Question

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(A) Find the population in 2001.
(B) What would be its population in 2005?

Answer 1

(A) Given,
Population in the year 2003 = 54,000
54,000 = (Population in 2001) ${(1+\frac{5}{100})}^2$
54,000 = (Population in 2001) ${(1+\frac{1}{10})}^2$

54,000 = (Population in 2001) ${\left(\frac{21}{20}\right)}^2$
Population in 2001 $=54,000\times {\left(\frac{20}{21}\right)}^2=48,979.59$
$\therefore$ The population in the year 2001 was approximately 48,980.

(B) Given,
Population in the year 2003 = 54,000
54,000 = (Population in 2001) ${(1+\frac{5}{100})}^2$
54,000 = (Population in 2001) ${(1+\frac{1}{10})}^2$

Population in 2005 $=54,000{\left(\frac{21}{20}\right)}^2$
Population in 2005 = 59,535
$\therefore$ The population in the year 2005 was 59,535.