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Problems on H.C.F and L.C.M

Problems on H.C.F and L.C.M
There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and of the last three is 1001. Find the fourth number?

13
11
7
5

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.

35
22
48
52

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Required number =  H.C.F. of (132 - 62), (237 - 132) and (237 - 62)                                      =  H.C.F. of 70, 105 and 175 = 35.

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?

5054
5035
5016
5073

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Find the greatest number which, while dividing 19, 83 and 67, gives a remainder of 3 in each case?

16
19
18
17

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

17
21
24
31

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.Now 60/2 = 30Adding one bell at the starting it will 30+1 = 31

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:

1745
364
777
1045

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.

MORE MCQ ON Problems on H.C.F and L.C.M

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