Question 1208988
.
An air rescue plane averages 300 miles per hour in still air. 
It carries enough fuel for 5 hours of flying time. 
If, upon takeoff, it encounters a head wind of 30 mi/hr, how far can it fly 
and return safely? (Assume that the wind remains constant.)
~~~~~~~~~~~~~~~~~~~~~


<pre>
Let d be the maximum one way distance, in miles.


The rate of the plane (relative the ground) against the wind is 300-30 = 270 mils per hour.

The time to fly the distance of d miles is  {{{d/270}}} hours.



The rate of the plane (relative the ground) with the wind is 300+30 = 330 mils per hour.

The time to fly the distance of d miles is  {{{d/330}}} hours.


The total time flying is  {{{d/270}}} + {{{d/330}}}.


The time equation is

    {{{d/270}}} + {{{d/330}}} = 5  hours.


To solve, multiply all the terms by  270*330.  You will get

    330d + 270d = 5*270*330

    600d = 445500

       d = 445500/600 = 742.5.


<U>ANSWER</U>.  The maximum one-way distance to fly is  742.5 miles.
</pre>

Solved.