Problems on H.C.F and L.C.M The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 777 1045 1745 364 777 1045 1745 364 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
Problems on H.C.F and L.C.M 1095/1168 when expressed in simplest form is: 17/26 25/26 15/16 13/16 17/26 25/26 15/16 13/16 ANSWER DOWNLOAD EXAMIANS APP
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: 1440 1449 1439 1459 1440 1449 1439 1459 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case 4 13 7 9 4 13 7 9 ANSWER DOWNLOAD EXAMIANS APP
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? 1 minute 48 seconds 4 minute 48 seconds 2 minute 48 seconds 3 minute 48 seconds 1 minute 48 seconds 4 minute 48 seconds 2 minute 48 seconds 3 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.
Problems on H.C.F and L.C.M Product of two co-prime numbers is 117. Their L.C.M should be: equal to their H. 117 F Cannot be calculated 1 equal to their H. 117 F Cannot be calculated 1 ANSWER DOWNLOAD EXAMIANS APP
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