Mathematics Advanced • Year 12 • Module 7 • Lesson 1
Introduction to Financial Mathematics
Practise HSC-style writing on simple and compound interest, including a structured extended response.
1. Short-answer questions
1.1 Calculate the total amount and the total interest earned when $6,500 is invested for 4 years at 4.8% p.a. (a) simple interest and (b) compounded annually. Give answers to the nearest cent. 3 marks Band 3
1.2 $15,000 is invested at 7.2% p.a. compounded monthly for 3 years. Calculate, to the nearest cent, (a) the final balance and (b) the total interest earned. Show the periodic rate and the number of periods. 3 marks Band 3-4
1.3 Two products are advertised at the same headline rate of 6.0% p.a. for a $20,000 investment.
(a) Product P uses simple interest. Product Q uses compound interest, compounded annually.
(b) Find each product's final value after 10 years and the dollar difference.
(c) State, with one line of justification, the smallest whole number of years for which Q exceeds P. 4 marks Band 4
2. Extended response
2.1 An investor has $30,000 to invest for 5 years. Three products are available, all advertised at 6.0% per annum.
Product X: 6.0% p.a. simple interest.
Product Y: 6.0% p.a. compounded annually.
Product Z: 6.0% p.a. compounded monthly.
(a) Calculate the final balance of each product to the nearest cent. Show all working including periodic rate and number of periods for Product Z.
(b) Rank the products from highest to lowest final balance and justify the ranking using the structure of each formula.
(c) Explain in 2-3 sentences why the headline rate "6.0% p.a." is insufficient to compare these products, referencing the relationship between r, n, and total return. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — Product X simple calculation correct: A = 30,000(1 + 0.06 × 5) = $39,000.
• 1 mark — Product Y compound annually correct: A = 30,000(1.06)⁵ to nearest cent.
• 1 mark — Product Z monthly: explicit r = 0.005, n = 60, evaluation to nearest cent.
Part (b) — 2 marks
• 1 mark — correct ranking Z > Y > X.
• 1 mark — justification references that compounding adds interest-on-interest (Y beats X) and more-frequent compounding adds more interest cycles per year (Z beats Y).
Part (c) — 2 marks
• 1 mark — explicitly states that the headline (nominal) rate hides the compounding frequency.
• 1 mark — references the rate-period matching rule (r and n must share a time unit) and/or the effective annual rate concept.
Your response:
Stuck on (c)? Reference the lesson's "Critical Rule" callout: r and n must share the same time unit.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $6,500 at 4.8% for 4 years (3 marks)
Sample response. (a) Simple: A = 6,500(1 + 0.048 × 4) = 6,500 × 1.192 = $7,748.00; interest = $1,248.00. (b) Compound: A = 6,500(1.048)⁴ = 6,500 × 1.20996 = $7,940.47; interest = $1,440.47.
Marking notes. 1 mark — correct simple amount with substitution shown. 1 mark — correct compound amount with substitution shown. 1 mark — both interest figures (subtraction from P). A bald answer of $7,748 and $7,940.47 with no substitution scores 2/3.
1.2 — $15,000 at 7.2% monthly for 3 years (3 marks)
Sample response. Periodic rate r = 0.072/12 = 0.006 per month; number of periods n = 3 × 12 = 36. (a) A = 15,000(1.006)³⁶ = 15,000 × 1.239915 = $18,598.73. (b) Interest = 18,598.73 − 15,000 = $3,598.73.
Marking notes. 1 mark — explicit r = 0.006 and n = 36. 1 mark — correct final balance to nearest cent. 1 mark — correct interest. A common error is using r = 0.072 and n = 36 (mismatched units), which gives a wildly incorrect figure — 0/2 for the calculation portion.
1.3 — Simple vs compound at 6.0% on $20,000 (4 marks)
Sample response. (a)/(b) P (simple, 10 yr): A = 20,000(1 + 0.06 × 10) = 20,000 × 1.60 = $32,000.00. Q (compound annually, 10 yr): A = 20,000(1.06)¹⁰ = 20,000 × 1.790847 = $35,816.95. Difference = Q − P = $3,816.95 in favour of Q. (c) Test successive n: at n = 1, P = $21,200, Q = $21,200 (equal); for any n ≥ 2, (1.06)ⁿ exceeds 1 + 0.06n because the binomial expansion of (1.06)ⁿ has positive higher-order terms. So Q exceeds P for the first time at n = 2 years (Q = $22,472 vs P = $22,400).
Marking notes. 1 mark — correct P final value. 1 mark — correct Q final value. 1 mark — correct dollar difference, in favour of Q. 1 mark — correct first year Q wins (n = 2) with a one-line justification. Common error: students claim "Q always exceeds P from n = 1" — but they are equal at n = 1.
2.1 — Three products at 6.0% (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Final balances after 5 years on $30,000.
Product X (simple). A = 30,000(1 + 0.06 × 5) = 30,000 × 1.30 = $39,000.00. [1 mark]
Product Y (compound annually). A = 30,000(1.06)⁵ = 30,000 × 1.338226 = $40,146.77. [1 mark]
Product Z (compound monthly). r = 0.06/12 = 0.005 per month; n = 5 × 12 = 60. A = 30,000(1.005)⁶⁰ = 30,000 × 1.348850 = $40,465.49. [1 mark]
(b) Ranking. Z > Y > X. [1 mark]
Justification. Y beats X because each year's interest is added to the principal and itself earns interest the next year — the (1 + r)ⁿ factor exceeds 1 + rn for n ≥ 2 by exactly the higher-order binomial terms. Z beats Y because each annual 6% is split into twelve 0.5% slices that compound on each other within the year, producing extra "interest-on-interest" cycles. [1 mark]
(c) Why "6.0% p.a." alone is not enough. The headline (nominal) rate hides the compounding frequency. Without knowing how often interest is added, you cannot compute the actual annual return — Product Z's true effective rate is (1 + 0.06/12)¹² − 1 ≈ 6.17%, not 6.00%. A fair comparison requires either (i) the effective annual rate, or (ii) the rate-period-matched calculation r/n and nt — both of which respect the critical rule that r and n must share the same time unit. [2 marks]
Total: 7/7.
Band descriptors for marker.
Band 3: Two of three balances correct, no ranking justification or only restates rather than explains. ≈ 2-3 marks.
Band 4: All three balances correct, correct ranking but justification is vague ("compound is better"). ≈ 4-5 marks.
Band 5: Correct calculations and ranking with a structural justification (mentions (1+r)ⁿ vs 1+rn or the compounding-frequency effect). Part (c) restates rather than connects to nominal vs effective. ≈ 5-6 marks.
Band 6: Full calculations to the cent, ranking justified by formula structure, and part (c) explicitly references the rate-period matching rule and/or the effective annual rate concept with a numeric illustration. 7/7.