Question

I0​ =Initial Population= 16,000
Population after 2 years = I_{2}I2​ = 17640
The formula for increase or decrease in population is given by:
\begin{lgathered}I_{2}=I_{0}[1 +\frac{R}{100}]^t\\\\ 17,640= 16,000[ 1 +\frac{R}{100}]^2\end{lgathered}I2​=I0​[1+100R​]t17,640=16,000[1+100R​]2​
\begin{lgathered}\frac{1764}{1600}=[ 1 +\frac{R}{100}]^2\\\\ (\frac{42}{40})^2=[ 1 +\frac{R}{100}]^2\\\\ \frac{42}{40}=1 +\frac{R}{100}\\\\ \frac{2}{40}=\frac{R}{100}\\\\ R= \frac{200}{40} \\\\ R= 5\end{lgathered}16001764​=[1+100R​]2(4042​)2=[1+100R​]24042​=1+100R​402​=100R​R=40200​R=5​
Rate of increase= 5 %
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Answer 1

Yes 5% is in crease in population