A building worth ₹ 1,33,100 is constructed on land worth ₹ 72,900. After how many years will the value of both be the same if land appreciates at 10% p.a. and building depreciates at 10% p.a.?

### Concept & Equation This problem combines **Compound Appreciation** and **Compound Depreciation**. We need to track the future value of two separate assets changing at different rates (one positive, one negative) and find the point in time ($n$ years) when their values are equal. * Formula for Appreciation (Land): $A_1 = P_1\left(1 + \frac{R_1}{100}\right)^n$ * Formula for Depreciation (Building): $A_2 = P_2\left(1 - \frac{R_2}{100}\right)^n$ We set $A_1 = A_2$ and solve for $n$. ### Step-by-step Solution * **Given:** * Initial Land Value ($P_1$) = ₹ $72,900$ * Land Appreciation Rate ($R_1$) = $+10\%$ p.a. * Initial Building Value ($P_2$) = ₹ $1,33,100$ * Building Depreciation Rate ($R_2$) = $-10\%$ p.a. * **Setting up the Equation:** * Future value of Land = $72900 \times \left(1 + \frac{10}{100}\right)^n = 72900 \times (1.1)^n$ * Future value of Building = $133100 \times \left(1 - \frac{10}{100}\right)^n = 133100 \times (0.9)^n$ * Since their values become equal: $$72900 \times (1.1)^n = 133100 \times (0.9)^n$$ * **Calculation:** * Rearrange to group the exponential terms on one side and constants on the other: $$\frac{(1.1)^n}{(0.9)^n} = \frac{133100}{72900}$$ * Simplify both sides: $$\left(\frac{1.1}{0.9}\right)^n = \frac{1331}{729}$$ * Multiply the numerator and denominator of the fraction inside the parenthesis by $10$: $$\left(\frac{11}{9}\right)^n = \frac{1331}{729}$$ * Recognize the cubes of the numbers. You should know that $11^3 = 1331$ and $9^3 = 729$: $$\left(\frac{11}{9}\right)^n = \left(\frac{11}{9}\right)^3$$ * Equating the exponents: $$n = 3$$ ### Exam Strategy & Shortcut In competitive exams, numbers like $1331$ and $729$ are massive hints. They are highly recognizable perfect cubes ($11^3$ and $9^3$). When you see $+10\%$ and $-10\%$, you are dealing with multipliers of $\frac{11}{10}$ and $\frac{9}{10}$. The ratio of their starting values is $\frac{133100}{72900}$, which instantly reduces to $\frac{1331}{729}$. Because this ratio perfectly matches the cubes of the rate multipliers ($\frac{11^3}{9^3}$), you can instantly conclude that $n = 3$ without writing out the full algebraic equation. ### Common Pitfall A common trap is accidentally swapping the rates (e.g., applying depreciation to the land and appreciation to the building). Always double-check which asset is increasing and which is decreasing. Buildings generally depreciate due to wear and tear, while land appreciates due to scarcity, which aligns with logical real-world principles. ### Final Answer **Therefore, the correct answer is 3.**