SOLUTION: At 8:00 AM a freight car left from city A to city B carrying citrus fruits. At the same time, a bike rider left from city B to city A. The freight car's speed was 32 km/hour greate

Question 636846: At 8:00 AM a freight car left from city A to city B carrying citrus fruits. At the same time, a bike rider left from city B to city A. The freight car's speed was 32 km/hour greater than the bike rider.
They drove the same trail and their speeds were constant. The freight car arrived to city B, waited there for 1/4 of an hour, and then started driving back to city A. On the way back, the car met the bike rider 38 kilometers away from from city A. The distance between the two cities is 64 kilometers. What was the speed of the bike rider?
Thanks so much for your time! I really appreciate it!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
At 8:00 AM a freight car left from city A to city B carrying citrus fruits.
At the same time, a bike rider left from city B to city A.
The freight car's speed was 32 km/hour greater than the bike rider.
They drove the same trail and their speeds were constant.
The freight car arrived to city B, waited there for 1/4 of an hour, and then
started driving back to city A.
On the way back, the car met the bike rider 38 kilometers away from from city A.
The distance between the two cities is 64 kilometers.
What was the speed of the bike?
:
Let s = speed of the bike
then
(s+32) = speed of the freight
:
A point 38 km from A would be 64-38 = 26 km from B
freight traveled 64 + 26 = 90 km when they met
bike traveled 26 km when they met
:
A time equation.
freight time = bike time
+ =
multiply by 4s(s+32) to clear the denominators, assemble a quadratic equation
:
solve with the quadratic formula, the positive solution
s = 11.125 km is the bike speed
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