Question

the population of a town is 20,000 . If the annual death rate is 2% and annual birth rate is 4%. Calculate the population of the town after 2 years?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Population of a town, P = 20000

Annual death rate = 2 %

Annual birth rate = 4 %

It implies, Annual growth rate of population, r = 2 %

Time period, n = 2 years

We know,

Population ( P' ) of the town after n years which increased at the rate of r % per annum having initial population P is given by

$\boxed{\sf{ \: \: \: P' \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: \: }}$

So, on substituting the values, we get

$\rm \: P' \: = \: 20000 {\bigg[1 + \dfrac{2}{100} \bigg]}^{2}$

$\rm \: P' \: = \: 20000 {\bigg[1 + \dfrac{1}{50} \bigg]}^{2}$

$\rm \: P' \: = \: 20000 {\bigg[\dfrac{50 + 1}{50} \bigg]}^{2}$

$\rm \: P' \: = \: 20000 {\bigg[\dfrac{51}{50} \bigg]}^{2}$

$\rm\implies \:P' = 20808$

So, it means, population of town after 2 years is 20808.

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Additional Information

1, Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\sf{ \: \: \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: \: }}$

2, Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{\sf{ \: \: \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: \: }}$

3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\sf{ \: \: \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: \: }}$

4. Population ( P' ) of the town after n years which decreased at the rate of r % per annum having initial population P is given by

$\boxed{\sf{ \: \: \: P' \: = \: P \: {\bigg[1 - \dfrac{r}{100} \bigg]}^{n} \: \: \: }}$

Answer 2

Answer:

Given :-

• The population of a town is 20000.
• The annual death rate is 2%.
• Annual birth rate is 4%.

To Find :-

• What is the population of the town after 2 years.

Formula Used :-

$\bigstar \: \: \: \: \sf\boxed{\bold{\pink{A =\: P\bigg\lgroup 1 + \dfrac{r}{100}\bigg\rgroup ^n}}}\: \: \: \: \bigstar$

where,

• P = Present Population
• r = Total rate of population
• n = Number of years

Solution :-

First, we have to find the total rate of population:

Given :

• Annual Birth Rate = 4%
• Annual Death Rate = 2%

Hence,

$\small \implies \sf Total\: Rate\: Of\: Population =\: Birth\: Rate\: - Death\: Rate$

$\small \implies \sf Total\: Rate\: Of\: Population =\: 4\% - 2\%$

$\small \implies \sf\bold{\purple{Total\: Rate\: Of\: Population =\: 2\%}}$

Hence, the total rate of population is 2% .

Now, we have to find the population of town after 2 years :

Given :

• Present population of town = 20000
• Total rate of population = 2%
• Number of years = 2 years

According to the question by using the formula we get,

$\small \dashrightarrow \bf Population\: after\: 2\: years =\: P\bigg\lgroup 1 + \dfrac{r}{100}\bigg\rgroup ^n$

$\small \mapsto \sf Population\: after\: 2\: years =\: 20000\bigg\lgroup 1 + \dfrac{2}{100}\bigg\rgroup ^2$

$\small \mapsto \sf Population\: after\: 2\: years =\: 20000\bigg\lgroup \dfrac{102}{100}\bigg\rgroup ^2$

$\small \mapsto \sf Population\: after\: 2\: years =\: 20000\bigg\lgroup \dfrac{102}{100} \times \dfrac{102}{100}\bigg\rgroup$

$\small \mapsto \sf Population\: after\: 2\: years =\: 20000\bigg\lgroup \dfrac{10404}{10000}\bigg\rgroup$

$\small \mapsto \sf Population\: after\: 2\: years =\: 2{\cancel{0000}} \times \dfrac{10404}{1\cancel{0000}}$

$\small \mapsto \sf Population\: after\: 2\: years =\: 2 \times 10404$

$\small \mapsto \sf\bold{\red{Population\: after\: 2\: years =\: 20808}}$

$\therefore$ The population of the town after 2 years is 20808 .