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1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

A. $1 \frac{13}{17}$ hours

B. $2 \frac{8}{11}$ hours

C. $3 \frac{9}{17}$ hours

D. $4 \frac{1}{2}$ hours

Solution:

\begin{aligned} Net part filled in 1 hour \\ = \frac{1}{5} + \frac{1}{6} - \frac{1}{12} = \frac{17}{60} \\ ∴ The tank will be full in \frac{60}{17} hours \\ i . e ., 3 \frac{9}{17} hours \end{aligned}

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