Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. Now long will it take to fill the empty cistern, if all of them are opened together?

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Solution

$\begin{matrix} (\frac{1}{10} + \frac{1}{8} - \frac{1}{15}) \times t = 1 \frac{12 + 15 - 8}{120} \times t = 1 \frac{19}{120} \times t = 1 t = \frac{120}{19} \end{matrix}$

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Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. Now long will it take to fill the empty cistern, if all of them are opened together?

(110+18−115)×t=112+15−8120×t=119120×t=1t=12019

$\begin{matrix} (\frac{1}{10} + \frac{1}{8} - \frac{1}{15}) \times t = 1 \frac{12 + 15 - 8}{120} \times t = 1 \frac{19}{120} \times t = 1 t = \frac{120}{19} \end{matrix}$