Problems on H.C.F and L.C.M Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. 2 4 6 8 2 4 6 8 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.
Problems on H.C.F and L.C.M A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally he was paid only Rs. 1073. For how many days did he actually work? 39 35 40 37 39 35 40 37 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The greatest number of four digits that have 144 for their HCF is? 9935 9936 9930 9903 9935 9936 9930 9903 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the least which when divided y 16,18,20, and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder. 17004 18000 18002 18004 17004 18000 18002 18004 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Let the least number of six digits,which when divided by 4,6,10 and 15 leaves in each case the same remainder of 2, be N. The sum of the digits in N is : 6 4 5 3 6 4 5 3 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 1745 777 364 1045 1745 777 364 1045 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
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