Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 2 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 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 Two numbers are in the ratio 3 : 4. If their L.C.M. is 96. what is sum of the numbers? 84 48 76 56 84 48 76 56 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case? 3 4 5 6 3 4 5 6 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Two numbers are in the ratio of 15 : 11. If their H.C. F. is 13, Find he numbers . 175 and 123 195 and 143 195 and 163 185 and 133 175 and 123 195 and 143 195 and 163 185 and 133 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the smallest number of five digits exactly divisible by 16,24,36 and 54. 10368 10568 10268 10468 10368 10568 10268 10468 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Smallest number of five digits is 10000. Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432, On dividing 10000 by 432,we get 64 as remainder. Required number = 10000 +( 432 – 64 ) = 10368.
Problems on H.C.F and L.C.M The smallest number when increased by " 1 " is exactly divisible by 12, 18, 24, 32 and 40 is: 1439 1459 1449 1440 1439 1459 1449 1440 ANSWER DOWNLOAD EXAMIANS APP
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