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

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) (Show Source): You can put this solution on YOUR website!
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
+ =
multiply by 5b(b+25)
5b(b+25)* + 5b(b+25)* = 5b(b+25)*
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
:
+ =
Reduce fraction
+ =
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