Ncert solutions

# Question 10

Q10) 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?

Solution:

(i) Here, A_{2003}=\ Rs.\ 54,000, R = 5%, n= 2 years

Population would be less in 2001 than 2003 in two years.

Here the population is increasing.

\therefore A_{2003}=P_{2001}\left(1+\frac{R}{100}\right)^n

\Rightarrow54000=P_{2001}\left(1+\frac{5}{100}\right)^2

\Rightarrow54000=P_{2001}\left(1+\frac{1}{20}\right)^2

\Rightarrow54000=P_{2001_{ }}\left(1+\frac{1}{20}\right)^2

\Rightarrow54000=P_{2001}\left(\frac{21}{20}\right)^2

\Rightarrow54000=P_{2001}\times\frac{21}{20}\times\frac{21}{20}

\Rightarrow P_{2001}=\frac{54000\times20\times20}{21\times21}

\Rightarrow P_{2001}=48980 (Approx)

(ii) According to question, population is increasing. Therefore population in 2005,

A_{2005}=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}

= 59,535

Hence population in 2005 would be 59,535.

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