Question

The population of Town A is 8615125 and that of town B is 8489664. The population town A and Town B are increasing at a rate of 2% and 2.5% anually. In how much time will the population of two towns will be equal

Answer 1

Answer:

3 years. Refer to the attachment.

Answer 2

$\text{Total population after n years } = P( 1 + \dfrac{r}{100})^n$

Town A:

Population now = 8615125

Growth Rate = 2%

Find the population in Town A after n years:

$\text{Total population after n years } = P( 1 + \dfrac{r}{100})^n$

$\text{Total population after n years } = 8615125( 1 + \dfrac{2}{100})^n$

$\text{Total population after n years } = 8615125( 1.02)^n$

Town B:

Population now = 8489664

Growth Rate = 2.5%

Find the population in Town B after n years:

$\text{Total population after n years } = P( 1 + \dfrac{r}{100})^n$

$\text{Total population after n years } = 8489664( 1 + \dfrac{2.5}{100})^n$

$\text{Total population after n years } = 8489664( 1.025)^n$

Solve n:

$8615125(1.02)^n = 8489664( 1.025)^n$

$\dfrac{(1.02)^n}{(1.025)^n} = \dfrac{8489664}{8615125}$

$\bigg( \dfrac{1.02}{1.025}\bigg)^n = \dfrac{8489664}{8615125}$

$\log \bigg( \dfrac{1.02}{1.025}\bigg)^n = \log \bigg(\dfrac{8489664}{8615125} \bigg)$

$n = \log \bigg(\dfrac{8489664}{8615125} \bigg) \div \log \bigg( \dfrac{1.02}{1.025}\bigg)$

$n = 3$

Answer: It will take 3 years for both the town to have equal population.