How does money grow or shrink over time, and how are loans and investments modelled?
Apply simple and compound interest, depreciation, and recurrence relations to financial situations.
Simple and compound interest, depreciation, effective rates, and recurrence relations for loans and investments in TCE Mathematics Applications.
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What this dot point is asking
Financial mathematics in Unit 3 is about putting a dollar value on time. Money invested earns interest; assets lose value through depreciation; loans accumulate balances. You must move fluently between formulas, recurrence relations, and worded problems.
Here is the principal (starting amount), is the interest or depreciation rate written as a decimal per period, and is the number of periods. The single most common slip is mismatching and : if a rate is "6% per annum compounding monthly", then the per-period rate is and counts months, not years.
Simple versus compound
Simple interest grows the balance by the same dollar amount each period, so a graph of value against time is a straight line. Compound interest pays interest on previously earned interest, so the balance grows by an increasing dollar amount each period and the graph curves upward (exponentially).
Recurrence relations
A recurrence relation defines each term from the previous one. This is the natural language of a spreadsheet or financial calculator.
For compound interest at rate per period:
For reducing-balance depreciation:
For a loan with fixed repayment each period:
The loan recurrence captures both ideas at once: interest is added, then the repayment is subtracted. When repayments exceed the interest charged, the balance falls toward zero.
Effective annual rate
To compare investments with different compounding frequencies, convert the nominal rate to an effective annual rate:
where is the number of compounds per year. For the monthly example, , or about . This is why monthly compounding beats annual at the same nominal rate.
When checking your work, ask whether the answer is sensible: a compound investment must exceed the matching simple-interest result, and a depreciating asset must always be worth less than it cost. If not, recheck your and .
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 TASC General Mathematics7 marksQuestions about finance models. a) Choose the most appropriate model from a list for: i. a person repaying a loan in fortnightly instalments where interest compounds monthly; ii. the book value of a farm machine decreasing by 15% of the cost price each year; iii. interest on a credit card calculated daily where each day's interest affects the next. b) 6% p.a. compounding monthly is equivalent to an effective rate of 6.17% p.a. Explain. c) Jackie's antique watch grows in value at 2.5% each year while inflation averages 3% p.a. Explain whether this is a good investment.Show worked answer →
a) (i, 1 mark; ii and iii, 2 marks) i. Reducing-balance loan (annuities / present value of annuity). ii. Straight-line depreciation, since a fixed percentage of the original cost price (not the book value) is lost each year. iii. Compound interest, because interest is added to the balance and earns further interest.
b) (2 marks) Compounding monthly means interest is added 12 times a year, and the interest already added itself earns interest. So (1 + 0.06/12)^12 = 1.0617, giving an effective annual rate of 6.17%, slightly above the 6% nominal rate. The extra 0.17% is the effect of compounding.
c) (2 marks) The watch grows at 2.5% but inflation is 3%, so its value rises more slowly than prices. In real (inflation-adjusted) terms it is losing buying power, so it is not a good investment. Note the watch's dollar value still increases, it just does not keep pace with inflation.
2024 TASC General Mathematics8 marksA business owner buys a ute for 800. a) Using the Unit Cost Depreciation method: i. Find R, the depreciation per km, to three decimal places. ii. Find the Book Value after 100000 km. For straight line depreciation V = -3214n + 45000; for reducing balance V = 45000(1 - 0.12)^n. c) A tax specialist advises using the method predicting more depreciation. Which method should they use if they keep the car for: i. 5 years; ii. 12 years?Show worked answer →
a) i. (1 mark) Unit cost rate R = (cost - scrap)/total units = (45000 - 800)/320000 = 44200/320000 = 0.138 per km (3 dp).
a) ii. (1 mark) Book Value = 45000 - R x km = 45000 - 0.138 x 100000 = 45000 - 13800 = $31200.
c) i. (about 1 mark) For 5 years compare the two equations. Straight line gives V = -3214(5) + 45000 = 23748. Reducing balance predicts the lower value, so it predicts more depreciation, choose reducing balance.
c) ii. (about 1 mark) For 12 years straight line gives V = -3214(12) + 45000 = 9705. Now straight line predicts the lower value, so choose straight line for the larger depreciation. The crossover happens because reducing balance falls fast early then flattens.
2024 TASC General Mathematics5 marksCasey pays 25.00 fee. Interest is 20.99% p.a. compounding daily. a) How much will Casey pay if they pay the total owing before the due date? b) How much will Casey pay in total if they neglect to make the first three payment dates, paying interest for 70 days when they finally pay?Show worked answer →
a) (1 mark) Paying before the due date means it is still interest free, so Casey pays exactly $2500.
b) (3-4 marks) Missing three payment dates adds three 75, which is added to the 2680.73 (approximately). So Casey pays about 2500.