Mathematics Advanced • Year 12 • Module 7 • Lesson 1
Introduction to Financial Mathematics
Build fluency with simple and compound interest calculations and rate-period matching.
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.
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.
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)
| Q | Scenario | Working & final amount A |
|---|---|---|
| 4.1 1 | P = $2,000, r = 5% p.a. simple, n = 4 years | |
| 4.2 1 | P = $2,000, r = 5% p.a. compound, n = 4 years | |
| 4.3 1 | P = $10,000, r = 3% p.a. compound, n = 10 years | |
| 4.4 1 | P = $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
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:
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.