SOLUTION: Howard lives 2 miles from school. When he rode his bicycle, he traveled 5 miles per hour faster than when he walked, and thus saved 25 minutes. What was his average rate of walking

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Question 624149: Howard lives 2 miles from school. When he rode his bicycle, he traveled 5 miles per hour faster than when he walked, and thus saved 25 minutes. What was his average rate of walking?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Howard lives 2 miles from school.
When he rode his bicycle, he traveled 5 miles per hour faster than when he walked, and thus saved 25 minutes.
What was his average rate of walking?
:
Let w = his av rate of walking
then
(w+5) = his av rate of riding
:
Since we are using mph, convert 25 min: 25/60 = 5/12 hrs
:
Write time equation; time = dist/speed
:
2%2Fw - 2%2F%28%28w%2B5%29%29 = 5%2F12
multiply by 12w(w+5) to clear the denominator, results:
12(w+5)*2 - 12w(2) = 5w(w+5)
:
24w + 120 - 24w = 5w^2 + 25w
A quadratic equation
0 = 5w^2 + 25w - 120
Simplify, divide by 5
w^2 + 5w - 24 = 0
Factors to
(w+8)(w-3) = 0
the positive solution
w = 3 mph is the walking speed
:
:
Check this out (8 mph is the biking speed)
2/3 - 2/8 =
8/12 - 3/12 = 5/12 which is 25 min