Loan Repayment with Changing Interest Rates
A student borrows $12,000 at an annual interest rate of 6% compounded monthly. The loan is to be repaid with equal monthly payments over 5 years. After 24 payments, the interest rate drops to 4.5% compounded monthly. The student continues with the same monthly payment amount. Determine the remaining balance after the rate change and the total interest paid over the life of the loan. Assume payments are made at the end of each month.
1. Calculate the original monthly payment.
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The monthly interest rate is ( i = \frac{0.06}{12} = 0.005 ). The number of payments is ( n = 5 \times 12 = 60 ). The present value of an annuity formula gives ( P = R \frac{1 - (1+i)^{-n}}{i} ), where ( P = 12000 ). Solving for ( R ): ( R = \frac{P \cdot i}{1 - (1+i)^{-n}} = \frac{12000 \times 0.005}{1 - (1.005)^{-60}} ). Compute ( (1.005)^{-60} \approx 0.74137 ), so denominator ( \approx 0.25863 ). Thus ( R \approx \frac{60}{0.25863} \approx 231.99 ). The monthly payment is approximately $232.00.
2. Find the outstanding balance after 24 payments, just before the interest rate changes.
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The outstanding balance after 24 payments is the present value of the remaining 36 payments at the original interest rate. ( B_{24} = R \frac{1 - (1+i)^{-36}}{i} ) with ( R = 231.99 ), ( i = 0.005 ). Compute ( (1.005)^{-36} \approx 0.83564 ), so ( 1 - 0.83564 = 0.16436 ). Then ( B_{24} \approx 231.99 \times \frac{0.16436}{0.005} \approx 231.99 \times 32.872 \approx 7625.00 ). Alternatively, use the amortization formula: ( B_{24} = P(1+i)^{24} - R \frac{(1+i)^{24} - 1}{i} ). Both give approximately $7,625.
3. After the rate change to 4.5% compounded monthly, with the same monthly payment, how many additional payments are needed to pay off the loan? What is the total interest paid?
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New monthly rate ( i' = \frac{0.045}{12} = 0.00375 ). The balance ( B_{24} = 7625.00 ). The payment remains ( R = 231.99 ). The number of additional payments ( m ) satisfies ( B_{24} = R \frac{1 - (1+i')^{-m}}{i'} ). Solve for ( m ): ( (1+i')^{-m} = 1 - \frac{B_{24} \cdot i'}{R} = 1 - \frac{7625 \times 0.00375}{231.99} \approx 1 - 0.12325 = 0.87675 ). Then ( -m \ln(1.00375) = \ln(0.87675) ), so ( m = \frac{-\ln(0.87675)}{\ln(1.00375)} \approx \frac{0.1315}{0.003743} \approx 35.13 ). So 36 additional payments are needed (the last one smaller). Total payments = 24 + 36 = 60. Total paid = 24×231.99 + 35×231.99 + final payment. Compute final payment: after 35 additional payments, balance = ( B_{24}(1+i')^{35} - R \frac{(1+i')^{35} - 1}{i'} ). Calculate ( (1.00375)^{35} \approx 1.1393 ). Balance before final payment ≈ 7625×1.1393 - 231.99×(1.1393-1)/0.00375 ≈ 8687.16 - 231.99×37.1467 ≈ 8687.16 - 8618.00 ≈ 69.16. So final payment ≈ $69.16. Total interest = total paid - principal = (24×231.99 + 35×231.99 + 69.16) - 12000 = (59×231.99 + 69.16) - 12000 = (13687.41 + 69.16) - 12000 = 13756.57 - 12000 = $1,756.57.