Quantitative Aptitude - Arithmetic Ability Questions

4.036/0.04 = 4036/40 = 100.9

From the given data,

12 children 16 days work,
One child’s one day work = 1/192.

8 adults 12 days work,
One adult’s one day’s work = 1/96.

Work done in 3 days = ((1/96) x 16 x 3) = 1/2

Remaining work = 1 – 1/2 = 1/2

(6 adults+ 4 children)’s 1 day’s work = 6/96 + 4/192 = 1/12

1/12 work is done by them in 1 day.

1/2 work is done by them in 12 x (1/2) = 6 days.

As the pipes are operating alternatively, thus their 2 minutes job is = 14 + 16 = 512 

In the next 2 minutes the pipes can fill another 512 part of cistern. 

Therefore, In 4 minutes the two pipes which are operating alternatively will fill 512 + 512 = 56Remaining part = 1 - 56 = 16 

Pipe A can fill 14 of the cistern in 1 minute 

Pipe A can fill 16 of the cistern in = 23 min 

Therefore, Total time taken to fill the Cistern  

4 + 23 minutes.

Prime factors of 

252 = 2 x 2 x 3 x 3 x 7; 

308 = 2 x 2 x 7 x 11; 

198 = 2 x 3 x 3 x 11

Therefore LCM of 252, 308 and 198 is 2 x 2 x 3 x 3 x 7 x 11 = 4 x 9 x 77 = 2772.

x - 35 = 80x/100 

=> x = 175 

=> 4x/5 = 4x175/5 = 140.

Required Length = H.C.F of 700 cm, 385 cm and 1295 c

                             = 35 cm.

Let daughter’s age = A and mother’s age = B
Given: 2A+B = 70 and A+2B = 95
Solving B, we will get B = 40.

Given total sum = Rs. 427
And given that 3 times A’s share, 4 times B’s share and 7 times C’s share are all equal.
=> 3A = 4B = 7C
But given
=> A + B + C = 427
Now express A and B in terms of C i.e
=> (7C/3) + (7C/4) + C = 427
=> C = 84