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A pool with a drain

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A pump fills an empty pool in $4$ hours, and an open drain alone empties a full pool in $12$ hours, both at constant rates. The pump starts on the empty pool; after $1$ hour someone opens the drain by mistake, and from then on both keep running. How many hours after the pump started is the pool full? Give the answer as a fraction in lowest terms.

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Hint

Treat the drain as a negative rate and handle the first hour separately.

Solution

In the first hour the pump alone fills $\frac{1}{4}$ of the pool.
Afterwards the net rate is $\frac{1}{4} - \frac{1}{12} = \frac{1}{6}$ of the pool per hour.
The remaining $\frac{3}{4}$ of the pool takes $\frac{3}{4} \times 6 = \frac{9}{2}$ hours.
The total is $1 + \frac{9}{2} = \frac{11}{2}$ hours, so the answer is $\frac{11}{2}$ .

Answer: $\frac{11}{2}$

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