Two pipes can separately fill a tank in 20 hr | vedclass

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(C) The rate of filling by both pipes in $1$ $hr$ is $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$ of the tank.Time tak

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$16$

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(C) The rate of filling by both pipes in $1$ $hr$ is $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$ of the tank.Time taken to fill $\frac{1}{3}$ of the tank is $\frac{1/3}{1/12} = 4$ $hrs$.After the tank is $\frac{1}{3}$ full,a leak develops. The leak removes $\frac{1}{3}$ of the water supplied by both pipes per hour.Net rate of filling after the leak = $(	ext{Rate of both pipes}) - (	ext{Leakage rate}) = \frac{1}{12} - (\frac{1}{3} 	imes \frac{1}{12}) = \frac{1}{12} - \frac{1}{36} = \frac{3-1}{36} = \frac{2}{36} = \frac{1}{18}$ of the tank per hour.The remaining part of the tank to be filled is $1 - \frac{1}{3} = \frac{2}{3}$.Time taken to fill the remaining $\frac{2}{3}$ part = $\frac{2/3}{1/18} = \frac{2}{3} 	imes 18 = 12$ $hrs$.Total time taken = $4$ $hrs + 12$ $hrs = 16$ $hrs$.

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