# Aptitude - Problems on H.C.F and L.C.M

### Exercise :: Problems on H.C.F and L.C.M - General Questions

11.

Find the lowest common multiple of 24, 36 and 40.

 A. 120 B. 240 C. 360 D. 480

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:

 A. 3 B. 13 C. 23 D. 33

Explanation:

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.

13.

 Reduce 128352 to its lowest terms. 238368

A.
 3 4
B.
 5 13
C.
 7 13
D.
 9 13

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
```

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

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.

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

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.