SOLUTION: An express and local train leave Gray’s Lake at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local, and arrives 1 hour ahead of it. Find the

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: An express and local train leave Gray’s Lake at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local, and arrives 1 hour ahead of it. Find the       Log On


   

Question 155059: An express and local train leave Gray’s Lake at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local, and arrives 1 hour ahead of it. Find the speed of each train.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: The information " leave Gray’s Lake at 3 P.M" is extra information that is not helpful. So we can ignore the "3 pm"

Let x = speed of local train

Let's set up the equation for the local train:

d=rt Start with the distance-rate-time equation

50=%28x%29t Plug in d=50 (the distance from the lake to Chicago) and r=x

50%2Fx=t Divide both sides by "x" to solve for "t"

So the time "t" can be represented by t=50%2Fx

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Now let's set up the equation for the express train:

d=rt Go back to the distance-rate-time equation

Since the "express travels twice as fast as the local", this means that the speed of the express train is 2x mph. Also, because the express train "arrives 1 hour ahead of" the local train, this means that the time of the express train is t-1 hours

50=2x%28t-1%29 Plug in d=50, r=2x and replace "t" with t-1


50=2x%2850%2Fx-1%29 Plug in t=50%2Fx


50=2x%2850%2Fx%29-2x%281%29 Distribute


50=100-2x Multiply


-50=-2x Subtract 100 from both sides


25=x Divide both sides by -2 to isolate "x"


So the answer is x=25.


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Answer:
So the speed of the local train is 25 mph and the speed of the express train is 50 mph (since it is twice the speed of the local train)