3.8 Population Growth and Decline

Figure: Population growth is compound interest wearing a different name.

Population problems are compound-style applications of percentage: each year’s growth or decline acts on the previous year’s population, not the original one. If population P grows at r% per annum:

P_n = P(1+(r)/(100))^n (growth for n years)

P_n = P(1-(r)/(100))^n (decline for n years)

When growth and decline alternate across different years, apply the successive percentage change formula year by year rather than a single power.

Worked Example 1: The population of a town is 40,000. It grows at 5% per annum. Find the population after 2 years. P₂ = 40,000 × (1.05)² = 40,000 × 1.1025 = 44,100.

Worked Example 2: The population of a village, currently 50,000, is declining at 4% per year. Find the population after 2 years. P₂ = 50,000 × (0.96)² = 50,000 × 0.9216 = 46,080.

Worked Example 3: The population of a city was 20,000. It increased by 10% in the first year and decreased by 5% in the second year. Find the population at the end of two years. Net % change (successive formula): 10 + (−5) + (10 × −5)/100 = 5 − 0.5 = 4.5%. Final population = 20,000 × 1.045 = 20,900.

Worked Example 4: The population of a town was 10,000 two years ago and is 12,100 today, growing at a constant rate of r% per annum. Find r. 10,000 × (1 + r/100)² = 12,100 → (1 + r/100)² = 12,100/10,000 = 1.21 → (1 + r/100) = √1.21 = 1.1 So r = 10%. This “work backward to find the rate” variant is common when a question gives population figures two years apart instead of directly stating the growth rate — always take the ratio of the two populations and find its square root (for a 2-year gap) to recover the annual rate.

Worked Example 5 (reverse growth, three-year gap): The population of a town, growing at a constant rate per annum, was 8,000 three years ago and is 10,648 today. Find the annual rate of growth. 8,000 × (1 + r/100)³ = 10,648 → (1 + r/100)³ = 10,648/8,000 = 1.331. Since 1.1³ = 1.1 × 1.1 × 1.1 = 1.21 × 1.1 = 1.331, we get (1 + r/100) = 1.1, so r = 10%. For an n-year gap, take the nth root of the population ratio rather than only the square root — the same idea as Worked Example 4, extended.

Worked Example 6 (reverse decline): The population of a village, declining at a constant annual rate, was 50,000 two years ago and is 38,720 today. Find the annual rate of decline. 50,000 × (1 − r/100)² = 38,720 → (1 − r/100)² = 38,720/50,000 = 0.7744. √0.7744 = 0.88 (since 0.88² = 0.7744), so (1 − r/100) = 0.88, giving r = 12%.