Two pipes A and B can fill a tank in 32 minutes and 48 minutes respectively. If both the pipes are opened simultaneously, after how much time B should be turned off so that the tank is full in 20 minutes?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 18 minutes
Pipe A fills tank in = 32 minutes
Rate A = $\frac{1}{32}$ of tank per minute
Similarly,
Rate B = $\frac{1}{48}$ of tank
First $x$ minutes when both pipes are open
$x\left(\frac{1}{32}+\frac{1}{48}\right)$
Only A is open for remaining $(20-x)$ minutes
$(20-x)\frac{1}{32}$
Adding both
$x\left(\frac{1}{32}+\frac{1}{48}\right)+(20-x)\frac{1}{32}=1$
$\frac{5}{96}x+(20-x)\frac{3}{96}=1$
$5x+60-3x=96$
$2x=36$
$x=18$
∴ Pipe B should be turned off after 18 minutes.