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: A.11115 B.15110 C.15130 D.15310 E.None of these

Last Answer : Answer – B (15110) Explanation – Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10 = 15120 – 10 = 15110

Description : The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A.68 B.98 C.180 D.364 E.None of these

Last Answer : Answer – D (364) Explanation – L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4 = 364.

Description : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Last Answer : Answer – D (16) Explanation – L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16