Two pipes A and B alone can fill the tank in 6 h and 4 h, respectively. If they work alternatively for one hour each and first pipe A works, then time taken by them to fill the tank is |
A) 4 h
B) 5 h
C) \[4\frac{1}{2}\,h\]
D) \[5\frac{1}{2}\,h\]
Correct Answer: B
Solution :
1 h work of pipe \[A=\frac{1}{6}\] |
and 1 h work of pipes \[B=\frac{1}{4}\] |
\[\therefore \] 2 h work of both pipes \[=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\] |
4 h work of both pipes \[=\frac{5}{12}\times 2=\frac{5}{6}\] |
\[\therefore \] Remaining work \[=1-\frac{5}{6}=\frac{1}{6}\] |
Now, it is turn of pipe A and so the remaining work will be completed in 1 h. |
Total time taken to fill the tank \[=4+1=5\,h\] |
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