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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 mi
Question

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?​

A.

120

B.

137

C.

107

D.

127

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 = 118\frac{1}{18}

Rate of B = 120\frac{1}{20}​​

Combined rate = 118120\frac{1}{18} - \frac{1}{20}​​

=201818×20 = \frac{20 - 18}{18 \times 20}​​

=2360=1180= \frac{2}{360} = \frac{1}{180}​​

So, when both pipes are open, the tank fills at a rate of 1180\frac{1}{180}​ of the tank per minute.

In 6 minutes, Pipe A fills = 6×118=618=136 \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 23 \frac{2}{3}​ of the tank with both pipes open:

The combined rate is 1180\frac{1}{180}​ of the tank per minute, so the time taken to fill 23\frac{2}{3}​ of the tank is:

Time = 231180\frac{\frac{2}{3}}{\frac{1}{180}}​​

=23×180=120 minutes= \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 = 18018=10 \frac{180}{18} = 10​​

Pipe B’s rate of emptying =18020 \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 = 1201\frac{120}{1 }​ = 120 minutes

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

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