Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
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
The greatest number when divide 147,185 and 251 leaving the remainders 3,5 and 11.Means the greatest number is not exactly dividing the given numbers. If we subtract the remainders from the respective numbers, then they are exactly divisible by the greatest numbers. The required greatest number = HCF of (147 - 3),(185-5) and ( 251-11) =HCF of 144,180 and 240 =12
EXAMIANS