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

Problems on H.C.F and L.C.M
252 can be expressed as a product of primes as:

2 x 2 x 2 x 3 x 7
2 x 3 x 3 x 3 x 7
2 x 2 x 3 x 3 x 7
3 x 3 x 3 x 3 x 7

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

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

Problems on H.C.F and L.C.M
The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1:4, then the larger of the two numbers is:

84
108
12
48

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

127
126
128
125

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Required number = H.C.F. of (1657 - 6) and (2037 - 5)= H.C.F. of 1651 and 2032 = 127.

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:

2
-1
-2
1

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
The least number which should be added to 2497 so that the sum is exactly divisible by 5,6,4 and 3 is :

13
3
33
23

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

101
185
111
107

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

 Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

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

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