SOLUTION: Walt and gail left a stable for a ride to the mountain lookout. On the way up, they averaged 3 mi/h. On the way back along way same trail, they averaged 5 ¼ mi/h, and the return tr

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Walt and gail left a stable for a ride to the mountain lookout. On the way up, they averaged 3 mi/h. On the way back along way same trail, they averaged 5 ¼ mi/h, and the return tr      Log On


   

Question 331740: Walt and gail left a stable for a ride to the mountain lookout. On the way up, they averaged 3 mi/h. On the way back along way same trail, they averaged 5 ¼ mi/h, and the return trip took 5/7 h less time than the outward trip. How long was the entire horseback ride?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Walt and gail left a stable for a ride to the mountain lookout.
On the way up, they averaged 3 mi/h.
On the way back along way same trail, they averaged 5 ¼ mi/h, and the return
trip took 5/7 h less time than the outward trip.
How long was the entire horseback ride?
:
Let d = entire length of the ride
Then
.5d = one way distance to lookout, and return distance
:
Write time equation; Time = dist/speed
:
Time up - Return time = 5/7 hr
%28.5d%29%2F3 - %28.5d%29%2F5.25 = 5%2F7
Multiply equation by 42 to get rid of the denominators, results
14(.5d) - 8(.5d) = 6(5)
:
7d - 4d = 30
:
3d = 30
d = 10 mi is the total distance
:
:
Check solution by finding the times
5/3 = 1.667 hr
5/5.25 = .952
----------------
differ = .715 ~ 5/7; confirms our solution