The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011?
Population of the city in $2009, \mathrm{P}=120000$
Rate of increase, $\mathrm{R}=6 \%$
Time, $\mathrm{n}=3$ years
Then the population of the city in the year 2010 is given by
Population $=P \times\left(1+\frac{R}{100}\right)^{n}$
$=120000 \times\left(1+\frac{6}{100}\right)^{1}$
$=120000 \times\left(\frac{100+6}{100}\right)$
$=120000 \times\left(\frac{106}{100}\right)$
$=120000 \times\left(\frac{53}{50}\right)$
$=2400 \times 53$
$=127200$
Therefore, the population of the city in 2010 is 127200 .
Again, $p$ opulation of the city in $2010, P=127200$
Rate of decrease, $R=5 \%$
Then the population of the city in the year 2011 is given by
Population $=P \times\left(1-\frac{R}{100}\right)^{n}$
$=127200 \times\left(1-\frac{5}{100}\right)^{1}$
$=127200 \times\left(\frac{100-5}{100}\right)$
$=127200 \times\left(\frac{95}{100}\right)$
$=127200 \times\left(\frac{19}{20}\right)$
$=6360 \times 19$
$=120840$
Therefore, the population of the city in 2011 is 120840 .