1. **Problem statement:** Alana made a minimum payment of 35 on June 01. After interest was applied, her new balance was 1845.77. We need to find the nominal interest rate compounded daily. 2. **Formula for daily compounding interest:** $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after interest, - $P$ is the principal (balance after payment), - $r$ is the nominal annual interest rate (decimal), - $n$ is the number of compounding periods per year (daily means $n=365$), - $t$ is the time in years. 3. **Calculate the principal after payment:** Let the balance before payment be $B$. After paying 35, the balance is $B - 35$. After one day of interest, the balance is 1845.77. 4. **Set up the equation for one day ($t=\frac{1}{365}$):** $$1845.77 = (B - 35)\left(1 + \frac{r}{365}\right)^{1}$$ 5. **We need $B$ to find $r$. Since $B$ is not given, assume $B = 1845.77$ after interest, so before interest it was $1845.77 / \left(1 + \frac{r}{365}\right)$. But we only have one equation and two unknowns. The problem implies the payment was made before interest, so the balance before interest is $B - 35$. 6. **Rearranged:** $$\frac{1845.77}{B - 35} = 1 + \frac{r}{365}$$ 7. **Assuming the balance before payment is $B = 1845.77 + 35 = 1880.77$ (approximate), then:** $$\frac{1845.77}{1880.77 - 35} = \frac{1845.77}{1845.77} = 1$$ This is trivial, so we need to consider the interest applied on the balance after payment, meaning the balance before interest is $1845.77 / (1 + r/365)$. 8. **Instead, solve for $r$ using:** $$1845.77 = (1845.77 - 35)\left(1 + \frac{r}{365}\right)$$ $$1845.77 = 1810.77\left(1 + \frac{r}{365}\right)$$ 9. **Divide both sides:** $$\frac{1845.77}{1810.77} = 1 + \frac{r}{365}$$ $$1.0193 = 1 + \frac{r}{365}$$ 10. **Subtract 1:** $$0.0193 = \frac{r}{365}$$ 11. **Multiply both sides by 365:** $$r = 0.0193 \times 365 = 7.04$$ 12. **Convert to percentage:** $$r = 7.04\%$$ nominal annual interest rate compounded daily. --- **B) Equivalent annually-compounding rate:** 1. Use the formula for equivalent rates: $$\left(1 + \frac{r_{daily}}{365}\right)^{365} = 1 + r_{annual}$$ 2. Substitute $r_{daily} = 0.0704$: $$\left(1 + \frac{0.0704}{365}\right)^{365} = 1 + r_{annual}$$ 3. Calculate: $$\left(1 + 0.0001929\right)^{365} \approx e^{0.0704} = 1.0729$$ 4. So, $$r_{annual} = 1.0729 - 1 = 0.0729 = 7.29\%$$ --- **C) Balance on August 01 after payments:** 1. Timeline: - June 01: payment 35, balance 1845.77 after interest - July 01: payment 35 - August 01: payment 35 2. Calculate balance on July 01 before payment: $$B_{July} = (1845.77 - 35)\left(1 + \frac{0.0704}{365}\right)^{30}$$ 3. Calculate: $$B_{July} = 1810.77 \times \left(1 + 0.0001929\right)^{30} = 1810.77 \times 1.0058 = 1821.3$$ 4. After July 01 payment: $$1821.3 - 35 = 1786.3$$ 5. Calculate balance on August 01 before payment: $$B_{Aug} = 1786.3 \times \left(1 + 0.0001929\right)^{31} = 1786.3 \times 1.0060 = 1796.9$$ 6. After August 01 payment: $$1796.9 - 35 = 1761.9$$ --- **D) Three equal payments on Sept 01, Oct 01, Nov 01 to settle debt:** 1. Let each payment be $x$. 2. Balance on Sept 01 before payment: $$B_{Sept} = 1761.9 \times \left(1 + 0.0001929\right)^{31} = 1761.9 \times 1.0060 = 1772.5$$ 3. After Sept 01 payment: $$1772.5 - x$$ 4. Balance on Oct 01 before payment: $$\left(1772.5 - x\right) \times 1.0060$$ 5. After Oct 01 payment: $$\left(1772.5 - x\right) \times 1.0060 - x$$ 6. Balance on Nov 01 before payment: $$\left[\left(1772.5 - x\right) \times 1.0060 - x\right] \times 1.0060$$ 7. After Nov 01 payment, balance is zero: $$\left[\left(1772.5 - x\right) \times 1.0060 - x\right] \times 1.0060 - x = 0$$ 8. Simplify: $$\left(1772.5 - x\right) \times 1.0060^2 - x \times 1.0060 - x = 0$$ 9. Calculate $1.0060^2 = 1.0120$: $$1772.5 \times 1.0120 - x \times 1.0120 - x \times 1.0060 - x = 0$$ 10. Combine $x$ terms: $$1772.5 \times 1.0120 - x(1.0120 + 1.0060 + 1) = 0$$ 11. Calculate: $$1793.6 - x(3.018) = 0$$ 12. Solve for $x$: $$x = \frac{1793.6}{3.018} = 594.3$$ --- **E) Total interest paid since May 01:** 1. Initial balance on May 01 is unknown, but assume it was $B_0$. 2. Total payments made: - June 01: 35 - July 01: 35 - August 01: 35 - Sept 01, Oct 01, Nov 01: 3 payments of 594.3 each 3. Total payments: $$35 + 35 + 35 + 3 \times 594.3 = 105 + 1782.9 = 1887.9$$ 4. Final balance after Nov 01 is 0. 5. Interest paid = total payments - initial balance. 6. Initial balance before June 01 payment was approximately: $$1845.77 + 35 = 1880.77$$ 7. Interest paid: $$1887.9 - 1880.77 = 7.13$$ --- **Final answers:** - Nominal daily compounded rate: **7.04%** - Equivalent annual compounded rate: **7.29%** - Balance on August 01 after payment: **1761.9** - Three equal payments to settle debt: **594.3** each - Total interest paid since May 01: **7.13**