In a tank four taps of equal efficiency are filled on equal height intervals. The 1st tap is at the base of the tank and the 4th tap is at \(\frac{3}{4}\)th of height of the tank. Then calculate in how much time the whole tank will empty by all the pipes if the 1st tap alone can empty the tank in 16 hours ?

8\(\frac{1}{3}\) hrs

Let the total capacity of the tank is = 16 ltr

Efficiency of each piper (A : B  :  C : D) = 1 : 1 : 1 : 1

ATQ,

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

For next 4 ltrs. ⇒ Pipe B + C + D work with total efficiency 3.

For next 4 ltrs. ⇒ Pipe C + D work with the efficiency 2.

For last 4 ltrs. ⇒ Pipe D works with the efficiency 1.

Total time to empty = \(\frac{4}{4}\) + \(\frac{4}{3}\) + \(\frac{4}{2}\) + \(\frac{4}{1}\) 
= \(\frac{12 + 16 + 24 + 48}{12}\) 

= \(\frac{100}{12}\) = 8\(\frac{1}{3}\) hrs