Three pipes, D, E and F, can fill a tank in 6 min, 8 min and 12 min, respectively. All the pipes are opened simultaneously, and then pipes D and E are closed 3 minutes before tank is full. In how much time will the tank be full?

5 min

D+ = 6 min, E+ = 8 min, F+ = 12 min,

⇒ F worked for 3 min alone = 2 x 3 = 6 units,

⇒ Remaining = 24 - 6 =18 units,

⇒ D + E + F completed 18 units in = \(\frac{18}{4+3+2}\) = \(\frac{18}{9}\) = 2 min.

Therefore, total time taken to fill the tank = 3 + 2 = 5 minutes.