Aptitude - Pipes and Cistern - Discussion

Given,

First pipe takes 20h.
Second pipe take 24h.

ALL together take 15h.

Assuming that capacity of tank L. C. M [20, 24, 15] = 120.

First pipe can fill 6u/m.
Second pipe fill 5u/m.

All three fill 8u/m -------> 1.

First and second together fill in 11m/u----------> 2.

From 1 and 2 we get;

Third pipe fills the tank in 3u/m that is 120/3 = 40u of total fill.

But given 3 gallons are empty.

3 * 40 = 120gallons (capacity of tank).

120 gallon is waste water, which drains out from the tank in 15 min.

2 pipes can fill a tank in X and Y times respectively. How to calculate the time taken to fill the tank if both the pipes a open together?

How can we solve this problem by the unitary method?

Let the capacity of the tank or pump X gallons.
The first pipe can fill the tank in 20 min.
The second pipe can fill the tank in 24 min.
1st pipe 20 min. fill X gallon. So, First pipe in one min can fill X/20 gallons.
So, the second pipe can fill in X/24 gallons.
And the third pipe flows out 3 gallons per minute.
So, X/20 + X/24 - 3 equal X/15 therefore X equal 120 gallons.

The part of tank filled in 1 min by all pipes=(1/20)+(1/24)-(1/x).
= (44x-480)/480x.

The time required to fill 1 tank.
= 480x/(44x-480) minutes.

Where, x = time in minutes required to empty the tank by the waste pipe.
But all pipes fill the total tank in 15 minutes.

So, 15 = 480x/(44x-480).
x = 40 minutes.

The capacity of tank = discharge per minute * time.
= 3 * 40,
= 120 gallons.

How did 11/120?

Your logic is most acceptable or easy to me, Thanks @Rabi.