SOLUTION: Elliot jogs for 10 miles and then walks for another 10 miles. He jogs 2.5 miles per hour faster than he walks. The entire trip of 20 miles takes 6 hours. Find his jogging speed an

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Elliot jogs for 10 miles and then walks for another 10 miles. He jogs 2.5 miles per hour faster than he walks. The entire trip of 20 miles takes 6 hours. Find his jogging speed an      Log On

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Question 382200: Elliot jogs for 10 miles and then walks for another 10 miles. He jogs 2.5 miles per hour faster than he walks. The entire trip of 20 miles takes 6 hours.
Find his jogging speed and his walking speed.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Elliot jogs for 10 miles and then walks for another 10 miles. He jogs 2.5 miles per hour faster than he walks. The entire trip of 20 miles takes 6 hours.
Find his jogging speed and his walking speed.
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Jog DATA:
distance = 10 miles; rate = x+2.5 mph ; time = 10/(x+2.5) hr.
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Walk DATA:
distance = 10 miles ; rate = x mph ; time = 10/x hr.
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Equation:
time + time = 6 hrs.
10/(x+2.5) + 10/x = 6
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10x + 10(x+2.5) = 6x(x+2.5)
20x + 25 = 6x^2 + 15x
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6x^2 - 5x -25 = 0
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x = [5 +- sqrt(25-4*6*-25)]/12
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x = [5 +- sqrt(625)]/12
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x = [5+- 25]/12
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Positive solution:
x = 30/12 = 2.5 mph (walking rate)
x+2.5 = 5 mph (jogging rate)
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Cheers,
Stan H.
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