From two holes atop a covered aquarium certain amount of water can evaporate in 6 minutes,however one hole takes 5 minutes more two evaporate the same amount of water than the other.find the time taken by each whole to evaporate the same amount of water?
1/x + 1/(x+5) = 1/6
To solve this problem, we'll use the concept of rates. Let's assume that the rate of water evaporated from the first hole is x amount per minute.
Since the first hole takes 5 minutes less than the second hole to evaporate the same amount of water, we can say that the rate of water evaporated from the second hole is (x - 1) amount per minute.
Now, let's calculate the total amount of water evaporated in 6 minutes.
For the first hole:
Amount evaporated = Rate × Time = x × 6 = 6x
For the second hole:
Amount evaporated = Rate × Time = (x - 1) × (6 + 5) = (x - 1) × 11
Since we know that the same amount of water evaporates from both holes in 6 minutes, we can set up an equation:
6x = 11(x - 1)
Solving this equation will give us the value of x, which is the rate of water evaporated from the first hole. Once we find x, we can calculate the rate from the second hole as (x - 1).
Let's solve the equation:
6x = 11x - 11
11 - 6 = 11x - 6x
5 = 5x
x = 1
Now that we have x = 1, we can substitute it back to find the rates of evaporation from each hole.
Rate of first hole = x = 1 amount per minute
Rate of second hole = x - 1 = 1 - 1 = 0 amount per minute
Therefore, the first hole takes 1 minute to evaporate the same amount of water as the second hole. The second hole does not evaporate any water in any amount of time, as its rate is 0.
If the faster time is x, then
1/x + 1/(x+5) = 1/6