SOLUTION: An amateur cyclist is training for a road race. He rode the first 27-mile portion of his workout at a constant rate. He then reduced his speed by 3 mph for the remaining 18-mile co

Algebra ->  Average -> SOLUTION: An amateur cyclist is training for a road race. He rode the first 27-mile portion of his workout at a constant rate. He then reduced his speed by 3 mph for the remaining 18-mile co      Log On


   



Question 1174525: An amateur cyclist is training for a road race. He rode the first 27-mile portion of his workout at a constant rate. He then reduced his speed by 3 mph for the remaining 18-mile cool-down portion of the workout. Each portion of the workout took equal time. Find the cyclist's rate during the first portion and his rate during the cool-down portion.
First Portion: mph
Cool-Down: mph

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
d=s*t
27=s*t
18=(s-3)*t, since the time for each part is the same
t=27/s
t=18/(s-3)
so 27/s=18/(s-3)
cross-multiply and 27s-81=18s
9s=81
s=9 mph initially for 3 hours to do 27 miles
s-3=6 mph for 3 hours to do 18 miles