Pipe A can fill a tank in 18 minutes, while pipe B can empty the completely filled tank in 20 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 20 minutes.

Pipe A is opened first, and after 6 minutes, Pipe B is also opened.

Solution:

Rate of A = $$\frac{1}{18}$$

Rate of B = $$\frac{1}{20}$$​​

Combined rate = $$\frac{1}{18} - \frac{1}{20}$$​​

​$$= \frac{20 - 18}{18 \times 20}$$​​

​$$= \frac{2}{360} = \frac{1}{180}$$​​

So, when both pipes are open, the tank fills at a rate of $$\frac{1}{180}$$​ of the tank per minute.

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

Therefore, after 6 minutes,  of the tank is still empty.

Time to fill the remaining $$\frac{2}{3}$$​ of the tank with both pipes open:

The combined rate is $$\frac{1}{180}$$​ of the tank per minute, so the time taken to fill $$\frac{2}{3}$$​ of the tank is:

Time = $$\frac{\frac{2}{3}}{\frac{1}{180}}$$​​

​$$= \frac{2}{3} \times 180 = 120 \text{ minutes}$$​​

Thus, the remaining tank will be completely filled in 120 minutes after Pipe B is also opened 

Alternate Solution: 

LCM of 18 and 20: 180 

So, the total capacity of the tank is 180 units.

Pipe A’s rate of filling = $$\frac{180}{18} = 10$$​​

Pipe B’s rate of emptying =$$\frac{180}{20 }$$​ = 9

Pipe A works for 6 minutes = 10 ×6 = 60 units

After 6 minutes, 60 units have been filled, so 120 units remain.

Net rate when both pipes are open = 10 − 9 =1 

Time to fill remaining = $$\frac{120}{1 }$$​ = 120 minutes

Total time to fill the tank completely = 6 minutes + 120 minutes = 126 minutes