| 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? |
A) 30 min
B) 25 min
C) 45 min
D) 35 min
E) None of these
Correct Answer: C
Solution :
| According to the question, |
| Double the \[2.5\min =5\min \] |
| Now, part filled in 5 min\[=5\times \left( \frac{1}{12}+\frac{1}{15} \right)\] |
| \[=5\times \left( \frac{5+4}{60} \right)=5\times \frac{9}{60}=\frac{3}{4}\] |
| Part emptied in 1 min when P, R and M, all are opened. |
| \[=\frac{1}{6}-\left( \frac{1}{12}+\frac{1}{15} \right)=\frac{1}{6}-\left( \frac{5+4}{60} \right)=\left( \frac{1}{6}-\frac{3}{20} \right)=\frac{1}{60}\] |
| One-sixtieth part is emptied in 1 min. |
| \[\therefore \]Three-fourth part will be emptied in |
| \[60\times \frac{3}{4}=15\times 3=45\,\,\min \] |
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