DRDO Multi tasking staff Model paper 10

Let the capacity of the tank be 1 ltr.
Therefore, part filled by pipe A in 1 hr = 1/4 ltr/hr
Part filled by pipe B in 1 hr = 1/2 ltr/hr
Let the time for which pipe B remained active be t hr.
For first 3 hours pipe A worked alone and for next t hours, A and B both were active.
⇒ Tank will be full when 1 ltr is filled
⇒ 3 × ¼ + t × (1/4 + 1/2) = 1
⇒ 3/4 + 3t/4 = 1
⇒ 3t/4 = ¼
⇒ t = 1/3 hr = 20 minutes.
Therefore, tank will be full at 8hr + 20min = 8:20 pm