Population Problem Analysis

This question asks us to find the population of a city two years ago, given its current population and a consistent annual growth rate. This is a classic problem involving compound growth, similar to compound interest calculations.

Here's what we know:

• Current Population ($P_n$): 1,10,250
• Annual Growth Rate (r): 5%
• Time Period (n): 2 years

We need to find the population 2 years ago ($P_0$).

Growth Rate Formula Explained

The formula for population growth (or compound growth) is used to calculate the future value of a quantity when it grows at a constant percentage rate over time. The formula is:

$$P_n = P_0 \left(1 + \frac{r}{100}\right)^n$$

Where:

• $\mathbf{P_n}$ is the population after 'n' years (the current population).
• $\mathbf{P_0}$ is the initial population (the population 'n' years ago).
• $\mathbf{r}$ is the annual rate of increase (in percent).
• $\mathbf{n}$ is the number of years.

In our case, we need to find $P_0$. So, we can rearrange the formula to solve for $P_0$:

$$P_0 = \frac{P_n}{\left(1 + \frac{r}{100}\right)^n}$$

Past Population Calculation Step-by-Step

Let's substitute the given values into the rearranged formula to find the population 2 years ago:

• Current Population ($P_n$) = 1,10,250
• Rate of increase (r) = 5%
• Time (n) = 2 years

Step 1: Convert the percentage rate to a decimal.

$$r = 5\% = \frac{5}{100} = 0.05$$

Step 2: Calculate the growth factor.

$$1 + \frac{r}{100} = 1 + 0.05 = 1.05$$

Step 3: Raise the growth factor to the power of 'n' (number of years).

$$\left(1 + \frac{r}{100}\right)^n = (1.05)^2$$

$$ (1.05)^2 = 1.05 \times 1.05 = 1.1025$$

Step 4: Substitute all values into the formula for $P_0$.

$$P_0 = \frac{P_n}{\left(1 + \frac{r}{100}\right)^n}$$

$$P_0 = \frac{110250}{1.1025}$$

Step 5: Perform the division to find $P_0$.

$$P_0 = 100000$$

Summary of Calculations

Therefore, the population of the city 2 years ago was 1,00,000.