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A vessel has three pipes connected to it , two to supply liquid and one to draw liquid. The first alone can fill the vessel in $$4\frac{1}{2}$$ hours, the second in 3 hours and the third can empty it in $$1\frac{1}{2}$$ hours. If all the pipes are opened simultaneously when the vessel is half full, how soon will it be emptied?
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& {\text{Net part filled in 1 hour}} \cr
& = \frac{2}{3} - \left( {\frac{2}{9} + \frac{1}{3}} \right) \cr
& = \left( {\frac{2}{3} - \frac{5}{9}} \right) \cr
& = \frac{1}{9} \cr
& \therefore \,\frac{1}{9}\,:\,\frac{1}{2}\,::\,1\,:\,x \cr
& {\text{or}}\,\,\,x = \left( {\frac{1}{2} \times 9} \right) = 4\frac{1}{2}{\text{ hours}} \cr
& {\text{So, the tank will be emptied in}} \cr
& {\text{ = }}4\frac{1}{2}{\text{ hours}} \cr} $$
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