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SAGeneral MathematicsSyllabus dot point

How does a loan balance fall as we make regular repayments against compounding interest?

Model a reducing-balance loan with a recurrence relation, track the balance after each repayment, and analyse the effect of changing the repayment.

How to model a reducing-balance loan with a recurrence, build an amortisation schedule splitting each payment into interest and principal, and see how repayment size changes the loan term.

Generated by Claude Opus 4.77 min answer

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  1. What this dot point is asking
  2. How a reducing-balance loan works
  3. The amortisation schedule
  4. Analysing changes to the repayment
  5. The final repayment

What this dot point is asking

You must model the loan with a recurrence, build an amortisation schedule, and analyse how changing the repayment affects the term and total interest.

How a reducing-balance loan works

Interest is charged only on the amount still owed. Each period, interest is added to the balance, then the repayment is taken off. The recurrence is:

Bn+1=Bn(1+i)βˆ’RB_{n+1} = B_n(1 + i) - R

where BnB_n is the balance after nn repayments, ii is the interest rate per period (decimal) and RR is the regular repayment.

The amortisation schedule

An amortisation schedule splits each repayment into the interest charged that period and the principal repaid (the rest of the payment).

  • Interest for the period =iΓ—Bn= i \times B_n (the balance at the start of the period).
  • Principal repaid == repayment Rβˆ’R - interest.
  • New balance =Bnβˆ’= B_n - principal repaid.

Analysing changes to the repayment

The repayment RR must exceed the first period's interest, or the balance will grow instead of shrink. Increasing the repayment clears the loan in fewer periods and reduces the total interest paid; decreasing it lengthens the term and raises total interest.

The final repayment

The last repayment is usually smaller than the rest, because only the remaining balance plus its final period's interest is owed. To find it, take the second-last balance, add one period's interest, and that total is the final payment.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 SACE Stage 22 marksKeisha took out a home loan of 650000over20yearsataninterestrateof2.88650 000 over 20 years at an interest rate of 2.88% per annum, compounded monthly. Show that the minimum monthly repayment was approximately 3570.
Show worked answer β†’

Use the reducing-balance (amortisation) model and solve for the repayment R.

Monthly rate i = 2.88% / 12 = 0.24% per month = 0.0024.
Number of payments n = 20 times 12 = 240.
Present value (loan) PV = 650 000, future value = 0.

R = PV times i times (1 + i)^n / ((1 + i)^n - 1)
= 650 000 times 0.0024 times (1.0024)^240 / ((1.0024)^240 - 1)
= approximately $3570 per month.

Award 1 mark for the correct monthly rate and number of payments, and 1 mark for arriving at approximately $3570. Using a financial solver with PV = 650000, n = 240, i = 0.0024, FV = 0 gives the same result.

2023 SACE Stage 22 marksGrace took out a home loan of 356000over25yearsat5.11356 000 over 25 years at 5.11% per annum, compounded monthly. Show that Grace would pay approximately 2100 per month in loan repayments.
Show worked answer β†’

Apply the reducing-balance repayment model and solve for R.

Monthly rate i = 5.11% / 12 = 0.42583% per month = 0.0042583.
Number of payments n = 25 times 12 = 300.
Loan PV = 356 000, FV = 0.

R = PV times i times (1 + i)^n / ((1 + i)^n - 1)
= 356 000 times 0.0042583 times (1.0042583)^300 / ((1.0042583)^300 - 1)
= approximately $2100 per month.

Award 1 mark for the correct monthly rate and 300 payments, and 1 mark for showing the repayment is about $2100. A financial solver with PV = 356000, n = 300, i = 0.0042583, FV = 0 confirms this.

2019 SACE Stage 21 marksDakota's 25-year home loan of 350000at4.18350 000 at 4.18% per annum compounded monthly has a minimum monthly repayment of approximately 1880. Calculate the total cost of the loan.
Show worked answer β†’

The total cost of the loan is the repayment multiplied by the number of repayments.

Number of repayments n = 25 times 12 = 300.
Total cost = 1880 times 300 = $564 000.

So Dakota repays about 564000intotaloverthelifeoftheloan(comparedwiththe564 000 in total over the life of the loan (compared with the 350 000 borrowed, the extra roughly 214000isinterest).Awardthesinglemarkformultiplyingthemonthlyrepaymentby300togetapproximately214 000 is interest). Award the single mark for multiplying the monthly repayment by 300 to get approximately 564 000.

2021 SACE Stage 21 marksA reducing-balance loan with an offset account has a balance owing of 351700after2years.Atthistime351 700 after 2 years. At this time 15 000 is deposited into the previously empty offset account. State the new balance that interest will be calculated on as soon as the $15 000 is deposited into the offset account.
Show worked answer β†’

An offset account reduces the balance on which interest is charged by the amount held in the offset account, without actually reducing the loan principal.

New balance for interest = balance owing - offset amount
= 351 700 - 15 000 = $336 700.

So interest is calculated on 336700whilethe336 700 while the 15 000 sits in the offset account. Award the mark for the subtraction giving 336700.Theactualloanbalanceisstill336 700. The actual loan balance is still 351 700; only the interest-bearing portion changes.