Interest and Depreciation — Exam Practice
Integrate simple interest, compound interest and depreciation in HSC-style multi-part problems — including loan comparisons, investment decisions and mixed extended response questions.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Interest and depreciation questions in the HSC love to combine multiple concepts in a single scenario. A business might borrow money at compound interest to buy equipment that depreciates — and you could be asked about both in the same question. Or an investment question might compare simple versus compound interest, then ask which option a rational person should choose and why.
Before you start: if you had $10,000 to invest for 5 years, what would you want to know before choosing a product? Without calculating — write your gut feeling. We'll revisit this at the end of the lesson.
The key is recognising which formula applies — and the signal is always in how the rate is described.
Simple interest / "flat rate" / "interest on the principal only" → $I = Prn$, $A = P + I$.
Compound interest / "compounds annually/monthly" / "interest on the balance" → $A = P(1+r)^n$. Adjust $r$ and $n$ to match the compounding period.
Straight-line depreciation / "fixed amount per year" / gives a dollar figure → $S = V_0 - Dn$.
Declining balance depreciation / "depreciates at X% per annum" / "reducing balance" → $S = V_0(1-r)^n$.
Key facts
- Signal words that identify each of the four formulas
- How to adjust $r$ and $n$ for non-annual compounding
- That total depreciation = $V_0 - S$ for both depreciation methods
- That total interest = $A - P$ for both interest methods
Concepts
- Why a lower nominal rate can still cost more (different compounding)
- Why showing $(1 \pm r)^n$ as a separate line preserves method marks
- How to identify and compare options from a borrower's vs investor's perspective
Skills
- Identify the correct formula from question wording
- Compare investment/loan options and state the difference with a conclusion
- Find depreciation rate $r$ by rearranging $S = V_0(1-r)^n$
- Combine investment and depreciation in a single multi-part problem
The first step in any Interest and Depreciation question is identifying which formula applies — and the signal is always in how the rate is described.
| Question says… | Use this formula |
|---|---|
| "simple interest", "flat rate", "interest on the principal only" | I = Prn |
| "compound interest", "compounds annually/monthly", "interest on the balance" | A = P(1+r)^n |
| "straight-line", "fixed amount per year", gives a dollar depreciation figure | S = V₀ − Dn |
| "declining balance", "reducing balance", "depreciates at X% per annum" | S = V₀(1−r)^n |
What to write in your book
- Read for signal words: flat rate / fixed dollar → SL or SI; compounding / percentage rate → compound interest or declining balance.
- For compound interest with non-annual compounding: divide rate by periods per year, multiply time by periods per year.
- Always label $P$ (or $V_0$), $r$, $n$ before substituting — one line, three values.
Quick check: Which wording most strongly signals that you should use $A = P(1+r)^n$?
Comparison questions require calculating a quantity under two different conditions and identifying which is better — for an investor, higher final amount is better; for a borrower, lower total repayment is better.
The most common comparison formats:
- Simple vs compound interest on the same investment — compound always wins for $n > 1$, but you must quantify the difference.
- Two compound interest products with different rates and/or compounding frequencies — calculate $A$ for each and compare.
- Flat rate vs reducing balance loans — flat rate applies interest to the original principal throughout (like simple interest); compare total repayment amounts.
- Investment vs depreciation — e.g. "is it better to invest or buy equipment?" Calculate both and compare net positions.
For all comparisons: use identical time periods, calculate both fully, state a conclusion with the dollar difference, and identify who benefits.
What to write in your book
- Comparison structure: calculate Option A fully → calculate Option B fully → subtract → state which is better and by how much.
- Investor perspective: higher $A$ is better. Borrower perspective: lower total repayment is better.
- Always use the same time period for both options before comparing.
True or false: For a borrower comparing two loans, the option with the lower nominal interest rate is always cheaper.
Worked examples · 4 problems, reveal step by step
Aiko needs to borrow $18,000 for 3 years. Lender A offers simple interest at 7.5% per annum. Lender B offers compound interest at 6.8% per annum compounding annually. (a) Calculate the total amount repaid under each lender. (b) Which lender should Aiko choose, and by how much is it cheaper?
$P = \$18{,}000$, $r = 0.075$, $n = 3$
$I = Prn = \$18{,}000 \times 0.075 \times 3 = \$4{,}050.00$
$\therefore$ Total repaid (A) $= \$18{,}000 + \$4{,}050 = \$22{,}050.00$
$P = \$18{,}000$, $r = 0.068$, $n = 3$
$(1.068)^3 = 1.21819\ldots$
$\therefore A = \$18{,}000 \times 1.21819 = \$21{,}927.36$
Saving $= \$22{,}050.00 - \$21{,}927.36 = \$122.64$
$\therefore$ Aiko should choose Lender B — it is cheaper by $\$122.64$ over 3 years.
A business spends $28,000 on a new machine. The machine depreciates at 15% per annum declining balance. Simultaneously, the business invests $28,000 in a term deposit earning 4.2% per annum compounding semi-annually. (a) What is the machine's value after 4 years? (b) What is the investment worth after 4 years? (c) What is the difference between the investment value and the machine's value?
$V_0 = \$28{,}000$, $r = 0.15$, $(1-r) = 0.85$, $n = 4$
$(0.85)^4 = 0.52201\ldots$
$\therefore S = \$28{,}000 \times 0.52201 = \$14{,}616.18$
Semi-annual compounding: $r = 0.042 \div 2 = 0.021$ per half-year; $n = 4 \times 2 = 8$
$(1.021)^8 = 1.18088\ldots$
$\therefore A = \$28{,}000 \times 1.18088 = \$33{,}064.65$
Difference $= \$33{,}064.65 - \$14{,}616.18 = \$18{,}448.48$
$\therefore$ The investment is worth $\$18{,}448.48$ more than the machine after 4 years.
A boat purchased for $54,000 has a salvage value of $29,160 after 4 years of declining balance depreciation. Find the annual depreciation rate.
$\$29{,}160 = \$54{,}000 \times (1-r)^4$
$(1-r)^4 = \dfrac{\$29{,}160}{\$54{,}000} = 0.54$
$1 - r = (0.54)^{1 \div 4} = (0.54)^{0.25}$
$(0.54)^{0.25} = 0.8573\ldots$
$r = 1 - 0.8573 = 0.1427$
$\therefore$ Annual depreciation rate $= 0.1427 \times 100 = \mathbf{14.27\%}$ per annum
Nina has $12,000 to invest for 4 years. Product A offers 5.1% per annum compounding annually. Product B offers 4.9% per annum compounding monthly. Which product gives the larger final amount, and by how much?
$A_A = \$12{,}000(1.051)^4$
$(1.051)^4 = 1.220143\ldots$
$A_A = \$12{,}000 \times 1.220143 = \$14{,}641.72$
$r = 0.049 \div 12 = 0.004083333\ldots$ per month
$n = 4 \times 12 = 48$ months
$A_B = \$12{,}000(1.004083333\ldots)^{48} = \$14{,}592.50$
Difference $= \$14{,}641.72 - \$14{,}592.50 = \$49.22$
$\therefore$ Product A gives the larger final amount by $\$49.22$.
What to write in your book
- Show $(1 \pm r)^n$ as a separate line before multiplying by $P$ or $V_0$ — preserves method marks.
- Finding $r$ (declining balance): rearrange to $(1-r)^n = S/V_0$, take $n$th root, subtract from 1, convert to %.
- Always end comparison answers with a conclusion sentence stating which is better and by how much.
Fill the gap: $25,000 is invested at 6% per annum compounding annually for 5 years using Option A (simple interest 6%) and Option B (compound 5.6%). Option A gives $A = \$$.00 and Option B gives $A = \$32{,}829.15$, so Option B is better by $\$$.
Common errors · the 3 traps that cost marks
What to write in your book
- Extended response layout: (1) write formula, (2) list variables, (3) show $(1 \pm r)^n$ on its own line, (4) state final answer with units and conclusion.
- Reliable sequence for mixed problems: identify the model → label variables → calculate each part separately → compare → conclude.
- Error carried forward: if you make an arithmetic mistake but continue correctly, the marker can still award method marks — only if the working is clearly shown.
Match each problem type to its recommended approach:
Quick-fire practice · 5 calculations
$20,000 is borrowed at 8% per annum simple interest for 4 years. Find the total amount repaid.
$15,000 is invested at 3.6% per annum compounding quarterly for 3 years. Find the final amount.
Equipment worth $44,000 depreciates at 16% per annum declining balance. Find its value after 3 years.
An asset bought for $32,000 is worth $20,480 after 2 years of declining balance depreciation. Find the annual depreciation rate.
Compare: $10,000 invested for 2 years at 5% p.a. simple interest vs 4.8% p.a. compounding monthly. Which is better and by how much?
Top 3 list: Name THREE things you should always show in an extended response Interest or Depreciation question to maximise your marks.
Look back at what you wrote in the Think First section. The signal for which formula to use is always in the question wording — "simple interest" / "flat rate" → $I = Prn$; "compounds" → $A = P(1+r)^n$; "declining balance" → $S = V_0(1-r)^n$. You'd want to know the interest rate, the compounding frequency, any fees, and the time period before choosing a product.
What has changed? What did you get right? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 1. A savings account invests $14,500 at 5.3% per annum compounding annually for 3 years. Calculate: (a) the final amount, and (b) the interest earned. (3 marks)
SA 2. A business buys equipment for $36,000. It depreciates at 18% per annum declining balance. Find: (a) its value after 5 years, and (b) the total depreciation. (3 marks)
SA 3. Compare two investment options for $20,000 over 4 years: Option A is simple interest at 6.1% per annum; Option B is compound interest at 5.7% per annum compounding annually. Which option is better, and by how much? (4 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $I = 20000 \times 0.08 \times 4 = \$6{,}400$; $A = \$26{,}400$ · 2: $r = 0.036/4 = 0.009$; $n = 12$; $A = 15000(1.009)^{12} = \$16{,}695.86$ · 3: $(1-0.16) = 0.84$; $S = 44000(0.84)^3 = \$26{,}094.14$ · 4: $(1-r)^2 = 20480/32000 = 0.64$; $1-r = 0.8$; $r = 20\%$ · 5: SI: $A = 10000(1 + 0.05 \times 2) = \$11{,}000$; CI: $r = 0.048/12 = 0.004$, $n = 24$, $A = 10000(1.004)^{24} = \$11{,}003.07$; CI better by $\$3.07$
SA 1 (3 marks): $A = 14500(1.053)^3$ [1] $= 14500 \times 1.16779\ldots = \$16{,}932.48$ [1]. Interest $= 16932.48 - 14500 = \$2{,}432.48$ [1].
SA 2 (3 marks): $(1-r) = 0.82$ [1]. $S = 36000(0.82)^5 = 36000 \times 0.36950\ldots = \$13{,}300.76$ [1]. Total depreciation $= 36000 - 13300.76 = \$22{,}699.24$ [1].
SA 3 (4 marks): Option A: $I = 20000 \times 0.061 \times 4 = \$4{,}880$; $A = \$24{,}880.00$ [1 method + 1 answer]. Option B: $A = 20000(1.057)^4 = 20000 \times 1.24831 = \$24{,}966.18$ [1 method + 1 answer]. Option B is better by $\$24{,}966.18 - \$24{,}880.00 = \$86.18$ [1 conclusion].
Five timed questions across all four formulas. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering questions on Financial Mathematics. Pool: lessons 1–14.
Module 3 Complete!
You've worked through all 14 lessons covering Earning Money, Managing Money, and Interest & Depreciation — the full MS-F1 Financial Mathematics syllabus for Year 11.
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