A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:6 hours10 hours15 hours30 hours
Question
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:6 hours10 hours15 hours30 hours
Solution
Let's denote the rates at which the pipes can fill the tank as follows:
- Pipe 1 fills the tank at a rate of 1/A per hour
- Pipe 2 fills the tank at a rate of 1/B per hour
- Pipe 3 fills the tank at a rate of 1/C per hour
From the problem, we know that:
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The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. This can be written as 1/A + 1/B = 1/C.
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The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. This can be written as B = A - 5 and B = C + 4.
Substituting B = A - 5 and B = C + 4 into the first equation, we get:
1/A + 1/(A - 5) = 1/(A - 9)
Solving this equation for A, we find that A = 15. Therefore, the first pipe can fill the tank in 15 hours.
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