Question

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. Find the population in 2001.

• A. 49,500
• B. 48,980
• C. 51,960
• D. 52,085

Answer 1

The correct option is B

48,980

Let $A_{2001}$ be the population of the place in 2001.

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
Population would be less in 2001 than 2003 in two years.
$\therefore A_{2003}=A_{2001}{(1+\frac{R}{100})}^n$
$\Rightarrow 54000=A_{2001}{(1+\frac{5}{100})}^2\Rightarrow 54000=A_{2001}{(1+\frac{1}{20})}^2$
$\Rightarrow 54000=A_{2001}{\left(\frac{21}{20}\right)}^2\Rightarrow 54000=A_{2001}\times \frac{21}{20}\times \frac{21}{20}$

$\Rightarrow A_{2001}=\frac{54000\times 20\times 20\times 20}{21\times 21}$
$\Rightarrow A_{2001}=48,980\left(approx\right)$