Aptitude - Problems on H.C.F and L.C.M
Overview Exercise "I never think of the future. It comes soon enough."
- Albert Einstein
6.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Answer: E
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).
7.
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
Answer: A
Explanation:
Let the numbers be 3x , 4x and 5x .
Then, their L.C.M. = 60x .
So, 60x = 2400 or x = 40.
Hence, required H.C.F. = 40.
8.
The G.C.D. of 1.08, 0.36 and 0.9 is:
Answer: C
Explanation:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
9.
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Answer: D
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:
Answer: B
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.
ab = 3.
Greater number = 111.