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Two pipes A and B can fill a tank in 20 minutes and 10 minutes respectively. Both pipes A and B are opened together for some time and then pipe B is turned off. If the tank is filled in 15 minutes, then find after how many minutes pipe B is turned off?

Solution

✅ Correct Option: 2

Pipe A fills the tank in 20 minutes, so its rate is 120\dfrac{1}{20} of the tank per minute.

Pipe B fills the tank in 10 minutes, so its rate is 110\dfrac{1}{10} of the tank per minute.


When both pipes work together, their combined rate is:

120+110\dfrac{1}{20} + \dfrac{1}{10}

=120+220= \dfrac{1}{20} + \dfrac{2}{20}

=320= \dfrac{3}{20} of the tank per minute


Let pipe B work for tt minutes before being turned off.

During these tt minutes, both pipes work together and fill:

t×320t \times \dfrac{3}{20} of the tank

After pipe B is turned off, pipe A works alone for (15−t)(15 - t) minutes and fills:

(15−t)×120(15 - t) \times \dfrac{1}{20} of the tank


The total work equals 1 full tank:

t×320+(15−t)×120=1t \times \dfrac{3}{20} + (15 - t) \times \dfrac{1}{20} = 1

3t20+15−t20=1\dfrac{3t}{20} + \dfrac{15 - t}{20} = 1

3t+15−t20=1\dfrac{3t + 15 - t}{20} = 1

2t+1520=1\dfrac{2t + 15}{20} = 1

2t+15=202t + 15 = 20

2t=52t = 5

t=2.5t = 2.5

Therefore, pipe B is turned off after 2.5 minutes.

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