Question 199712
Since distance = speed x time,  time = speed/distance +or- wind speed/distance

(+or- depending on whether it is in the same direction as plane, in which case it would be +, or opposite direction and -)

time = (speed +or- wind speed)/distance

(abbreviations used:  time: t, speed in still air: s, wind speed: ws, distance: d)

Let's look at the first scenario: t = 4, ws = -40.  Therefore,  since t = (s - ws)/d, we get this equation:

4 = (s - 40)/d  ----) this has 2 variable, which means we will need a system of equations to solve it.

Let's look at the second part of the question to get the second equation.  Since t=(s+ws)/d,  we get this:

3 = (s + 40)/d

Now we have our two equations. We can use algebra to solve for s, the wind speed in still air.

4d = s-40 , so  d = (s-40)/4  using eq 1.   Also, 3d = s+40, so d = (s+40)/3  using eq. 2

Now we have 2 equations solving for d. Since distance stays the same, we can set these equal to each other to solve for the wind speed in still air:

(s-40)/4 = (s+40)/3 

3(s-40) = 4(s+40)

3s - 120 = 4s + 160

3s - 280 = 4s

Therefore, s = 280 km/h  (ignore -, since only asking for speed, not direction)