1. **State the problem:** Olivia and Sadie each take out a loan of 2000 at 6.5% compound interest per year. Olivia pays back 1000 two years after the loan, Sadie pays back 1000 ten years after the loan. We need to find how much more money Sadie owes than Olivia. 2. **Formula for compound interest:** The amount owed after $t$ years is given by $$A = P\left(1 + \frac{r}{100}\right)^t$$ where $P$ is the principal, $r$ is the annual interest rate, and $t$ is time in years. 3. **Calculate Olivia's remaining debt:** - Olivia pays back 1000 two years after the loan, so the loan grows for 2 years first: $$A_2 = 2000\left(1 + \frac{6.5}{100}\right)^2 = 2000 \times 1.065^2$$ - Calculate $1.065^2 = 1.134225$ - So, $$A_2 = 2000 \times 1.134225 = 2268.45$$ - Olivia pays back 1000, so remaining debt after payment is $$2268.45 - 1000 = 1268.45$$ - This remaining debt continues to compound for 1 more year (3 years total - 2 years before payment = 1 year after payment): $$A_{3} = 1268.45 \times 1.065 = 1351.18$$ 4. **Calculate Sadie's remaining debt:** - Sadie does not pay anything until 10 years after the loan, so the loan grows for 10 years: $$A_{10} = 2000 \times 1.065^{10}$$ - Calculate $1.065^{10} \approx 1.877455$ - So, $$A_{10} = 2000 \times 1.877455 = 3754.91$$ - Sadie pays back 1000, so remaining debt after payment is $$3754.91 - 1000 = 2754.91$$ 5. **Calculate how much more Sadie owes than Olivia:** $$\text{Difference} = 2754.91 - 1351.18 = 1403.73$$ **Final answer:** Sadie owes 1403.73 more than Olivia.