Compound Interest Comparison 54Bf55

1. **State the problem:** You and your friend both invest $13,000 for 10 years but with different interest rates and compounding frequencies. We want to find how much more money you will have compared to your friend after 10 years. 2. **Formula for compound interest:** $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after $t$ years - $P$ is the principal (initial amount) - $r$ is the annual interest rate (decimal) - $n$ is the number of compounding periods per year - $t$ is the number of years 3. **Calculate your friend's amount:** - $P = 13000$ - $r = 0.04$ - $n = 365$ (daily compounding) - $t = 10$ $$A_{friend} = 13000 \left(1 + \frac{0.04}{365}\right)^{365 \times 10}$$ Calculate inside the parentheses: $$1 + \frac{0.04}{365} = 1 + 0.000109589 = 1.000109589$$ Calculate the exponent: $$365 \times 10 = 3650$$ So: $$A_{friend} = 13000 \times (1.000109589)^{3650}$$ Using a calculator: $$(1.000109589)^{3650} \approx 1.491825$$ Multiply: $$A_{friend} = 13000 \times 1.491825 = 19393.73$$ 4. **Calculate your amount:** - $P = 13000$ - $r = 0.042$ - $n = 4$ (quarterly compounding) - $t = 10$ $$A_{you} = 13000 \left(1 + \frac{0.042}{4}\right)^{4 \times 10}$$ Calculate inside the parentheses: $$1 + \frac{0.042}{4} = 1 + 0.0105 = 1.0105$$ Calculate the exponent: $$4 \times 10 = 40$$ So: $$A_{you} = 13000 \times (1.0105)^{40}$$ Calculate the power: $$(1.0105)^{40} \approx 1.491825$$ Multiply: $$A_{you} = 13000 \times 1.491825 = 19393.73$$ 5. **Find the difference:** $$\text{Difference} = A_{you} - A_{friend} = 19393.73 - 19393.73 = 0.00$$ 6. **Conclusion:** After 10 years, you will have approximately the same amount as your friend, so the difference is $0.00$. **Note:** The amounts are very close because the slightly higher interest rate compounded quarterly nearly matches the lower rate compounded daily.