Aptitude - Pipes and Cistern - Discussion

Thanks @Krishna.

The first tank will take 3hrs to fill in half of the tank.

4 tanks will take 6/4=1.5hrs to fill the full tank.
but as half of the tank is already filled, it will take 0.75hr=45mins for the rest half tank.

One tap can fill the whole cistern in = 6 hr,
And half(1/2)tank can fill in = 3 hr.

Remaining (1/2)tank will filled with the 4 pipes(1+3) = 3hr = 180min,
4 pipes = 180/4 = 45min.

So the final answer is (1half+2nd half) = 3hr + 45min,
3hr and 45min.

Thanks @vicky.

Simple (3hour/1pipe)+(3hour/4pipe).
(First hf tank)+(sec half tank) = 3(3/4) = in hours then = 3(3*60)/4 = 3hr 45min.

Tank filled in 1hr =6units.
For half an hour =3units.
Remaining 3units.
Now 3 more taps.....+previous 1 tap =4taps.
Efficiency =3/4.
Total =3*3/4(60)=3hr 45min.

Here we are assuming rate =100
As per the statement,
A
R=100 h=? Capacity =300
=300/100=3.

A+B+C+D =300/400=0.75=45min

=> 3hr +45 min = 3hr 45.

@ Sravanthi :

only one tap fill the half tank. So, the remaining four should fill the remaining half tank.
Hence. 1 - 1/2 = 1/2.
It's should not be 1-2/3.

Time taken by one tap to fill half of the tank = 3 hours.
Part filled by one tap in 1 hour = 1/6,
Part filled by 4 taps together in 1 hour = 4 × 1/6 = 2/3.

Total part filled by one tap + total part filled by 4 taps = 1.
(3)×(1/6) + (x)×(2/3) = 1.
[where x is the time taken by 4 taps to fill the tank].
Solving the equation we get, x = 3/4.
Now,
The Total time taken to fill the tank = 3 hours + 3/4 hours = 15/4 hours, which can also be written as 3(3/4) hours, which is 3 hours 45 minutes.
So, the correct option is (b) 3 hours 45 minutes.

If 1 tap takes 6 hours --> 2 taps takes 6/2=3 (half time as fill faster).

If 2 taps takes 6/2=3 (half time)---4 taps takes 3/2=1.5 (i.e 1:30 min).

1:30 min= 60 min +30 min= 90 min.

90 min (time taken by 4 tapes to fill full tank).

But left is only half tank.

So 90/2 = 45 min.

Now total time = 3hour+45 min.