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The question asks to determine the population of a village two years ago if the current population is 2800 and the population increases at a rate of 6% per year.

### Solution:
This is a case of compound growth. The formula for the population after \( t \) years of growth is:

\[
P_{\text{current}} = P_{\text{initial}} \times (1 + r)^t
\]

Where:
- \( P_{\text{current}} \) is the current population.
- \( P_{\text{initial}} \) is the population 2 years ago (the value we want to find).
- \( r \) is the annual growth rate (6% or 0.06).
- \( t \) is the time in years (2 years).

Rearranging the formula to solve for \( P_{\text{initial}} \):

\[
P_{\text{initial}} = \frac{P_{\text{current}}}{(1 + r)^t}
\]

Substituting the values:

\[
P_{\text{initial}} = \frac{2800}{(1 + 0.06)^2} = \frac{2800}{(1.06)^2}
\]

Calculating:

\[
P_{\text{initial}} = \frac{2800}{1.1236} \approx 2492.04
\]

So, the population two years ago was approximately 2492, which rounds to **2500**.

### Answer: 
The correct option is **D) 2500**.

Would you like more details on any part of the calculation? Here are some related questions:

1. How would the formula change if the growth rate was compounded monthly instead of yearly?
2. What is the formula for exponential decay, and how would it apply in a population decrease scenario?
3. If the growth rate was 8% instead of 6%, what would the population have been two years ago?
4. Can you calculate the population three years ago instead of two using the same method?
5. How would the result change if the population increased at 6% for one year and 4% the next year?

**Tip:** When using compound growth or decay formulas, always check the units for time and rate to ensure they match, e.g., both in years or months.

The population of a village increases at the rate of 6% per year. What was the population 2 years ago if the present population is 2800?

Learn how to calculate the population of a village two years ago given a 6% annual growth rate and the present population of 2800. This problem explores compound growth and exponential functions, ideal for students in Grades 10-12.

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