Question

the population of a town 2yr ago was 62500 due to migration every year it decrease by 4% what is the present population ​

Answer 1

Answer:

57600 members

Step-by-step explanation:

population before one year :

=>62500-4/100×62500

=>60000 members

population this year :

=>60000-4/100×60000

=>57600 members

Answer 2

☆ To Find :

The Present Population of the town.

☆ We Know :

Depreciation :

$\purple{\sf{\underline{V = V_{0}\left(1 - \dfrac{r}{100}\right)^{n}}}}$

Where,

• V = Present Population
• V⁰ = Population Before n years
• n = Time
• r = Rate of Depreciation

Solution :

☆ Given :

• Time = 2 years
• Population Before 2 years = 62500
• Rate of Depreciation = 4% p.a.

☆ Calculation :

Using the formula and substituting the values in it ,we get :

$\purple{\sf{V = V_{0}\left(1 - \dfrac{r}{100}\right)^{n}}} \\ \\ \\ \\ \implies \sf{V = 62500 \times \left(1 - \dfrac{4}{100}\right)^{2}} \\ \\ \\ \\ \implies \sf{V = 62500 \times \left(\dfrac{100 - 4}{100}\right)^{2}} \\ \\ \\ \\ \implies \sf{V = 62500 \times \left(\dfrac{96}{100}\right)^{2}} \\ \\ \\ \\ \implies \sf{V = 62500 \times \dfrac{96}{100} \times \dfrac{96}{100}} \\ \\ \\ \\ \implies \implies \sf{V = 62500 \times \dfrac{24}{25} \times \dfrac{24}{25}} \\ \\ \\ \\ \implies \sf{V = 2500 \times 24 \times \dfrac{24}{25}} \\ \\ \\ \\ \implies \sf{V = 100 \times 24 \times 24} \\ \\ \\ \\ \implies \sf{V = 100 \times 576} \\ \\ \\ \\ \implies \sf{V = 57600} \\ \\ \\ \\ \therefore \purple{\sf{V = 57600}}$

Hence , The present Population is ₹ 57600.

Additional information :

• Growth =

$\sf{V = V_{0}\left(1 + \dfrac{r}{100}\right)^{n}}$

• Growth for two Rate =

$\sf{V = V_{0}\left(1 + \dfrac{r_{1}}{100}\right)\left(1 + \dfrac{r_{2}}{100}\right)}$