The current population of a town is 50,000. If it is growing at the rate of 6% compounded every year, what will be the population of the town after 2 years?
- AOption (a)
56180
- BOption (b)
56280
- COption (c)
56000
- DOption (d)
56080
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Correct answer
56180
Why this answer is correct
Shortcut Trick
Net percentage growth for 2 years at 6% rate = 6 + 6 + (6 × 6)/100 = 12.36%.
Current Population = 50,000.
Total Growth = 12.36% of 50,000 = 0.1236 × 50,000 = 6,180.
Population after 2 years = 50,000 + 6,180 = 56,180.
∴ The correct answer is 56180.
Alternate Method
Given:
Current Population (P) = 50,000
Growth Rate (R) = 6% per annum
Time Period (n) = 2 years
Formula Used:
Population after n years = P × (1 + R/100)n
⇒ Population after 1 year = 50,000 × (1 + 6/100)
⇒ Population after 1 year = 50,000 × 1.06 = 53,000
⇒ Population after 2 years = 53,000 × (1 + 6/100)
⇒ Population after 2 years = 53,000 × 1.06
⇒ Population after 2 years = 56,180
∴ The correct answer is 56180.
Additional Information
Successive Percentage Change
When a value increases by x% and then y% successively, the net percentage change is given by [x + y + (xy/100)]%.
Depreciation Formula
If the population decreases at a rate R, the population after n years is calculated as P × (1 − R/100)n.
Compound Growth Concept
Growth in population, bacteria, or compound interest follows the same mathematical model of exponential growth.