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?
a) $$4\frac{1}{2}$$ hours
b) $$5\frac{1}{2}$$ hours
c) $$6\frac{1}{2}$$ hours
d) None of these

Answer: a
Explanation:
$$\eqalign{ & {\text{Net part filled in 1 hour}} \cr & = \frac{2}{3} – \left( {\frac{2}{9} + \frac{1}{3}} \right) \cr & = \left( {\frac{2}{3} – \frac{5}{9}} \right) \cr & = \frac{1}{9} \cr & \therefore \,\frac{1}{9}\,:\,\frac{1}{2}\,::\,1\,:\,x \cr & {\text{or}}\,\,\,x = \left( {\frac{1}{2} \times 9} \right) = 4\frac{1}{2}{\text{ hours}} \cr & {\text{So, the tank will be emptied in}} \cr & {\text{ = }}4\frac{1}{2}{\text{ hours}} \cr} $$