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Loan Payments & Amortization - Finite Mathematics

  1. A home buyer is choosing between two 30-year fixed-rate mortgages for $400,000. Option A has a 6.0% annual rate with no points. Option B has a 5.7% annual rate but requires paying 2 points at closing (where 1 point is 1% of the loan amount). Considering the total of all monthly payments plus the upfront cost of points, how much more will the borrower pay in total with Option A compared to Option B over the full 30-year term?
    Answer: $18,975 For Option A (i=0.005,n=360i=0.005, n=360): Monthly payment RA=$400,0000.0051−(1.005)−360≈$2,398.20R_A = \$400,000 \frac{0.005}{1 - (1.005)^{-360}} \approx \$2,398.20. Total cost is 360×$2,398.20=$863,352360 \times \$2,398.20 = \$863,352. For Option B (i=0.00475,n=360i=0.00475, n=360): Monthly payment RB=$400,0000.004751−(1.00475)−360≈$2,323.27R_B = \$400,000 \frac{0.00475}{1 - (1.00475)^{-360}} \approx \$2,323.27. Total of payments is 360×$2,323.27=$836,377.20360 \times \$2,323.27 = \$836,377.20. The cost of points is 0.02×$400,000=$8,0000.02 \times \$400,000 = \$8,000. Total cost for Option B is $836,377.20+$8,000=$844,377.20\$836,377.20 + \$8,000 = \$844,377.20. The difference is $863,352−$844,377.20=$18,974.80\$863,352 - \$844,377.20 = \$18,974.80.
  2. An entrepreneur plans to borrow money to start a business and can afford to make payments of $2,000 per month for 10 years. Bank A offers a loan at a 9% annual interest rate. Bank B offers a loan at an 8.4% annual interest rate. How much more can the entrepreneur borrow from Bank B than from Bank A, assuming monthly compounding for both loans?
    Answer: $5,992 This problem requires calculating the present value (PP) for each loan scenario and finding the difference. The number of payments is n=10×12=120n = 10 \times 12 = 120. For Bank A: R=$2000R=\$2000, iA=0.09/12=0.0075i_A = 0.09/12 = 0.0075. PA=$20001−(1.0075)−1200.0075≈$157,983.33P_A = \$2000 \frac{1 - (1.0075)^{-120}}{0.0075} \approx \$157,983.33. For Bank B: R=$2000R=\$2000, iB=0.084/12=0.007i_B = 0.084/12 = 0.007. PB=$20001−(1.007)−1200.007≈$163,975.31P_B = \$2000 \frac{1 - (1.007)^{-120}}{0.007} \approx \$163,975.31. The difference is PB−PA=$163,975.31−$157,983.33=$5,991.98P_B - P_A = \$163,975.31 - \$157,983.33 = \$5,991.98. Distractor A uses an annual rate of 8% for Bank B. Distractor B is the difference in total payments if rates were simple interest on a fixed principal. Distractor C is the difference in total interest paid based on a hypothetical common principal.
  3. A loan of $10,000 is being repaid with 48 monthly payments at an annual interest rate of 12% compounded monthly. The calculated monthly payment is rounded to the nearest cent for billing purposes. What is the amount of the 48th and final payment?
    Answer: $263.24 First, find the unrounded monthly payment. With P=10000P=10000, i=0.12/12=0.01i=0.12/12=0.01, and n=48n=48, R=100000.011−(1.01)−48≈$263.338R = 10000 \frac{0.01}{1 - (1.01)^{-48}} \approx \$263.338. This is rounded up to a regular payment of Rpmt=$263.34R_{pmt} = \$263.34. Because the payment is rounded up, the final payment will be slightly less. The easiest way to find the adjustment is to compare the future value of the loan to the future value of the payments. FVloan=P(1+i)n=10000(1.01)48≈$16,122.26FV_{loan} = P(1+i)^n = 10000(1.01)^{48} \approx \$16,122.26. FVpmts=Rpmt(1+i)n−1i=263.34(1.01)48−10.01≈$16,122.36FV_{pmts} = R_{pmt} \frac{(1+i)^n - 1}{i} = 263.34 \frac{(1.01)^{48} - 1}{0.01} \approx \$16,122.36. The future value of the payments is $$0.10 higher, meaning the borrower has overpaid by $0.10 after 48 payments. Therefore, the final payment should be adjusted down by this amount: $263.34−$0.10=$263.24\$263.34 - \$0.10 = \$263.24.
  4. A small business loan of $75,000 is structured as a balloon mortgage. It is amortized over 30 years with monthly payments at a 7.5% annual interest rate, but the entire remaining balance is due in a single balloon payment at the end of the 10th year. What is the amount of this balloon payment?
    Answer: $66,134.18 The balloon payment is the outstanding balance of the loan at the end of the 10th year. First, calculate the monthly payment as if it were a 30-year loan. P=$75,000P=\$75,000, n=360n=360, i=0.075/12=0.00625i=0.075/12=0.00625. R=$75,0000.006251−(1.00625)−360≈$524.41R = \$75,000 \frac{0.00625}{1 - (1.00625)^{-360}} \approx \$524.41. Next, find the balance after 10 years (120 payments). The number of remaining payments in the original amortization schedule is 360−120=240360 - 120 = 240. The balloon payment is the present value of these remaining 240 payments. B120=$524.411−(1.00625)−2400.00625≈$66,134.18B_{120} = \$524.41 \frac{1 - (1.00625)^{-240}}{0.00625} \approx \$66,134.18. Distractor A is the monthly payment. Distractor B is a linear reduction (75000×20/3075000 \times 20/30). Distractor C miscalculates the remaining term as 15 years instead of 20.
  5. A student can afford a monthly payment of $450 for a car loan. If the loan terms are 6 years at a 7.2% annual interest rate compounded monthly, what is the maximum loan amount the student can afford, rounded to the nearest dollar?
    Answer: $26,253 This requires calculating the present value (PP) of an ordinary annuity. The monthly payment is R=$450R=\$450, the monthly interest rate is i=0.072/12=0.006i = 0.072/12 = 0.006, and the number of payments is n=6×12=72n = 6 \times 12 = 72. The formula for present value is P=R1−(1+i)−niP = R \frac{1 - (1+i)^{-n}}{i}. P=$4501−(1.006)−720.006≈$450×58.340=$26,253P = \$450 \frac{1 - (1.006)^{-72}}{0.006} \approx \$450 \times 58.340 = \$26,253. Distractor A uses the future value formula. Distractor B is the sum of all payments (450×72450 \times 72). Distractor D calculates the principal assuming it was a simple interest loan.
  6. A couple secures a 30-year mortgage for $350,000 with a fixed annual interest rate of 6%, compounded monthly. Assuming they make only the required monthly payments for the entire term, what is the total amount of interest they will pay over the life of the loan?
    Answer: $405,435 First, calculate the monthly payment, RR. The principal is P=$350,000P = \$350,000, the monthly interest rate is i=0.06/12=0.005i = 0.06 / 12 = 0.005, and the number of payments is n=30×12=360n = 30 \times 12 = 360. The payment formula is R=Pi1−(1+i)−nR = P \frac{i}{1 - (1+i)^{-n}}. R=$350,0000.0051−(1.005)−360≈$2,098.43R = \$350,000 \frac{0.005}{1 - (1.005)^{-360}} \approx \$2,098.43. Next, calculate the total amount paid over the 30 years: R×n=$2,098.43×360=$755,434.80R \times n = \$2,098.43 \times 360 = \$755,434.80. Finally, subtract the principal to find the total interest paid: $755,434.80−$350,000=$405,434.80\$755,434.80 - \$350,000 = \$405,434.80.
  7. A company borrows $50,000 for equipment, to be repaid over 5 years with monthly payments at a 9% annual interest rate compounded monthly. For the first payment made, which of the following correctly states the amounts applied to interest and principal?
    Answer: $375.00 for interest and $662.92 for principal. First, calculate the interest portion of the first payment. The balance is the full principal, P=$50,000P = \$50,000. The monthly rate is i=0.09/12=0.0075i = 0.09/12 = 0.0075. Interest for month 1 = P×i=$50,000×0.0075=$375.00P \times i = \$50,000 \times 0.0075 = \$375.00. Next, calculate the full monthly payment, RR, with n=60n=60: R=$50,0000.00751−(1.0075)−60≈$1,037.92R = \$50,000 \frac{0.0075}{1 - (1.0075)^{-60}} \approx \$1,037.92. The principal portion of the first payment is the remainder: R−Interest1=$1,037.92−$375.00=$662.92R - \text{Interest}_1 = \$1,037.92 - \$375.00 = \$662.92. Distractor B represents the split for the final payment. Distractor C is an average split over the loan term. Distractor D reverses the correct interest and principal amounts.
  8. Five years ago, a family took out a 30-year mortgage for $400,000 at a 7.2% annual rate. Today, they refinance the remaining balance for a new 20-year term at a 5.4% annual rate. What will their new monthly payment be? (All rates are compounded monthly).
    Answer: $2,347.85 This is a two-step problem. First, find the outstanding balance on the original loan. Original loan: P=$400,000P = \$400,000, n=360n=360, i=0.072/12=0.006i=0.072/12=0.006. The monthly payment was R=$2,714.49R = \$2,714.49. After 5 years (60 payments), 360−60=300360-60=300 payments remain. The balance is the present value of these 300 payments: B60=$2,714.491−(1.006)−3000.006≈$376,966.50B_{60} = \$2,714.49 \frac{1 - (1.006)^{-300}}{0.006} \approx \$376,966.50. Second, calculate the new payment for the refinanced loan. This balance becomes the new principal. New loan: Pnew=$376,966.50P_{new} = \$376,966.50, nnew=20×12=240n_{new}=20 \times 12=240, inew=0.054/12=0.0045i_{new}=0.054/12=0.0045. Rnew=$376,966.500.00451−(1.0045)−240≈$2,347.85R_{new} = \$376,966.50 \frac{0.0045}{1 - (1.0045)^{-240}} \approx \$2,347.85.
  9. A business takes a $120,000 loan for 10 years at an 8.4% annual rate compounded monthly. What is the outstanding balance on the loan immediately after the 60th payment is made?
    Answer: $72,454.49 First, calculate the monthly payment, RR. Here, P=$120,000P=\$120,000, i=0.084/12=0.007i = 0.084/12 = 0.007, and n=10×12=120n = 10 \times 12 = 120. R=$120,0000.0071−(1.007)−120≈$1,481.04R = \$120,000 \frac{0.007}{1 - (1.007)^{-120}} \approx \$1,481.04. The outstanding balance after 60 payments is the present value of the remaining 120−60=60120 - 60 = 60 payments. B60=R1−(1+i)−(n−k)i=$1,481.041−(1.007)−600.007≈$72,454.49B_{60} = R \frac{1 - (1+i)^{-(n-k)}}{i} = \$1,481.04 \frac{1 - (1.007)^{-60}}{0.007} \approx \$72,454.49. Distractor A assumes the principal is paid down linearly. Distractor B incorrectly subtracts the total of 60 payments from the original principal. Distractor C represents the principal paid off (P−B60P - B_{60}), not the remaining balance.
  10. A car is purchased for $45,000. The buyer makes a 20% down payment and finances the remaining amount with a 5-year loan at a 4.8% annual interest rate, compounded monthly. What is the total amount of interest paid over the life of the loan?
    Answer: $4,582 First, determine the loan principal. The down payment is 0.20×$45,000=$9,0000.20 \times \$45,000 = \$9,000. The principal PP is $45,000−$9,000=$36,000\$45,000 - \$9,000 = \$36,000. Next, calculate the monthly payment. Here, i=0.048/12=0.004i=0.048/12=0.004 and n=5×12=60n=5 \times 12=60. R=$36,0000.0041−(1.004)−60≈$676.37R = \$36,000 \frac{0.004}{1 - (1.004)^{-60}} \approx \$676.37. Total payments are 60×$676.37=$40,582.2060 \times \$676.37 = \$40,582.20. Total interest is total payments minus principal: $40,582.20−$36,000=$4,582.20\$40,582.20 - \$36,000 = \$4,582.20. Distractor A calculates the interest using the full $45,000 price as the principal. Distractor B is the monthly payment. Distractor D is the total amount paid.
  11. A person has a $250,000, 30-year mortgage with a 4.5% annual interest rate, compounded monthly. The monthly payment is $1266.71. On their 12th payment, they include an extra one-time principal payment of $5,000. Approximately how many months is the loan term reduced by as a result of this single extra payment?
    Answer: 13 months First, find the loan balance after 12 regular payments. There are 360−12=348360-12=348 payments remaining. The balance is the present value of these payments: B12=$1266.711−(1.00375)−3480.00375≈$246,312.73B_{12} = \$1266.71 \frac{1 - (1.00375)^{-348}}{0.00375} \approx \$246,312.73. After the extra $5,000 payment, the new principal is Pnew=$246,312.73−$5,000=$241,312.73P_{new} = \$246,312.73 - \$5,000 = \$241,312.73. Now, find how many payments nnewn_{new} of $1266.71 are needed to pay off this new principal: 241,312.73=1266.711−(1.00375)−nnew0.00375241,312.73 = 1266.71 \frac{1 - (1.00375)^{-n_{new}}}{0.00375}. Solving for nnewn_{new} gives nnew=−ln⁡(1−241312.73×0.003751266.71)ln⁡(1.00375)≈335n_{new} = -\frac{\ln(1 - \frac{241312.73 \times 0.00375}{1266.71})}{\ln(1.00375)} \approx 335 months. The original remaining term was 348 months. The new term is 335 months. The reduction is 348−335=13348 - 335 = 13 months.

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