Question

The population of a town was 3.600 three years back. It is 4,800 right now. What will be the population
three years hence if the growth rate of population has been compounded annually with a constant
rate?​

Answer 1

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

• Population of town 3 yrs back =3600.
• At present , population = 4800

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

• The population  three years hence if the growth rate of population has been compounded annually with a constant  rate?​

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

We know :

$\mapsto \pink{\sf A=P(1+\frac{r}{100} )^n}$

Acc to question ,

Acc to population 3 yrs back :

$\implies \sf 4800 = 3600(1+\frac{r}{100} )\\\\\\\implies \sf 48 / 36 = (1+\frac{r}{100} )^3\\\\\\\implies \sf 4/3 = (1+\frac{r}{100} )^3....(1)$

Acc to population from now :

$\implies \sf 4800 (1+\frac{r}{100} )^3$

From equation (1) ,

$:\implies \sf (1+\frac{r}{100} )^3$

Therefore,

$\implies \sf 4800 (4/3)\\\\\implies \sf \underline{6400.}$

Hence, required answer is 6400.

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