Problems on H.C.F and L.C.M

11. Find the lowest common multiple of 24, 36 and 40.

A. 120
B. 240
C. 360
D. 480

Correct Answer: C. 360

Answer Explanation:

2 | 24 – 36 – 40
——————–
2 | 12 – 18 – 20
——————–
2 | 6 – 9 – 10
——————-
3 | 3 – 9 – 5
——————-
| 1 – 3 – 5

L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

12. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A. 3
B. 13
C. 23
D. 33

Correct Answer: C. 23

Answer 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.

13

Correct Answer: Option C

Answer Explanation:

lowest terms

14. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A. 1677
B. 1683
C. 2523
D. 3363

Correct Answer: B. 1683

Answer Explanation:

L.C.M. of 5, 6, 7, 8 = 840.
So, the required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
So, Required number = (840 x 2 + 3) = 1683.

15. 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. After what time will they again at the starting point?

A. 26 minutes and 18 seconds
B. 42 minutes and 36 seconds
C. 45 minutes
D. 46 minutes and 12 seconds

Correct Answer: D. 46 minutes and 12 seconds

Answer Explanation:

L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

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