Financial Mathematics — Worked Solutions (HSC Maths Advanced)
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Worked examples for HSC Maths Advanced 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 Advanced — 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.
How to use these
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 — Compound interest
Question
$\$8000$ is invested at $5\%$ per annum, compounded annually. Find the value of the investment after $4$ years, to the nearest cent.
Solution
Use $A = P(1+r)^n$ with $P = 8000$, $r = 0.05$, $n = 4$.
$A = 8000(1.05)^4$.
$(1.05)^4 = 1.21550625$, so $A = 8000 \times 1.21550625 = 9724.05$.
$A = \$9724.05$. Keep full accuracy until the final rounding — early rounding costs cents and marks.
Compound interest grows by the same factor each period, so after $n$ years $A = P(1+r)^n$, where $r$ is the rate as a decimal.
Here $P = 8000$, $r = 5\% = 0.05$, and $n = 4$ years, so the growth factor each year is $1.05$.
$A = 8000(1.05)^4$. Working out $(1.05)^4 = 1.21550625$, we get $A = 8000 \times 1.21550625 = 9724.05$.
So after $4$ years the investment is worth $\$9724.05$. Notice we round only at the very end, to the nearest cent as asked.
$A = P(1+r)^n$; $P = 8000$, $r = 0.05$, $n = 4$.
- $A = 8000(1.05)^4$
- $(1.05)^4 = 1.21550625$
- $A = 8000 \times 1.21550625 = 9724.05$
$A = \$9724.05$.
Where the marks go
- 1 mark: Correct compound-interest setup $A = 8000(1.05)^4$
- 1 mark: Correctly evaluates the growth factor / expression
- 1 mark: Correct value $\$9724.05$ to the nearest cent
Key idea
Compound interest uses $A = P(1+r)^n$ with $r$ as a decimal and $n$ the number of compounding periods.
Example 2 — Arithmetic sequence
Question
An employee earns $\$52\,000$ in their first year, and their salary increases by $\$2500$ each year. Find their total earnings over the first $10$ years.
Solution
This is an arithmetic series: $a = 52000$, $d = 2500$, $n = 10$.
Use $S_n = \tfrac{n}{2}\big[2a + (n-1)d\big]$.
$S_{10} = \tfrac{10}{2}\big[2(52000) + 9(2500)\big] = 5\big[104000 + 22500\big] = 5(126500) = 632500$.
Total $= \$632\,500$. Identify $a$, $d$ and $n$ first — guessing the formula wastes time.
Each year the salary goes up by a fixed amount, $\$2500$, so the yearly salaries form an arithmetic sequence with first term $a = 52000$ and common difference $d = 2500$. "Total earnings" means we sum the first $10$ terms.
The sum of an arithmetic series is $S_n = \tfrac{n}{2}\big[2a + (n-1)d\big]$, which works because we're averaging the first and last terms and multiplying by how many there are.
Substituting $n = 10$: $S_{10} = \tfrac{10}{2}\big[2(52000) + 9(2500)\big] = 5\big[104000 + 22500\big] = 5 \times 126500 = 632500$.
So over the first $10$ years the employee earns a total of $\$632\,500$.
Arithmetic: $a = 52000$, $d = 2500$, $n = 10$.
- $S_n = \tfrac{n}{2}\big[2a + (n-1)d\big]$
- $S_{10} = 5\big[104000 + 22500\big] = 5(126500)$
- $= 632500$
Total $= \$632\,500$.
Where the marks go
- 1 mark: Identifies $a = 52000$, $d = 2500$ and an arithmetic series
- 1 mark: Correct substitution into $S_n = \tfrac{n}{2}[2a + (n-1)d]$
- 1 mark: Correct total $\$632\,500$
Key idea
A fixed annual increase gives an arithmetic series; sum it with $S_n = \tfrac{n}{2}[2a + (n-1)d]$.
Example 3 — Geometric series
Question
A person deposits $\$1000$ at the start of each year into an account, and each deposit is $5\%$ larger than the previous one. Find the total amount deposited over $6$ years, to the nearest dollar.
Solution
The deposits form a geometric sequence: $a = 1000$, $r = 1.05$, $n = 6$.
Use $S_n = \dfrac{a(r^n - 1)}{r - 1}$.
$S_6 = \dfrac{1000(1.05^6 - 1)}{0.05}$. Now $1.05^6 = 1.340095641$, so $1.05^6 - 1 = 0.340095641$.
$S_6 = \dfrac{1000(0.340095641)}{0.05} = \dfrac{340.095641}{0.05} = 6801.91$.
Total $\approx \$6802$. Note this is total deposits, not an account balance with interest.
Each deposit is $5\%$ bigger than the last, so the deposits multiply by the same factor $1.05$ each time — that's a geometric sequence with first term $a = 1000$ and common ratio $r = 1.05$. We want the sum of $6$ of them.
The geometric-series sum is $S_n = \dfrac{a(r^n - 1)}{r - 1}$.
Substituting: $S_6 = \dfrac{1000(1.05^6 - 1)}{1.05 - 1}$. Computing $1.05^6 = 1.340095641$, the numerator is $1000(0.340095641) = 340.095641$, and dividing by $0.05$ gives $6801.91$.
Rounded to the nearest dollar, the total deposited is $\$6802$.
Geometric: $a = 1000$, $r = 1.05$, $n = 6$.
- $S_n = \dfrac{a(r^n - 1)}{r - 1}$
- $1.05^6 = 1.340095641$
- $S_6 = \dfrac{1000(0.340095641)}{0.05} = 6801.91$
Total $\approx \$6802$.
Where the marks go
- 1 mark: Identifies a geometric series with $a = 1000$, $r = 1.05$
- 1 mark: Correct substitution into $S_n = \dfrac{a(r^n - 1)}{r - 1}$
- 1 mark: Correct total $\approx \$6802$
Key idea
Deposits that grow by a fixed percentage form a geometric series, summed with $S_n = \dfrac{a(r^n - 1)}{r - 1}$.
Example 4 — Future value of an annuity
Question
At the end of each year, $\$2000$ is deposited into an account earning $6\%$ per annum, compounded annually. Find the value of the annuity at the end of $5$ years, to the nearest cent.
Solution
Future value of an ordinary annuity: $FV = R\,\dfrac{(1+i)^n - 1}{i}$, with $R = 2000$, $i = 0.06$, $n = 5$.
$FV = 2000 \cdot \dfrac{(1.06)^5 - 1}{0.06}$. Now $(1.06)^5 = 1.3382255776$, so the numerator is $0.3382255776$.
$\dfrac{0.3382255776}{0.06} = 5.63709296$, and $FV = 2000 \times 5.63709296 = 11274.19$.
$FV = \$11\,274.19$. Or sum the geometric series of grown deposits — same answer. Keep full precision until the end.
Each $\$2000$ deposit sits in the account and earns compound interest until the end. The first deposit (end of year 1) earns interest for $4$ years, the next for $3$, and so on, with the last deposit earning none — so the final values form a geometric series. The neat formula that adds them up is the future-value-of-an-annuity formula: $FV = R\,\dfrac{(1+i)^n - 1}{i}$.
Here $R = 2000$ (the regular payment), $i = 0.06$ (the rate per period as a decimal), and $n = 5$ periods.
$FV = 2000 \cdot \dfrac{(1.06)^5 - 1}{0.06}$. Working out $(1.06)^5 = 1.3382255776$, the bracket is $0.3382255776$, and dividing by $0.06$ gives $5.63709296$.
Finally $FV = 2000 \times 5.63709296 = 11274.19$. So the annuity is worth $\$11\,274.19$ after $5$ years.
Future value of annuity: $FV = R\,\dfrac{(1+i)^n - 1}{i}$; $R = 2000$, $i = 0.06$, $n = 5$.
- $(1.06)^5 = 1.3382255776$
- bracket $= 0.3382255776$
- $\div 0.06 = 5.63709296$
- $\times 2000 = 11274.19$
$FV = \$11\,274.19$.
Where the marks go
- 1 mark: Correct annuity future-value formula $FV = R\dfrac{(1+i)^n - 1}{i}$
- 1 mark: Correct values substituted ($R = 2000$, $i = 0.06$, $n = 5$)
- 1 mark: Correctly evaluates $(1.06)^5$ and the expression
- 1 mark: Correct future value $\$11\,274.19$ to the nearest cent
Key idea
The future value of a stream of equal end-of-period deposits is $FV = R\,\dfrac{(1+i)^n - 1}{i}$ — a geometric series in disguise.
Example 5 — Depreciation
Question
A machine is purchased for $\$30\,000$ and depreciates by $12\%$ of its value each year (declining-balance method). Find its value after $3$ years, to the nearest dollar.
Solution
Declining balance: $S = V_0(1 - r)^n$ with $V_0 = 30000$, $r = 0.12$, $n = 3$.
$S = 30000(0.88)^3$. Now $0.88^3 = 0.681472$, so $S = 30000 \times 0.681472 = 20444.16$.
$S \approx \$20\,444$. The base is $(1 - r) = 0.88$, not $0.12$ — that's the classic slip.
Under declining-balance depreciation the asset loses a fixed percentage of its current value each year, so it keeps $100\% - 12\% = 88\%$ of its value annually. That makes it a geometric decay: $S = V_0(1 - r)^n$.
Here $V_0 = 30000$, $r = 0.12$ so the retained fraction is $1 - 0.12 = 0.88$, and $n = 3$.
$S = 30000(0.88)^3$. Since $0.88^3 = 0.681472$, we get $S = 30000 \times 0.681472 = 20444.16$.
Rounded to the nearest dollar, the machine is worth about $\$20\,444$ after $3$ years.
Declining balance: $S = V_0(1 - r)^n$; $V_0 = 30000$, $r = 0.12$, $n = 3$.
- retained fraction $1 - 0.12 = 0.88$
- $0.88^3 = 0.681472$
- $S = 30000 \times 0.681472 = 20444.16$
$S \approx \$20\,444$.
Where the marks go
- 1 mark: Correct declining-balance setup $S = 30000(0.88)^3$
- 1 mark: Correctly evaluates $(0.88)^3$
- 1 mark: Correct value $\approx \$20\,444$ to the nearest dollar
Key idea
Declining-balance depreciation uses $S = V_0(1 - r)^n$, keeping a fraction $(1 - r)$ of the value each period.
Frequently asked questions
Step-by-step solutions to exam-style questions on Financial Mathematics in HSC Maths Advanced, 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 Advanced syllabus for Financial Mathematics, using the methods and notation expected in the exam.