There are 3 inlet pipes. The first 2 pipes working simultaneously can fill a pool in the same time as the third pipe alone, while the second pipe alone fills the pool 5 hours faster than the first and 4 hours slower than the third. In what time will second and third pipe working together fill the empty pool?
Question
There are 3 inlet pipes. The first 2 pipes working simultaneously can fill a pool in the same time as the third pipe alone, while the second pipe alone fills the pool 5 hours faster than the first and 4 hours slower than the third. In what time will second and third pipe working together fill the empty pool?
Solution
Let's denote the time it takes for the first pipe to fill the pool as x hours.
According to the problem, the second pipe fills the pool 5 hours faster than the first pipe, so it takes (x-5) hours.
The third pipe fills the pool 4 hours faster than the second pipe, so it takes (x-5-4) = (x-9) hours.
The problem also states that the first two pipes working together can fill the pool in the same time as the third pipe alone. This means that the sum of their rates equals the rate of the third pipe.
We know that the rate is the reciprocal of the time, so we can write this equation as 1/x + 1/(x-5) = 1/(x-9).
Solving this equation for x, we get x = 15 hours.
So, the second pipe takes (15-5) = 10 hours to fill the pool, and the third pipe takes (15-9) = 6 hours.
If the second and third pipes work together, their combined rate is 1/10 + 1/6 = 4/15 pools per hour.
Therefore, the time it takes for the second and third pipes to fill the pool together is the reciprocal of their combined rate, or 15/4 = 3.75 hours.
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