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: 216 443 548 389 216 443 548 389 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 The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is 1008 1015 1022 1032 1008 1015 1022 1032 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is : 1 2 -2 -1 1 2 -2 -1 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocal of the numbers is equal to: 601/55 55/601 11/120 120/11 601/55 55/601 11/120 120/11 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, 11 : 31 am 12 : 31 am 10 : 31 am 9 : 31 am 11 : 31 am 12 : 31 am 10 : 31 am 9 : 31 am ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 60 and 62 seconds is 1860 seconds1860/60 = 31 minutesThey will beep together at 10:31 a.m.Sometimes questions on red lights blinking comes in exam, which can be solved in the same way
Problems on H.C.F and L.C.M Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder. 18004 18002 18000 17004 18004 18002 18000 17004 ANSWER DOWNLOAD EXAMIANS APP
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