An aircraft flew 5 hours with the wind. The return trip took 6 hours against the wind. If the speed of the plane in still air is 250 miles per hour more than the speed of the​ wind, find the wind speed and the speed of the plane in still air.

Answer 1

Let's call the speed of the wind "w" (in miles per hour) and the speed of the plane in still air "p" (in miles per hour).

Since the plane flew 5 hours with the wind, the distance it traveled can be expressed as 5(p+w). Similarly, the distance traveled in 6 hours against the wind can be expressed as 6(p-w).

As per the problem, the speed of the plane in still air is 250 miles per hour more than the speed of the wind, which can be written as p = w + 250.

We can now solve the problem using these three equations:

1. Distance with the wind: 5(p+w) = D
2. Distance against the wind: 6(p-w) = D
3. Speed in still air: p = w + 250

Since the distance is the same in both cases, we can equate equation 1 and equation 2:
5(p+w) = 6(p-w)

Simplifying this equation, we get:
5p + 5w = 6p - 6w
11w = p

Now, we can substitute the value of p from equation 3 into this equation to solve for w:
11w = w + 250
10w = 250
w = 25

Substituting the value of w into equation 3, we get:
p = w + 250
p = 25 + 250
p = 275

Therefore, the wind speed is 25 miles per hour and the speed of the plane in still air is 275 miles per hour.