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 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 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: 127 126 128 125 127 126 128 125 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 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 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: 101 185 111 107 101 185 111 107 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.
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