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Arithmetic Aptitude :: Problems on H.C.F and L.C.M
Find the lowest common multiple of 24, 36 and 40.
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.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Answer : Option C
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.
Reduce 128352 238368 to its lowest terms.
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
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:
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.
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.
D.
46 minutes and 12 seconds
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.
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
Answer : Option C
Explanation :
Other number =[ 11 × 7700 275 ] = 308
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
Answer : Option B
Explanation :
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
----------------------------
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
----------------------------
Required number = (1260 � 2) | 2 - 3 - 7 - 5
= 630.
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
Answer : Option D
Explanation :
Let the numbers be 3x and 4x . Then, their H.C.F. = x . So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Answer : Option B
Explanation :
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
252 can be expressed as a product of primes as:
Answer : Option A
Explanation :
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
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