Question 143133: An express and local train leaves Allen's point at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of each train. i came up with 75 mph for the express and 58 for the local, but this doesn't seem right.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! AT LEAST YOU TRIED!!!
ACTUALLY YOU CAN CHECK YOUR ANSWER, AS FOLLOWS:
TIME IT TOOK LOCAL=50/58= 0.862 HR
TIME IT TOOK EXPRESS=50/75=0.66667 HR
THERE SHOULD BE AN HOURS DIFFERENCE AND THERE'S NOT
LOOK AT THIS APPROACH:
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r= rate of the local
then 2r=rate of the express
Time it tool local to make the trip=50/r
Time it tool express to make the trip=50/2r
Now we are told that the express arrives 1 hour ahead of the local, so:
50/r -1 =50/2r multiply each term by 2r
50*2-2r=50 or
100-2r=50 subtract 100 from each side
100-100-2r=50-100 collect like terms
-2r=-50 divide each side by -2
r=25 mph------------------------------speed of local
2r=2*25=50 mph----------------------------speed of the express
CK
time it took local=50/25=2hrs
Local arrived at 5 PM
time it took express =50/50=1hr
express arrives at 4 PM
Hope this helps----ptaylor
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