Problems on H.C.F and L.C.M What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? 630 540 420 770 630 540 420 770 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 12, 18, 21, 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5 = 630.
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 869 4320 427 859 869 4320 427 859 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M Find the largest number which divides 62,132,237 to leave the same reminder None of these 35 25 30 None of these 35 25 30 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Trick is HCF of (237-132), (132-62), (237-62)= HCF of (70,105,175) = 35
Problems on H.C.F and L.C.M The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: 23 21 24 22 23 21 24 22 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37.Number to be added = (60 - 37) = 23.
Problems on H.C.F and L.C.M Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case 4 13 9 7 4 13 9 7 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: 10 20 40 30 10 20 40 30 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the numbers be 2x and 3x.Then, their L.C.M. = 6x.So, 6x = 48 or x = 8The numbers are 16 and 24.Hence, required sum = (16 + 24) = 40.
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