SOLUTION: You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average

Question 241510: You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average speed on the return trip?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average speed on the return trip?
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Service Call DATA:
distance = 56 miles ; rate = x mph ; time = d/r = 56/x hrs
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Return trip DATA:
distance = 56 miles ; rate = (x+8) mph ; time = 56/(x+8) hrs
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Equation:
go time - return time = 1/6 hr
56/x - 56/(x+8) = 1/6
1/x - 1/(x+8) = 1/336
Multiply thru by 336x(x+8) to get:
336(x+8) - 336x = x(x+8)
336*8 = x^2+8x
x^2 + 8x - 2688 = 0
(x-48)(x+56) = 0
Positive solution:
x = 48 mph (rate on the service call)
x+8 = 56 mph (rate on the return)
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Cheers,
Stan H.

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