Financial Mathematics - TCE Mathematics Applications (Tasmania)

Syllabus dot point

“Apply simple and compound interest, depreciation, and recurrence relations to financial situations.”

TCEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

You need to model how a single sum of money changes over time using simple interest (linear growth), compound interest (geometric growth), and depreciation (geometric decay), and express each as a recurrence relation.

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.

Core formulas

Simple interest: $I = Prn$ , so the value after $n$ periods is $A = P(1 + rn)$ .

Compound interest: $A = P(1 + r)^n$ , where $r$ is the rate per compounding period.

Flat-rate (straight-line) depreciation: $A = P(1 - rn)$ .

Reducing-balance depreciation: $A = P(1 - r)^n$ .

Here $P$ is the principal (starting amount), $r$ is the interest or depreciation rate written as a decimal per period, and $n$ is the number of periods. The single most common slip is mismatching $r$ and $n$ : if a rate is "6% per annum compounding monthly", then the per-period rate is $r = 0.06/12 = 0.005$ and $n$ counts months, not years.

Compounding period

The interval at which interest is calculated and added to the balance. Common periods are annually, semi-annually (twice a year), quarterly (4), monthly (12), and weekly (52).

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).

Compound interest over different compounding periods

We invest $\$8000$ at a nominal $6\%$ per annum for $3$ years and compare the final balance when interest is compounded annually versus monthly.

Step 1: Identify the formula and variables

The compound interest formula is $A = P(1 + r)^n$ , where $r$ is the rate per compounding period and $n$ is the number of periods. The key is that $r$ and $n$ must use the same time unit.

Step 2: Compute the balance compounded annually

Compounding once per year means the full annual rate applies each period, so $r = 0.06$ and $n = 3$ :

A = 8000(1.06)^3 = 8000 \times 1.191016 = \$9528.13

Step 3: Compute the balance compounded monthly

Compounding monthly means 12 periods per year. We divide the annual rate by 12 to get the monthly rate and multiply the years by 12 to count months: $r = 0.06/12 = 0.005$ and $n = 3 \times 12 = 36$ :

A = 8000(1.005)^{36} = 8000 \times 1.196680 = \$9573.44

Step 4: Compare the results

Monthly compounding earns $\$45.31$ more than annual compounding ( $\$9573.44$ versus $\$9528.13$ ) because interest is added more frequently, so earlier interest payments themselves earn interest sooner.

Final answers: Compounded annually: $\$9528.13$ ; compounded monthly: $\$9573.44$ .

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 $r$ per period:

A_{n+1} = A_n(1 + r), \quad A_0 = P

For reducing-balance depreciation:

A_{n+1} = A_n(1 - r), \quad A_0 = P

For a loan with fixed repayment $d$ each period:

A_{n+1} = A_n(1 + r) - d, \quad A_0 = \text{amount borrowed}

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.

Reducing-balance depreciation

A car bought for $\$32\,000$ depreciates at $15\%$ per year using the reducing-balance method. We need to find its value after $4$ years.

Step 1: Choose the correct formula

Reducing-balance depreciation keeps a percentage of the remaining value each period, so the value shrinks by a fixed factor rather than a fixed dollar amount. The formula is $A = P(1 - r)^n$ , where $P$ is the purchase price, $r$ is the annual depreciation rate as a decimal, and $n$ is the number of years.

Step 2: Substitute the known values

Here $P = 32000$ , $r = 0.15$ (because $15\% = 0.15$ ), and $n = 4$ :

A = 32000(1 - 0.15)^4 = 32000(0.85)^4

Step 3: Evaluate

A = 32000 \times 0.52200625 = \$16\,704.20

The car retains only $0.85^4 \approx 52.2\%$ of its original value, reflecting how each year's loss is a smaller dollar amount than the year before.

Final answer: After 4 years the car is worth approximately $\$16\,704.20$ , a loss of $\$15\,295.80$ .

Effective annual rate

To compare investments with different compounding frequencies, convert the nominal rate to an effective annual rate:

r_{\text{eff}} = \left(1 + \frac{r_{\text{nom}}}{k}\right)^k - 1

where

k

is the number of compounds per year. For the

6\%

monthly example,

r_{\text{eff}} = (1.005)^{12} - 1 = 0.061678

, or about

6.17\%

. This is why monthly compounding beats annual at the same nominal rate.

Reading the question

Underline the words "per annum", "per month", "compounding", and "flat rate" before you start. They tell you which formula to use and how to set $r$ and $n$ .

Mixing up the rate and the period

Students often use $r = 0.06$ with $n = 36$ for a 3-year monthly-compounding problem. That treats each month as if it earned a full year of interest, giving a wildly inflated answer. Always divide the annual rate by the number of compounds per year, and count $n$ in the same units.

Choosing the right depreciation model

TASC problems often ask you to compare straight-line (flat-rate) and reducing-balance depreciation, or to choose a unit-cost model. Straight-line subtracts a fixed dollar amount each year, so the book value falls along a straight line and can in principle reach zero. Reducing-balance multiplies by a fixed factor less than one each year, so it falls quickly at first then flattens and never quite reaches zero. Unit-cost depreciation links the loss to usage: the rate per unit is $R = \dfrac{\text{cost} - \text{scrap value}}{\text{total units of life}}$ , and the book value after $u$ units is $V = \text{cost} - Ru$ .

Comparing straight-line and reducing-balance

A machine costs $\$20\,000$ . Compare straight-line at $\$2500$ per year with reducing-balance at $15\%$ per year over $3$ years.

Straight-line

$V = 20000 - 2500 \times 3 = 20000 - 7500 = \$12\,500$ .

Reducing-balance

$V = 20000(1 - 0.15)^3 = 20000 \times 0.85^3 = 20000 \times 0.614125 = \$12\,282.50$ .

Reducing-balance gives the lower value here because the early years lose more. Over a long enough horizon the straight-line value drops below the reducing-balance value, which is why the two methods cross.

Real versus nominal growth

An asset that grows at a rate below inflation is losing purchasing power even though its dollar value rises. Compare the growth rate with the inflation rate: if the growth rate exceeds inflation the investment gains in real terms, if it is below inflation it loses real value, and if it equals inflation it merely holds its real worth. This is why TASC finance questions ask you to judge an investment against an inflation figure rather than just checking that the dollar amount went up.

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 $r$ and $n$ .

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks

a) Choose the most appropriate model 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 watch grows at

2.5\%

each year while inflation averages

3\%

p.a. Is this a good investment?

Show worked answer →

a) (i, 1 mark; ii and iii, 2 marks) i. Reducing-balance loan (present value of an 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 $\left(1 + \frac{0.06}{12}\right)^{12} = 1.0617$ , giving an effective annual rate of $6.17\%$ , slightly above the $6\%$ nominal rate. The extra $0.17\%$ is the compounding effect.

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 loses buying power, so it is not a good investment. Its dollar value still rises, it just does not keep pace with inflation.

Practice8 marks

A business owner buys a ute for

\$45\,000

, expected to last

320\,000

km with a scrap value of

\$800

. a) Using unit-cost depreciation: i. Find

R

, the depreciation per km (3 dp). ii. Find the book value after

100\,000

km. For straight line

V = 45\,000 - 3214n

; for reducing balance

V = 45\,000(1 - 0.12)^n

. c) A tax specialist wants the method predicting more depreciation. Which method if they keep the car for i. 5 years; ii. 12 years?

Show worked answer →

a) i. (1 mark) Unit-cost rate $R = \frac{\text{cost} - \text{scrap}}{\text{total units}} = \frac{45000 - 800}{320000} = \frac{44200}{320000} = 0.138$ per km (3 dp).

a) ii. (1 mark) Book value $= 45000 - R \times \text{km} = 45000 - 0.138 \times 100000 = 45000 - 13800 = \$31\,200$ .

c) i. (1 mark) For 5 years: straight line $V = 45000 - 3214(5) = \$28\,930$ ; reducing balance $V = 45000(0.88)^5 = \$23\,748$ . Reducing balance predicts the lower value, so it predicts more depreciation; choose reducing balance.

c) ii. (1 mark) For 12 years: straight line $V = 45000 - 3214(12) = \$6432$ ; reducing balance $V = 45000(0.88)^{12} = \$9705$ . Now straight line predicts the lower value, so choose straight line. The crossover happens because reducing balance falls fast early then flattens.

Practice5 marks

Casey pays

\$2500

for a holiday on a credit card. Purchases are interest free until the first payment is due; a missed payment incurs a

\$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 in total if they miss 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) (4 marks) Missing three payment dates adds three $\$25$ late fees $= \$75$ , added to the $\$2500$ owing before interest is charged. Apply daily compound interest for 70 days at a daily rate of $\frac{0.2099}{365}$ : amount $= 2575 \times \left(1 + \frac{0.2099}{365}\right)^{70} = 2575 \times 1.04106 = \$2680.73$ (approximately). So Casey pays about $\$2681$ , well above the original $\$2500$ .

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC