SOLUTION: A cyclist leaves town a heading east at the same time another cyclist leaves town B ,60 kilometers away heading west on the same highway the east bound cyclist can average 25 kph t

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Question 1007435: A cyclist leaves town a heading east at the same time another cyclist leaves town B ,60 kilometers away heading west on the same highway the east bound cyclist can average 25 kph the other cyclist can average only 15 kph the cyclist ride until they meet how long does it take?

Found 2 solutions by macston, fractalier:
Answer by macston(5194) (Show Source):
You can put this solution on YOUR website!
The cyclist approach one another at:
.
15kph+25kph=40kph
.
To travel 60 km:
.
60km/40kph=1.5 hours
.
ANSWER: They will meet in 1.5 hours.

Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website!
The distance between them is 60 km.
The rate at which these cyclists close that distance is 15+25 = 40 kph.
Of course, distance is rate times time, or d=rt.
Then t = d/r = 60/40 = 1.5 hours.