In a tank four equal efficiency's taps are fitted on equal intervals and the 4th pipe is at \(\frac{3}{4}\) of the height of the tank. If one tap can empty the tank in 12 hours then find in how much time the whole tank will empty?

6 h 15 min

For 1st 3 ltrs. ⇒ Pipe A + B +C + D work with total efficiency 4.

For next 3 ltrs. ⇒ Pipe B + C + D work with total efficiency 3.

For next 3 ltrs. ⇒ Pipe C + D work with the efficiency 2.

For last 3 ltrs. ⇒ Pipe D works with the efficiency 1.

Total time to empty = \(\frac{3}{4}\) + \(\frac{3}{3}\) + \(\frac{3}{2}\) + \(\frac{3}{1}\)

                                = \(\frac{3 + 4 + 6 + 12}{4}\) = \(\frac{25}{4}\)

                                =  6\(\frac{1}{4}\) = 6 hours 15 minutes