Two pipes can fill a tank in 15 min and 20 min respectively. Due to some blockage in pipes, only \(\frac{1}{2}\) water and \(\frac{2}{3}\) water can flow through first and second pipe respectively. After removing that blockage, the tank filled in 5 minutes. After how much time the blockage was removed?

6 min 15 sec

A(+)   :   B(-)

 15    :    20    Time

  4     :     3     Efficiency

Total work = 60         (LCM of  15 and 20)

In last five minutes, the work done will be 5 x 7 = 35

Remaining work = 60 -  35 = 25

Due to blockage, efficiency of pipe A is  \(\frac{4}{2}\) = 2 and that of pipe B is  \(\frac{6}{3}\) = 2

Time taken to fill 25 ltrs. =  \(\frac{work}{efficiency}\) =  \(\frac{25}{4}\) = 6 \(\frac{1}{4}\) = 6 min 15 seconds

Hence,

The blockage was removed after 6 min 15 sec.