Pipe A takes 4/5 of the time required by pipe B to fill an empty tank individually. When an outlet pipe C is also opened simultaneously with pipes A and B, it takes 4/5 more time to fill the empty tank than it takes when only pipe A and pipe B are opened together. If it takes 40 hours to fill the tank when all the three pipes are opened simultaneously, in what time (in hours) will pipe C empty the full tank, operating alone?

Question

Accepted Answer

Let the time taken by pipe B to fill the tank be T hours.
Then, time taken by pipe A = (4/5)T hours.

Rates of filling:
Pipe A's rate = 1 / ((4/5)T) = 5/(4T)
Pipe B's rate = 1 / T

When pipes A and B work together, their combined rate = 5/(4T) + 1/T = (5/4T) + (4/4T) = 9/(4T)

Let the time taken by pipes A and B together to fill the tank be t hours.
So, t = 1 / (combined rate) = 1 / (9/(4T)) = 4T/9

Given that when pipe C is also opened, it takes 4/5 more time than t to fill the tank.
So, time taken with A, B, and C together = t + (4/5)t = (9/5)t

Given this time is 40 hours:
(9/5)t = 40
=> t = (40 * 5) / 9 = 200 / 9 hours

Recall t = 4T/9, so:
4T/9 = 200/9
=> 4T = 200
=> T = 50 hours

Now, rate of pipe A = 5/(4T) = 5/(4*50) = 5/200 = 1/40
Rate of pipe B = 1/T = 1/50

Rate of A + B = 1/40 + 1/50 = (5/200) + (4/200) = 9/200

Let rate of pipe C (emptying) = 1/x (since it empties, rate is negative)

Given that with pipe C open, time taken is 40 hours, so rate = 1/40

So, (rate of A + rate of B - rate of C) = 1/40
=> 9/200 - 1/x = 1/40

Multiply both sides by 200x:
9x - 200 = 5x

=> 9x - 5x = 200
=> 4x = 200
=> x = 50 hours

Therefore, pipe C empties the tank in 50 hours.

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