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. 8 4 6 2 8 4 6 2 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 H.C.F of 4 * 27 * 3125, 8 * 9 * 25 * 7 and 16 * 81 * 5 * 11 * 49 is: 540 180 360 1260 540 180 360 1260 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The L.C.M. of two numbers is 45 times their H.C.F. If one of the numbers is 125 and the sum of H.C.F. and L.C.M. is 1150, the other number is: 225 235 215 220 225 235 215 220 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M he maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is: 1911 1001 910 91 1911 1001 910 91 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: 120 40 200 80 120 40 200 80 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 3x, 4x and 5x.Then, their L.C.M. = 60x.So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40).Hence, required H.C.F. = 40.
Problems on H.C.F and L.C.M The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 0, 62,98 and 130 as remainders respectively is . 15120 15210 11115 15110 15120 15210 11115 15110 ANSWER DOWNLOAD EXAMIANS APP
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