Pipe A can fill a tank in 18 minutes, while pipe B can empty the completely filled tank in 27 minutes. Initially, pipe A is opened and after 6 minutes pipe B is also opened. In how much time (in minutes) will the remaining tank be filled completely?

Given:

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

Time taken by Pipe B to empty the tank = 27 minutes

Pipe A is opened for 6 minutes initially.

Then Pipe B is also opened.

Calculations:

LCM of 18 and 27 = 54 units (Let the total capacity of the tank be 54 units)

Efficiency of Pipe A (filling) = \(\frac{54}{18}\) = 3 units/minute

Efficiency of Pipe B (emptying) = \(\frac{54}{27}\) = 2 units/minute

Work done by Pipe A in the first 6 minutes = 6 minutes × 3 units/minute

⇒ Work done = 18 units

Remaining capacity to be filled = Total capacity - Work done by Pipe A

⇒ Remaining capacity = 54 - 18

⇒ Remaining capacity = 36 units

When Pipe B is also opened, the net efficiency = Efficiency of A - Efficiency of B

⇒ Net efficiency = 3 - 2

⇒ Net efficiency = 1 unit/minute

Time taken to fill the remaining tank = Remaining capacity / Net efficiency

⇒ Time taken = \(\frac{36}{1}\)

⇒ Time taken = 36 minutes

∴ The remaining tank will be filled completely in 36 minutes.