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?

Correct option is D

Rate of a pipe = $$\frac{1}{\text{time taken to fill the tank}}$$​​

Time to fill remaining part = $$\frac{\text{Remaining part}}{\text{Combined rate of P and Q}} ​$$​​

Rate of P = $$\frac{1}{60}$$​ ​tank per minute.

Rate of  Q = $$\frac{1}{40}$$​ ​tank per minute.

Part filled by Q in 30 min. = $$30 \times \frac{1}{40} = \frac{30}{40} = \frac{3}{4}$$​​

Remaining part of the tank to be filled = $$1 - \frac{3}{4} = \frac{1}{4}.$$​​

Combined rate P + Q = $$\frac{1}{60} + \frac{1}{40} = \frac{2}{120} + \frac{3}{120} = \frac{5}{120} = \frac{1}{24}$$​​

Thus, the tank will be filled in 6 minutes more after Pipe P is turned on.

Efficiency of P = $$\frac{120}{60}$$​ = 2

Efficiency of Q = $$\frac{120}{40}$$​ = 3