| Three taps are fitted to a cistern. The empty cistern is filled by the first and second taps in 3 and 4 h, respectively. The full cistern is emptied by the third tap in 5 h. If all three taps are opened simultaneously, then the empty cistern will be filled up in [SSC (CGL) 2013] |
A) \[1\frac{14}{23}h\]
B) \[2\frac{14}{23}h\]
C) \[2\,\,\text{h}\,\,40\min \]
D) \[1\,\,\text{h}\,\,56\min \]
Correct Answer: B
Solution :
| Given, time taken by first tap to fill the cistern \[=3\,\,\text{h}\] |
| Part of cistern filled by first tap in \[1\,\,\text{h}=\frac{1}{3}\] ... (i) |
| (if a pipe fills a tank in x h, then the part of tank filled in \[1\,\,\text{h}=\frac{1}{x}\]) |
| Similarly, part of cistern filled by second tap in \[1\,\,\text{h}=\frac{1}{4}\] (ii) |
| and part of cistern emptied by third tap in \[1\,\,\text{h}=\frac{1}{5}\] (iii) |
| Now, part of cistern filled by all taps In 1 h |
| \[=\frac{1}{3}+\frac{1}{4}-\frac{1}{5}=\frac{20+15-12}{60}=\frac{23}{60}\] |
| Hence, all three taps can fill the cistern in \[\frac{60}{23}\]or\[2\frac{14}{23}h.\] |
| (If a pipe fills \[\frac{1}{x}\] part of the tank in 1 h, then the time taken by the pipe to fill the full tank \[=x\,\,h\]) |
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