SOLUTION: Two hikers leave town A at the same time and walk by two different routes to Town B. The average speed of one hiker is 3 mph faster than the other hiker. The slower hiker reaches t
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Question 999968: Two hikers leave town A at the same time and walk by two different routes to Town B. The average speed of one hiker is 3 mph faster than the other hiker. The slower hiker reaches town B 15 minutes (1/4 hour) earlier than the faster, becuase the route taken by the faster hiker is 30 miles long, while the route taken by the slower hiker is only 21 miles long. What is the average rate of each hiker? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let speed of slower hiker be x
speed of faster is x=3
Time taken by slower hiker = 21/x
time taken by faster hiker = 30/(x+3)
Time taken by faster hiker - time taken by slower hiker = 1/4
multiply equation by 4x(x+3)
30*4x -21*4(x+3) = 4x(x+3)/4
120x-84x-252 = x^2+3x
x^2 +3x+84x-120x+252=0
x^2-33x+252=0
x^2-21x-12x+252=0
x(x-21)-12(x-21)=0
(x-12)(x-21)=0
x=12 OR 21
