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.
120
240
360
480
Answer: Option
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:
3
13
23
33
Answer: Option
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
3
4
5
13
7
13
9
13
Answer: Option
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:
1677
1683
2523
3363
Answer: Option
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 ?
26 minutes and 18 seconds
42 minutes and 36 seconds
45 minutes
46 minutes and 12 seconds
Answer: Option
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.