Loan Balance D11C14

1. **State the problem:** Zandile makes monthly payments of $1800$ for a loan with an annual interest rate of $11.918\%$ compounded monthly, over $8$ years. We want to find the remaining loan balance immediately after the $4^{th}$ payment of the $4^{th}$ year. 2. **Identify key values:** - Monthly payment, $P = 1800$ - Annual interest rate, $r_{annual} = 11.918\% = 0.11918$ - Monthly interest rate, $r = \frac{0.11918}{12} = 0.0099315$ - Total number of payments, $N = 8 \times 12 = 96$ - Number of payments made until $4^{th}$ payment of $4^{th}$ year: $n = (4-1) \times 12 + 4 = 40$ 3. **Formula for loan balance after $n$ payments:** $$ B_n = P \times \frac{1 - (1 + r)^{-(N - n)}}{r} $$ This formula calculates the present value of the remaining payments after $n$ payments have been made. 4. **Calculate the remaining balance:** Calculate $N - n = 96 - 40 = 56$ Calculate $(1 + r)^{-(N - n)} = (1 + 0.0099315)^{-56}$ First, calculate $1 + r = 1.0099315$ Calculate $1.0099315^{56} \approx e^{56 \times \ln(1.0099315)}$ Calculate $\ln(1.0099315) \approx 0.009883$ So, $56 \times 0.009883 = 0.5535$ Then, $1.0099315^{56} \approx e^{0.5535} \approx 1.739$ Therefore, $(1 + r)^{-56} = \frac{1}{1.739} \approx 0.5751$ Now substitute into the formula: $$ B_n = 1800 \times \frac{1 - 0.5751}{0.0099315} = 1800 \times \frac{0.4249}{0.0099315} $$ Calculate the fraction: $$ \frac{0.4249}{0.0099315} \approx 42.78 $$ Multiply by $1800$: $$ B_n = 1800 \times 42.78 = 77004 $$ This seems too large, so let's check the calculation carefully. 5. **Recalculate carefully:** Calculate $\frac{0.4249}{0.0099315}$: $$ \frac{0.4249}{0.0099315} = 42.78 $$ Multiply by $1800$: $$ 42.78 \times 1800 = 77004 $$ This is the present value of remaining payments in months, so the amount is $77,004$, which is too high for a loan with $1800$ monthly payments. This suggests a misinterpretation: the formula used is for the present value of an annuity, but the loan amount is not given, so we cannot find the remaining balance directly without the original loan amount. 6. **Find the original loan amount first:** The original loan amount $L$ is the present value of all $96$ payments: $$ L = P \times \frac{1 - (1 + r)^{-N}}{r} $$ Calculate $(1 + r)^{-N} = (1.0099315)^{-96}$ Calculate $\ln(1.0099315) = 0.009883$ Calculate $96 \times 0.009883 = 0.9488$ Calculate $1.0099315^{96} = e^{0.9488} = 2.582$ So, $$ (1 + r)^{-96} = \frac{1}{2.582} = 0.3871 $$ Calculate numerator: $$ 1 - 0.3871 = 0.6129 $$ Calculate fraction: $$ \frac{0.6129}{0.0099315} = 61.72 $$ Calculate original loan amount: $$ L = 1800 \times 61.72 = 111096 $$ 7. **Calculate remaining balance after $n=40$ payments:** Remaining balance formula: $$ B_n = L \times (1 + r)^n - P \times \frac{(1 + r)^n - 1}{r} $$ Calculate $(1 + r)^n = (1.0099315)^{40}$ Calculate $40 \times 0.009883 = 0.3953$ Calculate $e^{0.3953} = 1.485$ Calculate $B_n$: $$ B_n = 111096 \times 1.485 - 1800 \times \frac{1.485 - 1}{0.0099315} $$ Calculate first term: $$ 111096 \times 1.485 = 164,900.76 $$ Calculate numerator of second term: $$ 1.485 - 1 = 0.485 $$ Calculate fraction: $$ \frac{0.485}{0.0099315} = 48.85 $$ Calculate second term: $$ 1800 \times 48.85 = 87,930 $$ Calculate remaining balance: $$ B_n = 164,900.76 - 87,930 = 76,970.76 $$ This is still very high, indicating a misunderstanding. 8. **Use amortization formula for remaining balance:** The remaining balance after $n$ payments is also given by: $$ B_n = P \times \frac{1 - (1 + r)^{-(N - n)}}{r} $$ Calculate $N - n = 96 - 40 = 56$ Calculate $(1 + r)^{-56} = 0.5751$ (from step 4) Calculate numerator: $$ 1 - 0.5751 = 0.4249 $$ Calculate fraction: $$ \frac{0.4249}{0.0099315} = 42.78 $$ Calculate remaining balance: $$ B_n = 1800 \times 42.78 = 77,004 $$ This matches previous result. 9. **Interpretation:** The remaining balance after $40$ payments is approximately $77,004$. This is the amount Zandile still owes immediately after the $4^{th}$ payment of the $4^{th}$ year. 10. **Final answer:** $$ \boxed{77004.00} $$