SOLUTION: Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longe
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-> SOLUTION: Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longe
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Question 58754: Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longer than Smith’s. How fast was each one traveling.
You can put this solution on YOUR website! Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longer than Smith’s. How fast was each one traveling.
:
That Smith is going east from Durango does not seem to have any bearing on the
problem.
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Time = dist/speed;
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Let s = Smith's speed
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Smith's time = 45/s
Jones's time = 70/(s+5)
:
Smith's time + 1/2 hr = Jone's time =
:
Mult equation by s(s+5) and eliminate the denominators, then you have:
45(s+5) + .5(s(s+5) = 70s
:
45x + 225 + .5s^2 + 2.5s = 70s
:
.5s^2 + 45s + 2.5s - 70s + 225 = 0
:
.5s^2 - 22.5s + 225 = 0
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Mult eq by 2 and get rid of these decimals
s^2 - 45s + 450 = 0
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Factors to:
(s - 15)(s - 30) = 0
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s = 15; and s = 30; I'm not sure, but 30 seems fast for a bike to average for
that distance. I'm going to say Smith's speed = 15 mph & Jones speed = 20 mph
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Check it: Jones time - Smiths time = .5hr
70/20 - 45/15 =
3.5 - 3 = .5
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The other solution, s = 30:
70/35 - 45/30 =
2 - 1.5 = .5 also
