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


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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.

Questions and Answers

Q 1 - H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 & 16 x 81 x 5 x 11 49 is?

A - 360

B - 540

C - 180

D - 1260

Answer : C

Explanation

4 x 27 x 3125 = 22 x 33 x 55; 8 x 9 x 25 x 7 = 23 x 32 x 52 x 7;
16 x 81 x 5 x 11 x 49 = 24 x 34 x 5 x 72 x 11.
Therefore H.C.F = 180.

Q 2 - H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 24 x 3 5 x 52 x 72?

A - 23 x 3 5 x 52 x 72

B - 22 x 3 5 x 72

C - 22 x 5 x 72

D - 25 x 5 2 x 72

Answer : B

Explanation

3240 = 23 x 32 x 5; 3600 = 24 x 32 52
H.C.F = 36 = 22 x 32.
Since H.C.F is the product of the lowest powers of common factors so the the third number must have 22 x 32 as its factor
Since L.C.M is the product of highest powers of common prime factors so the third number must have 35 x 72 as its factor
Therefore third number = 22 x 35 x 72

Q 3 - The least multiple of 7, which leaves a remainder 4, when divided by 6,9,15 and 18 is?

A - 630

B - 364

C - 360

D - 540

Answer : A

Explanation

L.C.M  of 6,9,15 and 18 is 90.
Let required number be 90k + 4, which is a multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4. 
 Therefore Required number 90 x 4 + 4 = 364.

Q 4 - The H.C.F and L.C.M of the two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1:4, then the larger number of the twi is?

A - 84

B - 48

C - 58

D - 80

Answer : A

Explanation

Let the number be z and 4z. Then z x 4z = 84 x 21 
 = z2 = 84 x 214 = z = 21
Hence larger number is 4z = 4 x 21 = 84

Q 5 - The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and pencils?

A - 91

B - 910

C - 1001

D - 1911

Answer : B

Explanation

Required number of students = H.C.F of 1001 and 910 = 91.

Q 6 - Find the H.C.F. of 2/3, 2/5 and 2/7

A - 3/105

B - 2/105

C - 5/105

D - 7/105

Answer : B

Explanation

H.C.F. of given fractions =  (H.C.F.of denominators)/(L.C.M.of numerators)=2/105
(  H.C.F.of numerators    = 2  L.C.M.of denominators =105)

Q 7 - Three number are in the ratio of 5: 6: 7 and their L.C.M. is 2100. Their H.C.F. is:

A - 40

B - 30

C - 20

D - 10

Answer : D

Explanation

Let the numbers be 5X, 6X and 7X
Then, their L.C.M. = 210X.
So, 210X = 2100 or X = 10.
The numbers are (5 x 10), (6 x 10) and (7 x 10).Hence, required H.C.F. = 10.

Q 8 - If two numbers are greater than 13 and the H.C.F of two numbers be 13, L.C.M 273, then the sum of the numbers is:

A - 286

B - 130

C - 288

D - 290

Answer : B

Explanation

Let the number be 13 a and 13 b, where a and b are co-primes.
Then, 13a * 13b=   (13* 273)  ⇒ab= 21
Two co-primes with product 21are 3 and 7.
∴ numbers are (13*3, 13*7) i.e , 39 and 91.
Their sum = (39+91) = 130

Q 9 - The H.C.F of three numbers is 12 and they are in the ratio of 1:2:3, find the numbers.

A - 6, 12, 18

B - 10, 20 ,30

C - 12, 24, 36

D - 24, 48, 72

Answer : C

Explanation

Let the numbers be a, 2a , 3a . Then , their H.C.F is a.
∴  a =12  and hence , the numbers  are 12, 24, 36.

Q 10 - If the product of two co-primes is 117 then their L.C.F should be

A - 1

B - 117

C - equal to their H.C.F

D - cannot be calculated

Answer : B

Explanation

H.C.F of co-primes=1
H.C.F * L.C.M = Their product = 117
∴ 1* L.C.M = 117 ⇒L.C.M = 117


aptitude_hcf_lcm.htm

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