Ex.8.3 Q10 Comparing Quantities Solutions - NCERT Maths Class 8

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Question

The population of a place increased to \(\rm{}\,54,000\) in \(\rm{}\,2003\) at a rate of \(5\%\) per annum

(i) find the population in \(2001.\)

(ii) what would be its population in \(2005\)?

Text Solution

What is known?

Population in \(2003\), Time Period and Rate of population growth

What is unknown?

Population in \(2005\) and \(2001\)

Reasoning:

\(\begin{align}A = P \left( 1 + \frac{r}{100} \right)^{\rm{n}}\end{align}\)

\(P=\rm{}\, 54000 \) in the year \(2003\)

\(N =2\) years

\(R =5\%\) p.a. compounded annually

Steps:

(i) Population in the year \(2001\)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ 54000 &= P\left( {{1 + }\frac{{5}}{{{100}}}} \right)^2  \\ 54000 &= P\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2}  \\ 54000 &= P\left( {\frac{{21}}{{20}}} \right)^2  \\ 54000 &= P \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ 
  P &= 54000 \times \frac{{400}}{{441}} \\ P &= 48979.6 \\ \end{align}\] 

The population in \(2001\) \(= \rm{}\,48980\).

(ii) Population in the year \(2005\)

\[\begin{align}A& = P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\ & = 54000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{2}  \\ & = 54000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2}  \\ & = 54000\left( {\frac{{21}}{{20}}} \right)^2  \\ &= 54000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ & = 54000 \times \frac{{441}}{{400}} \\ &= 135 \times 441 \\ &= 59535 \\ \end{align}\]

The population in \(2005 \) \(=\rm{} 59535.\)

The population in \(2001\) \(=\rm{} 48980\).

The population in \(2005\) \(= \rm{}\,59535\).