Question

A town's population is increased at a constant annual rate from 40000 in 2006 to 42436 in 2008. a.) find the annual percentage increase. b.)find the population size in 2007.

Answer 1

Step-by-step explanation:

(a) Let annual percentage increase be 'r'= %

Population in 2006 , x= 40000

Population in 2008, y = 42436

Time period n =2 years

we know that

$y = x(1 + \frac{r}{100} ) {}^{2}$

$= > 42436 = 40000(1 + \frac{r}{100}) {}^{2}$

$= > \frac{42436}{40000} = (1 + \frac{r}{100} ) {}^{2}$

$= > 1 + \frac{r}{100} = \sqrt{ \frac{42436}{40000} }$

$= > 1 + \frac{r}{100} = \frac{206}{200}$

$= > \frac{r}{100} = \frac{206}{200} - 1$

$= > \frac{r}{100} = \frac{6}{200}$

$= > r = \frac{6}{200} \times 100$

=> r =3

Hence the annual percentage increase of population is equal to 3%

(b) Now population in 2007 = x(1+ 3/100)

= 40000×103/100

= 400×103

= 41200