Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously,after how much time it should be closed s that the tank is full in 18 minutes ?

8 minutes

Time taken by pipe A $=24$ minutes.

Time taken by pipe B $=32$ minutes.

Rate of A $=\frac{1}{24}$ tank per minute.

Rate of B $=\frac{1}{32}$ tank per minute.

Combined rate of A and B:

$\frac{1}{24}+\frac{1}{32}=\frac{4+3}{96}=\frac{7}{96}$ tank per minute.

Let both pipes be opened for $x$ minutes.

Work done in $x$ minutes:

$x\cdot\frac{7}{96}$

Remaining time $=18-x$ minutes.

After closing one pipe, only pipe A works.

Work done by A in remaining time:

$(18-x)\cdot\frac{1}{24}$

Total work $=1$ tank:

$\frac{7x}{96}+\frac{18-x}{24}=1$

Multiply throughout by $96$:

$7x+4(18-x)=96$

$7x+72-4x=96$

$3x=24$

$x=8$

final answer: both pipes should be kept open for $8$ minutes