| A cistern has three pipes A, B and C. A and B can fill it in 3 and 4 h, respectively and C can empty it in 1 h. If the pipes are opened at 3 pm, 4 pm and 5 pm, respectively on the same day, then cistern will be empty at |
A) 7: 12 pm
B) 7: 15 pm
C) 7: 10 pm
D) 7: 18 pm
Correct Answer: A
Solution :
| 1 h work of pipe \[A=\frac{1}{3}\] |
| 1 h work of pipe \[B=\frac{1}{4}\] |
| 1 h work of pipe \[C=-\frac{1}{1}\] |
| Total part fill by A and B in \[2h=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}\] |
| and 1 h work of A, B and C \[=\frac{1}{3}+\frac{1}{4}-1=-\frac{5}{12}\] |
| i.e. \[\frac{5}{12}\] part of tank can he emptied in 1 h. |
| \[\therefore \] Time taken to empty \[\frac{11}{12}\text{part}=\frac{11}{12}\times \frac{12}{5}=\frac{11}{5}h\] |
| = 2 h 12 min |
| \[\therefore \] Required time \[\text{=}\,\,\text{5}\,\text{pm+2}\,\text{h}\,\text{12 min}\] |
| = 7: 12 pm |
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