2.10 Population Growth and Depreciation as CI Applications

Population growth and machine/asset depreciation are simply CI and “negative CI” in disguise — refer back to your Percentage chapter for the successive percentage-change logic; here is the direct link to the CI formula.

Population growth (increases every year, like CI): Population after T years = P(1+(R)/(100))^T

Depreciation (decreases every year — same structure but with a minus sign): Value after T years = P(1-(R)/(100))^T

Solved Example 2.10.1: The population of a town is 20,000 and grows at 10% per annum. Find the population after 2 years.

Solution: Population = 20000 × (1.1)² = 20000 × 1.21 = 24,200.

Solved Example 2.10.2: A machine worth Rs. 50,000 depreciates at 10% p.a. Find its value after 2 years.

Solution: Value = 50000 × (0.9)² = 50000 × 0.81 = Rs. 40,500.

Solved Example 2.10.3 (mixed growth/depreciation): A machine’s value increases by 10% in the first year but then depreciates by 10% in the second year. Find the net percentage change over 2 years.

Solution: Net multiplying factor = 1.1 × 0.9 = 0.99, i.e., a net decrease of 1% — never simply “0% change,” a classic percentage-chapter trap that reappears constantly in CI-styled questions.

Solved Example 2.10.4: The population of a village increases at 5% p.a. If the population after 2 years is 44,100, find the initial population (edge case: working backwards from the final value).

Solution: 44100 = P × (1.05)² = P × 1.1025. So P = 44100/1.1025 = 40,000. This is the exact same “divide to undo the multiplying factor” idea used for present-value installment problems in Section 2.9(b).

Solved Example 2.10.5: A machine bought for Rs. 62,500 is valued at Rs. 40,000 after 2 years due to depreciation. Find the rate of depreciation per annum.

Solution: 40000 = 62500 × (1 − R/100)². So (1 − R/100)² = 40000/62500 = 0.64, giving 1 − R/100 = 0.8, so R/100 = 0.2. Rate of depreciation = 20% p.a.