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Pipe A and B can fill a tank in 15 hours and 20 hours respectively. Pipe C can empty the tank in 25 hours. Pipes A, B and C works together but pipe C is closed after 10 hours , then time taken by pipe A and B to fill the remaining tank is:

Solution

✅ Correct Option: 2

Rate of Pipe A =115= \dfrac{1}{15} tank per hour

Rate of Pipe B =120= \dfrac{1}{20} tank per hour

Rate of Pipe C =−125= -\dfrac{1}{25} tank per hour (negative since it empties)


Combined rate when all three pipes work together:

115+120−125\dfrac{1}{15} + \dfrac{1}{20} - \dfrac{1}{25}

Finding LCM of 15, 20, and 25 gives 300.

20300+15300−12300\dfrac{20}{300} + \dfrac{15}{300} - \dfrac{12}{300}

=23300= \dfrac{23}{300} tank per hour


Work done in first 10 hours with all three pipes:

23300×10\dfrac{23}{300} \times 10

=230300= \dfrac{230}{300}

=2330= \dfrac{23}{30} of the tank


Remaining work for pipes A and B:

1−23301 - \dfrac{23}{30}

=730= \dfrac{7}{30} of the tank


Combined rate of pipes A and B only:

115+120\dfrac{1}{15} + \dfrac{1}{20}

Finding LCM of 15 and 20 gives 60.

460+360\dfrac{4}{60} + \dfrac{3}{60}

=760= \dfrac{7}{60} tank per hour


Time taken by A and B to fill remaining tank:

Time=WorkRate\text{Time} = \dfrac{\text{Work}}{\text{Rate}}

=7/307/60= \dfrac{7/30}{7/60}

=730×607= \dfrac{7}{30} \times \dfrac{60}{7}

=6030= \dfrac{60}{30}

=2= 2 hours

Therefore, pipes A and B take 2 hours to fill the remaining tank.

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