1. **Problem 1a:** Lulu borrows 240,000 with 12% interest compounded monthly and pays 5,300 monthly. Find the number of payments. Step 1: Identify variables: - Principal $P=240,000$ - Monthly interest rate $i=\frac{12\%}{12}=1\%=0.01$ - Monthly payment $R=5,300$ Step 2: Use the annuity formula for present value: $$P=R \times \frac{1-(1+i)^{-n}}{i}$$ Step 3: Substitute values: $$240,000=5,300 \times \frac{1-(1.01)^{-n}}{0.01}$$ Step 4: Simplify: $$\frac{240,000 \times 0.01}{5,300}=1-(1.01)^{-n}$$ $$\frac{2,400}{5,300}=1-(1.01)^{-n}$$ $$0.45283=1-(1.01)^{-n}$$ Step 5: Solve for $(1.01)^{-n}$: $$(1.01)^{-n}=1-0.45283=0.54717$$ Step 6: Take natural logarithm: $$-n \ln(1.01)=\ln(0.54717)$$ Step 7: Calculate: $$-n \times 0.00995 = -0.6025$$ Step 8: Solve for $n$: $$n=\frac{0.6025}{0.00995} \approx 60.56$$ Step 9: Number of payments is approximately 61 months. 2. **Problem 1b:** Find the final payment if made one month after the last 5,300 payment. Step 1: Calculate balance after 60 payments: $$B=R \times \frac{1-(1+i)^{-(n-1)}}{i}$$ $$B=5,300 \times \frac{1-(1.01)^{-60}}{0.01}$$ Step 2: Calculate $(1.01)^{-60} \approx 0.54703$ Step 3: Calculate balance: $$B=5,300 \times \frac{1-0.54703}{0.01}=5,300 \times 45.297=240,059.1$$ Step 4: Add one month interest: $$Final = B \times (1+i) = 240,059.1 \times 1.01 = 242,459.7$$ Step 5: Subtract last payment: $$Final\ payment = 242,459.7 - 5,300 = 237,159.7$$ 3. **Problem 2:** Present value 270,000, semi-annual payment 15,600, 8% compounded semi-annually. Find number of payments. Step 1: Variables: - $P=270,000$ - $R=15,600$ - $i=\frac{8\%}{2}=4\%=0.04$ Step 2: Use annuity formula: $$270,000=15,600 \times \frac{1-(1.04)^{-n}}{0.04}$$ Step 3: Simplify: $$\frac{270,000 \times 0.04}{15,600}=1-(1.04)^{-n}$$ $$0.6923=1-(1.04)^{-n}$$ Step 4: Solve for $(1.04)^{-n}$: $$(1.04)^{-n}=1-0.6923=0.3077$$ Step 5: Take natural logarithm: $$-n \ln(1.04)=\ln(0.3077)$$ Step 6: Calculate: $$-n \times 0.03922 = -1.1787$$ Step 7: Solve for $n$: $$n=\frac{1.1787}{0.03922} \approx 30.05$$ Step 8: Number of payments is approximately 30. 4. **Problem 3a:** Margarita borrows 500,000, pays 18,500 quarterly, interest 10.5% compounded quarterly. Find number of full payments. Step 1: Variables: - $P=500,000$ - $R=18,500$ - $i=\frac{10.5\%}{4}=2.625\%=0.02625$ Step 2: Use annuity formula: $$500,000=18,500 \times \frac{1-(1.02625)^{-n}}{0.02625}$$ Step 3: Simplify: $$\frac{500,000 \times 0.02625}{18,500}=1-(1.02625)^{-n}$$ $$0.7095=1-(1.02625)^{-n}$$ Step 4: Solve for $(1.02625)^{-n}$: $$(1.02625)^{-n}=1-0.7095=0.2905$$ Step 5: Take natural logarithm: $$-n \ln(1.02625)=\ln(0.2905)$$ Step 6: Calculate: $$-n \times 0.0259 = -1.236$$ Step 7: Solve for $n$: $$n=\frac{1.236}{0.0259} \approx 47.7$$ Step 8: Number of full payments is 48. 5. **Problem 3b:** Find final payment if made on day of last 18,500 payment. Step 1: Calculate balance after 47 payments: $$B=18,500 \times \frac{1-(1.02625)^{-47}}{0.02625}$$ Step 2: Calculate $(1.02625)^{-47} \approx 0.2905$ Step 3: Calculate balance: $$B=18,500 \times \frac{1-0.2905}{0.02625}=18,500 \times 26.97=499,945$$ Step 4: Final payment: $$Final = P - B = 500,000 - 499,945 = 55$$ **Final answers:** 1a) 61 payments 1b) Final payment about 237,160 2) 30 payments 3a) 48 payments 3b) Final payment about 55