question_answer

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:

(a) Here A = 54000, R = 5%, n = 2 year,

P = ?

A = P                        \[{{\left( 1+\frac{R}{100} \right)}^{n}}\]

54000=P                                            \[{{\left( 1+\frac{5}{100} \right)}^{2}}\]

= P                                               \[{{\left( \frac{21}{20} \right)}^{2}}\]

\[P=\frac{54000\times 20\times 20}{21\times 21}\]

= 48980 (approx.)

Population in 2001 was 48980.

(b) Here, P = 54000, R = 5% p.a., n = 2 years

A = P                                            \[{{\left( 1+\frac{R}{100} \right)}^{n}}\]

= 54000                                            \[{{\left( 1+\frac{5}{100} \right)}^{2}}\]

\[=\text{ }54000\text{ }\times \frac{21}{20}\times \frac{21}{20}\]

= 59535

Hence, population in 2005 would be 59535.