Worked Solutions
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.
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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.