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Reducing-Balance Loans: Repayment Tables and Extra Repayments — Answers

Answer sheet with worked solutions for the Year 13 financial mathematics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 13 · Mathematics · Financial Mathematics

Reducing-Balance Loans: Repayment Tables and Extra Repayments — Answers

Part A: The repayment table

In a reducing-balance loan, interest is charged each period on the balance still owing, then the repayment is subtracted. A repayment table tracks each period: opening balance, interest = balance × r, principal repaid = repayment − interest, closing balance = opening balance + interest − repayment.

As a recurrence relation: Vₙ₊₁ = (1 + r)Vₙ − M, with V₀ = the amount borrowed.

Worked example. $5000 at 12% p.a. compounded monthly (r = 0.01), repaid at $300 per month. Month 1: interest = $50.00, principal = $250.00, balance = $4750.00. Month 2: interest = $47.50, principal = $252.50, balance = $4497.50.

1.A loan of $12 000 is charged 9% p.a. interest compounded monthly. What is the monthly interest rate?

Answer: 0.75% per month

  1. 9% ÷ 12 = 0.75% per month (r = 0.0075)
2.The loan of $12 000 is repaid at $500 per month. Complete the first row of the repayment table: interest for month 1, principal repaid and closing balance.

Answer: Interest $90.00; principal $410.00; balance $11 590.00

  1. Interest = 12 000 × 0.0075 = $90.00
  2. Principal repaid = 500 − 90 = $410.00
  3. Closing balance = 12 000 + 90 − 500 = $11 590.00
3.Continue the table for months 2 and 3 (round each interest amount to the nearest cent).

Answer: Month 2: interest $86.92, balance $11 176.92. Month 3: interest $83.83, balance $10 760.75

  1. Month 2: interest = 11 590 × 0.0075 = $86.925 ≈ $86.92; principal = 500 − 86.92 = $413.08; balance = 11 590 − 413.08 = $11 176.92
  2. Month 3: interest = 11 176.92 × 0.0075 = $83.83 (83.8269); principal = 500 − 83.83 = $416.17; balance = 11 176.92 − 416.17 = $10 760.75
4.Which recurrence relation models the balance Vₙ of this loan after n months?
  1. AVₙ₊₁ = 1.0075Vₙ − 500, V₀ = 12 000
  2. BVₙ₊₁ = 1.09Vₙ − 500, V₀ = 12 000
  3. CVₙ₊₁ = Vₙ − 500, V₀ = 12 000
  4. DVₙ₊₁ = 0.9925Vₙ − 500, V₀ = 12 000

Answer: Vₙ₊₁ = 1.0075Vₙ − 500, V₀ = 12 000

Each month the balance grows by the monthly rate 0.75% (multiply by 1.0075), then the $500 repayment is subtracted.

5.In a reducing-balance loan with equal repayments, the interest part of each repayment gets smaller over time while the principal part gets larger.

Answer: True

Interest is charged on the balance owing, which falls every month, so less of each equal repayment goes to interest and more goes to reducing the balance.

Part B: Repayments, total interest and the balance owing

The regular repayment for a loan of P over n periods at rate r per period is M = [frac Pr/1 − (1 + r)[sup −n]]. Round M to the cent, then total interest = M × n − P.

The balance owing after k repayments is Vₖ = P(1 + r)[sup k] − M × [frac (1 + r)[sup k] − 1/r]: what the loan would have grown to, minus what the repayments have grown to.

Worked example. $10 000 at 12% p.a. compounded monthly over 1 year: M = = $888.49, and total interest = 888.49 × 12 − 10 000 = $661.88.

6.A car loan of $25 000 is taken at 7.2% p.a. compounded monthly over 5 years. Find the monthly repayment.

Answer: $497.39

  1. r = 0.072 ÷ 12 = 0.006; n = 60
  2. M = = = $497.39
7.Find the total interest paid on the car loan.

Answer: $4843.40

  1. Total repaid = 497.39 × 60 = $29 843.40
  2. Total interest = 29 843.40 − 25 000 = $4843.40
8.Find the balance owing on the car loan after 2 years (24 repayments).

Answer: $16 061.23

  1. Vₖ = P(1 + r)k − M ×
  2. V₂₄ = 25 000 × 1.00624 − 497.39 ×
  3. = 28 859.68… − 12 798.45… = $16 061.23
  4. After 40% of the term, about 64% of the original amount is still owed, because early repayments are mostly interest.
9.If the car loan were taken over 10 years instead of 5 years, the monthly repayment would be halved.

Answer: False

Over 120 months, M = = $292.85, which is more than half of $497.39, because interest is charged for longer. The total interest also rises, to 292.85 × 120 − 25 000 = $10 142.

Part C: Number of repayments, extra repayments and rate changes

To find how many repayments are needed, set Vₙ = 0 and solve: (1 + r)n (M − Pr) = M, so n = [frac ln(M ÷ (M − Pr))/ln(1 + r)]. Round up: the last repayment is smaller and equals the remaining balance plus one period of interest.

Extra repayments and lump sums reduce the balance early, so every later period is charged less interest. A rate rise increases the repayment needed to finish on time.

Worked example. $10 000 at 12% p.a. compounded monthly, repaid at $500 per month: n = = 22.4…, so 23 repayments (22 full ones and a smaller final one).

10.$18 000 is borrowed at 6% p.a. compounded monthly and repaid at $400 per month. How many repayments are needed, and how much is the final (smaller) repayment?

Answer: 52 repayments; final repayment $42.41

  1. r = 0.005. Solve 0 = 18 000 × 1.005n − 400 ×
  2. 1.005n (400 − 90) = 400, so 1.005n = and n = = 51.1…
  3. So 51 full repayments and a 52nd smaller one: 52 repayments.
  4. Balance after 51 repayments = $42.20; with one more month of interest the final repayment is 42.20 × 1.005 = $42.41.
11.A home loan of $300 000 is taken at 6% p.a. compounded monthly over 25 years. Find the monthly repayment and the total interest.

Answer: $1932.90; total interest $279 870

  1. r = 0.005; n = 300
  2. M = = = $1932.90
  3. Total interest = 1932.90 × 300 − 300 000 = 579 870 − 300 000 = $279 870
12.The borrower in the previous question pays $2200 per month instead. How long does the loan take to repay, and about how much interest is saved?

Answer: 230 repayments (19 years and 2 months); about $74 751.87 saved

  1. 1.005n (2200 − 1500) = 2200, so n = = 229.6…
  2. 229 full repayments plus a smaller final repayment of $1318.13: 230 repayments, or 19 years and 2 months instead of 25 years.
  3. Total paid = 2200 × 229 + 1318.13 = $505 118.13, so interest = $205 118.13.
  4. Interest saved = 279 870 − 205 118.13 = $74 751.87
13.If the rate on the $300 000 loan over 25 years were 6.5% p.a. compounded monthly instead of 6%, how much more would the monthly repayment be?

Answer: $92.72 more

  1. M = = $2025.62
  2. Increase = 2025.62 − 1932.90 = $92.72 per month
14.Which change reduces the total interest on a loan the most, assuming each is affordable?
  1. AIncreasing the regular repayment
  2. BExtending the term of the loan
  3. CMaking the same repayment but switching to a higher interest rate
  4. DDelaying the first repayment

Answer: Increasing the regular repayment

Paying more each period reduces the balance faster, so less interest is charged. A longer term, a higher rate or a delay all leave more money owing for longer.

Part D: Extended problem

Jordan borrows $40 000 for a car at 8.4% p.a. compounded monthly, to be repaid in equal monthly repayments over 4 years.

(a) Find the monthly repayment.

(b) Complete the first two rows of the repayment table (interest, principal repaid and closing balance).

(c) Find the balance owing after 2 years.

(d) Find the total interest Jordan would pay if the loan runs its full term.

(e) After 2 years Jordan makes a lump-sum payment of $5000 and keeps paying the same monthly repayment. How many more repayments are needed, and how many fewer is this than planned?

(f) About how much interest does the lump sum save?

15.Answer parts (a) to (f), showing full working. Give dollar amounts to the nearest cent.[12 marks]

Model answer: (a) r = 0.007, n = 48: M = = $984.04. (b) Month 1: interest = 40 000 × 0.007 = $280.00, principal = 984.04 − 280 = $704.04, balance = $39 295.96. Month 2: interest = 39 295.96 × 0.007 = $275.07, principal = $708.97, balance = $38 586.99. (c) V₂₄ = 40 000 × 1.00724 − 984.04 × = $21 670.37. (d) Total interest = 984.04 × 48 − 40 000 = 47 233.92 − 40 000 = $7233.92. (e) New balance = 21 670.37 − 5000 = $16 670.37; n = = 18.1, so 18 full repayments and a final repayment of $94.23: 19 more repayments, 5 fewer than the 24 remaining. (f) Without the lump sum: 24 × 984.04 = $23 616.96 still to pay. With it: 5000 + 18 × 984.04 + 94.23 = $22 806.95. Interest saved ≈ $810.01.

Success criteria:

  • (a) correct r, n and repayment $984.04 (2 marks)
  • (b) two correct rows of the table (2 marks)
  • (c) balance after 24 months $21 670.37 (2 marks)
  • (d) total interest $7233.92 (1 mark)
  • (e) 19 more repayments with a final smaller repayment, and 5 fewer than planned (3 marks)
  • (f) compares total remaining outlay to find about $810 saved (2 marks)
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