Question

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. What would be its population in 2005?

• A. 59,535
• B. 57,892
• C. 62,899
• D. 67,134

Answer 1

The correct option is A

59,535

Let $A_{2005}$ be the population of the place in 2005.

Let $A_{2003}$ be the population of the place in 2003. Then $A_{2003}$ = 54,000

Let R be the rate(per annum) at which the population increased. Then R = 5%

Let n = 2 years

According to the question, population is increasing .

$\therefore A_{2005}=A_{2003}{(1+\frac{R}{100})}^n$
$=54000{(1+\frac{5}{100})}^2$

$=54000{(1+\frac{1}{20})}^2$
$=54000{\left(\frac{21}{20}\right)}^2$

$=54000\times \frac{21}{20}\times \frac{21}{20}$
$=59,535$
Hence population in 2005 would be 59,535.