<PRE><B>Question 507497</B>
A plane flew on a mission and returned to its base.
 It took fifteen minutes less time to get to its destination than it did to get back.
 The average speed on the trip out was 220 mph.
 The average speed on the way back was 200 mph.
 How long did the trip out take?
:
change 15 min to .25 hrs
:
Let t = time for trip out
then
(t+.25) = time for trip back
:
Write a distance equation (dist out and back are the same)
:
220t = 200(t+.25)
220t - 200t + 50
220t - 200t = 50
20t = 50
t = 50/20
t = 2.5 hrs for the trip out
;
:
:
Check this by finding the distances
220(2.5) = 550 mi
200(2.75) =550 mi; confirms our solution of t = 2.5</PRE>