1. An amateur cyclist training for a road race rode the first 20 mile

portion of his workout at a constant rate. For the nest 16 mile

cool down portion of his workout he reduced his speed by 2 miles

per hour. Each portion of the workout took the same amount of

time. Find the cyclist's rate during the first portion and his rate

during his cool down.

Answer: the cyclist's speed during the first portion is 10 mph and his speed during the cool down portion is 8 mph.

Step-by-step explanation:

Let x represent the cyclist's rate during the first portion.

The amateur cyclist training for a road race rode the first 20 mile

portion of his workout at a constant rate.

Time = distance/speed

Time taken to ride through the first portion is

20/x

For the next 16 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. It meas that his speed during the cool down portion is (x - 2) mph. Time taken to ride through the cool down portion is

16 / (x - 2)

Since each portion of the workout took the same time, it means that

20/x = 16 / (x - 2)

Cross multiplying, it becomes

20 (x - 2) = 16x

20x - 40 = 16x

20x - 16x = 40

4x = 40

x = 40/4

x = 10 mph

His speed during the cool down portion is

10 - 2 = 8 mph