Problems on H.C.F and L.C.M Find the least number which when divide by 2, 3, 4, 5 and 6 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 7 leaves no remainder? 119 210 154 126 119 210 154 126 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number 420 360 180 110 420 360 180 110 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Product of 2 numbers = product of their HCF and LCM144 * x = 12 * 5040x = (12*5040)/144 = 420
Problems on H.C.F and L.C.M The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: 15110 15210 11115 15120 15110 15210 11115 15120 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the largest number of four digits which is exactly divisible by 27,18,12,15 10720 8720 11720 9720 10720 8720 11720 9720 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP LCM of 27-18-12-15 is 540.After dividing 9999 by 540 we get 279 remainder.So answer will be 9999-279 = 9720
Problems on H.C.F and L.C.M The least number which when increased by 5 is divisible by each one of 24, 32, 36 and 54, is 427 869 4320 859 427 869 4320 859 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to: 11/120 55/601 601/55 120/11 11/120 55/601 601/55 120/11 ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS