Question

The population of a town increases by 20% every
year. If its present population is 2,16,000. Find its
population (i) after 2 years, (ii) 2 years ago.​

Answer 1

Answer:

• (i) After 2 years = 311040
• (ii) 2 years ago = 138240

Explanation:

Given,

Population of a town is 2,16,000

Every year it increases by 20%

Population after 2 years is given by :-

$\odot \: { \underline{ \boxed{ \sf \: p \bigg(1 + \dfrac{r}{100} \bigg)^{n} }}}$

Where,

• p(population) = 216000
• r(rate) = 20%
• n(time) = 2 years.

From the given data,

$= \sf 216000\bigg(1 + \dfrac{20}{100} \bigg)^{2}$

$= \sf 216000\bigg( 1 + \dfrac{1}{5} \bigg)^{2}$

$= \sf 216000\bigg( \dfrac{6}{5} \bigg)^{2}$

$= \sf 216000\bigg( \dfrac{6}{5} \bigg)^{2}$

$= \sf 216000\bigg( \dfrac{36}{25} \bigg)$

$= \sf \: \cancel{216000 }\times \dfrac{36}{ \cancel{25}}$

$= \sf \: 8640 \times 36$

$= \sf \: 311040$

∴ Population after 2 years would be 311040.

Now,

Population 2 years ago is given by :-

$\odot \: { \underline{ \boxed{ \sf \: p \bigg(1 - \dfrac{r}{100} \bigg)^{n} }}}$

Where,

• p(population) = 21600
• r(rate) = 20%
• n(time) = 2 years.

$\sf = 216000\bigg(1 - \dfrac{20}{100} \bigg)^{2}$

$\sf = 216000\bigg(1 - \dfrac{1}{5} \bigg)^{2}$

$\sf = 216000\bigg(\dfrac{4}{5} \bigg)^{2}$

$\sf = 216000\bigg(\dfrac{16}{25} \bigg)$

$\sf = \cancel{ 216000 }\times \dfrac{16} {\cancel{ {25} }}$

$\sf = 8640 \times 16$

$\sf = 138240$

∴ 2 years ago the population was 138240

_____________________

Formula used :

$\longrightarrow \: \bf \: p \bigg(1 + \dfrac{r}{100} \bigg)^{n}$

Where, p denotes the principal, r is the rate and n is the time.