SOLUTION: A cyclist traveled 60 mi at a constant rate before reducing the speed by 5 mph. Another 40 mi was traveled at the reduced speed. The total time for the 100-mile trip was 8 h. Find
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A cyclist traveled 60 mi at a constant rate before reducing the speed by 5 mph. Another 40 mi was traveled at the reduced speed. The total time for the 100-mile trip was 8 h. Find
Log On
Question 931101: A cyclist traveled 60 mi at a constant rate before reducing the speed by 5 mph. Another 40 mi was traveled at the reduced speed. The total time for the 100-mile trip was 8 h. Find the rate during the first 60 mi. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! 60 miles at speed x mph
40 miles at (x-5) mph
Total time = 8 hours
time first leg + time second leg = 8
t=d/r
60/x + 40/(x-5) = 8
multiply equation by x(x-5)
60(x-5)+40x = 8x(x-5)
60x-300+60=8x^2-40x
8x^2-140x+300=0
/4
2x^2-35x+75=0
2x^2-30x-5x+75=0
2x(x-15)-5(x-15)=0
(x-15)(2x-5)=0
x= 15 OR x=5/2
5/2 is not possible
so speed = 15 mph