<PRE><B>Question 1034493</B>
Mr. Johns drives a semitrailer. On one trip to a city 275 mi from his home, he traveled 30 min longer going than he did returning along the same route because of construction work. How long did it take him to make the trip each way if his rate going was 5 mi/hr slower than his rate returning? How fast did he travel each way? 
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Going DATA:
dist = 275 ; time = t+(1/2) hr ; rate = 275/(2t+1)/2 = 550/(2t+1) mph
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Return DATA:
dist = 275 ; time = t hrs ; rate = 275/t mph
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Equation::
return rate - rate = 5 mph
275/t - 550/(2t+1) = 5
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275(2t+1) - 550t = 5t(2t+1)
550t + 275 - 550t = 10t^2 + 5t
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10t^2 + 5t - 275 = 0
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2t^2 + t - 137.5 = 0
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t = 8.045 hrs (time to return)
t + 0.5 = 8.545 hrs (time to Go)
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Cheers,
Stan H.
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