**Question:**

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?

**Solution:**

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 .