Mathematics Standard • Year 12 • Module 7 • Lesson 2
Compound Interest — Past-Paper Style
Practise HSC-style short answers and one structured extended response on compound interest with full marking criteria.
1. Short-answer questions
1.1 $12,000 is invested at 5.4% p.a. compounded quarterly. Calculate the value of the investment after 6 years, to the nearest cent. 3 marks Band 3
1.2 A bank offers two savings products:
Product X: 5.8% p.a. compounded monthly.
Product Y: 5.85% p.a. compounded annually.
(a) Calculate the effective annual rate of Product X to 2 decimal places of a percent.
(b) State which product is the better choice for a saver, with a one-line reason. 3 marks Band 3-4
1.3 $8,000 is invested at 6% p.a. compounded monthly. After how many complete years will the investment first exceed $12,000?
Show working (you may use a trial-and-error / list-the-values approach without using a calculus log technique). 4 marks Band 4
2. Extended response
2.1 Hana receives $15,000 from her grandparents on her 18th birthday and wants to invest it for 10 years. She is comparing three options.
Option A: 5.6% p.a. simple interest for 10 years.
Option B: 5.6% p.a. compounded annually for 10 years.
Option C: 5.4% p.a. compounded monthly for 10 years.
(a) Calculate the final value of Option A.
(b) Calculate the final value of Option B.
(c) Calculate the final value of Option C and its effective annual rate (to 2 d.p. of a percent).
(d) Rank the three options from best to worst for Hana, and write a one-sentence justification that names the winner and the dollar gap to the next-best option. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct simple-interest final value using A = P(1 + rn).
Part (b) — 1 mark
• 1 mark — correct compound-annual final value using A = P(1 + r)^n.
Part (c) — 3 marks
• 1 mark — correct r/k = 0.0045 and kn = 120 stated or used.
• 1 mark — correct final value using A = P(1 + r/k)^(kn).
• 1 mark — correct effective annual rate to 2 d.p. of a percent.
Part (d) — 2 marks
• 1 mark — correct ranking of all three options.
• 1 mark — clear conclusion sentence naming the best option and the dollar gap to the next-best.
Your response:
Stuck on (d)? Once the three final amounts are calculated, sort them in descending order and write a sentence with "Option ___ pays the most, by $___ over Option ___".How did this worksheet feel?
What I'll revisit before next class:
1.1 — $12,000 at 5.4% quarterly for 6 years (3 marks)
Sample response.
r/k = 0.054 / 4 = 0.0135.
kn = 4 × 6 = 24.
A = 12,000 × (1.0135)^24 = 12,000 × 1.380429 = $16,565.15.
Marking notes. 1 mark — correct r/k and kn. 1 mark — correct substitution into A = P(1 + r/k)^(kn). 1 mark — correct final answer to the nearest cent. A bare $16,565.15 with no working scores 1/3.
1.2 — Effective rate comparison (3 marks)
(a) Sample response. Effective_X = (1 + 0.058/12)^12 − 1 = (1.004833)^12 − 1 = 1.05957 − 1 = 0.05957 = 5.96% p.a.
(b) Sample response. Since 5.96% (Product X effective) > 5.85% (Product Y), Product X is better for a saver, earning a higher effective annual return.
Marking notes. (a) 1 mark — correct r/k and kn substituted. 1 mark — correct effective rate to 2 d.p. (b) 1 mark — correct comparison sentence identifying Product X as better with a numerical reason.
1.3 — When does $8,000 first exceed $12,000 at 6% monthly? (4 marks)
Sample response.
Use A = 8,000 × (1.005)^(12n) and test integer years:
n = 6: A = 8,000 × (1.005)^72 = 8,000 × 1.43204 = $11,456.30 (under $12,000)
n = 7: A = 8,000 × (1.005)^84 = 8,000 × 1.52037 = $12,162.96 (over $12,000) ✓
Therefore the investment first exceeds $12,000 after 7 complete years.
Marking notes. 1 mark — correct formula stated. 1 mark — value at n = 6 (under target). 1 mark — value at n = 7 (over target). 1 mark — explicit conclusion: 7 years. Answers giving 6 years (rounding down) lose the conclusion mark — the wording is "first exceed".
2.1 — Three investment options for $15,000 over 10 years (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Option A (simple interest 5.6%).
A_A = 15,000 × (1 + 0.056 × 10) = 15,000 × 1.56 = $23,400.00. [1 mark — correct simple-interest final value.]
(b) Option B (compound annual 5.6%).
A_B = 15,000 × (1.056)^10 = 15,000 × 1.722571 = $25,838.56. [1 mark — correct compound-annual value.]
(c) Option C (compound monthly 5.4%).
r/k = 0.054/12 = 0.0045. kn = 12 × 10 = 120. [1 mark — correct r/k and kn.]
A_C = 15,000 × (1.0045)^120 = 15,000 × 1.713590 = $25,703.85. [1 mark — correct final value.]
Effective rate = (1.0045)^12 − 1 = 1.05536 − 1 = 0.05536 = 5.54% p.a. [1 mark — correct effective rate.]
(d) Ranking and recommendation.
Best → worst: B ($25,838.56) > C ($25,703.85) > A ($23,400.00). [1 mark — correct ranking.]
Conclusion: Hana should choose Option B (5.6% p.a. compounded annually), which gives $25,838.56 — $134.71 more than Option C and $2,438.56 more than Option A over the 10 years. [1 mark — clear conclusion sentence naming Option B and the dollar gap to the next-best option.]
Total: 7/7.
Band descriptors for marker.
Band 3: Options A and B correctly evaluated, Option C attempted but with wrong r/k or wrong kn, no effective rate, no ranking. ≈ 3 marks.
Band 4: All three numerical values correct including Option C, but effective rate omitted or wrong; no ranking sentence. ≈ 5 marks.
Band 5: Full numerical solution with correct effective rate, ranking listed but conclusion sentence is bare (e.g. just "B is best") without naming the dollar gap. ≈ 6 marks.
Band 6: Complete: three correct final values, correct effective rate, ranking, AND a clear conclusion sentence that names Option B and the dollar gap to the next-best option. 7/7.