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

Problems on H.C.F and L.C.M
Reduce 4128/4386 to its lowest terms

13/17
14/13
16/17
13/18

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Find the smallest number of five digits exactly divisible by 16,24,36 and 54.

10568
10268
10368
10468

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Smallest number of five digits is 10000.       Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,       On dividing 10000 by 432,we get 64 as remainder.    Required number = 10000 +( 432 – 64 ) = 10368.

Problems on H.C.F and L.C.M
Find the largest number of four digits exactly divisible by 12,15,18 and 27.

6720
8720
9720
7720

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The Largest number of four digits is 9999.       Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.      On dividing 9999 by 540,we get 279 as remainder .   Required number = (9999-279) = 9720.

Problems on H.C.F and L.C.M
A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally he was paid only Rs. 1073. For how many days did he actually work?

40
37
35
39

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Which of the following has the most number of divisors?

182
101
176
99

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.Option B, 182 = 1 x 2 x 7 x 13. Option C, 99 = 1 x 3 x 3 x 11. Option D, 101 = 1 x 101.Divisors of 99 are 1, 3, 9, 11, 33, 99.Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176. Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.Hence, 176 have the most number of divisors.

Problems on H.C.F and L.C.M
Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case

505
404
606
707

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

  3 |  6 - 7  - 8  - 9 - 12-----------------------------  4 | 2  - 7  - 8  - 3 - 4  ---------------------------   2 | 2  - 7  - 2  - 3 - 1 --------------------------     | 1  - 7  - 1  - 3 - 1 L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.Hence required number = (504 +1) = 505   

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