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Problems on H.C.F and L.C.M

Problems on H.C.F and L.C.M
The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number

360
180
110
420

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Product of 2 numbers = product of their HCF and LCM144 * x = 12 * 5040x = (12*5040)/144 = 420

Problems on H.C.F and L.C.M
What will be the LCM of 8, 24, 36 and 54

112
345
444
216

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

LCM of 8-24-36-54 will be2*2*2*3*3*3 = 216

Problems on H.C.F and L.C.M
Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .

3683
4683
2683
1683

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

L.C.M. of 5,6,7,8 = 840.    Required number is of the form 840k + 3     Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 X 2 + 3)=1683

Problems on H.C.F and L.C.M
Find the largest number of four digits exactly divisible by 12,15,18 and 27.

9720
7720
8720
6720

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The Largest number of four digits is 9999.       Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.      On dividing 9999 by 540,we get 279 as remainder .   Required number = (9999-279) = 9720.

Problems on H.C.F and L.C.M
L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

1
2
-2
-1

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Find the HCF of 54, 288, 360

17
18
16
19

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Lets solve this question by factorization method.18=2×3²,288=2×2×2×2×2×3²,360=2³×3²×5So HCF will be minimum term present in all three, i.e.2×3²=18

MORE MCQ ON Problems on H.C.F and L.C.M

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