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Tap X can fill a pot in 6 hours and tap Y can fill it in 8 hours. If both the taps are opened simultaneously, but after two hours tap X is closed, then in how many hours will tap Y take to fill the remaining part of the pot?1. 2 hours2. \(3\frac{1}{3}\) hours3. \(2\frac{2}{3}\)hours4. 4 hours |
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Answer» Correct Answer - Option 2 : \(3\frac{1}{3}\) hours Given: Tap X can fill a pot in 6 hours and tap Y can fill it 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 the pot filled in 2 hours by tap X and Y ⇒ \(2{\rm{}}\left( {\frac{1}{6}{\rm{}} + {\rm{}}\frac{1}{8}} \right){\rm{}} = {\rm{}}2{\rm{}}\left( {\frac{{4{\rm{\;}} + {\rm{\;}}3}}{{24}}} \right){\rm{}} = {\rm{}}\frac{7}{{12}}\) Remaining part = \(1 - \frac{7}{{12}}{\rm{}} = {\rm{}}\frac{5}{{12}}\) Time taken by tap Y in filling \(\frac{5}{{12}}\) part ⇒ \(\frac{5}{{12}} \times 8{\rm{}} = {\rm{}}\frac{{10}}{3}{\rm{}} = {\rm{}}3\frac{1}{3}\) hours ∴ The required time is \(3\frac{1}{3}\) hours. |
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