VCE General Mathematics Units 3 & 4 · Exam practice · 17 October 2026
A finance question often looks like a calculator exercise. The decisive step happens earlier: deciding what the balance means and when interest and payments occur. Once that is clear, the recurrence relation, amortisation table and final repayment all tell the same story.
This guide focuses on reducing-balance loans and regular deposits, within the recursion and financial modelling content described in VCAA’s current General Mathematics support materials. All situations and questions below are original, simplified mathematical models. They are not financial advice or actual loan offers.
1. Define the balance before touching the calculator
Write a sentence such as: “Bn is the loan balance immediately after the nth monthly repayment.” Then B0 is the amount borrowed before the first interest charge and repayment.
If the periodic interest rate is i as a decimal and the repayment A occurs at the end of each period, the model is:
B0 = P, Bn+1 = (1 + i)Bn − A.
The opening balance earns or incurs interest first; the payment is then subtracted. For a nominal rate of 6% per annum with monthly compounding, i = 0.06 ÷ 12 = 0.005, so the multiplier is 1.005. A period is one month, not one year.
Be careful with an effective annual rate: dividing it by 12 does not give the equivalent monthly compound rate. If an effective annual rate is r, the equivalent monthly rate satisfies (1 + i)12 = 1 + r. Read the rate description before choosing a conversion.
2. Worked example: build an amortisation table
A fictional loan starts at $12,000. Interest is 6% per annum, compounded monthly, with repayments of $350 at the end of each month. Assume no fees and a constant rate. Let Bn be the balance after n repayments.
B0 = 12,000, Bn+1 = 1.005Bn − 350.
For month 1, interest = 0.005 × 12,000 = $60. The repayment reduces principal by 350 − 60 = $290. Therefore B1 = 12,000 − 290 = $11,710.
Month 2 uses the new opening balance: interest = 0.005 × 11,710 = $58.55. Principal reduction = 350 − 58.55 = $291.45, and B2 = $11,418.55.
| Payment | Interest ($) | Principal reduction ($) | Closing balance ($) |
|---|---|---|---|
| 1 | 60.00 | 290.00 | 11,710.00 |
| 2 | 58.55 | 291.45 | 11,418.55 |
| 3 | 57.09 | 292.91 | 11,125.64 |
The interest portion decreases because it is calculated on a smaller balance. Since the repayment stays fixed, the principal-reduction portion increases. A common mistake is to subtract the entire repayment from the principal and forget that some of it pays the interest.
Rounding convention: these examples retain unrounded balances internally and round displayed money to cents. If a question gives a cent-rounded table or explicitly requires interest to be rounded each period, follow that convention consistently instead.
3. Worked example: the final repayment is usually different
Continue the same loan until it is paid off. A recurrence table or financial solver gives:
- Balance after 36 regular repayments: $592.529560…
- Balance after 37 regular repayments: $245.492208…
That second balance is after payment 37. It still incurs one month of interest before the next scheduled payment. The amount owing then is:
1.005 × 245.492208… = $246.719669… ≈ $246.72.
There are 37 full payments of $350, followed by a final payment of $246.72 in month 38. Subtracting another $350 would produce a negative model balance. That negative number indicates an overpayment; it is not the final amount to pay.
Total interest = total repayments − amount borrowed = 37 × 350 + 246.719669… − 12,000 = $1,196.72, to the nearest cent.
Notice the counting: 38 payment dates does not mean 38 equal payments. Likewise, do not use the balance immediately after payment 37 as the final payment without adding the last interest charge.
4. Worked example: regular deposits change the sign
An account starts with $5,000 and earns a nominal 4.8% per annum, compounded monthly. Another $200 is deposited at the end of each month. Find the balance immediately after deposit 24.
Define Sn as the balance after n monthly deposits. The monthly rate is 0.048 ÷ 12 = 0.004, giving S0 = 5,000 and Sn+1 = 1.004Sn + 200.
Check the first step: S1 = 1.004 × 5,000 + 200 = $5,220. A recurrence table or financial solver gives S24 = $10,530.156518… ≈ $10,530.16.
Total contributions are 5,000 + 24 × 200 = $9,800, so interest earned is $730.16. The full final balance is not all interest: most of it is money contributed.
If deposits were made at the beginning of each month, each new deposit would also earn interest that month. With the balance defined immediately before the next beginning-of-month deposit, the update would be Sn+1 = 1.004(Sn + 200). The timing changes the model, even when the monthly rate and deposit are identical.
5. Check technology against the first period
For a time-value-of-money solver, match the number of periods, nominal annual percentage rate, payment frequency, compounding frequency and payment timing to the question. Cash inflows and outflows normally need opposite signs. A borrowed amount received now and repayments paid later should therefore have opposite signs.
Before trusting a long calculation, reproduce month 1 manually. If the calculator disagrees, inspect its settings. Reset old values rather than assuming an earlier question used the same frequency. VCAA’s approved-technology page lists the permitted technology for the 2026 examinations; follow your school’s instructions and the examination rules.
6. Four practice questions
Try these before reading the solutions. All payments occur at the end of each stated period unless specified otherwise.
- A $6,000 loan charges a nominal 7.2% per annum, compounded monthly. Repayments are $150 monthly. Write the recurrence and find the balance after two repayments.
- For L0 = 8,000 and Ln+1 = 1.01Ln − 70, determine whether the loan balance initially increases or decreases. Explain using the first interest charge.
- Immediately after the last regular repayment, a loan balance is $184.26. The monthly interest rate is 0.6%. Find the amount required at the next payment date to clear it, to the nearest cent.
- An investment starts with $2,000, earns 0.5% interest per month and receives $100 at each month’s end. Find the balance after two deposits and the interest earned over those two months.
7. Worked solutions
Question 1: match the monthly rate
i = 0.072 ÷ 12 = 0.006. Thus L0 = 6,000 and Ln+1 = 1.006Ln − 150. The first balance is 6,000 × 1.006 − 150 = 5,886. The next is 5,886 × 1.006 − 150 = 5,771.316, or $5,771.32.
Question 2: a payment can be too small
The first interest charge is 0.01 × 8,000 = $80. The $70 payment does not cover it, so the balance increases by $10 to $8,010. For this initial balance and rate, a payment must exceed $80 to reduce the principal in the first period. A payment of exactly $80 would leave the balance unchanged under this fixed model.
Question 3: include the final interest
The payoff is 184.26 × 1.006 = 185.36556, or $185.37. Paying $184.26 at the next scheduled date would omit the interest accumulated since the previous payment.
Question 4: separate contributions from growth
S0 = 2,000 and Sn+1 = 1.005Sn + 100. Therefore S1 = 2,110 and S2 = 2,220.55. Contributions total $2,200, so the interest earned is $20.55.
Your last check before moving on
- Have I defined what n counts and when the balance is measured?
- Does the interest multiplier use the rate for one period?
- Does the payment sign and timing match the scenario?
- Have I distinguished the total paid from the interest paid?
- Have I checked the last payment separately and rounded as instructed?
For official practice, use VCAA’s General Mathematics examination papers, assessment guides and reports. This is an independent Loav revision resource and is not endorsed by VCAA.
