Two pipe A and B can fill a tank in 20 hours and 16 hours respectively. Pipe B alone is kept open for 1/4th of time and both pipe are kept open for the remaining time. What is the time required to fill the tank ?

10 hours

The correct answer is option (4) : 10 hours

Part of tank filled by both the pipes in 1 hr $=\frac{1}{20}+\frac{1}{16}=\frac{9}{80}$

Let the time required to the fill the tank be t hrs, then

$\frac{t}{4}×\frac{1}{16}+\frac{3t}{4}×\frac{9}{80}=1$

$⇒\frac{5t+27t}{320}=1$

$32t = 320 $

$t = 10$

Hence, the required to fill the tank is 10 hours.