A tank can be filled by pipe A in 2 h and pipe B in 6 h. At 10 am pipe A was opened. At what time will tank be filled if pipe B is opened at 11 am? [SSC (CGL) 2013] |
A) 12: 45 am
B) 5 pm
C) 11: 48 am
D) 12 pm
Correct Answer: C
Solution :
Given, tank filled by A in 2h and tank filled by B in 6 h. |
Then, part of the tank filled in 1 h by pipe A \[=\frac{1}{2}\] |
Part of the tank filled in 1 h by pipe B \[=\frac{1}{6}\] |
According to the question, |
Pipe A was opened at 10 am and till, 11 am, pipe A alone worked. So, part of the tank filed between 10 am to 11 am by pipe A \[=\frac{1}{2}\] |
\[\therefore \]Remaining work \[=1-\frac{1}{2}=\frac{1}{2}\] |
Now, remaining work will be done by both pipes |
(A + B). |
Tank filled by (A + B) working together |
\[=\frac{xy}{x+y}\] [where, \[x=2,\]\[y=6\]] |
\[=\frac{2\times 6}{6+2}=\frac{12}{8}=\frac{3}{2}h\] |
Hence, half tank filled by (A + B) |
\[=\frac{3}{2\times 2}h=\frac{3}{4}h=\frac{3}{4}\times 60=45\,\,\min \] |
Hence, the tank will be filled at 11: 45 am. |
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