SOLUTION: 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 tra
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: 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 tra
Log On
Question 91981: 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.
Thanks Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
Let r=speed(rate) of freight train
Then r+25=speed of passenger train
distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r
time it takes freight train=200/r
time it takes passenger train=325/(r+25)
Now we are told that the above two times are the same. Soooooooooo
200/r=325/(r+25) multiply both sides by r(r+25) to get rid of fractions
200(r+25)=325r get rid of parens
200r+5000=325r subtract 5000 and also 325r from both sides
200r-325r+5000-5000=325r-325r-5000 collect like terms
-125r=-5000 divide both sides by -125
r=40 mi/hr------------------------speed of freight train
r+25=40+25=65 mi/hr---------------------speed of passenger train
CK
200/40=325/65
5=5