SOLUTION: Smith bicycled 45 miles goin east from Durango and Jones bicycled 70 miles. Jones averaged 5 miles per hou more than Smith, and his trip took on-half hour longer than Smith's. How
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Question 124875: Smith bicycled 45 miles goin east from Durango and Jones bicycled 70 miles. Jones averaged 5 miles per hou more than Smith, and his trip took on-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! 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 on-half hour longer than Smith's. How fast was each one traveling?
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Let s = Smith's speed
It says,"Jones averaged 5 miles per hour more than Smith," therefore:
(s+5) = Jones' speed
:
Write a time equation: Time =
Smiths time = Jones time - half an hour +
Multiply equation by 2s(s+5) to get rid of the denominators
2s(s+5)* + 2s(s+5)*
:
Cancel out the denominators and you have:
2(s+5)*45 = 2s*70 - s(s+5)
:
90s + 450 = 140s - s^2 - 5s
:
s^2 + 5s + 90s - 140s + 450 = 0
:
s^2 - 45s + 450 = 0; Arranged as a quadratic equation
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This factors to
(s - 15) (s - 30) = 0
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s = 15, or s = 30;
Check and you will see that either solution will work in the original equation
:
Probably s = 15 is more reasonable
