Exercise: Problems On Hcf And Lcm
Questions for: Problems On Hcf And Lcm
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
Answer: D
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.
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
Answer: B
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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| Reduce | 128352 | to its lowest terms. |
| 238368 |
A:
| 3 |
| 4 |
B:
| 5 |
| 13 |
C:
| 7 |
| 13 |
D:
| 9 |
| 13 |
Answer: C
128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 รท 18336 7
Therefore, ------ = -------------- = --
238368 238368 รท 18336 13
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The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
A:
3
B:
13
C:
23
D:
33
Answer: C
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.
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Find the lowest common multiple of 24, 36 and 40.
A:
120
B:
240
C:
360
D:
480
Answer: C
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.
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