SOLUTION: Science and medicine. A passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the s

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Science and medicine. A passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the s      Log On


   



Question 76754: Science and medicine. A passenger train can travel 325 mi in the same time a
freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed of each. How do I do this?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

Science and medicine. A passenger train can 
travel 325 mi in the same time a freight train
takes to travel 200 mi. If the speed of the 
passenger train is 25 mi/h faster than the 
speed of the freight train, find the speed of
each. How do I do this?

Make this DRT-chart:

                  DISTANCE   RATE       TIME
Passenger train     
  Freight train   

Fill in the two distances, 325 and 200 miles

                  DISTANCE   RATE       TIME
Passenger train     325      
  Freight train     200       


Let x = the speed of the freight train.  So
fill that in:


                  DISTANCE   RATE       TIME
Passenger train     325       
  Freight train     200        x       

>>...the speed of the passenger train is 25 mi/h 
faster than the speed of the freight train...<<

So the speed of the passenger train = x+25.
Fill that in:

                  DISTANCE   RATE       TIME
Passenger train     325      x+25      
  Freight train     200        x       


Now fill in the TIMEs using TIME = DISTANCE/RATE

                  DISTANCE   RATE       TIME
Passenger train     325      x+25      325%2F%28x%2B25%29 
  Freight train     200        x        200%2Fx

Now that the chart is all filled in, look for the
fact that you haven't used.  The words

>>...in the same time...<<

tell us the two times are equal, so our equation
is

       325%2F%28x%2B25%29+=+200%2Fx

Can you solve that equation?:

Answer = 40 mph for the freight train, and
x+25 or 65 for the passenger train.

Edwin