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

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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

Correct Answer: C. 111

Answer Explanation:

Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
So, 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).
So, The 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

Correct Answer: A. 40

Answer Explanation:

Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
So, 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

Correct Answer: C. 0.18

Answer Explanation:

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

Correct Answer: B. 2

Answer Explanation:

Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
So, 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

Correct Answer: D. 364

Answer Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.
Let the 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.
So, Required number = (90 x 4) + 4 = 364.

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