The population of a town is increased by 15% in the first year and 8% in the second year, but decreased by 12 `(1)/(2)%` in the third year. If the polulation at the end of the third year is 26.082. find the original polulation of the towm.

Options

A. 30000
B. 25000
C. 24000
D. 28000

Answer

To find the original population of the town, we will denote the original population as \( x \).

### Step-by-step Solution:

1. **First Year Increase**:
The population increases by 15% in the first year.
\[ \text{Population after 1st year} = x + 0.15x = 1.15x \]

2. **Second Year Increase**:
The population then increases by 8% in the second year.
\[ \text{Population after 2nd year} = 1.15x + 0.08(1.15x) = 1.15x \times 1.08 = 1.242x \]

3. **Third Year Decrease**:
The population decreases by \( 12 \frac{1}{2} \% \) in the third year, which is equivalent to \( \frac{25}{2} \% \) or \( 12.5\% \).
\[ \text{Population after 3rd year} = 1.242x - 0.125(1.242x) = 1.242x \times (1 - 0.125) = 1.242x \times 0.875 \]

4. **Calculate the Population after 3rd Year**:
\[ \text{Population after 3rd year} = 1.242x \times 0.875 = 1.08675x \]

5. **Setting up the Equation**:
We know that the population at the end of the third year is 26,082.
\[ 1.08675x = 26,082 \]

6. **Solving for \( x \)**:
\[ x = \frac{26,082}{1.08675} \]
Now, we calculate \( x \):
\[ x \approx 24,000 \]

### Conclusion:
The original population of the town is approximately **24,000**.