| Two pipes A and B can fill a tank in 36 min and 45 min, respectively. Another pipe C can empty the tank in 30 min. First A and B are opened. After 7 min, C is also opened. The tank is filled up in [SSC (10+2) 2014] |
A) 39 min
B) 46 min
C) 40 min
D) 46 min
Correct Answer: A
Solution :
| Work done by pipes A and B in 1 min |
| \[=\frac{1}{36}+\frac{1}{45}=\frac{5+4}{180}=\frac{9}{180}=\frac{1}{20}\] |
| Work done by pipes A and B in 7 min \[=\frac{7}{20}\] |
| Left work \[=1-\frac{7}{20}=\frac{13}{20}\] |
| Now, work done by pipes A, B and C in 1 min |
| \[=\frac{1}{36}+\frac{1}{45}-\frac{1}{30}=\frac{5+4-6}{180}=\frac{3}{180}=\frac{1}{60}\] |
| \[\therefore \]Work done by A, B and C in x min = Left work |
| \[\Rightarrow \] \[x\times \frac{1}{60}=\frac{13}{20}\]\[\Rightarrow \]\[x=\frac{13}{20}\times 60=39\min \] |
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