A tank can be filled by pipe $p_{1}$ in 3 hours and by pipe $P_{2}$ in $5$ hours. When the tank is full, it can be drained by pipe $P_{3}$ in $4$ hours. If the tank is initially empty and all three pipes are open, how many hours will it take to fill up the tank?

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Solution

We can extend the result to cover three pipes. Only thing we have to keep in mind is that $P_{3}$ drains the tank and we have to use negative sign for this.
If $t$ is the time taken to fill up the tank when all the three pipes are open, then
$\frac{1}{t} = \frac{1}{3} + \frac{1}{5} - \frac{1}{4} = \frac{20 + 12 - 15}{60} = \frac{17}{60}$ .
Hence, $t = \frac{60}{17}$ . Hence it takes $3 \frac{9}{17}$ hours to fill up the tank.

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A tank can be filled by pipe p1 in 3 hours and by pipe P2 in 5 hours. When the tank is full, it can be drained by pipe P3 in 4 hours. If the tank is initially empty and all three pipes are open, how many hours will it take to fill up the tank?

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