Quantitative Aptitude

General Questions

Quantitative Aptitude Exercise Mode

General Questions

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Question 1
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A
4
B
5
C
6
D
8
Correct Answer: Option A

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question 2
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A
276
B
299
C
322
D
345
Correct Answer: Option C

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.

Question 3
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A
9000
B
9400
C
9600
D
9800
Correct Answer: Option C

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

Question 4
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A
4
B
7
C
9
D
13
Correct Answer: Option A

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

     = H.C.F. of 48, 92 and 140 = 4.

Question 5
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A
4
B
10
C
15
D
16
Correct Answer: Option D

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1
2

= 16 times.

Question 6
Find the lowest common multiple of 24, 36 and 40.

A
120
B
240
C
360
D
480
Correct Answer: Option 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.
Question 7
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
Correct Answer: Option 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.

Question 8
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
Correct Answer: Option 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.

Question 9
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
Correct Answer: Option 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.

Question 10

Reduce 128352 to its lowest terms.
238368

A
3/4
B
5/13
C
7/13
D
9/13
Correct Answer: Option 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 ÷ 18336    
 ------  =  -------------- 
 238368     238368 ÷ 18336    

= 7/13
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