work problem

[inkarea]

The wind is blowing at an average rate of 10 miles per hour. Riding with the wind, a bicyclist can cycle 75 miles in the same amount of time it takes to cycle 15 miles against the wind. What is the bicyclist's average rate in calm air?

I think I need to set this up as r=d/t but I haven't been able to set it up to get the correct answer. Thanks for any help!

 

[soroban]

Hello, inkarea!

The wind is blowing at an average rate of 10 miles per hour.
Riding with the wind, a bicyclist can cycle 75 miles in the same amount of time it takes to cycle 15 miles against the wind.
What is the bicyclist's average rate in calm air?

I think I need to set this up as r=d/t . . . . . um. not quite

\(\displaystyle \text{The problem compares }times\quad\hdots\quad\text{We need: }\:t \:=\:\frac{d}{r}\)

\(\displaystyle \text{Let }r\text{ = rate of the bike in calm air.}\)


\(\displaystyle W\!ith\text{ the wind, its rate is: }\:r+10\,\text{ mph.}\)
. . . . . . . . . . . . \(\displaystyle \text{To go 75 miles, it takes: }\:\frac{75}{r+10}\text{ hours.}\)

\(\displaystyle Against\text{ the wind, its rate is: }\:r-10\,\text{ mph.}\)
. . . . . . . . . . . . \(\displaystyle \text{To go 15 miles, it takes: }\:\frac{15}{r-10}\text{ hours.}\)


\(\displaystyle \text{These two times are equal.}\)
. . \(\displaystyle \text{There is our equation! }\quad\hdots\quad \frac{75}{r+10} \;=\;\frac{15}{r-10}\)


\(\displaystyle \text{Now solve for }r.\)