The present population of a village is 8,820. Twenty years ago, it was 8,000.…

When a quantity grows at a constant percentage rate every fixed time period, its value after n such periods follows the compound-growth relation: Final value = Initial value × (1 + r)n, where r is the growth rate per period as a decimal — the same relation used for compound interest.

1. The population 20 years before the present (two 10-year periods earlier) was 8,000, and the present population is 8,820, so the number of growth periods is n = 20 ÷ 10 = 2.

2. Apply the compound-growth relation with n = 2: 8,000 × (1 + r)2 = 8,820.

3. Divide both sides by 8,000: (1 + r)2 = 8,820 ÷ 8,000 = 1.1025.

4. Take the square root of both sides: 1 + r = √1.1025 = 1.05.

5. Subtract 1 from both sides: r = 0.05, i.e., the rate of increase is 5% per 10-year period.

Cross-check: starting from 8,000 and applying a 5% increase twice — 8,000 × 1.05 = 8,400 after the first 10 years, then 8,400 × 1.05 = 8,820 after the second 10 years — reproduces the given present population exactly, confirming the computed rate.

So the population increases at 5% per 10-year period.