Question 299199
.
An airplane took 4 hours to fly 1200 miles agianst a headwind. The return trip with the wind took 3 hours. 
Find the speed of the plane in still air and the speed of the wind. 
What is the speed of the {{{highlight(cross(wind))}}} <U>plane</U> in still air? What is the wind speed?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;To the visitor: 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"against a headwind" is not a good term.  Say simply "against the wind".



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Notice that I edited your post to make sense from nonsense.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution by @mananth uses the "zigzag" logic, which is not recommended.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;To teach in a right way, I use a straightforward logic without zigzags. It is the style, which is recommended.



<pre>
Let the rate of the plane in still air be x miles per hour.
Let the rate of the wind be y miles per hour.


The effective rate of the plane against the wind is  (x-y).
According to the problem, it is 

    x - y = {{{1200/4}}} = 300 mph.      (the distance divided by the time)


The effective rate of the plane with the wind is  (x+y).
According to the problem, it is 

    x + y = {{{1200/3}}} = 400 mph.      (the distance divided by the time)


Thus you have these two equations to find x and y

    x + y = 400,    (1)

    x - y = 300.    (2)


To find x, add equations (1) and (2).  You will get

    2x = 400 + 300 = 700,  x = 700/2 = 350.


To find y, subtract equation (2) from equation (1).  You will get

    2y = 400 - 300 = 100,  y = 100/2 = 50.


<U>ANSWER</U>.  The rate of the plane in still air is 350 mph.

         The rate of the wind is 50 mph.
</pre>

Solved by a standard way using a straightforward logic.