| A swimming pool has 3 drain pipes. The first two pipes A and B, operating simultaneously can empty the pool in half the time that C, the 3rd pipe, alone takes to empty it. Pipe A, working alone, takes half the time taken by pipe B. Together they take 6 h 40 min to empty the pool. Time taken by pipe A to empty the pool. Time taken (in hours) by pipe A to empty the pool, is [SSC (CGL) 2012] |
A) 15
B) 10
C) 30
D) 7
Correct Answer: A
Solution :
| Let time taken by pipe \[B=2x\,\,h\] |
| Part of the tank emptied by B in \[1\,\,h=\frac{1}{2x}\] |
| Time taken by pipe \[A=x\,\,h\] |
| Similarly, by A in \[1\,\,h=\frac{1}{x}\] |
| \[\therefore \]Time taken by pipe \[C=\frac{2}{\frac{1}{2x}+\frac{1}{x}}\] |
| \[=\frac{2}{\frac{1+2}{2x}}=\frac{4x}{3}h\] |
| Now, part of the tank filled by all together |
| \[\frac{1}{x}+\frac{1}{2x}+\frac{3}{4x}=\frac{1}{6+\frac{2}{3}}\] |
| \[\Rightarrow \] \[\frac{4+2+3}{4x}=\frac{1}{6+\frac{2}{3}}\] |
| \[\Rightarrow \] \[\frac{4+2+3}{4x}=\frac{3}{20}\] |
| \[\Rightarrow \] \[9\times 20=4x\times 3\] |
| \[\therefore \] \[x=\frac{9\times 20}{4\times 3}=15\] |
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