Question

Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4% p.a.

Answer 1

- Hint:- Get the value of present population (P), rate of increase (R) and time n. Substitute these values in the formula to find the increase in population. Simplify the expression to get the required answer.

Complete step-by-step answer: -
It is said that presently there are 12 lakh people in a city. After 2 years the rate of increase is 4%. We need to find the population after 2 years.
Thus we can give the present population as, P = 12 lakh.
The rate of increase can be given as, R = 4%.
Now, the time period, n =2.
Thus we can use the formula to calculate the increase in population over the year.
Population \[=P{{\left( 1+\dfrac{R}{100} \right)}^{n}}\]
Now let us substitute P = 12 lakh \[=12\times {{10}^{5}}=12,00,000\], R = 4% and n = 2.
\[\therefore \]Population after 2 years \[=12\times {{10}^{5}}{{\left( 1+\dfrac{4}{100} \right)}^{2}}\]
\[\begin{align}
  & =12\times {{10}^{5}}{{\left( 1+0.04 \right)}^{2}}=12\times {{10}^{5}}{{\left( 1.04 \right)}^{2}} \\
 & =12\times {{10}^{5}}\times 1.04\times 1.04 \\
 & =12.9792\times {{10}^{5}} \\
 & =12,97,920 \\
\end{align}\]
Thus we got the population of the town after 2 years as 12,97,920.
Thus the increase in the population = 1297920 – 1200000 = 97920.
Thus in a gap of 2 years, the population increased by 97920.
\[\therefore \]The population of the city after 2 years = 1297920.

Note:-The formula for getting the increase in population \[=P{{\left( 1+\dfrac{R}{100} \right)}^{n}}\].
If we need to find the depreciation or decrease in the population, then we use the formula,
population \[=P{{\left( 1-\dfrac{R}{100} \right)}^{n}}\].