question_answer 26)

                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 ?

Answer:

                (i) Let the population in 2001 be P.                                           R = 5% p.a.                    \[n=2\] years \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}=P{{\left( 1+\frac{5}{100} \right)}^{2}}\] \[=P{{\left( 1+\frac{1}{20} \right)}^{2}}\,=P{{\left( \frac{21}{20} \right)}^{2}}\] According to the question, \[P{{\left( \frac{21}{20} \right)}^{2}}\,=54,000\] \[\Rightarrow \]               \[P=54,000{{\left( \frac{20}{21} \right)}^{2}}\] \[=54,000\times \frac{20}{21}\times \frac{20}{21}\] = 48, 980 (approx.) \[=54-000\times 21\times 21\] = 48,980 (approx.) Hence, the population in 2001 was 48,980 (approx.) (ii) P = 54, 000 R = 5% p.a. \[n=2\] years \[\therefore \]  \[A=P{{\left( 1+\frac{R}{100} \right)}^{n}}\] \[=54,000{{\left( 1+\frac{5}{100} \right)}^{2}}\] \[=54,000{{\left( 1+\frac{1}{20} \right)}^{2}}\] \[=54,000{{\left( \frac{21}{20} \right)}^{2}}\] \[=54,000\times \frac{21}{20}\times \frac{21}{20}=59,535\] Hence, the population in 2005 would be 59,