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Tap A and Tap B can fill a cistern into 6 hours and 2 hours respectively. Tap C can empty it in 3 hours. If all three taps are open alternatively 1 hour but start with A, then the whole tank will fill in how many hours?1. 11 hours2. 10 hours3. 7 hours4. 9 hours5. None of these |
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Answer» Correct Answer - Option 5 : None of these Given: Tap A can fill a tank in = 6 hours Tap B can fill a tank in = 2 hours Tap C can empty a tank in = 3 hours Calculation: Let the total capacity of cistern be LCM (6, 2, 3) = 6 litres Now, Tank filled by A in 1 hour = 6/6 ⇒ 1 litre/hour Tank filled by B in 1 hour = 6/2 ⇒ 3 litre/hour Tank empty by C in 1 hour = 6/3 ⇒ (– 2) litre/hour Tank filled by A, B & C in (1 + 1 +1) hour = (1 + 3 – 2) ⇒ 2 litre/3 hour In the first 3 hour tank filled by 2 litre In next 1 hour by tap A tank filled by 1 litre ⇒ In (3 + 1) = 4 hour tank filled by (2 + 1) = 3 litre In next 1 hour by tap B tank filled by 3 litre ⇒ In (4 + 1) = 5 hour tank filled by (3 + 3) = 6 litre ⇒ In 5 hour the tank become fill ∴ If all three taps are open alternatively 1 hour but start with A, then the whole tank will fill in 5 hours.
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