Question

The current population of a town is 10,000. If the population of the town is increased by 10% every year, then the population of the town after three years will be
A) 13,310
B) 13,500
C) 14,000
D) 14,500

Answer 1

Hint:
The problem can be solved by treating the problem as a problem of compound interest. The formula to calculate the amount after \[t\] years, with \[r\] rate compound added for a principle \[p\] is $p{\left( {1 + r} \right)^t}$

Complete step by step solution:
The problem can be solved by treating the problem as of compound interest. The population of the town is increasing at the compound rate of 10% every year, as the 10% of population increment is done on the present population, assuming no people are leaving the town.

The amount after \[t\] years of time with compound interest rate of \[r\] per year on the initial amount of \[p\] is given by $p{\left( {1 + r} \right)^t}$.
The value of \[r\]should be in the range of 0 to 1, where 0 denotes 0% of rate and 1 denotes 100% as the rate.

In the problem statement, we are given the rate of increase of population to be 10% every year. Therefore \[r\] is $\dfrac{{10}}{{100}}$or 0.1.
Substituting the 0.1 for \[r\], 3 for \[t\] and 10000 for \[p\] in the formula $p{\left( {1 + r} \right)^t}$ for amount after \[t\] years with \[r\] rate of compound interest added on initial amount \[p\]
$10000{\left( {1 + 0.1} \right)^3}$
We can solve the above equation to find the population of the town after 3 years.
$10000 \times {1.1^3} = 13310$
Thus the population of the town after 3 years with initial population of 10000 and 10% rate of increase every year is 13310.

Thus option A is the correct answer.

Note:
The amount after \[t\] years of time with compound interest rate of \[r\] per year on the initial amount of \[p\] is given by $p{\left( {1 + r} \right)^t}$. Alternatively the population of the town can be solved by finding the population after every year iteratively.