Question 292296: i know the basic setup of the problem, but i don't know how to simplify the 4 hours and 30 minutes, i always come up with long decimal answers. please help.
two cars leave a city on the same road, one driving 12 mph faster than the other. After 4 hours, the car traveling faster stops for lunch. After 4 hours and 30 minutes the car traveling slower stops for lunch. Assuming that the person in the faster car is still eating lunch, the cars are now 24 miles apart. How fast is each car driving?
Answer by ptaylor(2198)   (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 one car
and r+12=rate of the other car
After 4 hours the car travelling faster has travelled 4(r+12)mi
After 4.5 hours the car travelling slower has travelled 4.5r mi
Now we are told that the difference between these distances is 24 mi, so:
4(r+12)-4.5r=24
4r+48-4.5r=24 subtract 48 from each side
4r-4.5r=24-48
-0.5r=-24
r=48 mph----------------rate of slower car
r+12=48+12=60 mph---rate of faster car
CK
48*4.5=216 mi
60*4=240 mi
240-216=24 mi
hope this helps--ptaylor
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