Problems on H.C.F and L.C.M Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively. 1494 1294 1394 1194 1494 1294 1394 1194 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6. Required number = (L.C.M. of 20,25,35,40) – 6 =1394.
Problems on H.C.F and L.C.M Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? 4 10 16 15 4 10 16 15 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30 + 1 = 16 times. 2
Problems on H.C.F and L.C.M Find the lowest common multiple of 24, 36 and 40. 120 360 240 480 120 360 240 480 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
Problems on H.C.F and L.C.M Find the greatest number that will divided 43,91 and 183 so as to leave the same remainder in each case 9 4 7 13 9 4 7 13 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? 4 minute 48 seconds 1 minute 48 seconds 3 minute 48 seconds 2 minute 48 seconds 4 minute 48 seconds 1 minute 48 seconds 3 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.
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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