Quantitative Aptitude - Arithmetic Ability Questions

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.

The ratio of fees collected from B.Tech : MBA = 4x * 25y : 5x * 16y 

                                                                  = 100xy : 80xy 

                                                                  = 5xy : 4 xy  = 5k : 4k

The amount collected only from MBA students = 49×1.62 lakh = Rs. 72,000

Efficiency of kaushalya = 5%

Efficiency of kaikeyi  = 4%

Thus, in 10 days working together they will complete only 90% of the work.

          [(5+4)*10] =90

Hence, the remaining work will surely done by sumitra, which is 10%.

Thus, sumitra will get 10% of Rs. 700, which is Rs.70

4.036/0.04 = 4036/40 = 100.9

x - 35 = 80x/100 

=> x = 175 

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

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.

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

                             = 35 cm.

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