L.C.M and H.C.F : Aptitude Test

1.

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

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

Answer: Option 'C'

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.

2.

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

   A.) 5
   B.) 6
   C.) 7
   D.) 4

Answer: Option 'D'

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

3.

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.) 40
   C.) 38
   D.) 48

Answer: Option 'B'

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.

4.

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.) 4
   B.) 8
   C.) 12
   D.) 16

Answer: Option 'D'

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30/2 + 1 = 16 times

5.

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.) 46 minutes and 12 seconds
   C.) 42 minutes and 36 seconds
   D.) 45 minutes

Answer: Option 'B'

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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