Question

the population of a village is 64000 if the population increases at the rate of 2.5 % per annum find the population at the end of 3 years​

Answer 1

$\huge{\mathfrak{\red{\underline{\underline{........Solution........}}}}}$

Given:

=> Initial population of village (P) = 64000

=> Rate (r) = 2.5%

=> Time (t) = 3 years.

To Find:

=> Population at the end of 3 years.

Formula used:

$\sf{\implies P\bigg(1+\dfrac{r}{100}\bigg)^{t}}$

Now, put the value in the formula.

$\sf{\implies P\bigg(1+\dfrac{r}{100}\bigg)^{t}}$

$\sf{\implies 64000\bigg(1+\dfrac{2.5}{100}\bigg)^{3}}$

$\sf{\implies 64000\bigg(\dfrac{100+2.5}{100}\bigg)^{3}}$

$\sf{\implies 64000\bigg(\dfrac{102.5}{100}\bigg)^{3}}$

$\sf{\implies 64000(1.025)^{3}}$

$\sf{\implies 64000\times 1.076890625}$

$\large{\boxed{\boxed{\pink{\sf{\implies 68921}}}}}$

Hence, population at the end of 3 years is 68921.

Answer 2

$\bf{\large{\underline{\underline{\rm{Answer:}}}}}$

The population of village after 3 years will be 68.921.

$\bf{\large{\underline{\underline{\rm{Explaination:}}}}}$

$\bf{\large{\underline{\sf{Given:}}}}$

• Population of a village is 64,000.

Population increases at the rate of 2.5% every year.

Time period = 3 years

$\bf{\large{\underline{\sf{To\:find:}}}}$

The population of the village after 3 years.

$\bf{\large{\underline{\sf{Solution:}}}}$

Let the initial population of the village be P = 64000

Let the rate at which the population increases evry year be r = 2.5%

Let the time period be t = 3 years

$\bf{\underline{\boxed{\large{\rm{Formula\:to\:be\:used:}}}}}$

$\rightarrow$$\bf\large\rm{P}$ $\bf\large({1+\frac{r}{100})^t}$

Block in the values,

$\rightarrow$ $\bf{64000}$ $\bf({1\:+\frac{2.5}{100})^3}$

$\rightarrow$$\bf{64000}$ $\bf({\frac{100\:+\:2.5}{100})^3}$

$\rightarrow$ $\bf{64000}$ $\bf\large({\frac{102.5}{100})^3}$

$\rightarrow$$\bf{64000}$ $\bf\large{\times1.025}$

$\rightarrow$$\bf{64000\times\:1.025\times\:1.025\times\:1.025}$

$\rightarrow$$\bf{64000\times\:1.050625\:\times1.025}$

$\rightarrow$$\bf{64000\times\:1.076890625}$

$\rightarrow$$\bf\large\rm{68,921}$