Problems on H.C.F and L.C.M Find the least multiple of 23, which when divided by 18, 21 and 24 leaves remainders 7, 10 and 13 respestively. 3036 3024 3002 3013 3036 3024 3002 3013 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: 443 389 548 216 443 389 548 216 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 Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together. 17 24 31 21 17 24 31 21 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.Now 60/2 = 30Adding one bell at the starting it will 30+1 = 31
Problems on H.C.F and L.C.M The largest four-digit number which when divided by 4, 7 or 13 leaves a remainder of 3 in each case,is : 9893 9831 8739 9834 9893 9831 8739 9834 ANSWER DOWNLOAD EXAMIANS APP
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? 485 391 431 381 485 391 431 381 ANSWER DOWNLOAD EXAMIANS APP
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: 101 185 111 107 101 185 111 107 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.
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