Worked Solutions
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.
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.
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