Mathematics Advanced • Year 12 • Module 7 • Lesson 1

Introduction to Financial Mathematics

Build fluency with simple and compound interest calculations and rate-period matching.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the formulas:

Simple interest total amount: A = ____________________

Compound interest total amount: A = ____________________

Q1.2 An account pays 7.2% p.a. compounded monthly. State the periodic rate r and the number of periods n for a 5-year term.

r = ____________    n = ____________

Q1.3 In one sentence, state the critical rule that links the rate r and the number of periods n in any compound-interest calculation.

Stuck? Revisit lesson § Formula Reference and § Compounding Periods.

2. Worked example, Maya's monthly-compounded investment

Follow every line. Each step has a short reason.

Problem. Maya invests $8,000 in an account paying 6% p.a. compounded monthly. Find the balance after 4 years.

Step 1, Convert the annual rate to a periodic (monthly) rate.

r = 0.06 ÷ 12 = 0.005

Reason: monthly compounding means the rate per period is the annual rate divided by 12.

Step 2, Convert years to total compounding periods.

n = 4 × 12 = 48 periods

Reason: r and n must share a time unit; both are now monthly.

Step 3, Substitute into A = P(1 + r)ⁿ.

A = 8,000 × (1.005)⁴⁸

A = 8,000 × 1.270489

Step 4, Evaluate and round to the nearest cent.

A ≈ $10,163.91

Reason: money answers in HSC questions are quoted to the nearest cent.

Conclusion. Maya's balance after 4 years is $10,163.91, an increase of $2,163.91 in interest.

3. Faded example, fill in the missing steps

Tom borrows $12,000 at 5.4% p.a. simple interest for 3.5 years. Fill in each blank line. 4 marks

Step 1, Identify the values:

P = $______________,   r = ______________ (as a decimal),   n = ______________ years

Step 2, Substitute into A = P(1 + rn):

A = 12,000 × (1 + ______ × ______)

Step 3, Evaluate inside the bracket:

A = 12,000 × ______________

Step 4, Compute the final amount:

A = $______________

Conclusion. Tom must repay $______________, of which $______________ is interest.

Stuck? Revisit lesson § Simple Interest, Example.

4. Graduated practice, compute the final amount

Show the substitution and the final amount (to nearest cent) for each. Assume compounding is annual unless stated. Use the same time units for r and n.

Foundation, single-step substitution (4 questions)

QScenarioWorking & final amount A
4.1 1P = $2,000, r = 5% p.a. simple, n = 4 years
4.2 1P = $2,000, r = 5% p.a. compound, n = 4 years
4.3 1P = $10,000, r = 3% p.a. compound, n = 10 years
4.4 1P = $500, r = 4% p.a. simple, n = 6 months (convert n first)

Standard, typical HSC difficulty (6 questions)

Show working in the space below each part, at least one substitution line and one evaluation line.

4.5 $15,000 is invested at 7.2% p.a. compounded monthly for 3 years. Find the final balance and the total interest.    2 marks

4.6 $6,500 is invested at 4.8% p.a. for 4 years. Compare the final amounts under (a) simple interest and (b) compound interest (compounded annually).    2 marks

4.7 A term deposit of $25,000 is offered at 5.4% p.a. compounded quarterly. Find the balance after 5 years.    2 marks

4.8 $4,000 grows at 6% p.a. compounded semi-annually for 7 years. Find the final amount.    2 marks

4.9 Sara puts $1,200 in an account paying 3.65% p.a. compounded daily. Find the balance after 2 years. (Use 365 days per year.)    2 marks

4.10 $9,000 is invested for 6 years at 5.5% p.a. compounded annually. State (a) the final amount, (b) the total interest earned, and (c) the percentage growth over the term.    2 marks

Extension, combine concepts (2 questions)

4.11 Two products both advertise "6% p.a." Product X: simple interest. Product Y: compound interest compounded annually. By how many dollars do they differ on a $20,000 deposit after 15 years? Show both calculations and the difference.    3 marks

4.12 A loan of $50,000 is taken at 4.8% p.a. compounded monthly. Without finding the final amount, explain in 2-3 lines why the effective annual cost is higher than 4.8% and use the formula (1 + r/n)ⁿ − 1 to compute the effective rate.    3 marks

Stuck on 4.12? The formula appears in Lesson 3, but you have all the pieces from rate-period matching here.

5. Self-check the easy 3

Tick the first three once you have checked the method works.

How did this worksheet feel?

What I'll revisit before next class:

Answers, Do not peek before attempting

Q1.1, Formulas

Simple: A = P(1 + rn).   Compound: A = P(1 + r)ⁿ.

Q1.2, Monthly rate and periods

r = 0.072 ÷ 12 = 0.006.   n = 5 × 12 = 60.

Q1.3, Critical rule

The rate r and the number of periods n must use the same time unit. If interest compounds monthly, divide the annual rate by 12 and multiply the number of years by 12.

Q3, Faded example: Tom's loan

P = $12,000; r = 0.054; n = 3.5 years.
A = 12,000 × (1 + 0.054 × 3.5) = 12,000 × (1 + 0.189) = 12,000 × 1.189 = $14,268.00. Interest = $2,268.00.

Q4.1, Simple, P = 2,000, r = 5%, n = 4

A = 2,000(1 + 0.05 × 4) = 2,000 × 1.20 = $2,400.00. Interest = $400.

Q4.2, Compound, P = 2,000, r = 5%, n = 4

A = 2,000(1.05)⁴ = 2,000 × 1.21550625 = $2,431.01. Interest = $431.01 (about $31 more than simple).

Q4.3, Compound, P = 10,000, r = 3%, n = 10

A = 10,000(1.03)¹⁰ = 10,000 × 1.343916 = $13,439.16.

Q4.4, Simple, n = 6 months

Convert n to years: n = 0.5. A = 500(1 + 0.04 × 0.5) = 500 × 1.02 = $510.00.

Q4.5, $15,000 at 7.2% p.a. monthly for 3 years

r = 0.072/12 = 0.006; n = 36. A = 15,000(1.006)³⁶ = 15,000 × 1.239915 = $18,598.73. Total interest = $3,598.73.

Q4.6, $6,500 at 4.8% p.a. for 4 years

(a) Simple: A = 6,500(1 + 0.048 × 4) = 6,500 × 1.192 = $7,748.00.
(b) Compound: A = 6,500(1.048)⁴ = 6,500 × 1.20996 = $7,940.47. Compound exceeds simple by $192.47.

Q4.7, $25,000 at 5.4% p.a. quarterly for 5 years

r = 0.054/4 = 0.0135; n = 20. A = 25,000(1.0135)²⁰ = 25,000 × 1.307985 = $32,699.63.

Q4.8, $4,000 at 6% p.a. semi-annually for 7 years

r = 0.06/2 = 0.03; n = 14. A = 4,000(1.03)¹⁴ = 4,000 × 1.512590 = $6,050.36.

Q4.9, $1,200 at 3.65% p.a. daily for 2 years

r = 0.0365/365 = 0.0001; n = 730. A = 1,200(1.0001)⁷³⁰ = 1,200 × 1.075660 = $1,290.79.

Q4.10, $9,000 at 5.5% p.a. for 6 years

(a) A = 9,000(1.055)⁶ = 9,000 × 1.378843 = $12,409.59.
(b) Interest = 12,409.59 − 9,000 = $3,409.59.
(c) Percentage growth = 3,409.59 ÷ 9,000 × 100 = 37.88% over 6 years.

Q4.11, Simple vs compound at 6% for 15 years on $20,000

Simple: A = 20,000(1 + 0.06 × 15) = 20,000 × 1.90 = $38,000.00.
Compound: A = 20,000(1.06)¹⁵ = 20,000 × 2.396558 = $47,931.16.
Difference = 47,931.16 − 38,000 = $9,931.16. Compound earns nearly $10,000 more because each year's interest itself earns interest.

Q4.12, Effective rate on a 4.8% monthly loan

The advertised rate 4.8% p.a. is split into 12 monthly slices of 0.4%. Each slice is added to the balance and then earns interest the following month, so the actual annual cost exceeds 4.8%. Using the effective-rate formula:
reff = (1 + 0.048/12)¹² − 1 = (1.004)¹² − 1 = 1.04907 − 1 = 4.907% p.a.
The borrower effectively pays about 0.11 percentage points more than the advertised nominal rate.