A tank is being filled by two taps, M and N. M fills the tank completely in 18 hours and N fills it in 9 hours. A leakage at the bottom of the tank can empty the completely filled tank in 27 hours. How long will it take to fill the empty tank completely if both taps are opened while the leakage continues?
\(7\frac{5}{7}\) hours
In 1 hour tap M fills \(\frac{1}{18}\), tap N fills \(\frac{1}{9}\), and the leakage empties \(\frac{1}{27}\) of the tank.
Net part filled in 1 hour \(=\frac{1}{18}+\frac{1}{9}-\frac{1}{27}\).
Taking the LCM of 18, 9 and 27 as 54: \(\frac{3}{54}+\frac{6}{54}-\frac{2}{54}=\frac{7}{54}\) of the tank per hour.
Time to fill the whole tank \(=\frac{54}{7}=7\frac{5}{7}\) hours.
Hence, the tank is filled completely in \(7\frac{5}{7}\) hours.
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