Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively while a third pipe C can empty the full tank in 20 minutes. All the three pipes are opened in the beginning but pipe C is closed 6 minutes before the tank is filled. In what time will the tank is full?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7 minutes
In one minute, A fills \( \frac{1}{12} \), B fills \( \frac{1}{15} \) and C empties \( \frac{1}{20} \) of the tank.
Let the tank be full after t minutes. A and B work all t minutes; C works only (t − 6) minutes.
\( \displaystyle \frac{t}{12}+\frac{t}{15}-\frac{t-6}{20}=1 \)
Multiply by 60, the LCM of 12, 15 and 20:
\( 5t+4t-3(t-6)=60 \)
\( 6t+18=60 \)
\( t=7\ \text{minutes} \)
Shortcut: Check Option (3): in 7 minutes A and B fill
\( \displaystyle 7\left(\frac{1}{12}+\frac{1}{15}\right)=7\times\frac{3}{20}=\frac{21}{20} \)
and C, open for 1 minute, empties \( \frac{1}{20} \). Net:
\( \displaystyle \frac{21}{20}-\frac{1}{20}=1 \) tank.