Problems on H.C.F and L.C.M The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: 308 508 208 408 308 508 208 408 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Other number = (11 x 7700/275) = 308.
Problems on H.C.F and L.C.M What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? 420 630 770 540 420 630 770 540 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 12, 18, 21, 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5 = 630.
Problems on H.C.F and L.C.M The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: 48 36 18 24 48 36 18 24 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.So, the numbers 12 and 16.L.C.M. of 12 and 16 = 48.
Problems on H.C.F and L.C.M The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: 268 689 1015 432 268 689 1015 432 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = (L.C.M. of 12,16, 18, 21, 28) + 7= 1008 + 7= 1015
Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 2 x 2 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 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.
EXAMIANS