<PRE><B>Question 181007</B>
A B747 aircraft flew 6 hours with the wind. The return trip took 7 hours against
 the wind.If the speed of the plane in still air is 13 times the speed of the 
wind. Find the wind speed and the speed of the plane in still air
:
let s = speed of the wind
then
13s = speed of the plane in still air
:
The distance of both trips are the same. Write a distance equation:
With dist = against dist
6(13s + s) = 7(13s - s)
:
6(14s) = 7(12s)
84s = 84s
:
There is no unique solution. Any 13:1 speed:wind relationship will work
a few
Plane, wind, dist
650, 50, 4200 mi
:
390, 30, 2520 mi
:
260, 20, 1680 mi
:
The fact that it is a Boeing 747 may be a clue, find the cruising speed of
that and start from there</PRE>