RRB NTPC - L.C.M and H.C.F : Aptitude Test

1.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 

   A.) 114
   B.) 294
   C.) 364
   D.) 194

Answer: Option 'C'

Answer: Option 'C'
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 × 4) + 4 = 364.

2.

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.) 267
   B.) 318
   C.) 308
   D.) 279

Answer: Option 'C'

Other number = (11 x 7700/275) = 308.

3.

Four prime numbers are arranged in ascending order according to their magnitude. Product of first Three is 20,677 and the product of last three 33,263 

   A.) 23
   B.) 29
   C.) 31
   D.) 37

Answer: Option 'D'

37

4.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

   A.) 28
   B.) 32
   C.) 40
   D.) 64

Answer: Option 'C'

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.

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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