1. **State the problem:** You want to compare two loans to see which one results in less interest paid. 2. **Given data:** - Loan 1: 8.6% annual interest compounded monthly - Loan 2: 9.8% annual interest compounded monthly 3. **Formula for compound interest:** $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$ - $P$ is the principal (initial loan amount) - $r$ is the annual interest rate (decimal) - $n$ is the number of compounding periods per year - $t$ is the time in years 4. **Important note:** Since the principal $P$ and time $t$ are the same for both loans, we only need to compare the growth factors: $$\left(1 + \frac{r}{n}\right)^{nt}$$ 5. **Calculate monthly interest rates:** - Loan 1 monthly rate: $\frac{8.6}{100 \times 12} = 0.0071667$ - Loan 2 monthly rate: $\frac{9.8}{100 \times 12} = 0.0081667$ 6. **Compare growth factors for 1 year ($t=1$):** - Loan 1: $$\left(1 + 0.0071667\right)^{12} = (1.0071667)^{12}$$ - Loan 2: $$\left(1 + 0.0081667\right)^{12} = (1.0081667)^{12}$$ 7. **Calculate values:** - Loan 1: $$1.0071667^{12} \approx 1.0896$$ - Loan 2: $$1.0081667^{12} \approx 1.1039$$ 8. **Interpretation:** - Loan 1 grows to about 1.0896 times the principal after 1 year - Loan 2 grows to about 1.1039 times the principal after 1 year 9. **Conclusion:** You will pay less interest with the loan at 8.6% annual interest compounded monthly because the amount owed grows less over the same period. **Final answer:** Choose the loan with 8.6% annual interest compounded monthly to pay less interest.