HCF and LCM Quiz Set 009

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Question 1

Which is the smallest number, which, when divided by 21 and 80, leaves the remainder 4?

 A

1684.

 B

1686.

 C

1683.

 D

1687.

Soln.
Ans: a

The LCM of 21 and 80 = 1680, is the smallest number, which, when divided by either of them leaves the remainder 0. Now adding, 1680 + 4 = 1684 is the required number.


Question 2

Which is the smallest number, which, when divided by 9 and 65, leaves the remainder 7?

 A

592.

 B

594.

 C

591.

 D

595.

Soln.
Ans: a

The LCM of 9 and 65 = 585, is the smallest number, which, when divided by either of them leaves the remainder 0. Now adding, 585 + 7 = 592 is the required number.


Question 3

Which is the smallest number, which, when divided by 21 and 75, leaves the remainder 3?

 A

528.

 B

530.

 C

527.

 D

531.

Soln.
Ans: a

The LCM of 21 and 75 = 525, is the smallest number, which, when divided by either of them leaves the remainder 0. Now adding, 525 + 3 = 528 is the required number.


Question 4

Which is the least number which must be added to 21937 so that it becomes divisible by 21, 95 and 77?

 A

8.

 B

9.

 C

7.

 D

10.

Soln.
Ans: a

The LCM of 21, 95 and 77 = 21945. The lcm when divided by either of these numbers leaves the remainder 0. The minimum number to be added is 21945 - 21937 = 8.


Question 5

The sum of LCM and HCF of two numbers is 232. If LCM is 57 times the HCF, then the product of the two numbers is?

 A

912.

 B

914.

 C

913.

 D

915.

Soln.
Ans: a

Let L be the LCM, and H the HCF. Then H + L = 232, and L = 57H. Solving these for H, we get H = $232/{57 + 1}$ = 4, and L = 228. The product of the two numbers is equal to the product of the lcm and hcf = 4 × 228 = 912.


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This Blog Post/Article "HCF and LCM Quiz Set 009" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2019-08-18.

Posted by Parveen(Hoven),
Aptitude Trainer


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