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

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

6. 

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A. 101
B. 107
C. 111
D. 185

Answer: Option C

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.


7. 

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A. 40
B. 80
C. 120
D. 200

Answer: Option A

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.


8. 

The G.C.D. of 1.08, 0.36 and 0.9 is:

A. 0.03
B. 0.9
C. 0.18
D. 0.108

Answer: Option C

Explanation:

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

H.C.F. of given numbers = 0.18.


9. 

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. 4

Answer: Option B

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).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.


10. 

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

A. 74
B. 94
C. 184
D. 364

Answer: Option D

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.





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