Financial Mathematics — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 financial mathematics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Standard 2 — Financial Mathematics. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
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Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Reducing-balance loan
Question
Priya borrows \$8000 on a reducing-balance loan charged at 1% per month. She repays \$700 at the end of each month.
(a) Using a recurrence relation, find the amount still owing after she has made 2 repayments. Give your answer to the nearest cent.
(b) Explain why the amount of interest charged decreases each month.
Solution
(a) Each month, add 1% interest then subtract the \$700 repayment. The recurrence is $A_{n} = 1.01\,A_{n-1} - 700$, with $A_0 = 8000$.
Month 1: $A_1 = 1.01(8000) - 700 = 8080 - 700 = 7380$.
Month 2: $A_2 = 1.01(7380) - 700 = 7453.80 - 700 = 6753.80$.
So she owes \$6753.80 after 2 repayments.
(b) Interest is charged on the balance, and the balance shrinks each month, so 1% of a smaller number is less interest. Quote the recurrence — examiners want the multiplier and the repayment shown explicitly.
(a) On a reducing-balance loan, each month two things happen: interest is added, then a repayment is taken off. Interest of 1% means multiplying by $1.01$, and the repayment is $-700$. That gives the recurrence $A_n = 1.01\,A_{n-1} - 700$ starting from $A_0 = 8000$.
Let's step through it. After month 1: $A_1 = 1.01 \times 8000 - 700 = 8080 - 700 = 7380$. After month 2: $A_2 = 1.01 \times 7380 - 700 = 7453.80 - 700 = 6753.80$. So she owes \$6753.80.
(b) Here's the why: interest is calculated on the current amount owing. Because each repayment lowers that balance, the 1% is applied to a smaller and smaller amount, so the interest portion falls every month — which is exactly what "reducing balance" describes.
(a) Recurrence $A_n = 1.01\,A_{n-1} - 700$, $A_0 = 8000$.
- $A_1 = 1.01(8000) - 700 = 7380$
- $A_2 = 1.01(7380) - 700 = 6753.80$
Owing $= \$6753.80$.
(b) Interest is 1% of the balance; the balance falls each month, so the interest charged falls too.
Where the marks go
- 1 mark: Correct recurrence relation $A_n = 1.01A_{n-1} - 700$
- 1 mark: Correct balance after 1 repayment ($7380)
- 1 mark: Correct balance after 2 repayments (\$6753.80)
- 1 mark: Valid explanation linking falling balance to falling interest
Key idea
A reducing-balance loan applies interest to the current balance then subtracts a repayment; as the balance falls, so does the interest charged.
Example 2 — Depreciation
Question
A café buys a coffee machine for \$9500. It depreciates by the declining-balance method at 15% per year.
(a) Find the salvage value of the machine after 4 years, to the nearest dollar.
(b) The café will replace the machine once its value first drops below \$4000. After how many full years does this happen?
Solution
(a) Declining balance: $S = V_0(1 - r)^n$ with $V_0 = 9500$, $r = 0.15$, $n = 4$.
$S = 9500(0.85)^4 = 9500 \times 0.52200625 \approx \$4959$.
(b) Need $9500(0.85)^n < 4000$, i.e. $(0.85)^n < 0.42105$.
$0.85^5 = 0.4437$ (still above), $0.85^6 = 0.3771$ (below). So it first drops below \$4000 after **6 years**. Don't round the rate — use $0.85$ exactly through the powers.
(a) Declining-balance depreciation keeps a fixed percentage of the value each year, so we keep $100\% - 15\% = 85\%$, i.e. multiply by $0.85$ each year. The formula is $S = V_0(1 - r)^n = 9500(0.85)^4$. Computing $0.85^4 \approx 0.5220$, we get $S \approx \$4959$.
(b) We want the value to fall below \$4000, so $9500(0.85)^n < 4000$, which rearranges to $(0.85)^n < 0.42105$. Testing whole years: at $n = 5$, $0.85^5 \approx 0.4437$ (not yet below); at $n = 6$, $0.85^6 \approx 0.3771$ (below). So it happens after 6 full years. We test whole years because the café only checks once a year.
(a) $S = V_0(1-r)^n$:
- $S = 9500(0.85)^4 = 9500 \times 0.52201 \approx \$4959$
(b) Need $9500(0.85)^n < 4000 \Rightarrow (0.85)^n < 0.42105$.
- $0.85^5 \approx 0.4437$ (above)
- $0.85^6 \approx 0.3771$ (below)
- First below \$4000 after 6 years
Where the marks go
- 1 mark: Correct declining-balance setup $9500(0.85)^n$
- 1 mark: Correct salvage value after 4 years (\$4959)
- 1 mark: Correctly identifies 6 years for value below \$4000
Key idea
Declining-balance depreciation multiplies by $(1 - r)$ each year, so the value follows $V_0(1-r)^n$ — test whole years to find when it crosses a threshold.
Frequently asked questions
Step-by-step solutions to exam-style questions on Financial Mathematics in HSC Maths Standard 2, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Standard 2 syllabus for Financial Mathematics, using the methods and notation expected in the exam.