The population of town increases by 5% every year. If the present population is 9261. the population 3 years ago was:

Options

A. 5700
B. 6000
C. 7500
D. 8000

Explanation

To find the population of the town 3 years ago, we can use the formula for compound interest, which in this case applies to population growth. The formula is:

\[ A = P \left(1 + \frac{r}{100}\right)^n \]

Where:
- \( A \) = the amount (present population)
- \( P \) = the principal (population 3 years ago)
- \( r \) = rate of increase (5% in this case)
- \( n \) = number of years (3 years)

Given:
- Present population \( A = 9261 \)
- Rate of increase \( r = 5\% \)
- Number of years \( n = 3 \)

### Step 1: Set up the equation
We need to rearrange the formula to solve for \( P \):
\[ P = \frac{A}{\left(1 + \frac{r}{100}\right)^n} \]

### Step 2: Substitute the known values
Substituting the values we have:
\[ P = \frac{9261}{\left(1 + \frac{5}{100}\right)^3} \]

### Step 3: Calculate \( 1 + \frac{5}{100} \)
Calculate \( 1 + \frac{5}{100} \):
\[ 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \]

### Step 4: Raise to the power of 3
Now raise \( 1.05 \) to the power of 3:
\[ (1.05)^3 = 1.157625 \]

### Step 5: Substitute back into the equation
Now substitute this back into the equation for \( P \):
\[ P = \frac{9261}{1.157625} \]

### Step 6: Perform the division
Now perform the division:
\[ P \approx 8000 \]

### Conclusion
The population of the town 3 years ago was approximately **8000**.

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