SOLUTION: An express and local train leave GraysLake 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 the local. Find t

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: An express and local train leave GraysLake 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 the local. Find t      Log On

Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo .
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 193131: An express and local train leave GraysLake 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 the local. Find the speed of each train.
Answer by jonvaliente(64) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=no of hours it takes for the local train
x-1=no of hours it takes for the express train
.
Since speed=distance/time,
speed of the local train = 50/x
speed of the express train =50/(x-1)
If the express train is twice as fast as the local train then:
2%2A%2850%2Fx%29=50%2F%28x-1%29
100%2Fx=50%2F%28x-1%29
Multiplying both sides by x*(x-1), we get:
100%2A%28x-1%29=50%2Ax
100x-100=50x
Subtracting 50x from both sides, we get:
50x-100=0
Adding 100 on both sides, we get:
50x=100
Diving both sides by 50 we get:
x=2
So, the local train takes 2 hours to get to Chicago and the express only takes 1 hour to get there.
speed of the local train = 50/2 = 25 mph
speed of the express train = 50/1 = 50 mph