Problems on H.C.F and L.C.M Product of two co-prime numbers is 117. Their L.C.M should be: F equal to their H. Cannot be calculated 117 1 F equal to their H. Cannot be calculated 117 1 ANSWER DOWNLOAD EXAMIANS APP
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 1015 689 432 268 1015 689 432 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 Express 1168/1095 in simple form? 16/15 19/18 18/17 17/16 16/15 19/18 18/17 17/16 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP To express any fraction in simple (small form), we have to find the HCF of numerator and denominator. 1095)1168(1 1095 73)1095(15 1095 ——— 0 HCF of 1095 and 1168 is 73Therefore (1168/73)/(1095/73) = 16/15
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: 24 48 18 36 24 48 18 36 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 What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? 540 420 630 770 540 420 630 770 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 Let the least number of six digits,which when divided by 4,6,10 and 15 leaves in each case the same remainder of 2, be N. The sum of the digits in N is : 6 5 4 3 6 5 4 3 ANSWER DOWNLOAD EXAMIANS APP
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