Question 191433: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was each one traveling?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was each one traveling?
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Smith DATA:
distance = 45 mi ; time = x hrs ; rate = d/t = 45/x mph
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Jones DATA:
distance = 70 mi ; time (x+0.5) hrs ; rate = d/t = 70/(x+(1/2)) = mph
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Equation:
Jones' rate - Smith's rate = 5
(45/x) - (70/((2x+1)/2) = 5
(9/x) - 28/(2x+1) = 1
9(2x+1) -28x = x(2x+1)
2x^2 + 30x = 18x+9
2x^2 + 12x - 9 = 0
x = [-12 +- sqrt(144 -4*2*-9)]/4
x = [-12 + 14.7]
x = 2.7 hrs
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Smith rate = 45/x = 45/2.7 = 16 2/3 mph
Jones rate = 16 2/3 + 5 = 21 2/3 mph
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Cheers,
Stan H.
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