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H.C.F & L.C.M. - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to H.C.F & L.C.M.. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - Reduce ¹²⁸³⁵²⁄₂₃₈₃₆₈ to its lowest terms.

Answer : A

Explanation

Q 2 - The ratio of two numbers is 3:4 and their H.C.F is 4. Their L.C.M is?

Answer : C

Explanation

Let the numbers be 3x and 4x.
Then, their H.C.F = z
So, z = 4 
 So, the numbers are 12 and 16. L.C.M of 12 and 16 = 48.

Q 3 - The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above conditions is?

Answer : A

Explanation

Let the numbers be 33a and 33b. then 33a + 33b = 528
a + b = 16.
Now, co-primes with product 16 are (1,15) (3,13) (5,11) and (7,9)
So, the required numbers are (33 x 1, 33 x 15) (33 x 3, 33 x 13) (33 x 5, 33 x 11) (33 x 7, 33 x 9)
the number of such pairs is 4.

Q 4 - Three numbers are in the ratio 3:4:5 and their L.C.M is 2400. Their H.C.F is?

Answer : D

Explanation

Let the numbers be 3z, 4z, and 5z. then their L.C.M = 60z 
 So, 60z = 2400 or z = 40.
Therefore the numbers are (3 x 40), (4 x 40), (5 x 40).
Hence the required H.C.F = 40.

Q 5 - L.C.M of two prime numbers a and b(a>b) is 161. The value of 3b - a is?

Answer : B

Explanation

H.C.F of two prime numbers is 1. Product of numbers = (1 x 161) = 161
Let the numbers be a and b. Then a x b = 161
now co-primes with product 161 are (1, 161) (7,23).
Since a and b are prime numbers and a > b, we have a = 23 and b = 7.
Therefore  3b - a = (3 x 7) - 23 = -2.

Q 6 - The L.C.M. of two numbers is 72. Their H.C.F. is 12. If one number is 12, the other is

Answer : D

Explanation

Let the other number be X
HCF*LCM=Product of two numbers
12*72=24*X
⇒X=(12*72)/24=36

Q 7 - X, Y and Z start at the same time in the same direction to run around a circular stadium. X completes a round in 126 seconds, Y in 154 seconds and Z in 99 seconds, all starting at the same point. After what time will they again at the starting point?

Answer : B

Explanation

L.C.M. of 126,154 and 99 = 1386.
So, X, Y and Z will again meet at the starting point in 1386 sec. i.e., 23 min. 06 sec.

Q 8 - If the sum of the H.C.F and L.C.F of two numbers is 680 and their L.C.M is 84 times the H.C.F. If one of the numbers is 56, the other number is:

Answer : D

Explanation

h+L = 680 and L = 84 h
∴ h+84 h = 680 ⇒ 85h ⇒680 ⇒h= 8
∴ L= (84*8) = 672
Now, h*L =56*x  ⇒ 8* 672=  56*x⇒ x  = (8*672)/56  = 96
∴ The other number is  96.

Q 9 - If the H.C.F of two numbers and their difference is 12 find the numbers.

Answer : D

Explanation

The difference of requisite numbers must be 12 and one must be divisible by 12 .
Hence, the required numbers are 84 and 96

Q 10 - If the capacity of two pots is 120 liters and 56 liters. Find the container capacity which can exactly measure the contents of the two pots, is :

Answer : C

Explanation

Required capacity = H.C.F of 120 L and 56 L.= 8 L = 8000cc

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