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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 hours
2. 10 hours
3. 7 hours
4. 9 hours
5. None of these

1 Answer

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Best 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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