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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%
- The tank fills at different rates with each tap:
- Tap A fills the tank in 10 hours (1/10 of the tank per hour).
- Tap B fills the tank in 20 hours (1/20 of the tank per hour).
- Tap C fills the tank in 25 hours (1/25 of the tank per hour).
- When all three taps are open:
- Combined rate = 1/10 + 1/20 + 1/25 = (10 + 5 + 4)/100 = 19/100 per hour.
- In the first 2 hours, they fill 2 * 19/100 = 38/100 of the tank.
- Tap C is then closed:
- Combined rate of A and B = 1/10 + 1/20 = 3/20 per hour.
- In the next 4 hours, they fill 4 * 3/20 = 12/20 = 60/100 of the tank.
- Together in 6 hours, 38/100 + 60/100 = 98/100 of the tank is filled.
- Remaining 1/50 of the tank is filled by tap A alone:
- Tap A fills 1/10 per hour, so it takes 1/5 hours more.
- Total work by tap A:
- In 6 hours, tap A fills 1/10 * 6 = 6/10.
- Plus the extra 1/50, tap A fills 6/10 + 1/50 = 31/50 of the tank.
- Calculate percentage contributed by tap A:
- (31/50) * 100 = 62%.
- Option 2 - 52% is correct.
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