A tap can fill a tank in 3 hours, and a leak can empty in 5 hours. If the tap and the leakage which is half closed are opened together, then find the time taken to fill the tank.

4\(\frac{2}{7}\) 7\(\frac{1}{2}\) 7\(\frac{2}{3}\) 6\(\frac{2}{3}\)

4\(\frac{2}{7}\)

Efficiency: ⇒ Tap - (half closed) leakage  = 5 - \(\frac{3}{2}\) = \(\frac{7}{2}\) Now → ⇒ The required time to fill the tank = \(\frac{15}{7}\) × 2 = \(\frac{30}{7}\) = 4\(\frac{2}{7}\)