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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 pipe B is turned off. If the tank is filled in 15 minutes, then after how many minutes pipe B is turned off?

Solution

Correct Option: 4

Pipe A fills the tank in 20 minutes, so in 1 minute, Pipe A fills 120\frac{1}{20} of the tank.

Pipe B fills the tank in 30 minutes, so in 1 minute, Pipe B fills 130\frac{1}{30} of the tank.


When both pipes work together, they fill per minute:

120+130\frac{1}{20} + \frac{1}{30}

=360+260= \frac{3}{60} + \frac{2}{60}

=560= \frac{5}{60}

=112= \frac{1}{12} of the tank


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

Work done when both pipes operate (for tt minutes):

=t×112=t12= t \times \frac{1}{12} = \frac{t}{12}

Work done when only Pipe A operates (for remaining time):

Remaining time =15t= 15 - t minutes

=(15t)×120=15t20= (15 - t) \times \frac{1}{20} = \frac{15 - t}{20}


Total work equals 1 full tank:

t12+15t20=1\frac{t}{12} + \frac{15 - t}{20} = 1

Multiplying by 60:

5t+3(15t)=605t + 3(15 - t) = 60

5t+453t=605t + 45 - 3t = 60

2t=152t = 15

t=7.5t = 7.5

Therefore, Pipe B is turned off after 7.5 minutes.

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