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Module 3 · L14 of 14 ~55 min MST-12-S2-02 ⚡ +90 XP available

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.

Today's hook, A business might borrow money at compound interest to buy equipment that depreciates, and you could be asked about both in the same question. The mathematics is never harder than what you've already practised, the challenge is recognising which formula to use, in what order, and making sure your answer actually responds to the question being asked.

0/5QUESTS

Get oriented

Recall the four formulas, meet the summary table and settle the key terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall, your gut answer first

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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.

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Formula Summary, all four formulas in one place

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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" → $S = V_0(1-r)^n$.

Simple/compound interest and both depreciation methods.

Signal words → formula choice. Write all variables before substituting.

"flat rate" → simple

Flat rate, fixed rate on original principal, simple interest → $I = Prn$. One formula, no compounding adjustment needed.

"compounds" → $A = P(1+r)^n$

Any compounding wording → adjust $r$ and $n$ to the compounding period before substituting. "Per annum" describes the quoted rate, not the period.

Label every variable

Write $P = \ldots$, $r = \ldots$, $n = \ldots$ before the formula. Prevents substitution errors and helps a marker follow your working.

What you'll master

Know

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

Understand

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

Can do

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

Key terms

Simple InterestInterest calculated only on the principal: $I = Prn$. Rate is constant and does not compound.

Compound InterestInterest calculated on the current balance (principal plus accumulated interest): $A = P(1+r)^n$.

Straight-Line DepreciationA fixed dollar amount $D$ is deducted from an asset's value each year: $S = V_0 - Dn$.

Declining BalanceA fixed percentage of the current value is deducted each year: $S = V_0(1-r)^n$.

Compounding PeriodThe frequency at which interest is calculated and added to the balance, annually, quarterly, monthly, etc.

Effective RateThe actual annual return after compounding more than once per year, always higher than the nominal rate.

Identify the right formula

Read a mixed question and decide which of the four formulas it is asking for.

Identifying the Right Formula in Mixed Questions

core concept

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

Must do: Label every variable before substituting into any formula. Write $P = \ldots$, $r = \ldots$, $n = \ldots$ as a list before the formula. This prevents substitution errors and makes your working easy to follow under exam pressure.

Common error: "Per annum" does not automatically mean annual compounding. A rate of 6% per annum compounding monthly still uses $r = 0.06 \div 12$ per month and $n$ in months. "Per annum" describes the quoted rate, compounding frequency is stated separately.

Mixed financial questions require choosing the correct formula: simple interest (I = Prn) for flat-rate scenarios; compound interest (A = P(1+r)^n) for reinvested interest; depreciation formulas for falling asset values.

Pause, copy the three-formula selector: flat-rate interest (no reinvestment) → I = Prn; interest reinvested each period → A = P(1+r)^n; falling asset value → S = V₀ − Dn (straight-line) or S = V₀(1−r)^n (declining balance) into your book.

Quick check: Which wording most strongly signals that you should use $A = P(1+r)^n$?

Compare investments and loans

Handle the questions that ask which option is better, and by how much.

Investment and Loan Comparison Questions

core concept

Selecting the right formula, I = Prn for flat-rate scenarios, A = P(1+r)^n for reinvested interest, and depreciation formulas for falling asset values, is step one. Comparing two financial products requires step two: express both options in the same metric (total interest paid, or final value) over the same time period, with rates converted to the same compounding frequency.

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.

Must do: State which option is better and by how much, a calculation without a conclusion loses the final mark. Write "Option A produces $\$X$ more than Option B over $n$ years."

Common error: For a borrower, the better option is lower total cost, not higher interest rate. Always calculate and compare totals rather than assuming the lower nominal rate is cheaper (different compounding frequencies can change the outcome).

Investment and loan comparison questions require converting all options to the same metric, usually total cost, total interest, or final value, before comparing. Convert rates to matching periods and check all conditions are equal.

Pause, copy the comparison principle: convert all options to the same metric (total interest or final value), the same time period, and the same compounding frequency before drawing any conclusion about which product is better into your book.

True or false: For a borrower comparing two loans, the option with the lower nominal interest rate is always cheaper.

Work four mixed examples

Follow four exam-style problems that switch between interest and depreciation.

Worked examples · 4 problems, reveal step by step

PROBLEM 1 · SIMPLE VS COMPOUND LOAN COMPARISON

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?

Lender A, simple interest
$P = \$18{,}000$, $r = 0.075$, $n = 3$
$I = Prn = \$18{,}000 \times 0.075 \times 3 = \$4{,}050.00$
$\therefore$ Total repaid (A) $= \$18{,}000 + \$4{,}050 = \$22{,}050.00$

Simple interest: flat rate applied to the original principal for 3 years.

PROBLEM 2 · COMBINED INVESTMENT AND DEPRECIATION

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?

Part (a), machine depreciation
$V_0 = \$28{,}000$, $r = 0.15$, $(1-r) = 0.85$, $n = 4$
$(0.85)^4 = 0.52201\ldots$
$\therefore S = \$28{,}000 \times 0.52201 = \$14{,}616.18$

Apply declining balance formula. The machine has lost almost half its value in 4 years at 15% per annum.

PROBLEM 3 · FINDING THE DEPRECIATION RATE

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.

Set up the equation
$\$29{,}160 = \$54{,}000 \times (1-r)^4$

Substitute the known values, $S = \$29{,}160$, $V_0 = \$54{,}000$, $n = 4$. Solve for $r$.

PROBLEM 4 · TWO COMPOUND INVESTMENT PRODUCTS

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?

Product A, annual compounding
$A_A = \$12{,}000(1.051)^4$
$(1.051)^4 = 1.220143\ldots$
$A_A = \$12{,}000 \times 1.220143 = \$14{,}641.72$

Annual compounding uses the annual rate directly, with $n = 4$ periods.

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 $\$$.

Avoid the mixed-question traps

Fix the compounding-period and units slips that cost marks under exam pressure.

Common errors · the 3 traps that cost marks

Trap 01

Not adjusting $r$ and $n$ for non-annual compounding

"4.8% per annum compounding quarterly" means $r = 0.048/4 = 0.012$ per quarter and $n$ in quarters. Using the annual rate and annual periods gives the wrong answer.

Trap 02

Rounding $(1 \pm r)^n$ too early

Do not round $(1.021)^8$ to 2 decimal places before multiplying by $P$. Rounding early introduces large errors in the final dollar amount. Keep all calculator decimal places until the final answer.

Trap 03

Giving a comparison without a conclusion

Calculate both values and then stop, without stating which is better, drops the final mark. Always write: "Option X is [better/cheaper] by $\$$[amount]."

Match each problem type to its recommended approach:

loan comparison
investment choice
unknown depreciation rate
mixed asset vs investment

write model, isolate power, solve for r
find depreciated value and invested value separately
calculate both totals, compare borrower cost
find final amount for each, identify larger return

Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

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.

Revisit your thinking

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?

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Practise mixed questions

Answer the question bank, then write full short-answer responses.

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 33 marks

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)

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ApplyBand 33 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)

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AnalyseBand 44 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)

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📖 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].

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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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← Lesson 13 · Depreciation Module 4 · Lesson 1 →

Module overview · Maths Standard