Problems on H.C.F and L.C.M The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 185 101 107 111 185 101 107 111 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 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.
Problems on H.C.F and L.C.M Find the greatest number which will divide 25, 73 and 97 as so to leave the same remainder in each case? 32 12 24 18 32 12 24 18 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Product of two co-prime numbers is 117. Their L.C.M should be: F equal to their H. 117 Cannot be calculated 1 F equal to their H. 117 Cannot be calculated 1 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: 443 548 389 216 443 548 389 216 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 room is 4 meters 37 cm long and 3 meters 23cm broad. It is required to pave the floor with minimum square slabs. Find the number of slabs required for this purpose? 381 391 431 485 381 391 431 485 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Reduce 4128/4386 to its lowest terms 14/13 16/17 13/18 13/17 14/13 16/17 13/18 13/17 ANSWER DOWNLOAD EXAMIANS APP