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: 40 120 200 80 40 120 200 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 Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder? 5035 5073 5016 5054 5035 5073 5016 5054 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the HCF of 54, 288, 360 18 17 19 16 18 17 19 16 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Lets solve this question by factorization method.18=2×3²,288=2×2×2×2×2×3²,360=2³×3²×5So HCF will be minimum term present in all three, i.e.2×3²=18
Problems on H.C.F and L.C.M The H.C.F of two numbers is 8. Which of the following can never be their L.C.M? 56 60 48 24 56 60 48 24 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: 308 208 408 508 308 208 408 508 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Other number = (11 x 7700/275) = 308.
Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 2 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 2 x 2 x 3 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, 252 = 2 x 2 x 3 x 3 x 7.
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