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

(ii) What would be its population in 2005?

(i) We are given: Population in 2003 = 54,000, Rate = 5% p.a.
Time = 2003 – 2001 = 2 years

\(Population\; in \;2003 = Population\; in \;2001\times(1+\frac{R}{100})^n\)

\(\Rightarrow 54,000=Population\; of \;2001\times(1+\frac{5}{100})^2\)

\(\Rightarrow 54,000=Population \;of \;2001\times(\frac{21}{20})^2\)

\(\Rightarrow 54,000=Population\; of \;2001\times(\frac{441}{400})\)

∴ Population of 2001= \(\frac{54,000\times 400}{441}\)

= \(\frac{21,6,00,000}{441}=48,979.59≈48,980\)

(ii) \(Population \;in \;2005= Population\;in\;2003\times(1+\frac{R}{100})^n\)

\( =54,000\times(1+\frac{5}{100})^2\)

\( =54,000\times(\frac{21}{20})^2\)

\( =54,000\times(\frac{441}{400})^2\)

\(=135\times441=59,535\)