The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is

### Concept & Exponential Growth To find when a sum more than doubles, we need to find the smallest integer value of $n$ for which the amount is greater than $2P$. $$P\left(1 + \frac{R}{100}\right)^n > 2P$$ ### Step-by-Step Solution * **Given:** Rate of interest $R = 20\%$. * We need: $P\left(1 + \frac{20}{100}\right)^n > 2P$ * Divide both sides by $P$: $\left(1 + \frac{1}{5}\right)^n > 2$ * $\left(\frac{6}{5}\right)^n > 2$, which means $(1.2)^n > 2$. * Let's evaluate powers of $1.2$: * For $n = 1$: $(1.2)^1 = 1.2$ * For $n = 2$: $(1.2)^2 = 1.44$ * For $n = 3$: $(1.2)^3 = 1.44 \times 1.2 = 1.728$ * For $n = 4$: $(1.2)^4 = 1.728 \times 1.2 = 2.0736$ * Since $2.0736 > 2$, the sum will be more than doubled in 4 years. ### Exam Strategy & Shortcut For 20%, you can use the Rule of 72 as a quick approximation for doubling time. $72 / 20 = 3.6$ years. Since the interest is compounded annually, we round up to the next complete year, which is 4. ### Common Pitfall Miscalculating the decimal multiplications (e.g., assuming $1.2^3$ is greater than 2) or confusing it with simple interest where it would take exactly 5 years to double. ### Final Answer Therefore, the correct answer is **4**.