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Pipe X can fill an empty pool in 6 hours and pipe Y in 8 hours. If both the pipes are opened and after two hours pipe X is closed, then how much time Y will take to fill the remaining tank?1. \(7\frac{1}{2}\) hours2. \({\rm{\;}}2\frac{2}{5}\) hours3. \(2\frac{1}{3}\) hours4. \(3\frac{1}{3}\) hours |
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Answer» Correct Answer - Option 4 : \(3\frac{1}{3}\) hours Given: Pipe X can fill an empty pool in 6 hours and pipe Y in 8 hours Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of pool filled by pipes X and Y in 2 hours ⇒ \(2{\rm{\;}}\left( {\frac{1}{6}{\rm{}} + {\rm{}}\frac{1}{8}} \right)\) ⇒ \(2{\rm{}}\left( {\frac{{4{\rm{\;}} + {\rm{\;}}3}}{{24}}} \right){\rm{}} = {\rm{}}\frac{7}{{12}}\) Remaining part = \(1 - \frac{7}{{12}}{\rm{}} = {\rm{}}\frac{5}{{12}}\) This part is filled by pipe Y. Required time = \(\frac{5}{{12}} \times 8\) ⇒ 10/3 hours = \(3\frac{1}{3}\) hours ∴ Y will take \(3\frac{1}{3}\) hours to fill the remaining tank. |
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