Problems on H.C.F and L.C.M Reduce 391/667 to lowest terms . 31/45 17/29 11/13 23/37 31/45 17/29 11/13 23/37 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP H.C.F. of 391 and 667 is 23.On dividing the numerator and denominator by 23, we get :391 = 391 ¸23 = 17667 667¸ 23 29
Problems on H.C.F and L.C.M The least number of five digits which is exactly divisible by 12, 15 and 18, is: 10020 10015 10080 10010 10020 10015 10080 10010 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 11720 8720 9720 10720 11720 8720 9720 10720 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 Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers are 12, 24, 36 10, 20, 30 5, 10, 15 4, 8, 12 12, 24, 36 10, 20, 30 5, 10, 15 4, 8, 12 ANSWER DOWNLOAD EXAMIANS APP
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 . 15210 15110 11115 15120 15210 15110 11115 15120 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: 1045 777 1745 364 1045 777 1745 364 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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