SOLUTION: A cyclist rode the first 22​-mile portion of his workout at a constant speed. For the 16​-mile cooldown portion of his​ workout, he reduced his speed by 33 miles

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Question 1038880: A cyclist rode the first 22​-mile portion of his workout at a constant speed. For the 16​-mile cooldown portion of his​ workout, he reduced his speed by 33 miles per hour. Each portion of the workout took the same time. Find the​ cyclist's speed during the first portion and find his speed during the cooldown portion.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

The formula we will use here is

D = r*t

where
D = distance
r = rate (aka speed)
t = time

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"A cyclist rode the first 22 mile portion of his workout at a constant speed"
means that we know D = 22, so

D = r*t
22 = r*t

Let's solve for t to get

t = 22/r

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"For the 16 mile cooldown portion of his? workout, he reduced his speed by 33 miles per hour"

the distance is now D = 16 and the speed goes from r to r-33

D = r*t
16 = (r-33)*t
t = 16/(r-33)

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We have these two equations

t = 22/r
t = 16/(r-33)

set the two right sides equal to each other and solve for r


22/r = 16/(r-33)
22(r-33) = 16r
22r-726 = 16r
22r-726-22r = 16r-22r
-726 = -6r
-6r = -726
-6r/(-6) = -726/(-6)
r = 121

So his initial speed is 121 miles per hour which seems to be way too fast to be realistic.

The speed during the cool-down period is 121 - 33 = 88 miles per hour which is also very unrealistic for a bicyclist.

There's a possibility there could be a typo in the problem.