Question 268621
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 and the wind remains constant, how far can the plane fly and then return safely? 
Is the rate simply 300-30? How do I find the distance traveled?

The time the plane can stay in the air is 5 hours. As you determined, the speed that the plane can fly against the wind is 300-30. But on a round trip the plane will fly back with a speed of 300+30. 

Let t be the time taken for the trip against the wind. Then the part of the trip with the wind will be 5-t since the whole trip will take 5 hours. Let's say the distance for each leg of the trip is d.  Then we have:

1.) d = rate*time = (300-30)*t  (against the wind)
2.) d = rate*time = (300+30)*(5-t) (with the wind)

Since both expressions are equal to d we have: 

270*t = 330*(5-t)

Solve the above for t then calculate the distance d by substituting this value for t in either 1.) or 2.) above.

Remember the TOTAL distance traveled is 2*d.