The population of a town was 3600 three years ago. It is 4800 at present. Assuming that the rate of growth has been constant and compounding annually, what will be the population three years from now?

Introduction / Context:
This question treats population growth in terms of compound interest. The population three years ago and now are given, and we are told that the growth rate has been constant and compounding annually. Using this, we can find the yearly growth factor and then apply it to project the population three years into the future.

Given Data / Assumptions:

• Population three years ago, P0 = 3600.
• Current population, P3 = 4800.
• Growth rate is constant with annual compounding.
• We must find population after another 3 years, that is, six years from the starting point.

Concept / Approach:

If the annual growth factor is k = 1 + r, then population after n years is P_n = P_0 * k^n. We are given P3 and P0, so P3 / P0 = k^3. From this we find k. Then P6 = P0 * k^6. However, since P6 = P3 * k^3, and k^3 is already known as P3 / P0, we can compute P6 directly as P6 = P3 * (P3 / P0).

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

The values 6000, 6500, and 6600 may seem reasonable but do not correspond to a constant compound growth that takes the town from 3600 to 4800 and then forward another 3 years. Only 6400 satisfies the compound growth conditions exactly.

Common Pitfalls:

Students sometimes treat the growth as linear and assume a fixed increase of 1200 every three years, which would give incorrect answers. Another error is to attempt to find the rate r explicitly, even though it is not necessary to compute k^3 and k^6 for this type of ratio problem.

Final Answer:

The population three years from now will be 6400.