Two water pumps are filling a pool. One of the pumps is high power and can fill the pool 5 hours before the other can do. However, they both working together can fill half of the pool in 3 hours. In how many hours the high power pump can fill the pool?
If the high power pump can fill it in x hours, then the slower pump takes x+5 hours. So, we have
1/x + 1/(x+5) = (1/2)/3
x = 10
To solve this problem, let's assign variables to represent the rates at which each pump fills the pool. Let x denote the rate of the slow pump (in pools per hour) and y denote the rate of the high power pump (in pools per hour).
Given that the high power pump takes 5 hours less than the slow pump to fill the pool, we can express their rates as follows:
Rate of slow pump: 1 pool / (x pools per hour)
Rate of high power pump: 1 pool / (y pools per hour)
We are also given that both pumps working together can fill half of the pool in 3 hours. Since the combined rate of the two pumps is the sum of their individual rates, we can set up the equation:
(1/x + 1/y) * 3 = 1/2
Simplifying the equation, we get:
3/x + 3/y = 1/2
Now, let's solve for y in terms of x:
3/x + 3/y = 1/2
Multiply both sides by 2xy to eliminate the denominators:
6y + 6x = xy
Rearrange the equation:
xy - 6y - 6x = 0
Factor out y:
y(x - 6) - 6x = 0
y(x - 6) = 6x
y = (6x) / (x - 6)
Since we want to find the rate at which the high power pump fills the pool, we need to find y when x = 5 (since the high power pump takes 5 hours less than the slow pump):
y = (6 * 5) / (5 - 6)
y = -30
However, since y represents the rate at which the pump fills the pool, it cannot be negative. This means that our initial assumption that the high power pump takes 5 hours less than the slow pump was incorrect.
Therefore, there is no valid solution to this problem based on the given information.