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?

Correct option is A

Given:

Pipe A fills the tank in 18 minutes

Pipe B empties the tank in 27 minutes

Pipe A runs alone for 6 minutes, then both A and B run together

Formula Used:

Work done = Rate × Time

Solution:

A’s rate =$$\frac{1}{18}$$​ tank/min

B’s rate =$$-\frac{1}{27}$$​ tank/min (since it empties)

Combined rate =$$\frac{1}{18} - \frac{1}{27} = \frac{3 - 2}{54} = \frac{1}{54}$$​ tank/min

In 6 minutes, A alone fills $$6 \times\frac{1}{18} = \frac{1}{3}$$​​​​ of the tank

Remaining = $$1 - \frac{1}{3} = \frac{2}{3}$$​​

Time to fill$$\frac{1}{3}$$​ tank at combined rate of$$\frac{1}{54}$$​​

​$$\frac{2}{3} \div \frac{1}{54} = \frac{2}{3} \times 54 = 36 \text{ minutes}$$

Alternate Method: 

Total Work = 54 Unit 

A efficiency = 3

A's 6 minute Work = 6 $$\times$$ 3 = 18 Unit

Remaining work = 54 - 18 = 36 unit 

Combined Efficiency = 3 - 2 = 1   

Remaining take time together = $$\frac{36}{1} = 36$$ minute