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Question

# A pipe can fill a cistern in 10 hours. Due to a leak in the bottom it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?

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Solution

## $\mathrm{When}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{leakage},\mathrm{the}\mathrm{pipe}\mathrm{can}\mathrm{fill}\mathrm{the}\mathrm{cistern}\mathrm{in}10\mathrm{hours}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{pipe}\mathrm{can}\mathrm{fill}\frac{1}{10}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{cistern}\mathrm{in}1\mathrm{hour}.\phantom{\rule{0ex}{0ex}}\mathrm{When}\mathrm{there}\mathrm{is}\mathrm{leakage},\mathrm{the}\mathrm{pipe}\mathrm{can}\mathrm{fill}\mathrm{the}\mathrm{cistern}\mathrm{in}12\mathrm{hours}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{leakage},\mathrm{the}\mathrm{pipe}\mathrm{can}\mathrm{fill}\frac{1}{12}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{cistern}\mathrm{in}1\mathrm{hour}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{in}\mathrm{one}\mathrm{hour},\mathrm{due}\mathrm{to}\mathrm{leakge},\left(\frac{1}{10}-\frac{1}{12}\right)\mathrm{th}\mathrm{or}\frac{1}{60}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{cistern}\mathrm{is}\mathrm{emptied}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{cistern}\mathrm{will}\mathrm{be}\mathrm{emptied}\mathrm{by}\mathrm{the}\mathrm{leakage}\mathrm{in}60\mathrm{hours}.$

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