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?

18 minutes

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.