SOLUTION: Sam rode his bike for 1 mile to a friend's house. Then his friend's mom drove him 12 miles at a rate that was 25 miles per hour faster than he was able to ride his bike. If the e

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Question 507878: Sam rode his bike for 1 mile to a friend's house. Then his friend's mom drove him 12 miles at a rate that was 25 miles per hour faster than he was able to ride his bike. If the entire trip took 3/5 of an hour, what was Sam's rate of speed on the bike?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Sam rode his bike for 1 mile to a friend's house.
Then his friend's mom drove him 12 miles at a rate that was 25 miles per hour faster than he was able to ride his bike.
If the entire trip took 3/5 of an hour, what was Sam's rate of speed on the bike?
:
Let b = bike speed
then
(b+25) = car speed
:
Write a time equation, time = dist/speed
:
Bike time + car time = 3/5 hr
1%2Fb + 12%2F%28%28b%2B25%29%29 = 3%2F5
multiply by 5b(b+25)
5b(b+25)*1%2Fb + 5b(b+25)*12%2F%28%28b%2B25%29%29 = 5b(b+25)*3%2F5
cancel out the denominators, resulting in
5(b+25) + 5b(12) = 3b(b+25)
5b + 125 + 60b = 3b^2 + 75b
65b + 125 = 3b^2 + 75b
:
Arrange as a quadratic equation on the right
0 = 3b^2 + 75b - 65b - 125
3b^2 + 10b - 125 = 0
:
you can use the quadratic formula to find b, but this will factor;
(3b+25)(b-5) = 0
:
the positive solution
b = 5 mph is the bike speed
:
:
Check this by finding that actual time of each
5 + 25 = 30 mph is the cars speed
:
1%2F5 + 12%2F30 =
Reduce fraction
1%2F5 + 2%2F5 = 3%2F5