SOLUTION: 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.
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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.
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
