Question 670733: 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 1 mph faster than the other hiker. The slower hiker reaches town B 1/2 hour earlier than the faster, because the route taken by the faster is 15 miles long, while the route taken by the slower is ten miles long. What is the average rate of each hiker?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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 1 mph faster than the other hiker. The slower hiker reaches town B 1/2 hour earlier than the faster, because the route taken by the faster is 15 miles long, while the route taken by the slower is 10 long. What is the average rate of each hiker?
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Faster hiker DATA:
rate = x+1 mph ; distance = 15 miles ; time = 15/(x+1) hrs
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Slower hiker DATA:
rate = x mph ; distance = 10 miles ; time = 10/x hrs
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Equation:
faster- slower = 1/2 hr
15/(x+1) - 10/x = 1/2
30x - 20(x+1) = x(x+1)
10x - 20 = x^2+x
x^2 - 9x + 20 = 0
(x-4)(x-5) = 0
x = 4 or x = 5 (possible rates of the slower hiker)
x+1 = 5 or x+1 = 6 (corresponding rates for the faster hiker)
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Cheers,
Stan H.
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