Question

Solve the given question in a detail manner:
The present population of a town is \[17640094\]. If the rate of growth in its population is \[5\% \] per annum, find:
a. Its population in \[2\] years hence.
b. Its population one year before.

Answer 1

Hint: Represent each value for a variable. Apply the given values in the formulas for the population after certain years and the population before certain years. Solve the equations and simplify them by arithmetic operations.

Complete step-by-step solution:
Let us assign a variable representation for every value.
Let \[{N_0}\] be the initial population,
Given initial population,
\[{N_0} = 17640094\]
Let \[r\] be the rate at which population changes
Given rate of population change,
\[r = 5\% \]
Let \[n\] be the number of years.
We have the formula for the population after \[n\] years.
Population after \[n\] years \[ = {N_ + } = {N_0}{\left( {1 + r} \right)^n}\]
We also have the formula for the population before \[n\] years.
Population before \[n\] years\[ = {N_ - } = \dfrac{{{N_0}}}{{{{\left( {1 + r} \right)}^n}}}\]
a. We should find the population after \[2\] years.
Here, \[n = 2\]
Since it is after \[n = 2\] years, we use the formula;
Population after \[n\] years \[ = {N_ + } = {N_0}{\left( {1 + r} \right)^n}\]
Substituting the value of \[n = 2\], we get;
Population after \[2\] years \[ = {N_2} = {N_0}{\left( {1 + r} \right)^2}\]
\[{N_0} = 17640094\]
\[r = 5\% \]
Substituting the values, we have;
$\Rightarrow$\[{N_2} = 17640094{\left( {1 + 5} \right)^2}\]
Adding the terms in the brackets, we get;
$\Rightarrow$\[{N_2} = 17640094{\left( 6 \right)^2}\]
By squaring the term, we get;
$\Rightarrow$\[{N_2} = 17640094 \times 36\]
Multiplying the terms, we get;
$\Rightarrow$\[{N_2} = {\text{635043384}}\]
Therefore, we have the population after \[2\] years.
The population after \[2\] years \[{\text{ = 63,50,43,384}}\]
b. We should find the population before \[1\] year.
Here, \[n = 1\]
Since it is after \[n = 1\] years, we use the formula;
Population after \[n\] years \[ = {N_ + } = {N_0}{\left( {1 + r} \right)^n}\]
Substituting the value of \[n = 1\], we get;
Population before \[n\] years\[ = {N_ - } = \dfrac{{{N_0}}}{{{{\left( {1 + r} \right)}^n}}}\]
\[{N_0} = 17640094\]
\[r = 5\% \]
Substituting the values, we have;
$\Rightarrow$\[{N_{ - 1}} = \dfrac{{17640094}}{{{{\left( {1 + 5} \right)}^1}}}\]
Adding the terms in the brackets, we get;
$\Rightarrow$\[{N_{ - 1}} = \dfrac{{17640094}}{{{{\left( 6 \right)}^1}}}\]
By squaring the term, we get;
$\Rightarrow$\[{N_{ - 1}} = \dfrac{{17640094}}{6}\]
Dividing the terms, we get;
$\Rightarrow$\[{{\text{N}}_{ - 1}}{\text{ = 2940015}}{\text{.666666667}}\]
Rounding off the term because it is a population, we get;
$\Rightarrow$\[{{\text{N}}_{ - 1}}{\text{ = 2940015}}\]
Therefore, we have the population before \[1\] years.
The population before \[1\] years \[{\text{ = 2940015}}\]

Note: The average rate of the population growth is the ratio between the increase in the population size and the population for that year which is multiplied by \[100\]. It is the rate at which the individuals increase in a given time period. It is expressed as a fraction. The rate is also calculated by the formula;
\[r = \dfrac{{b - d}}{N}\]
Where,
\[b = \]number of births
\[d = \] number of deaths