Problems on H.C.F and L.C.M Reduce 391/667 to lowest terms . 31/45 11/13 23/37 17/29 31/45 11/13 23/37 17/29 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 multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 364 1045 1745 777 364 1045 1745 777 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.
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
Problems on H.C.F and L.C.M The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: 208 308 508 408 208 308 508 408 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Other number = (11 x 7700/275) = 308.
Problems on H.C.F and L.C.M H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is 1260 540 180 360 1260 540 180 360 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The smallest number when increased by " 1 " is exactly divisible by 12, 18, 24, 32 and 40 is: 1449 1440 1459 1439 1449 1440 1459 1439 ANSWER DOWNLOAD EXAMIANS APP