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

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

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

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

$$=135\times441=59,535$$