SOLUTION: A passenger train and a freight train leave cities that are 300 miles apart and travel towards each other. The passenger train is travelling 15mph faster than the freight train. Fi

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Question 509742: A passenger train and a freight train leave cities that are 300 miles apart and travel towards each other. The passenger train is travelling 15mph faster than the freight train. Find the speed of each train if they pass each other after 4 hours.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=rate of freight train
Then r+15=rate of passenger train
distance freight train travels after 4 hours=r*4=4r
distance passenger train travels after 4 hours=(r+15)*4=4r+60
Now we know that the distance the freight train travels plus the distance the passenger train travels has to add up to 300 mi, sooooo
4r+4r+60=300
8r=240
r=30 mph-------------------------speed of freight train
r+15=30+15=45 mph---------------speed of passenger train
CK
4*30+4*45=300
120+180=300
300=300
Another way:
Let d=distance freight train travels in 4 hours
Then 300-d=distance passenger train travels in 4 hours
Let r=rate of freight train
Then r+15=rate of passenger train
4r=d or
4r-d=0------------------------------------------eq1
4(r+15)=300-d or
4r+d=240--------------------------------------eq2
add eq1 and eq2
8r=240
same as before
Hope this helps----ptaylor