1. **State the problem:** Simon invested an amount of money in a savings account with 0.5% compound interest per annum. After 3 years, the amount is 12180.90. We need to find the initial investment. 2. **Formula used:** The compound interest formula is $$A = P \left(1 + \frac{r}{100}\right)^n$$ where: - $A$ is the amount after $n$ years - $P$ is the principal (initial investment) - $r$ is the annual interest rate (in %) - $n$ is the number of years 3. **Substitute known values:** $$12180.90 = P \left(1 + \frac{0.5}{100}\right)^3$$ 4. **Simplify inside the parentheses:** $$1 + \frac{0.5}{100} = 1 + 0.005 = 1.005$$ 5. **Calculate the power:** $$1.005^3 = 1.005 \times 1.005 \times 1.005 = 1.015075125$$ 6. **Rewrite the equation:** $$12180.90 = P \times 1.015075125$$ 7. **Solve for $P$:** $$P = \frac{12180.90}{1.015075125}$$ 8. **Show cancellation step:** $$P = \frac{\cancel{12180.90}}{\cancel{1.015075125}}$$ (just indicating division) 9. **Calculate $P$:** $$P \approx 12000$$ **Final answer:** Simon invested approximately 12000 initially.