<PRE><B>Question 459758</B>
AN AIRCRAFT FLEW 4 HOURS WITH THE WIND. THE RETURN TRIP TOOK 5 HOURS AGAINST THE WIND. IF THE SPEED OF OF THE PLANE IN STILL AIR IS 384 MILES PER HOUR MORE THAN THE WIND SPEED, FIND THE SPEED OF THE WIND SPEED AND THE SPEED OF THE PLANE IN STILL AIR,

Let x = speed of plane in still air
    y = wind speed

Distance = Distance

4 (x+y) = 5(x-y)
4x +4y = 5x-5y
x = 9y
--------------------------
where x = 384+y

By substitution you get
9y = 384+y
8y = 384
y = 48
x = 384+48
x = 432

The wind speed is 48 miles per hour while the speed of plane in still air is 432 miles per hour.</PRE>