recall your d = rt, distance = rate * time
so... hmm if the plane has a still air speed of say "r", when it's going with the wind, is not really going "r" fast, is going "r + 40 ", because the wind is adding 40mph to its speed, is really moving.
and when the plane is going against the wind, is not really going "r" fast either, is going " r - 40 ", because the wind is eroding speed from it.
now, it went 2240 miles for say "t" hours, and it did 1920 miles for the same length of time, "t" hours.
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{with the wind}&2240&r+40&t\\
\textit{against the wind}&1920&r-40&t
\end{array}
\\\\\\
\begin{cases}
2240=(r+40)t\implies \frac{2240}{r+40}=\boxed{t}\\
1920=(r-40)t\\
----------\\
1920=(r-40)\left( \boxed{\frac{2240}{r+40}} \right)
\end{cases}
\\\\\\
1920=\cfrac{(r-40)2240}{r+40}\implies 1920(r+40)=(r-40)2240
\\\\\\
1920r+76800=2240r-89600\implies 166400=320r
\\\\\\
\cfrac{166400}{320}=r[/tex]
and surely you know how much that is.
