Question

The increase of population of town is 10%every year. If the present population is 100000 what will be the population in after 3 years

Answer 1

Given :

• The increase of population of town is 10% every year.
• The present population is 100000.

To find :

• What will be the population in after 3 years ?

Solution :

Formula Used :-

${\boxed{\bold{Uniform\:rate\: of\: growth=p(1+\dfrac{r}{100})^n}}}$

• Terms identification :
• p = Population
• r = Rate of increasing population.
• n = time (years)

Here,

• p = 100000
• r = 10%
• n = 3 years

After 3 years, the population will be ,

$\implies\sf{100000(1+\dfrac{10}{100})^3}$

$\implies\sf{100000(1+\dfrac{1}{10})^3}$

$\implies\sf{100000\times(\dfrac{11}{10})^3}$

$\implies\sf{100000\times\dfrac{11}{10}\times\dfrac{11}{10}\times\dfrac{11}{10}}$

$\implies\sf{100\times\:11\times\:11\times\:11}$

$\implies\sf{133100}$

Therefore,

After 3 years, the population will be 133100 .

______________________

Answer 2

$\huge \mathfrak{Answer:-}$

$\implies$ 133100

$\large\underline\mathrm{Given:-}$

• The increase of population of town is 10%every year.
• The present population is 100000.

$\large\underline\mathrm{To \: find}$

• what will be the population in after 3 years.

$\large\underline\mathrm{Using \: formula}$

• p = 100000
• r = 10%
• n = 3 years

$\large\underline\mathrm{Solution}$

$\implies$ 100000(1 + 10/100)3

$\implies$ 100000(1 + 1/10)3

$\implies$ 100000 × 11/10³

$\implies$ 100000 × 11/10 × 11/10 × 11/10

$\implies$ 100 × 11 × 11 × 11

$\implies$ 133100

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{After \: 3 \: year's, \: the \: population \: will \: be \: 133100.}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$