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Question 82862: Dear Tutor,
Thank-you for taking your time and looking at this problem. I wouldn't be botering you, but I am very stumped. Here it is:
Sally drives 720 miles to work and 720 miles back. Her average rate going to work is 12 miles per hour greater than returning. If the entire road trip took 35 hours, what was her average speed going and returning?
Thanks again for looking at this. I really need your help!
Sincerely, Victoria
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Sally drives 720 miles to work and 720 miles back. Her average rate going to work is 12 miles per hour greater than returning. If the entire road trip took 35 hours, what was her average speed going and returning?
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To-work DATA:
distance = 720 miles ; Rate = x+12 mph ; Time = d/r = 720/(x+12) hrs
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From-work DAA:
distance = 720 miles ; Rate = x mph ; Time = d/r = 720/x hrs
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EQUATION:
time + time = 35 hrs
720/(x+12) + 720/x = 35
Multiply thru by x(x+12) to get:
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720x + 720(x+12) = 35x(x+12)
1440x + 8640 = 35x^2 + 420x
35x^2-1020x-8640
7x^2-204x-1728=0
x = [204+-sqrt90000]/14
x = [204+-300]/14
x = 504/14 or -96/14
Positive solution: x = 36 (Rate driving from work)
x+12 = 48 (Rate driving to work)
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Cheers,
Stan H>
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