1. **State the problem:** Caroline and Qasim each invest 80000 in accounts with different interest rates and compounding frequencies. We want to find how much more money Caroline will have than Qasim after 10 years. 2. **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, $r$ is the annual interest rate (decimal), $n$ is the number of compounding periods per year, and $t$ is the number of years. 3. **Convert interest rates to decimals:** - Caroline's rate: $8 \frac{1}{8}\% = 8.125\% = 0.08125$ - Qasim's rate: $7 \frac{3}{4}\% = 7.75\% = 0.0775$ 4. **Calculate Caroline's amount:** - $P = 80000$ - $r = 0.08125$ - $n = 365$ (daily compounding) - $t = 10$ $$A_C = 80000 \left(1 + \frac{0.08125}{365}\right)^{365 \times 10}$$ 5. **Calculate Qasim's amount:** - $P = 80000$ - $r = 0.0775$ - $n = 4$ (quarterly compounding) - $t = 10$ $$A_Q = 80000 \left(1 + \frac{0.0775}{4}\right)^{4 \times 10}$$ 6. **Calculate Caroline's amount numerically:** $$A_C = 80000 \left(1 + \frac{0.08125}{365}\right)^{3650} = 80000 \left(1 + 0.0002226\right)^{3650}$$ Calculate the base: $$1 + 0.0002226 = 1.0002226$$ Raise to power: $$1.0002226^{3650} \approx e^{3650 \times 0.0002226} = e^{0.812} \approx 2.252$$ So, $$A_C \approx 80000 \times 2.252 = 180160$$ 7. **Calculate Qasim's amount numerically:** $$A_Q = 80000 \left(1 + \frac{0.0775}{4}\right)^{40} = 80000 \left(1 + 0.019375\right)^{40} = 80000 \times 1.019375^{40}$$ Calculate the base: $$1.019375^{40} = e^{40 \times \ln(1.019375)}$$ Calculate $\ln(1.019375) \approx 0.01919$ So, $$e^{40 \times 0.01919} = e^{0.7676} \approx 2.154$$ Therefore, $$A_Q \approx 80000 \times 2.154 = 172320$$ 8. **Find the difference:** $$\text{Difference} = A_C - A_Q = 180160 - 172320 = 7840$$ **Final answer:** Caroline will have approximately 7840 more than Qasim after 10 years.