<PRE><B>Question 1023707</B>
Sam was driving to Austin Texas on i-35 at a constant speed of 75 miles per hour with the wind. On his way home he traveled into the wind and maintain the constant speed. He drove 80 miles on the way there and decided to take a shortcut coming home and only drove 68 miles. If the total trip took 4 hours what was the speed of the wind? 
---------------------
With Wind DATA:
dist = 80 miles ; rate = 75+w ; time = 80/(75+w) hrs
---
Against Wind DATA:
dist = 68 miles ; rate = 75-w ; time = 68/(75-w) hrs
------------------
Equation:
time + time = 4 hrs
80/(75+w) + 68/(75-w) = 4 hrs
-------------------------------------
80(75-w) + 68(75+w) = 4(75^2-w^2)
----------------------------
80*75 - 80w + 68*75 + 68w = 4*75^2 - 4*w^2
----------------------------
148*75 - 12w = 4*75^2 - 4w^2
---------
4w^2 - 12w - 11400 = 0
-----
w^2 - 3w - 2850 = 0
-----------------------
wind = 54.9 mph
--------------
Cheers,
Stan H.
</PRE>