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

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Q 1 - 252 can be expressed as a product of primes as

A - 3 x 3 x 3 x 3 x 7

B - 2 x 2 x 3 x 3 x 7

C - 2 x 2 x 2 x 3 x 7

D - 2 x 3 x 3 x 3 x 7

Answer : B

Explanation

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

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 - 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?

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 the two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10 then their difference is?

Answer : D

Explanation

Let the numbers be z and (100 - z). Then, z (100 - z) = 5 x 495
= z² - 100z + 2475 = 0
(z - 55) (z - 45) = 0
z = 55 or z = 45 
 Therefore the numbers are 45 and 55. Required difference is (55 - 45) = 10

Q 5 - The H.C.F of two numbers is 8. Which of the following can never be their L.C.M?

Answer : A

Explanation

H.C.F of two numbers divide their L.C.M exactly. Clearly 8 is not a factor of 60.

Q 6 - Find the L.C.M. of 2³x11, 2⁵x3 and 2²x3x11.

Answer : D

Explanation

To find the L.C.M. of  2³x11, 2⁵x3 and 2²x3x11, choose the highest powers of prime factors i.e. 2⁵*3*11=1056.

Q 7 - Find the H.C.F. of 1296 and 2400

Answer : B

Explanation

1296=2⁴x3⁴ and 2400=2⁵x3x5²
HCF=2⁴x3=48

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.

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 - H.C.F and L.C.F of two numbers x and y are 3 and 105 respectively. If x+y=36 then figure out 1/x + 1/y.

Answer : D

Explanation

We have X *y =3 *105 ⇒ xy = 315
∴  x+y/xy =36/315  = 4/35 ⇒ 1/y+1/x = 4/35 ⇒ 1/x+1/y = 4/35.

Q 10 - The L.C.F of two prime numbers x and y is 161. If x > y then the value of (3y- x) is:

Answer : A

Explanation

161 = 7*23  , so, X = 23 and y= 7
∴ (3y-x)=  (3*7-23 )= -2

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