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**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 machine bought for ₹$80{,}000$ depreciates at $20\%$ per year. What is its value at the end of $3$ years?

A₹$38{,}400$
B₹$40{,}960$
C₹$45{,}000$
D₹$51{,}200$
Answer & Solution
Correct answer: B. ₹$40{,}960$
Depreciated value $= V(1 - R/100)^n = 80{,}000 \times (0.8)^3$. $(0.8)^3 = 0.512$, so $80{,}000 \times 0.512 = 40{,}960$. Option D (51,200) is the value after $2$ years (off-by-one); option A (38,400) is just $0.8^3 \times 75{,}000$ with the wrong base.
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