Pipes A and B can fill a tank in 18 hours and 27 hours, respectively, whereas pipe C can empty the full tank in 54 hours. All three pipes are opened together, but pipe C is closed after 12 hours. In how much time (in minutes) will the one-third of the remaining part of the tank be filled by A and B together?

Given:

Time taken by Pipe A to fill the tank = 18 hours

Time taken by Pipe B to fill the tank = 27 hours

Time taken by Pipe C to empty the tank = 54 hours

All three pipes are opened together for 12 hours.

Formula used:

Work done by a pipe in 1 hour = \(\dfrac{1}{\text{Time taken by pipe}}\)

Total work done in 1 hour by all three pipes together = Work done by A + Work done by B - Work done by C

Calculation:

Work done by A in 1 hour = \(\dfrac{1}{18}\)

Work done by B in 1 hour = \(\dfrac{1}{27}\)

Work done by C in 1 hour = \(\dfrac{1}{54}\)

Total work done by all three pipes in 1 hour:

⇒ \(\dfrac{1}{18} + \dfrac{1}{27} - \dfrac{1}{54} = \dfrac{3+2-1}{54} = \dfrac{4}{54} = \dfrac{2}{27}\)

Total work done in 12 hours:

⇒ \(\dfrac{2}{27} \times 12 = \dfrac{24}{27} = \dfrac{8}{9}\)

Remaining part of the tank after 12 hours = \(\dfrac{1}{9}\)

One-third of the remaining part = \(\dfrac{1}{3} \times \dfrac{1}{9} = \dfrac{1}{27}\)

Combined rate of A and B:

⇒ \(\dfrac{1}{18} + \dfrac{1}{27} = \dfrac{3+2}{54} = \dfrac{5}{54}\)

Time to fill \(\dfrac{1}{27}\) of the tank by A and B:

⇒ \(\dfrac{\dfrac{1}{27}}{\dfrac{5}{54}} = \dfrac{54}{27 \times 5} = \dfrac{2}{5} \text{ hours}\)

Convert to minutes:

⇒ \(\dfrac{2}{5} \times 60 = 24 \text{ minutes}\)

∴ The correct answer is 24 minutes.