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H.C.F and L.C.M Problems-Data Intensity 1 Multiple Questions and Answers.
Exercise Questions ::
H.C.F and L.C.M Problems
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. See More Information
Published by:Michael Daani
Answer: Option C
Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Hence: Number to be added = (60 - 37) = 23.
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Published by:Michael Daani
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 See More Information
Published by:Michael Daani
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? A. 1677 B. 1683 C. 2523 D. 3363
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Answer: Option B
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
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Published by:Michael Daani
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 ? A. 26 minutes and 18 seconds B. 42 minutes and 36 seconds C. 45 minutes D. 46 minutes and 12 seconds
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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.
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Published by:Michael Daani
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