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 ?
Answer & explanation
Correct answer: option 2
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