Problems on H.C.F and L.C.M The largest four – digit number which when divided by 4, 7 or 13 leaves a remainder of 3 in each case, is 9893 9831 8739 9834 9893 9831 8739 9834 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M LCM of 18 and 27 is: 89 91 85 54 89 91 85 54 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: 20 40 10 30 20 40 10 30 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 2x and 3x.Then, their L.C.M. = 6x.So, 6x = 48 or x = 8The numbers are 16 and 24.Hence, required sum = (16 + 24) = 40.
Problems on H.C.F and L.C.M The least number,which when divided by 48,60,72,108 and 140 leaves 38,50,62,98 and 130 as remainder respectively, is : 15210 15120 15110 11115 15210 15120 15110 11115 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Express 1168/1095 in simple form? 16/15 18/17 17/16 19/18 16/15 18/17 17/16 19/18 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 A, B and C start running around a circular stadium and complete one round in 27 s, 9 s and 36 s, respectively. In how much time will they meet again at the starting point? 4 minute 48 seconds 3 minute 48 seconds 1 minute 48 seconds 2 minute 48 seconds 4 minute 48 seconds 3 minute 48 seconds 1 minute 48 seconds 2 minute 48 seconds ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 27, 9 and 36 = 108 So they will meet again at the starting point after 108 sec. i.e., 1 min 48 sec.
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