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Amortization Schedules - Finite Mathematics

  1. A loan is amortized with constant monthly payments. Let PkP_k be the principal portion and IkI_k be the interest portion of the kk-th payment. Which statement best describes the monthly change in these values over the life of the loan?
    Answer: The amount by which PkP_k increases each month becomes progressively larger as the loan matures. This question tests the dynamics of amortization. The change in the principal portion from one payment to the next is given by the relationship Pk+1−Pk=Pk×iP_{k+1} - P_k = P_k \times i, where ii is the periodic interest rate. Since the principal portion PkP_k is always increasing over the life of the loan, the difference (Pk+1−Pk)(P_{k+1} - P_k) also increases. This means the principal portion grows by a larger and larger amount each month. Similarly, the change in the interest portion is Ik−Ik+1=Pk×iI_k - I_{k+1} = P_k \times i. As PkP_k increases, the amount by which the interest decreases each month also grows larger. A) and B) are incorrect because the change is not constant. The change depends on PkP_k, which is not constant. C) is incorrect because the amount of decrease in interest (Pk×iP_k \times i) gets larger, not smaller, as PkP_k increases. D) is correct. The increase in principal (Pk×iP_k \times i) gets larger as PkP_k increases.
  2. An amortization schedule shows that for payment 84, the principal portion is $623.45 and the interest portion is $876.55. If the monthly interest rate is 0.75%, what will be the principal portion of payment 85?
    Answer: $628.13 The monthly payment is $623.45 + $876.55 = $1,500.00. After payment 84, the remaining balance decreases by the principal portion: new balance = (old balance - $623.45). Since the interest portion of payment 84 was $876.55, the balance before payment 84 was $876.55 ÷ 0.0075 = $116,873.33. After payment 84, the balance is $116,873.33 - $623.45 = $116,249.88. For payment 85, the interest portion is $116,249.88 × 0.0075 = $871.87, so the principal portion is $1,500.00 - $871.87 = $628.13.
  3. A loan has equal monthly payments of $1,250. In month 36, the interest portion was $445.20. In month 37, the interest portion was $441.15. What is the monthly interest rate for this loan?
    Answer: 0.50% The principal portion of payment 36 was $1,250 - $445.20 = $804.80. This principal payment reduced the balance, causing the interest to decrease by $445.20 - $441.15 = $4.05 in payment 37. Since interest = balance × rate, and the balance decreased by $804.80, we have: $804.80 × rate = $4.05. Therefore, rate = $4.05 ÷ $804.80 = 0.005 = 0.50%.
  4. An amortization schedule shows that over payments 20-24, the total interest paid was $3,456 and the total principal paid was $1,544. If this represents exactly 5 payments of $1,000 each, what can be concluded about the trend in the interest portions?
    Answer: Interest portions are decreasing, but payment 20 had the largest interest portion In any standard amortization schedule, interest portions decrease over time as the principal balance decreases. Since we have 5 payments totaling $5,000 with $3,456 interest and $1,544 principal, the interest portions must be decreasing from payment 20 to 24. Payment 20 would have the largest interest portion and payment 24 the smallest. Option A misinterprets what the total amounts indicate about the trend. Option B is partially correct but identifies the wrong payment. Option C incorrectly assumes constant payments.
  5. In an amortization table, payment 48 has an interest portion of $567.89. The borrower makes an additional principal payment of $5,000 with payment 48. How will this affect the interest portion of payment 49, assuming the regular monthly payment amount stays the same?
    Answer: The interest portion will decrease by the monthly rate times $5,000 When you encounter amortization problems involving extra principal payments, remember that interest is always calculated on the outstanding loan balance at the beginning of each period. The key insight is understanding how additional principal payments affect future interest calculations. Here's what happens: When the borrower makes an extra $5,000 principal payment with payment 48, this reduces the outstanding loan balance by exactly $5,000 before payment 49 is calculated. Since interest for payment 49 is computed by multiplying the monthly interest rate by the new (reduced) balance, the interest portion will decrease by the monthly rate times $5,000. This makes answer C correct. Let's examine why the other options miss the mark. Answer A suggests the interest decreases by exactly $25.00, but this assumes a specific interest rate without any given information—there's no basis for this precise amount. Answer B proposes dividing $5,000 by remaining payments, but this confuses how interest calculation works; interest isn't distributed evenly across remaining payments. Answer D claims no change until payment 50, which ignores the fundamental principle that interest recalculates immediately when the principal balance changes. The mathematical relationship is: Interest reduction=Monthly rate×Extra principal payment\text{Interest reduction} = \text{Monthly rate} \times \text{Extra principal payment} Study tip: Remember that in amortization, interest always responds immediately to balance changes. When you see extra principal payments, look for answers that involve multiplying the additional payment by the interest rate—this direct relationship is a cornerstone of amortization mathematics.
  6. A homeowner has an adjustable-rate mortgage. For the first 60 payments, the rate was 4.2% APR, and after payment #60, the outstanding balance is $210,000. Just before payment #61 is due, the rate adjusts to 4.8% APR. The loan is then re-amortized over the remaining term, resulting in a new, higher monthly payment. What is the interest portion of payment #61?
    Answer: $840.00 The question asks for the interest portion of payment #61, which is calculated based on the outstanding balance after payment #60 and the newly adjusted interest rate. The information about re-amortization and the new payment amount is extra detail not needed to find the interest portion for this specific payment. 1. Identify the outstanding balance before payment #61: B60=$210,000B_{60} = \$210,000. 2. Identify the new annual interest rate: 4.8%. 3. Calculate the new monthly interest rate: inew=0.048/12=0.004i_{\text{new}} = 0.048 / 12 = 0.004. 4. Calculate the interest for the 61st period: I61=B60×inew=$210,000×0.004=$840.00I_{61} = B_{60} \times i_{\text{new}} = \$210,000 \times 0.004 = \$840.00. Distractors: A) This is the interest calculated using the old rate: $210,000×(0.042/12)=$210,000×0.0035=$735.00\$210,000 \times (0.042 / 12) = \$210,000 \times 0.0035 = \$735.00. This is a very common mistake. C) and D) These are plausible-looking dollar amounts that could result from miscalculations, for example, using the annual rate directly or confusing it with other rates.
  7. Let PkP_k be the principal portion of the kk-th payment for a loan being amortized with constant monthly payments at a periodic interest rate of ii. Which of the following expressions correctly relates Pk+1P_{k+1} to PkP_k?
    Answer: Pk+1=Pk(1+i)P_{k+1} = P_k(1+i) This question asks for the fundamental relationship between the principal portions of two consecutive payments. Let RR be the constant total payment, IkI_k be the interest portion of payment kk, and Bk−1B_{k-1} be the balance before payment kk. We know that Pk=R−IkP_k = R - I_k and Pk+1=R−Ik+1P_{k+1} = R - I_{k+1}. The interest portions are Ik=Bk−1iI_k = B_{k-1}i and Ik+1=BkiI_{k+1} = B_k i. The balance is reduced by the principal portion, so Bk=Bk−1−PkB_k = B_{k-1} - P_k. Substituting this into the expression for Ik+1I_{k+1}: Ik+1=(Bk−1−Pk)i=Bk−1i−Pki=Ik−PkiI_{k+1} = (B_{k-1} - P_k)i = B_{k-1}i - P_k i = I_k - P_k i. Now substitute this back into the expression for Pk+1P_{k+1}: Pk+1=R−(Ik−Pki)=(R−Ik)+PkiP_{k+1} = R - (I_k - P_k i) = (R - I_k) + P_k i. Since R−Ik=PkR - I_k = P_k, we have Pk+1=Pk+Pki=Pk(1+i)P_{k+1} = P_k + P_k i = P_k(1+i). Distractors represent common algebraic errors or misunderstandings: A) Confuses multiplication with addition. C) Incorrectly subtracts the interest effect. D) Incorrectly uses the total payment RR instead of the principal portion PkP_k to calculate the increase in principal.
  8. Two loans, Loan A and Loan B, are taken out for the same principal amount of $200,000. Loan A has a 15-year term at 6% APR. Loan B has a 30-year term at 5% APR. Both have monthly payments. Which of the following statements is true regarding the first payment for each loan?
    Answer: The interest portion is greater for Loan A, and the principal portion is also greater for Loan A. We need to analyze the first payment (k=1k=1) for each loan. The beginning balance for the first payment is the same for both loans: B0=$200,000B_0 = \$200,000. 1. Calculate the first month's interest for each loan: Loan A: monthly rate iA=0.06/12=0.005i_A = 0.06 / 12 = 0.005. Interest IA,1=$200,000×0.005=$1,000I_{A,1} = \$200,000 \times 0.005 = \$1,000. Loan B: monthly rate iB=0.05/12≈0.004167i_B = 0.05 / 12 \approx 0.004167. Interest IB,1=$200,000×(0.05/12)=$833.33I_{B,1} = \$200,000 \times (0.05/12) = \$833.33. So, the interest portion is greater for Loan A. 2. Calculate the total monthly payment for each loan using the amortization formula R=Pi(1+i)n(1+i)n−1R = P \frac{i(1+i)^n}{(1+i)^n - 1}: Loan A (n=180): RA=2000000.005(1.005)180(1.005)180−1≈$1,687.71R_A = 200000 \frac{0.005(1.005)^{180}}{(1.005)^{180} - 1} \approx \$1,687.71. Loan B (n=360): RB=200000(0.05/12)(1+0.05/12)360(1+0.05/12)360−1≈$1,073.64R_B = 200000 \frac{(0.05/12)(1+0.05/12)^{360}}{(1+0.05/12)^{360} - 1} \approx \$1,073.64. 3. Calculate the first month's principal for each loan: Loan A: PA,1=RA−IA,1=$1,687.71−$1,000=$687.71P_{A,1} = R_A - I_{A,1} = \$1,687.71 - \$1,000 = \$687.71. Loan B: PB,1=RB−IB,1=$1,073.64−$833.33=$240.31P_{B,1} = R_B - I_{B,1} = \$1,073.64 - \$833.33 = \$240.31. So, the principal portion is also greater for Loan A. Therefore, both the interest and principal portions of the first payment are greater for Loan A. Distractors are based on incorrect comparisons of these four values.
  9. Immediately after making payment #24, the borrower makes an additional lump-sum payment of $5,000 directly towards the principal. How much interest is saved on the very next scheduled payment (#25) as a result of this extra payment?
    Answer: $25.00 The interest saved on the next payment is the periodic interest rate applied to the amount of the extra principal payment. 1. First, determine the periodic interest rate, ii. We know the interest for payment #24 (I24=$300I_{24} = \$300) was calculated on the balance before that payment (B23B_{23}). 2. The balance before payment #24 is the ending balance after payment #24 plus the principal paid in payment #24: B23=B24+P24=$59,800+$200=$60,000B_{23} = B_{24} + P_{24} = \$59,800 + \$200 = \$60,000. 3. Now calculate the monthly interest rate: i=I24/B23=$300/$60,000=0.005i = I_{24} / B_{23} = \$300 / \$60,000 = 0.005. 4. The extra payment of X=$5,000X = \$5,000 reduces the principal upon which the next month's interest is calculated. The amount of interest saved is this extra payment multiplied by the monthly rate: Interest Saved =X×i=$5,000×0.005=$25.00= X \times i = \$5,000 \times 0.005 = \$25.00. Distractors: B) This is the principal portion of payment #24, an irrelevant value. C) This would be the result if the annual rate were 7.2% (i=0.006i=0.006), a common rate. 5000×0.006=$305000 \times 0.006 = \$30. D) This would be the result if the annual rate were 12% (i=0.01i=0.01). 5000×0.01=$505000 \times 0.01 = \$50. Students might guess a rate instead of calculating it.
  10. For a certain loan with constant monthly payments, the interest portion of payment #80 was $420.00 and the interest portion of payment #81 was $418.62. If the constant monthly payment is $650.00, what is the annual interest rate for the loan?
    Answer: 7.2% The key insight is that the decrease in interest paid from one period to the next is equal to the interest earned on the principal portion of the earlier payment. 1. First, calculate the principal portion of payment #80: P80=Total Payment−I80=$650.00−$420.00=$230.00P_{80} = \text{Total Payment} - I_{80} = \$650.00 - \$420.00 = \$230.00. 2. Next, find the decrease in interest from payment #80 to #81: ΔI=I80−I81=$420.00−$418.62=$1.38\Delta I = I_{80} - I_{81} = \$420.00 - \$418.62 = \$1.38. 3. This decrease, ΔI\Delta I, is the monthly interest on the principal P80P_{80} that was just paid off. So, ΔI=P80×i\Delta I = P_{80} \times i, where ii is the monthly interest rate. 4. Solve for ii: $1.38=$230.00×i  ⟹  i=$1.38/$230.00=0.006\$1.38 = \$230.00 \times i \implies i = \$1.38 / \$230.00 = 0.006. 5. Convert to annual rate: 0.006×12=0.072=7.2%0.006 \times 12 = 0.072 = 7.2\%.
  11. A borrower notices that in their amortization schedule, the principal portion increases by approximately $3.50 each month. If the current monthly payment is $975 and the current principal portion is $285, what will be the interest portion four payments later?
    Answer: $676.00 Currently: principal = $285, so interest = $975 - $285 = $690. If principal increases by $3.50 each payment, then four payments later: principal portion = $285 + 4($3.50) = $285 + $14 = $299. Therefore, interest portion = $975 - $299 = $676. Note that the $3.50 increase per payment is an approximation that works well for the middle portion of most amortization schedules where the rate of change is nearly linear.
  12. The outstanding balance on a 30-year (360-month) loan after 359 payments have been made is $1,125.40. The annual interest rate is 6.6% compounded monthly. What is the amount of the 360th and final payment needed to fully pay off the loan?
    Answer: $1,131.59 The final payment must cover the remaining outstanding balance (B359B_{359}) plus the interest that accrues on that balance during the final payment period. 1. Identify the outstanding balance: B359=$1,125.40B_{359} = \$1,125.40. 2. Calculate the monthly interest rate: i=0.066/12=0.0055i = 0.066 / 12 = 0.0055. 3. Calculate the interest for the final month: I360=B359×i=$1,125.40×0.0055=$6.1897≈$6.19I_{360} = B_{359} \times i = \$1,125.40 \times 0.0055 = \$6.1897 \approx \$6.19. 4. The final payment is the sum of the last balance and the interest on it: Final Payment =B359+I360=$1,125.40+$6.19=$1,131.59= B_{359} + I_{360} = \$1,125.40 + \$6.19 = \$1,131.59. Distractors: A) This is just the outstanding balance, neglecting the final month's interest. C) This might result from a calculation error, perhaps using an incorrect interest rate like 0.006. 1125.40×1.006=1132.151125.40 \times 1.006 = 1132.15. D) This might be the regular payment amount for the loan, but the final payment is typically slightly different to account for rounding over the life of the loan.

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