SOLUTION: im gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. How hast did Jim
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-> SOLUTION: im gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. How hast did Jim
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Question 53712: im gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. How hast did Jim go there and back?: Jim gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. What was Jims speed on the trip to the location and his speed on the trip home? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Jim gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. What was Jim's speed on the trip to the location and his speed on the trip home?
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My understanding is there are two one-way trips of 40 mi
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Let the initial speed = s; Then then the returning speed = (s-10)
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Time = Dist/speed
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Time out = Time back - 2 hrs
40/s = 40/(s-10) - 2
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Eliminate the denominators; mult equation by s(s-10); resulting in:
40(s-10) = 40s - 2(s(s-10))
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40s - 400 = 40s - 2s^2 + 20s
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+2s^2 + 40s - 40s - 20s - 400 = 0
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2s^2 - 20s - 400 = 0; our old friend the quadratic equation
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s^2 - 10s - 200 = 0; simplified, divided eq by 2
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Factors to:
(s-20) (s+10) = 0
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Positive solution: s = +20 mph out then 10 mph back:
:
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Check using time: 40/20 = 2hr out & 40/10 = 4 hr back (2 hours longer)
