Mathematics Advanced • Year 12 • Module 7 • Lesson 2
Compound Interest in Depth
Practise HSC-style writing on transposing A = P(1 + r)ⁿ — including a structured extended response with marking criteria.
1. Short-answer questions
1.1 An investment of $9,000 grows to $14,500 in 7 years under annual compound interest. Find the annual rate, to two decimal places, showing the transposed formula and substitution. 3 marks Band 3
1.2 A family wants $80,000 in 10 years. The available account pays 5.5% p.a. compounded annually. Find the single lump sum required today, to the nearest cent. 2 marks Band 3-4
1.3 $6,000 is invested at 4.5% p.a. compound interest. (a) Use logarithms to find the exact time, in years (to 2 dp), for the investment to double. (b) Verify with the Rule of 72 and state the percentage error. (c) Without further calculation, explain why your Rule-of-72 error is small. 4 marks Band 4
Stuck on 1.3(c)? Reference the Rule-of-72 derivation: ln 2 ≈ 0.72 and ln(1 + r) ≈ r for small r.2. Extended response
2.1 A house purchased in 2018 for $850,000 is sold in 2026 for $1,200,000. Throughout, assume annual compound growth at a fixed rate r.
(a) Find the annual growth rate r, to two decimal places.
(b) Project the value of the house at the end of 2030 (12 years from purchase) using this rate. Show the formula used.
(c) Using logarithms, find the year (from 2018) in which the house first exceeds $1,500,000 at the same rate. Give your answer as a year number from 2018 and as a calendar year.
(d) The seller in 2030 argues that "houses always grow at this rate." Critique that claim in 2-3 sentences using one mathematical and one real-world point. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — writes r = (A/P)1/n − 1 and substitutes the correct three values.
• 1 mark — evaluates to r ≈ 4.39% (4.40% acceptable).
Part (b) — 2 marks
• 1 mark — uses A = 850,000(1.0439)¹² with the rate from (a).
• 1 mark — final figure ≈ $1,422,860 (any answer in the band $1,420,000-$1,425,000 acceptable).
Part (c) — 2 marks
• 1 mark — writes n = ln(1,500,000/850,000) / ln(1.0439) and evaluates to ≈ 13.22 years.
• 1 mark — interprets correctly as "first exceeds in year 14 from 2018 = 2032".
Part (d) — 2 marks
• 1 mark — mathematical point (e.g., extrapolating a constant rate forever is the Rule-of-72-style assumption — produces unbounded exponential growth).
• 1 mark — real-world point (interest rates, demographic change, recession risk, council restrictions, etc.).
Your response:
Stuck on (d)? Pair "exponential growth never stops in the formula" with one real-world brake on house prices.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rate for $9,000 → $14,500 in 7 yr (3 marks)
Sample response. Transpose: r = (A/P)1/n − 1. Substitute: r = (14,500/9,000)1/7 − 1 = (1.6111)0.1429 − 1 = 1.07016 − 1 = 0.0702. Hence r ≈ 7.02% p.a.
Marking notes. 1 mark — correct transposition stated explicitly. 1 mark — correct substitution (especially the 1/n exponent). 1 mark — evaluation to 2 dp as a percentage. Common errors: subtracting 1 before taking the root, or writing the answer as 7.0% (1 dp) loses 0.5.
1.2 — Lump sum to reach $80,000 in 10 yr at 5.5% (2 marks)
Sample response. P = A / (1 + r)ⁿ = 80,000 / (1.055)¹⁰ = 80,000 / 1.708144 = $46,834.46.
Marking notes. 1 mark — correct formula with substitution. 1 mark — correct cent-level answer. A common error is computing 80,000 × 1.055⁻¹⁰ on the calculator without writing the formula — accept the final answer but deduct 0.5 for missing working.
1.3 — Doubling time at 4.5% (4 marks)
(a) Sample response. n = ln 2 / ln 1.045 = 0.6931 / 0.04402 = 15.75 years.
(b) Sample response. Rule of 72: 72 ÷ 4.5 = 16 years. Error = |16 − 15.75| / 15.75 × 100 ≈ 1.59%.
(c) Sample response. The Rule of 72 is most accurate around 8% because it uses 72 instead of 100 · ln 2 ≈ 69.3 (an over-estimate) and substitutes ln(1 + r) by r (an under-estimate); the two errors cancel near r ≈ 8% and are still small at 4.5% — within about 2%.
Marking notes. (a) 1 mark for the formula, 1 mark for the correct numeric answer. (b) 1 mark for the estimate and the percentage error. (c) 1 mark for citing both compensating approximations or for citing accuracy "between 4% and 12%" (acceptable from the lesson's wording).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Annual growth rate. Use r = (A/P)1/n − 1 with P = 850,000, A = 1,200,000, n = 8.
r = (1,200,000 / 850,000)1/8 − 1 = (1.4118)0.125 − 1 = 1.04393 − 1 ≈ 4.39% p.a. [2 marks — formula + numeric value.]
(b) Projected value at end of 2030. n = 12 years from purchase. A = 850,000(1.0439)¹² = 850,000 × 1.67395 ≈ $1,422,860.91. [2 marks — formula + numeric value.]
(c) First year exceeding $1,500,000. n = ln(1,500,000 / 850,000) / ln(1.0439) = ln 1.7647 / ln 1.0439 = 0.5681 / 0.04298 = 13.22 years. Since the house exceeds $1,500,000 part-way through year 14 from purchase, the first full year above $1.5 m is year 14 from 2018 = 2032. [2 marks — logarithm calculation + correct calendar year interpretation.]
(d) Critique. Mathematical: The model A = 850,000(1.0439)ⁿ grows without bound, so the seller's claim is the assumption that the historical 8-year compound rate continues forever — which the formula can compute but real markets cannot sustain (at 4.39% for 70 years, the house would be worth $17 million in 2088). Real-world: Australian house prices depend on interest rates, immigration, planning rules and recession cycles — none captured in a single-rate compound model; the 4.39% is a backward-looking average, not a forward-looking guarantee. [2 marks — one mathematical + one real-world point.]
Total: 8/8.
Band descriptors for marker.
Band 3: Calculations attempted, rate roughly correct, projection attempted but no logarithm step or no critique. ≈ 3-4 marks.
Band 4: (a), (b), and a numerical answer to (c) without interpreting "first year exceeds" correctly, weak critique. ≈ 5-6 marks.
Band 5: All four parts present, calendar-year answer correct in (c), critique gives one good point. ≈ 6-7 marks.
Band 6: Full working with formulas explicit at each step, calendar-year interpretation in (c), and critique pairing one mathematical assumption with one real-world brake on growth. 8/8.