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There are 3 taps, A, B and C, in a tank. These can fill the tank in 10 h, 20 h and 25 h, respectively. At first, all three taps are
opened simultaneously. After 2 h, tap C is closed and tap A and B keep running. After 4 h, tap B is also closed. The remaining
tank is filled by tap A alone. Find the percentage of work done by tap A itself.
75%
52%
72%
32%
- Tap A fills the tank in 10 hours (so, 1/10 per hour).
- Tap B fills the tank in 20 hours (so, 1/20 per hour).
- Tap C fills the tank in 25 hours (so, 1/25 per hour).
Let’s break it down:
- First 2 hours: All three are open, so work = 2 × (1/10 + 1/20 + 1/25)
= 2 × (0.1 + 0.05 + 0.04) = 2 × 0.19 = 0.38
- Next 2 hours: A and B are open, so work = 2 × (1/10 + 1/20) = 2 × 0.15 = 0.3
- Next 2 hours: Only A is open, so work = 2 × (1/10) = 0.2
- Total after 6 hours: 0.38 + 0.3 + 0.2 = 0.88 (88% filled)
Remaining: 1 - 0.88 = 0.12 (12% left)
- Tap A fills the remaining 12% alone. Time needed = 0.12 × 10 = 1.2 hours
- Total work by A:
- First 2 hours: 2 × (1/10) = 0.2
- Next 2 hours: 2 × (1/10) = 0.2
- Finally, 1.2 × (1/10) = 0.12
- Total = 0.2 + 0.2 + 0.2 + 0.12 = 0.72 (72%)
Option analysis:
- Option 1: 75% – Incorrect
- Option 2: 52% – Incorrect
- Option 3: 72% – Correct
- Option 4: 32% – Incorrect
By: santosh ProfileResourcesReport error
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