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:

11(¹³/₁₇) hours
22(⁸/₁₁) hours
33(⁹/₁₇) hours
44(¹/₂) hours

Answer:- 1

Explanation:-

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}

$\eqalign{ & {\text{Net}}\,{\text{part}}\,{\text{filled}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = \left( {\frac{1}{5} + \frac{1}{6} - \frac{1}{{12}}} \right) = \frac{{17}}{{60}} \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\frac{{60}}{{17}}{\text{hours}} \cr & i.e.,\,3\frac{9}{{17}}\,{\text{hours}} \cr}$