1. **Problem statement:** Hitanshi invests money at an annual compound interest rate of $x\%$. After 11 years, the investment doubles in value. We need to find $x$. 2. **Formula used:** The compound interest formula is: $$ A = P \left(1 + \frac{r}{100}\right)^t $$ where $A$ is the amount after time $t$, $P$ is the principal, $r$ is the annual interest rate in percent, and $t$ is the time in years. 3. **Given:** $A = 2P$ (investment doubles), $t = 11$ years. 4. Substitute values into the formula: $$ 2P = P \left(1 + \frac{x}{100}\right)^{11} $$ 5. Divide both sides by $P$: $$ \cancel{P} \times 2 = \cancel{P} \times \left(1 + \frac{x}{100}\right)^{11} \implies 2 = \left(1 + \frac{x}{100}\right)^{11} $$ 6. Take the 11th root of both sides: $$ 2^{\frac{1}{11}} = 1 + \frac{x}{100} $$ 7. Solve for $x$: $$ \frac{x}{100} = 2^{\frac{1}{11}} - 1 $$ $$ x = 100 \left(2^{\frac{1}{11}} - 1\right) $$ 8. Calculate the numerical value: $$ 2^{\frac{1}{11}} \approx 1.065 $$ $$ x \approx 100 \times (1.065 - 1) = 100 \times 0.065 = 6.5 $$ **Final answer:** The annual compound interest rate $x$ is approximately **6.5\%**.