Home › SSC CGL › Quantitative Aptitude › Percentages › **Percentages — formulas used here.** - Definiti…
**Percentages — formulas used here.** - Definition. $x\%$ means $\tfrac{x}{100}$. - Increase consumption. If price rises by $R\%$ and spending must stay constant, consumption falls by $\tfrac{R}{100+R} \times 100\%$. - Decrease consumption. If price falls by $R\%$ and spending must stay constant, consumption rises by $\tfrac{R}{100-R} \times 100\%$. - Growth. A population $P$ growing at $R\%$ a year becomes $P(1 + \tfrac{R}{100})^n$ after $n$ years, and was $P / (1 + \tfrac{R}{100})^n$, $n$ years ago. - Depreciation. A value $V$ depreciating at $R\%$ a year becomes $V(1 - \tfrac{R}{100})^n$ after $n$ years. - Reciprocal comparisons. If A is $R\%$ more than B, then B is $\tfrac{R}{100+R} \times 100\%$ less than A. If A is $R\%$ less than B, then B is $\tfrac{R}{100-R} \times 100\%$ more than A. **Question.** A town's current population is $50{,}000$ and grows at $10\%$ per year. What was its population $2$ years ago (to the nearest whole number)?
A$40{,}000$
B$41{,}322$
C$45{,}000$
D$48{,}000$
Answer & Solution
Correct answer: B. $41{,}322$
Population $n$ years ago is $\dfrac{P}{(1 + R/100)^n}$. Plugging in: $\dfrac{50{,}000}{(1.1)^2} = \dfrac{50{,}000}{1.21} \approx 41{,}322$.
Option A (40,000) is what you'd get by naively subtracting two flat $10\%$ chunks; the correct treatment uses compound discount.
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