Aptitude - Pipes and Cistern - Discussion

Work done by A in 1 min + Work done by B in 1 min = Work done by both in 1 min.

Any one tell me simplest way to solve this question?

Let it be 3x, 1x.

So, 4x = 1/36.

Then x = 144.

Let A be faster pipe & B be slower pipe.

We know that [b] Capacity = Work/Time [/b].

Assume the capacity of A as 3 unit/min (i.e. A pipe can fill three times more capacity than B as given in the question).

Then the capacity of B is 1 unit/min.

Total capacity (A+B) = 3+1= 4 unit/min.

If the pipes together (A+B) fill the tank in 36 min, then net work = capacity*time.

i.e Net work = 4*36 = 144 units.

Now the time taken by slower pipe B = Network/Capacity of B.

Time (B) = 144/1 = 144 min.

Similarly for A pipe = 144/3 = 48 min.

Ans: 144 min.

Thanks now I am understood.

I didn't understand the problem could you explain once.

Assume A = Fast pipe.

B = Slow pipe.

A takes x time. And B takes 3x time.

So, in 1 min 1/x + 1/3x = 1/36.

By solving 4x/3x^2 = 1/36.

Finally x = 4*36/3.

x = 48.

Time taken by slow pipe = 3x = 3*48 = 144 min.

I am getting a different answer,

Two pipe fills the tank in 36 min.

Let's take slower one = A.

Faster one as B.

So, B = A*3.

If B+A = 36.

Then B = 27 and A = 9.

If B alone has to fill the tank then it will fill in.

27min (of B) + 9/3 of A = 27+3 = 30 min.

And if go like this and A alone has to fill the tank then it would be:

30*3 = 90 min.

Why I this is not right someone Please explain?

P1 = 3P2 According to the question.
[P1 + P2] = 1/36.
[P2] = ?

P1 + P2 = 1/36.
3P2 + P2 = 1/36.
4P2 = 1/36.
P2 = 1/144 {1/36 * 4 = 144}.

==>P2 = 144min.

Actually A=3B because A pipe fill d tank 3 times fast as B pipe.
So A=3B but in time.

Then B = A/3
Let, A= x-----(1) assume i.e B=x/3---------(2)

Formula 1/A+1/B=1/36 substitute both in
1/x+1/x/3 = 1/36,
1/x+3/x = 1/36,
4/x = 1/36.
=>x=4*36 =>144.