Exercise :: Problems on H.C.F and L.C.M - General Questions
- Problems on H.C.F and L.C.M - Important Formulas
- Problems on H.C.F and L.C.M - General Questions
| 11. | Find the lowest common multiple of 24, 36 and 40. |
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Answer: Option C Explanation: 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360. |
| 12. | The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: |
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Answer: Option C Explanation: L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37.
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Number to be added = (60 - 37) = 23.| 13. |
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Answer: Option C Explanation: 128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ÷ 18336 7
Therefore, ------ = -------------- = --
238368 238368 ÷ 18336 13
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| 14. | The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: |
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Answer: Option B Explanation: L.C.M. of 5, 6, 7, 8 = 840.
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
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| 15. | A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? |
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Answer: Option D Explanation: L.C.M. of 252, 308 and 198 = 2772. So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec. |