SOLUTION: An express train and a local train leave Grays Lake at 3:00 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local and arrivies 1 hour ahead of th

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: An express train and a local train leave Grays Lake at 3:00 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local and arrivies 1 hour ahead of th      Log On


   

Question 237015: An express train and a local train leave Grays Lake at 3:00 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local and arrivies 1 hour ahead of the local. Find the speed of each train algebraically and show work. I believe the speed of the trains is 40 and 20 miles per hour respectivelly.. I'm having trouble formulating the equations
Thanks
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
An express train and a local train leave Grays Lake at 3:00 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local and arrivies 1 hour ahead of the local. Find the speed of each train algebraically and show work. I believe the speed of the trains is 40 and 20 miles per hour respectivelly.. I'm having trouble formulating the equations
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Express train DATA
rate = 2x mph ; distance = 50 miles ; time = d/r = 50/2x hrs
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Local train DATA:
rate = x mph ; distance = 50 miles ; time = 50/x hrs
------------------------
Equation:
local time - express time = 1 hr
50/x - 50/2x = 1
50/x - 25/x = 1
Multiply thru by "x":
50 - 25 = x
x = 25 mph (speed of the local train)
2x = 50 mph (speed of the express train)
========================================
Cheers,
Stan H.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


First let's test your theory.

We know that distance equals rate times time, or ,

So if the express train went 50 miles at 40 miles per hour, it would have taken hours. And it would have taken the local train @ 20mph hours. That means that the express train would have arrived an hour and 15 minutes ahead of the local, not 1 hour like the problem states. [BUZZZ!!!] Thank you for playing, we have some lovely parting gifts...

Let represent the rate of the local train. Then must be the rate of the express train. Let be the time it took the local train, then is the time it took the express train. The distance is the same 50 miles for both trains, so:

For the local train:

For the express:

Multiply both sides of the first equation by



Multiply both sides of the second equation by



Add 1 to both sides of the second equation and combine using the LCD of



Now we have two expressions for in terms of . Since :



Multiply both sides by



I'll let you take it from there.

John