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Two pipes A and B can fill a tank in 20 minutes and 30 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 time for which the pipe B works is:

Solution

✅ Correct Option: 2

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

Pipe B fills the tank in 30 minutes, so its rate is 130\frac{1}{30} tank per minute.


Let pipe B work for tt minutes.

Pipe A works for the entire 15 minutes.

Pipe B works for tt minutes, then is turned off.


Work done by Pipe A:

120×15\frac{1}{20} \times 15

=1520= \frac{15}{20}

=34= \frac{3}{4} tank


Work done by Pipe B:

130×t\frac{1}{30} \times t

=t30= \frac{t}{30} tank


Together they fill 1 complete tank:

34+t30=1\frac{3}{4} + \frac{t}{30} = 1

t30=1−34\frac{t}{30} = 1 - \frac{3}{4}

t30=14\frac{t}{30} = \frac{1}{4}

t=304t = \frac{30}{4}

t=7.5t = 7.5 minutes

t=712t = 7\frac{1}{2} minutes


Therefore, pipe B works for 7127\frac{1}{2} minutes.

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