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