1. **Problem Statement:** Suzana has two options for her education fund of 26,442. Option 1: Invest 26,442 at 4.25% interest compounded monthly and withdraw equal monthly payments for 4 years. Option 2: Receive 560 every month for 4 years from her parents. 2. **Formula for Option 1 (Annuity Withdrawal):** The formula for the monthly payment $P$ from a principal $PV$ with monthly interest rate $i$ over $n$ months is: $$P = PV \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ 3. **Calculate parameters:** - Annual interest rate $r = 4.25\% = 0.0425$ - Monthly interest rate $i = \frac{0.0425}{12} = 0.0035417$ - Number of months $n = 4 \times 12 = 48$ - Principal $PV = 26,442$ 4. **Calculate monthly payment for Option 1:** $$P = 26,442 \times \frac{0.0035417(1+0.0035417)^{48}}{(1+0.0035417)^{48} - 1}$$ Calculate $(1+0.0035417)^{48} \approx 1.1855$ $$P = 26,442 \times \frac{0.0035417 \times 1.1855}{1.1855 - 1} = 26,442 \times \frac{0.004198}{0.1855} = 26,442 \times 0.02263 = 598.6$$ So, monthly payment from investment is approximately 598.6. 5. **Compare with Option 2:** Parents give 560 monthly. 6. **Conclusion:** Since 598.6 > 560, Option 1 yields more monthly money. Suzana should choose Option 1. --- 7. **Problem b:** Laptop cost 3,250 with 10% down payment and 3.5% interest compounded monthly over 36 months. 8. **Calculate down payment:** $$\text{Down payment} = 0.10 \times 3,250 = 325$$ 9. **Loan amount:** $$PV = 3,250 - 325 = 2,925$$ 10. **Monthly interest rate:** $$i = \frac{3.5\%}{12} = 0.0029167$$ 11. **Number of payments:** $$n = 36$$ 12. **Monthly payment formula:** $$P = PV \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ Calculate $(1+0.0029167)^{36} \approx 1.1097$ $$P = 2,925 \times \frac{0.0029167 \times 1.1097}{1.1097 - 1} = 2,925 \times \frac{0.003236}{0.1097} = 2,925 \times 0.02952 = 86.3$$ Monthly payment is approximately 86.3. 13. **Total payment:** $$86.3 \times 36 = 3,106.8$$ 14. **Interest charged:** $$3,106.8 - 2,925 = 181.8$$ --- 15. **Problem c i):** After 30 payments, find remaining balance. 16. **Remaining balance formula:** $$B = P \times \frac{1 - (1+i)^{-(n-k)}}{i}$$ where $k=30$ payments made. Calculate $(1+i)^{-(n-k)} = (1.0029167)^{-6} \approx 0.9827$ $$B = 86.3 \times \frac{1 - 0.9827}{0.0029167} = 86.3 \times \frac{0.0173}{0.0029167} = 86.3 \times 5.93 = 511.5$$ Additional payment to settle loan immediately is approximately 511.5. 17. **Problem c ii):** Total interest paid if settled after 30 payments plus additional payment. Total paid = (30 payments \times 86.3) + 511.5 = 2,589 + 511.5 = 3,100.5 Interest paid = 3,100.5 - 2,925 = 175.5 --- 18. **Problem d:** Missed first 4 payments, pay all arrears on 5th payment. 19. **Calculate arrears with interest:** Each missed payment accrues interest for months until 5th payment. - 1st missed payment: 4 months interest - 2nd missed payment: 3 months interest - 3rd missed payment: 2 months interest - 4th missed payment: 1 month interest Calculate future value of each missed payment $P=86.3$: $$FV = P \times (1+i)^m$$ Sum: $$86.3 \times (1.0029167^4 + 1.0029167^3 + 1.0029167^2 + 1.0029167^1)$$ Calculate powers: $$1.0117 + 1.0087 + 1.0058 + 1.0029 = 4.0291$$ Total arrears: $$86.3 \times 4.0291 = 347.7$$ 20. **Add 5th payment:** $$86.3 + 347.7 = 434.0$$ Suzana's father should pay approximately 434.0 on the 5th payment to settle all arrears. **Final answers:** - a) Choose Option 1 with monthly payment 598.6 - b) Monthly payment 86.3, interest charged 181.8 - c i) Additional payment 511.5 - c ii) Total interest paid 175.5 - d) Payment on 5th month to settle arrears 434.0