Problems on H.C.F and L.C.M What is the least number which when divided by 6, 9, 12 and 18 leaves remainder 4 in each care? 56 40 36 30 56 40 36 30 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are? 800,500 700,400 820,369 820,360 800,500 700,400 820,369 820,360 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The traffic light at three different road crossing change after every 48 sec, 72 sec. and 108 sec. respectively . If they all change simultaneously at 8 : 20 : 00 hours, then at what time will they ag 5 : 27 : 12 hrs. 8 : 27 : 12 hrs. 7 : 27 : 12 hrs. 6 : 27 : 12 hrs. 5 : 27 : 12 hrs. 8 : 27 : 12 hrs. 7 : 27 : 12 hrs. 6 : 27 : 12 hrs. ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: 548 389 443 216 548 389 443 216 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = (L.C.M. of 12, 15, 20, 54) + 8= 540 + 8= 548.
Problems on H.C.F and L.C.M A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. Aft 45 minutes 26 minutes and 18 seconds 46 minutes and 12 seconds 42 minutes and 36 seconds 45 minutes 26 minutes and 18 seconds 46 minutes and 12 seconds 42 minutes and 36 seconds ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
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: 432 268 689 1015 432 268 689 1015 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = (L.C.M. of 12,16, 18, 21, 28) + 7= 1008 + 7= 1015
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