SOLUTION: A train traveled 120 miles from city A to city B and returned the same distance in a total time of 6 hours. If the train traveled 30 mph slower on the return trip, how fast did the
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Question 980069: A train traveled 120 miles from city A to city B and returned the same distance in a total time of 6 hours. If the train traveled 30 mph slower on the return trip, how fast did the train travel in each direction? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A train traveled 120 miles from city A to city B and returned the same distance in a total time of 6 hours.
If the train traveled 30 mph slower on the return trip, how fast did the train travel in each direction?
:
let s = train speed traveling from A to B
then
(s-30) = train speed from B to A
:
Write a time equation; time = dist/speed
:
A>B time + B>A time = 6 hrs + = 6
multiply thru by s(s-30)
s(s-30)* + s(s-30)* = 6s(s-30)
Cancel the denominators and you have
120(s-30) + 120s = 6s^2 - 180s
120s - 3600 + 120s = 6s^2 - 180s
240s - 3600 = 6s^2 - 180s
Combine to form a a quadratic equation on right
0 = 6s^2 - 180s - 240s + 3600
6s^2 - 420s + 3600 = 0
simplify, divide by 6
s^2 - 70s + 600 = 0
Factors to
(s-10)(s-60) = 0
s = 60 mph is the only reasonable answer, train speed A to B
then
60 - 30 = 30 mph the return train speed
:
:
See if that checks out, find the actual time each way
120/60 = 2 hrs
120/30 = 4 hrs
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total time: 6 hrs
