There are three taps A, B and C in the tank. These taps can fill the tank in 10 hours, 20 hours and 25 hours respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours, tap B is also closed. The remaining tank is filled by Tap A alone. Find the percentage of the work done by Tap A itself.
Question
There are three taps A, B and C in the tank. These taps can fill the tank in 10 hours, 20 hours and 25 hours respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours, tap B is also closed. The remaining tank is filled by Tap A alone. Find the percentage of the work done by Tap A itself.
Solution
Sure, let's solve this step by step:
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First, we need to find the rate at which each tap fills the tank. This is done by taking the reciprocal of the time it takes for each tap to fill the tank. So, Tap A fills the tank at a rate of 1/10 tank/hour, Tap B at 1/20 tank/hour, and Tap C at 1/25 tank/hour.
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For the first 2 hours, all three taps are open. So in one hour, they collectively fill 1/10 + 1/20 + 1/25 = 0.12 of the tank. Therefore, in 2 hours, they fill 2 * 0.12 = 0.24 or 24% of the tank.
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After 2 hours, Tap C is closed, and Taps A and B continue to fill the tank for another 4 hours. So in one hour, Taps A and B collectively fill 1/10 + 1/20 = 0.15 of the tank. Therefore, in 4 hours, they fill 4 * 0.15 = 0.6 or 60% of the tank.
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At this point, 24% + 60% = 84% of the tank is filled, and Tap A continues to fill the remaining 16% of the tank alone.
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Therefore, the percentage of the work done by Tap A itself is the work done in the first 2 hours (with Taps B and C), plus the work done in the next 4 hours (with Tap B), plus the work done alone. This is (0.12 * 2/3) * 2 + (0.15 * 1/2) * 4 + 0.16 = 0.08 + 0.3 + 0.16 = 0.54 or 54%.
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