# Aptitude - Problems on H.C.F and L.C.M

Exercise : Problems on H.C.F and L.C.M - General Questions

- Problems on H.C.F and L.C.M - Formulas
- Problems on H.C.F and L.C.M - General Questions

11.

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

Answer: Option

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:

Answer: Option

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.

Reduce | 128352 | to its lowest terms. |

238368 |

Answer: Option

Explanation:

128352) 238368 ( 1 128352 --------------- 110016 ) 128352 ( 1 110016 ------------------ 18336 ) 110016 ( 6 110016 ------- x ------- So, H.C.F. of 128352 and 238368 = 18336. 128352 128352 ÷ 18336 7 Therefore, ------ = -------------- = -- 238368 238368 ÷ 18336 13

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:

Answer: Option

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840*k + 3*

Least value of *k* for which (840*k* + 3) is divisible by 9 is *k* = 2.

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 ?

Answer: Option

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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