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 \(2005\)?

 Video Solution
Comparing Quantities
Ex 8.3 | Question 10

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 \(2005\) \(= \rm{}\,59535\).