SOLUTION: Set up a table and solve using an Algebraic Equation.
A passenger train can travel 245 miles in the same amount of time that it takes a freight train to travel 200 miles. If th
Question 1039533: Set up a table and solve using an Algebraic Equation.
A passenger train can travel 245 miles in the same amount of time that it takes a freight train to travel 200 miles. If the rate of the passenger train is 15 MPH faster than the rate of the freight train. Find train.
I have Let r=rate of the freight train
r+15 rate of the passenger train.
t= time
d=rt
Passenger train
245=r+15(t)
245=rt+15t
Freight Train:
200=rt
245=200+15t= 45=15t = 3
200=r(3)= 66.7
66.7 freight train
81.7 passenger train.
Is this correct ?
Thank you Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A passenger train can travel 245 miles in the same amount of time that it takes a freight train to travel 200 miles. If the rate of the passenger train is 15 MPH faster than the rate of the freight train.
-------------
r*t = 200
(r+15)*t = 245
------------
r*t = 200 --> t = 200/r
Sub for t
(r+15)*(200/r) = 245
(r+15)*200 = 245r
245r = 200r + 3000
45r = 3000
r = 66.67
==============
Looks right.
I have Let r=rate of the freight train
r+15 rate of the passenger train.
t= time
d=rt
Passenger train
245=r+15(t)
245=rt+15t
Freight Train:
200=rt
