This problem involves calculating the annual percentage rate at which a town's population grew over a period of three years. We are given the initial population, the final population, and the time duration. This type of problem is typically solved using the concept of compound growth, similar to how compound interest works.
The formula for population growth, assuming a constant annual growth rate, is:
$ P_t = P_0 \left(1 + \frac{R}{100}\right)^t $
Where:
Let's identify the values given in the question:
Now, we substitute these values into the population growth formula:
$ 1,06,480 = 80,000 \left(1 + \frac{R}{100}\right)^3 $
Our goal is to find the value of R. We can rearrange the equation to solve for it.
Divide both sides of the equation by the initial population ($P_0$):
$ \frac{1,06,480}{80,000} = \left(1 + \frac{R}{100}\right)^3 $
Simplify the fraction:
$ \frac{106480}{80000} = \frac{10648}{8000} $
We can simplify this fraction further by dividing both the numerator and the denominator by common factors. Dividing both by 8:
$ \frac{10648 \div 8}{8000 \div 8} = \frac{1331}{1000} $
So the equation becomes:
$ \frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3 $
To eliminate the exponent '3', we take the cube root of both sides of the equation:
$ \sqrt[3]{\frac{1331}{1000}} = \sqrt[3]{\left(1 + \frac{R}{100}\right)^3} $
We know that $11^3 = 1331$ and $10^3 = 1000$. Therefore:
$ \frac{11}{10} = 1 + \frac{R}{100} $
Converting the fraction to a decimal:
$ 1.1 = 1 + \frac{R}{100} $
Now, isolate the term $\frac{R}{100}$ by subtracting 1 from both sides:
$ 1.1 - 1 = \frac{R}{100} $
$ 0.1 = \frac{R}{100} $
Finally, multiply both sides by 100 to find the value of R:
$ R = 0.1 \times 100 $
$ R = 10 $
The value of R, representing the annual percentage growth rate of the town's population, is 10.