# 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 $$2005$$ $$= \rm{}\,59535$$.

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