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

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Questions and Answers

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

A - 7⁄13

B - 3⁄4

C - 5⁄13

D - 9⁄13

Answer : A

Explanation

Q 2 - The G.C.D of 1.08, 0.36 and 0.9 is?

A - 0.108

B - 0.9

C - 0.03

D - 0.18

Answer : B

Explanation

Given numbers are 1.08, 0.36 and 0.9. H.C.F of 108, 36 and 90 is 18.
H.C.F of given numbers is 0.18.

Q 3 - Let the least number of six digit, which when divided by 4,6,10 and 15, leaves in each case the same remainder of 2. The sum of the digits N is?

A - 5

B - 32⁄3

C - 4

D - 62⁄3

Answer : A

Explanation

Least number of 6 digits  is 100000. L.C.M of 4,6,10 and15 = 60
On dividing 100000 by 60, the remainder obtained is 40.
Therefore Least number of 6 digits divisible by 4,6,10, and 15 = 100000 + (60 - 40) = 100020.
Therefore N = (100020 + 2) = 100022.
Sum of digits in N = (1 + 2 + 2) = 5.

Q 4 - The L.C.M of two numbers is 48. The numbers are in the ratio 2:3. The sum of the number is?

A - 40

B - 45

C - 50

D - 60

Answer : A

Explanation

Let the numbers be 2z, and 3z. then their L.C.M = 6z 
 So, 6z = 48 or z = 8.
Therefore the numbers are 16 and 24
Hence the required sum = 40.

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

A - -1

B - -2

C - 1

D - 2

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 - Find the L.C.M. of 2/3, 2/5 and 2/7.

A - 2

B - 2/3

C - 2/5

D - 2/7

Answer : A

Explanation

L.C.M. of given fractions =  (L.C.M.of numerators)/(H.C.F.of denominators)=2/1=2
(L.C.M.of numerators = 2  H.C.F.of denominators =1)

Q 7 - The G.C.D. of 0.93, 0.60 and 0.75 is:

A - 0.03

B - 0.06

C - 0.09

D - 0.12

Answer : A

Explanation

Given numbers are 0.93, 0.60 and 0.75.   H.C.F. of 93, 60 & 75 is 3.
 H.C.F. of given numbers = 0.03

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 - Numbers which have 16 as their H.C.F and 136 as their L.C.M , we can definitely say that:

A - only one such pair exists

B - only two such pairs exist.

C - many such pairs exist

D - no such a pair exists

Answer : D

Explanation

H.C.F is always a factor of L.C.M
Since 16 does not divide 136 , so, no such pair exists.

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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