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.
Answer by ankor@dixie-net.com(22740) (Show Source): 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.
:
Time = dist/speed;
:
Let s = Smith's speed
:
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
:
Mult eq by 2 and get rid of these decimals
s^2 - 45s + 450 = 0
:
Factors to:
(s - 15)(s - 30) = 0
:
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
:
Check it: Jones time - Smiths time = .5hr
70/20 - 45/15 =
3.5 - 3 = .5
:
The other solution, s = 30:
70/35 - 45/30 =
2 - 1.5 = .5 also
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