Two towns, Centreville and Middletown, had the same population at the end \(2022.\)
The population of Centreville decreased by \(2.5\%\) from the end of \(2022\) to the end of \(2023.\) Then, the population increased by \(8.4\%\) from the end of \(2023\) to the end of \(2024.\)
The population of Middletown increased by \(r\%\), where \(r>0\), from the end of \(2022\) to the end of \(2023.\) Then, the population of Middletown increased by \((r+2)\%\) from the end of \(2023\) to the end of \(2024.\)
Surprisingly, the populations of both towns were the same again at the end of \(2024.\) Determine the value of \(r\), rounded to the nearest tenth.
Let \(p\) be the population of Centreville at the end of \(2022\). Since Centreville and Middletown have the same population size at the end of \(2022\), then \(p\) is also the population of Middletown at the end of \(2022.\)
The population of Centreville decreased by \(2.5\%\) in \(2023\), so the population at the end of \(2023\) was \[p -\frac{2.5}{100}p = \left(1-\frac{2.5}{100}\right)p = 0.975p\] The population of Centreville then increased by \(8.4\%\) during \(2024\), so the population at the end of \(2024\) was \[0.975p + \left(\frac{8.4}{100}\right)(0.975p)=\left(1 + \frac{8.4}{100}\right)(0.975p)= 1.084(0.975p) = 1.0569p\]
The population of Middletown increased by \(r\%\) in \(2023\), so the population at the end of \(2023\) was \[p + \frac{r}{100}p = \left(1+\frac{r}{100}\right)p\] The population of Middletown then increased by \((r+2)\%\) during \(2024\), so the population at the end of \(2024\) was \[\left(1+\frac{r}{100}\right)p + \frac{r+2}{100} \left(1+\frac{r}{100}\right)p = \left(1 +\frac{r}{100}\right)\left(1+\frac{r+2}{100}\right)p\]
Since the populations of Centreville and Middletown are equal at the end of \(2024\), we have \[\left(1+\frac{r}{100}\right)\left(1+\frac{r+2}{100}\right)p =1.0569p\] Dividing both sides by \(p>0\) and multiplying both sides by \(10\,000\) to clear fractions, we have \((100+r)(102+r)=10\,569.\) Thus, \(10\,200+202r+r^2=10\,569\), and so \(r^2+202r-369=0.\)
Using the quadratic formula, \(r = \dfrac{-202 \pm \sqrt{202^2 - 4(-369)}}{2} \approx 1.8, -203.8.\)
Since \(r>0\), we have \(r \approx 1.8\%\), correct to one decimal place.