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 - Find the H.C.F of 108, 288, and 360.

A - 36

B - 35

C - 40

D - 45

Answer : A

Explanation

108 = 23 x 33
288 = 25 x 32
360 = 23 x 32 x 5
Therefore H.C.F = 2 2 x 32 = 36

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 - 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?

A - 4

B - 6

C - 8

D - 12

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 - 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 - Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder in each case. then the sum of the digits in N is?

A - 5

B - 4

C - 6

D - 8

Answer : B

Explanation

N = H.C.F of (4665 - 1305), (6905 - 4665), (6905 - 1305)
= H.C.F of 3360, 2240 and 5600 = 1120.
Sum of digits in N = (1 + 1 + 2 + 0) = 4.

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 - A milk vendor has three kinds of milk: 68 litres, 119 litres and 153 litres. Find the least number of casks of equal size required to store all the milk without mixing.

A - 15

B - 20

C - 25

D - 30

Answer : B

Explanation

Size of the cask (G.C.D. of 68,119 and 153) =17
Number of casks =68/17+119/17+153/17=20

Q 8 - The L.C.M of two numbers is 12 times their H.C.F and sum of H.C.F and L.C.M is 403. Having one number as 93, find the other one.

A - 128

B - 124

C - 134

D - None of these

Answer : B

Explanation

Let L.C.M =x and H.C.F = y . Then,
X= 12y  and x+y = 403 ⇒12y+y = 403 ⇒13y= 403 ⇒y = 403/13 = 31
∴ x=   (12*31 ) = 372
Let the other number be z. then,
372*31 = 93*z ⇒ z = (372*31)/93  =124
Hence , the other number is 124 .

Q 9 - If ratio of two numbers is 2:3 and the product of their H.C.F and L.C.M is 33750, find the sum of the numbers:

A - 250

B - 325

C - 375

D - 425

Answer : C

Explanation

Let the numbers be 2x and 3x.  Then, H.C.F = x and L.C.M = 6x
∴ 6x*x = 33750  ⇒ x2 = 5625   ⇒ &sqrt;5625= 75
So, the numbers are 150  and 225 . Their sum is 375.

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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