Two pipes P and Q can individually fill a tank in 60 minutes and 40 minutes. If pipe Q alone is open for the first half an hour and then pipe P is also turned on, in how many minutes more will the tank get filled up?

Work = efficiency $$\times$$ time

Work=LCM(60,40)

       =120

Quantity filled by P in 1 hr =$$ \frac{120}{60} $$

                                       = 2

Quantity filled by Q in 1 hr =$$ \frac{120}{40}$$

                                          = 3

Efficiency of pipe P = 2

Efficiency of pipe Q = 3

Work done by pipe Q alone for the first half an hour = $$ 30 \times 3 $$

                                                                              = 90

Work done by pipe P and Q  for the remaining time t = $$ t \times (2+3) $$

                                                                               = $$ t \times 5 $$

Thus total work done, 

 $$ 90 + 5 \times t = 120 $$

Solving, t=6 minutes