1. **Problem statement:** A person invests 100000 at 10% per annum compounded annually, and the amount grows to 133100. We need to find: (i) Total compound interest earned. (ii) Number of years taken. (iii) Simple amount for the same rate and time. 2. **Formula for compound amount:** $$A = P \left(1 + \frac{r}{100}\right)^n$$ where $A$ is the amount, $P$ is the principal, $r$ is the annual interest rate, and $n$ is the number of years. 3. **Calculate total compound interest (i):** Compound interest $= A - P = 133100 - 100000 = 33100$ 4. **Calculate number of years (ii):** Given $A = 133100$, $P = 100000$, $r = 10$, solve for $n$: $$133100 = 100000 \left(1 + \frac{10}{100}\right)^n = 100000 (1.1)^n$$ Divide both sides by 100000: $$1.331 = (1.1)^n$$ Take natural logarithm: $$\ln(1.331) = n \ln(1.1)$$ $$n = \frac{\ln(1.331)}{\ln(1.1)} \approx \frac{0.286}{0.0953} \approx 3$$ So, the investment period is 3 years. 5. **Calculate simple amount (iii):** Simple interest formula: $$SI = \frac{P \times r \times t}{100} = \frac{100000 \times 10 \times 3}{100} = 30000$$ Simple amount $= P + SI = 100000 + 30000 = 130000$ **Final answers:** (i) Compound interest = 33100 (ii) Number of years = 3 (iii) Simple amount = 130000