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Two pipes can fill a cistern in 30 and 15h respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 5h extra are taken for cistern to be filled up. If the cistern is full, in what time would the leak empty it?

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Part of the cistern filled when the two pipes are opened simultaneously = \(\frac{1}{30}\)\(\frac{1}{15}\) = \(\frac{1+2}{30}\) = \(\frac{3}{30}\) = \(\frac{1}{10}\) i.e, The two pipes can fill the cistern together in 10 hours

Together with the leak, the cistern can be filled in (10 + 5) hours = 15 hours

Let the leak take x hours to empty the cistern. Then,

\(\frac{1}{10}\) - \(\frac{1}{X}\) = \(\frac{1}{15}\) \(\Rightarrow\) \(\frac{1}{X}\) = \(\frac{1}{10}\) - \(\frac{1}{15}\) = \(\frac{1}{30}\)

\(\therefore\) The leak takes 30 hours to empty the cistern

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