SOLUTION: Science and Nutrition. 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 faster than the spe

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Science and Nutrition. 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 faster than the spe      Log On


   

Question 94584: Science and Nutrition. 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 faster than the speed of the freight train, find the speed of each.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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 faster than the speed of the freight train, find the speed of each.
:
Let s = speed of the freight
Then
(s+25) = speed of the pass train
:
They are giving the time as the same for both trains, make a time equation from this fact:
Time = Dist/speed
:
325%2F%28%28s%2B25%29%29 = 200%2Fs
:
Cross multiply and you have:
:
325s = 200(s+25)
:
325s = 200s + 5000
:
325s - 200s = 5000
:
125s = 5000
:
s = 5000/125
:
s = 40 mph, speed of the freight
and
(s+25) = 65 mph, speed of the pass train
:
:
Check solutions, is the time the same?
200/40 = 325/65
5 = 5