Question
- Two pipes A and B can separately fill a cistern in 10 minutes and 15 minutes respectively. A person opens both the pipes together. When the cistern should have been full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when full?
Answer: Option B
Let the capacity of the cistern be C, the rate of pipe A be R1, and the rate of pipe B be R2.From the given information, we have:R1 = C/10 (since pipe A can fill the cistern in 10 minutes)R2 = C/15 (since pipe B can fill the cistern in 15 minutes)Let the rate of the waste pipe be R3 (in units of the cistern per minute).When all three pipes are open, the effective rate of filling the cistern is (R1 + R2 - R3) (in units of the cistern per minute).When the waste pipe is open, the effective rate of emptying the cistern is R3 (in units of the cistern per minute).When the waste pipe is closed, the effective rate of filling the cistern is (R1 + R2) (in units of the cistern per minute).
Using the above information, we can write the following equations:C = (R1 + R2 - R3) * t1 (since the cistern should have been full, but due to the waste pipe, it took t1 minutes to fill)C = (R1 + R2) * t2 (since the waste pipe was closed and it took t2 minutes to fill)C = R3 * t3 (since the waste pipe can empty the cistern, it takes t3 minutes to empty it)
We are given that t1 = t2 + 4 (since it took an additional 4 minutes to fill the cistern after the waste pipe was closed)Substituting the above equations, we get:(R1 + R2 - R3) * (t2 + 4) = (R1 + R2) * t2R1 + R2 - R3 = R1 + R2 - (C/t3)R3 = C/t3
Solving the above equations, we get t3 = 9 minutes.
Therefore, the correct option is B. 9 minutes.If you think the solution is wrong then please provide your own solution below in the comments section .
Was this answer helpful ?
Let the capacity of the cistern be C, the rate of pipe A be R1, and the rate of pipe B be R2.From the given information, we have:R1 = C/10 (since pipe A can fill the cistern in 10 minutes)R2 = C/15 (since pipe B can fill the cistern in 15 minutes)Let the rate of the waste pipe be R3 (in units of the cistern per minute).When all three pipes are open, the effective rate of filling the cistern is (R1 + R2 - R3) (in units of the cistern per minute).When the waste pipe is open, the effective rate of emptying the cistern is R3 (in units of the cistern per minute).When the waste pipe is closed, the effective rate of filling the cistern is (R1 + R2) (in units of the cistern per minute).
Using the above information, we can write the following equations:C = (R1 + R2 - R3) * t1 (since the cistern should have been full, but due to the waste pipe, it took t1 minutes to fill)C = (R1 + R2) * t2 (since the waste pipe was closed and it took t2 minutes to fill)C = R3 * t3 (since the waste pipe can empty the cistern, it takes t3 minutes to empty it)
We are given that t1 = t2 + 4 (since it took an additional 4 minutes to fill the cistern after the waste pipe was closed)Substituting the above equations, we get:(R1 + R2 - R3) * (t2 + 4) = (R1 + R2) * t2R1 + R2 - R3 = R1 + R2 - (C/t3)R3 = C/t3
Solving the above equations, we get t3 = 9 minutes.
Therefore, the correct option is B. 9 minutes.If you think the solution is wrong then please provide your own solution below in the comments section .
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution