Compound Interest 159Bf3

1. **Problem Statement:** Mr. Raul wants to have $1,650,000$ in 10 years. We need to find how much he should invest today in a bank that offers an interest rate of $1.25\%$ compounded semi-annually. 2. **Formula Used:** The formula for compound interest is: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where: - $A$ is the amount after time $t$ - $P$ is the principal (initial investment) - $r$ is the annual interest rate (decimal) - $n$ is the number of times interest is compounded per year - $t$ is the number of years 3. **Given Values:** - $A = 1,650,000$ - $r = 1.25\% = 0.0125$ - $n = 2$ (since compounded semi-annually) - $t = 10$ 4. **Rearranging the formula to find $P$:** $$ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} $$ 5. **Substitute the values:** $$ P = \frac{1,650,000}{\left(1 + \frac{0.0125}{2}\right)^{2 \times 10}} = \frac{1,650,000}{\left(1 + 0.00625\right)^{20}} = \frac{1,650,000}{(1.00625)^{20}} $$ 6. **Calculate the denominator:** $$ (1.00625)^{20} = 1.131408 $$ 7. **Calculate $P$:** $$ P = \frac{1,650,000}{1.131408} $$ 8. **Simplify with cancellation:** $$ P = \frac{\cancel{1,650,000}}{1.131408} = 1,458,263.16 $$ 9. **Final Answer:** Mr. Raul should invest approximately **1,458,263.16** today to have 1,650,000 in 10 years with 1.25% interest compounded semi-annually.