SOLUTION: A cyclist bikes at a constant speed for 18 miles. He then returns home at the same speed but takes a different route. His return trip takes one hour longer and is 23 miles. find hi
Algebra.Com
Question 509196: A cyclist bikes at a constant speed for 18 miles. He then returns home at the same speed but takes a different route. His return trip takes one hour longer and is 23 miles. find his speed.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
A cyclist bikes at a constant speed for 18 miles. He then returns home at the same speed but takes a different route. His return trip takes one hour longer and is 23 miles. find his speed.
=====================================
Let s = the cyclist's speed
Let t = the time for the 18 mile trip
The speed of the outbound trip = s = 18/t [1]
The speed of the return trip = s = 23/(t+1) [2]
Solve for t in [1], substitute into [2]:
s = 23/(18/s + 1)
Solve for s:
18 + s = 23
s = 5 mph
Check:
t = 18/5 = 3.6 h
t + 1 = 23/5 = 4.6 h
RELATED QUESTIONS
A cyclist bikes at a constant speed for 23 miles. He then returns home at the same speed... (answered by ewatrrr)
A cyclist bikes at a constant speed for 25 miles. He then returns home at the same speed... (answered by jorel1380)
A cyclist bikes at a constant speed for 20 miles. He then returns home at the same speed... (answered by KMST)
A cyclist bikes at a constant speed for 25 miles. He then returns home at the same speed (answered by MathLover1,MathTherapy)
A cyclyist bikes at a constant speed for 20 miles. He then returns home at the same... (answered by checkley75)
Cyclist bikes at a constant speed for 22 miles. He returns home at that speed but takes... (answered by josgarithmetic)
A cyclist bikes at a constant speed for 17 miles. He then returns home at the same speed... (answered by ankor@dixie-net.com)
a cyclist bikes at a constant speed for 19 miles.He then returns home at the same speed... (answered by josmiceli)
A cyclist bikes at a constant speed for 24 miles. He then returns home at the same speed (answered by Paul)