SOLUTION: I am looking for assistance with setting up this problem: A passenger train traveled to Schannesburg and back. The trip there took 20 hours and the trip back took 16 hours. It a

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Question 593715: I am looking for assistance with setting up this problem: A passenger train traveled to Schannesburg and back. The trip there took 20 hours and the trip back took 16 hours. It averaged 12 mph faster on the return trip then on the outbound trip. What was the passenger train's speed on the outbound trip?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A passenger train traveled to Schannesburg and back.
The trip there took 20 hours and the trip back took 16 hours.
It averaged 12 mph faster on the return trip then on the outbound trip.
What was the passenger train's speed on the outbound trip?
:
Let s = train speed outbound
then
(s+12) = train return speed
:
Write a distance equation: dist = time * speed
:
out dist = return dist
20s = 16(s+12)