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 ?
Answer & explanation
Correct answer: option 4
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.