A pipe P can fill a tank in 12 min and another pipe R can fill it in 15 min. But, the 3rd pipe M can empty it in 6 min. The 1st two pipes P and R are kept open for double the 2.5 min in the beginning and then the 3rd pipe is also opened. In what time is the tank emptied?
Correct Answer: Description for Correct answer:
According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min
= \( \Large 5 \times \left(\frac{1}{12}+\frac{1}{15}\right) = 5 \times \left(\frac{5+4}{60}\right) \)
= \( \Large 5 \times \frac{9}{60} = \frac{3}{4} \)
Part emptied in 1 min when R R and all are opened.
= \( \Large \frac{1}{6}- \left(\frac{1}{12}+\frac{1}{15}\right)=\frac{1}{6}- \left(\frac{5+4}{60}\right) \)
=\( \Large \left(\frac{1}{6}-\frac{3}{2}\right)=\frac{1}{60} \)
One-sixtieth part is emptied in 1 min.
Therefore, Three-fourth part will be emptied in\( \Large 60 \times \frac{3}{4}=15 \times 3 \) = 45 min.
Part of solved Pipes and Cisterns questions and answers :
>> Aptitude >> Pipes and Cisterns