| Pipe A can fill a tank in 4h and pipe B can fill it in 6h. If they are opened on alternate hours and if pipe A is opened first, then in how many hours the tank shall be full? [SSC (CGL) 2015] |
A) \[4\frac{1}{2}h\]
B) \[4\frac{2}{3}h\]
C) \[3\frac{1}{2}h\]
D) \[3\frac{1}{4}h\]
Correct Answer: B
Solution :
| Part filled by pipe A in \[1\,h=\frac{1}{4}\] |
| Part filled by pipe B in \[1\,h=\frac{1}{6}\] |
| Part filled by \[(A+B)\] in \[2\,h=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12}\] |
| \[\therefore \] Part filled by \[(A+B)\] in \[4h=\frac{5}{12}\times 2=\frac{5}{6}\] |
| Remaining part \[=1-\frac{5}{6}=\frac{1}{6}\] |
| Now, it is the turn of A. |
| Time taken by A to fill \[\frac{1}{6}\] part of the tank |
| \[=\frac{1}{6}\times 4=\frac{2}{3}h\] |
| \[\therefore \] Total time taken \[=\left( 4+\frac{2}{3} \right)h=4\frac{2}{3}h\] |
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