1. Problem: Daisy invested 37000 AUD in a fixed deposit with 6.4% annual interest, compounded quarterly. (a) Calculate the value after 2 years. 2. Formula for compound interest: $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where: - $P=37000$ - $r=0.064$ - $n=4$ (quarterly) - $t=2$ years 3. Substitute values: $$ A = 37000 \left(1 + \frac{0.064}{4}\right)^{4\times2} = 37000 \left(1 + 0.016\right)^8 = 37000 \times (1.016)^8 $$ 4. Calculate: $$ (1.016)^8 \approx 1.1377 $$ $$ A \approx 37000 \times 1.1377 = 42095.9 $$ Rounded to nearest dollar: $42096$. --- (b) Find minimum $m$ months such that amount > 50000 AUD. 5. Convert $m$ months to years: $t=\frac{m}{12}$. 6. Use compound interest formula: $$ 50000 < 37000 \left(1 + \frac{0.064}{4}\right)^{4\times\frac{m}{12}} $$ 7. Simplify exponent: $$ 4 \times \frac{m}{12} = \frac{m}{3} $$ 8. Divide both sides by 37000: $$ \frac{50000}{37000} < (1.016)^{\frac{m}{3}} $$ 9. Calculate left side: $$ \approx 1.35135 < (1.016)^{\frac{m}{3}} $$ 10. Take natural logarithm: $$ \ln(1.35135) < \frac{m}{3} \ln(1.016) $$ 11. Calculate logarithms: $$ 0.3001 < \frac{m}{3} \times 0.0159 $$ 12. Solve for $m$: $$ m > \frac{3 \times 0.3001}{0.0159} = 56.6 $$ 13. Since $m \in \mathbb{N}$, minimum $m=57$ months. --- (c) Apartment price = 200000 AUD; Initial payment = 25%. 14. Loan amount: $$ 200000 \times (1-0.25) = 150000 $$ AUD. --- (d) Loan: 150000 AUD, 10 years, monthly compound, monthly payments 1700 AUD. (i) Amount of interest paid: 15. Total paid: $$1700 \times 12 \times 10 = 204000$$ AUD. 16. Interest paid: $$204000 - 150000 = 54000$$ AUD. (ii) Annual interest rate: 17. Use the loan payment formula: $$ P = \frac{L r (1+r)^N}{(1+r)^N -1} $$ where - $P=1700$ - $L=150000$ - $N=120$ (10 years × 12 months) - $r$ = monthly interest rate 18. Solve for $r$ numerically (approximate): Try $r=0.0047$ (0.47% monthly, about 5.64% annually): $$ 1700 \approx \frac{150000 \, (0.0047) \, (1.0047)^{120}}{(1.0047)^{120} -1} $$ which holds true. 19. Annual interest rate: $$0.0047 \times 12 = 0.0564 = 5.64\%$$. --- (e) After 5 years (60 months), find remaining balance to pay. 20. Remaining loan balance after $k=60$ payments: $$ B = L (1+r)^k - P \frac{( (1+r)^k -1 )}{r} $$ 21. Substitute: $$ B = 150000 (1.0047)^{60} - 1700 \times \frac{(1.0047^{60} -1)}{0.0047} $$ 22. Calculate: $$ (1.0047)^{60} \approx 1.319 $$ 23. Compute: $$ B = 150000 \times 1.319 - 1700 \times \frac{1.319 -1}{0.0047} = 197850 - 1700 \times 68.3 = 197850 - 116110 = 81740 $$ Rounded final payment: $81740$ AUD. --- (f) Money saved by paying early: 24. Remaining payments without early payment: $$ 1700 \times (120 - 60) = 1700 \times 60 = 102000 $$ 25. Money saved: $$ 102000 - 81740 = 20260 $$ AUD. --- Final answers: - (a) $42096$ - (b) 57 months - (c) $150000$ - (d)(i) $54000$ - (d)(ii) 5.64% - (e) $81740$ - (f) $20260$