The population of a term increases by 6% annually . IF the persent population is 17490 . What was it a year ago ?

Answer

To find the population of a town a year ago, given that the population increases by 6% annually and the current population is 17,490, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between last year's population and this year's population**:
- If the population increases by 6%, it means that this year's population is 106% of last year's population.
- We can express this mathematically as:
\[ \text{Current Population} = \text{Last Year's Population} \times \left(1 + \frac{6}{100}\right) \]
- This simplifies to:
\[ \text{Current Population} = \text{Last Year's Population} \times 1.06 \]

2. **Set up the equation**:
- Let \( P \) be the population a year ago. Then, we can write:
\[ 17,490 = P \times 1.06 \]

3. **Solve for last year's population \( P \)**:
- To find \( P \), we need to divide both sides of the equation by 1.06:
\[ P = \frac{17,490}{1.06} \]

4. **Calculate the value**:
- Performing the division:
\[ P = \frac{17,490}{1.06} \approx 16,500 \]

5. **Conclusion**:
- Therefore, the population of the town a year ago was approximately 16,500.

### Final Answer:
The population of the town a year ago was **16,500**.

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