Mathematics Advanced • Year 12 • Module 7 • Lesson 6
Interest Rate Conversions & Multi-Stage Problems
Practise HSC-style writing on multi-stage compound interest, real-vs-nominal returns, and inflation adjustment.
1. Short-answer questions
1.1 $12,000 is invested at 4% p.a. for 3 years, then at 5.5% p.a. for 4 years. Find the final balance and the interest earned over the 7 years. 3 marks Band 3
1.2 The nominal return on an investment is 6.5% p.a. and inflation is 4.2% p.a. Find the exact real annual rate of return, to 4 decimal places, and state the percentage-point difference from the approximation r_real ≈ r_nom − inflation. 3 marks Band 3-4
1.3 $25,000 is placed in a term deposit at 5.2% p.a. for 6 years.
(a) Find the nominal balance after 6 years.
(b) Inflation averages 2.5% p.a. over the period. Find the real value of the balance in today's dollars and hence the real annual rate of return (to 2 dp). 4 marks Band 4
2. Extended response
2.1 Two cousins each receive a $40,000 inheritance and invest for 10 years.
Alex chooses Bank A: a flat 5% p.a. for 10 years.
Bea chooses Bank B: 8% p.a. for the first 2 years (a "bonus" rate), then 4% p.a. for the next 8 years.
Inflation averages 3% p.a. throughout the 10 years.
(a) Calculate both nominal balances and identify the winner in nominal terms.
(b) Calculate both real (today's-dollar) balances and the annualised real rate for each cousin (4 dp).
(c) Explain, in 2–3 sentences with reference to your numbers, why "headline" introductory rates often underperform a slightly lower flat rate, even when the headline rate looks more attractive. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — Alex's balance: A_A = 40,000 × (1.05)¹⁰ correctly evaluated.
• 1 mark — Bea's multi-stage line: A_B = 40,000 × (1.08)² × (1.04)⁸ correctly evaluated.
• 1 mark — explicit comparison and statement of winner (Alex).
Part (b) — 3 marks
• 1 mark — divides each nominal by (1.03)¹⁰ to get real values.
• 1 mark — uses the 10th-root formula to get each annualised real rate.
• 1 mark — both rates rounded correctly to 4 dp.
Part (c) — 2 marks
• 1 mark — references the geometric/factor-multiplication idea (rates do not average for compound interest).
• 1 mark — uses own numerical results to make the comparison concrete.
Your response:
Stuck on (c)? Think about what "8% for 2 years then 4% for 8 years" averages to numerically vs what its geometric mean does to the balance.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Multi-stage compound interest (3 marks)
Sample response. A = 12,000 × (1.04)³ × (1.055)⁴ = 12,000 × 1.12486 × 1.23882 = $16,723.10. Interest = 16,723.10 − 12,000 = $4,723.10.
Marking notes. 1 mark — sets up the multi-stage line correctly (correct exponents on correct rate). 1 mark — numerical evaluation correct to nearest cent. 1 mark — interest = A − P clearly stated.
1.2 — Exact real rate (3 marks)
Sample response. r_real = (1.065 / 1.042) − 1 = 1.02207 − 1 = 0.0221 (i.e. 2.21%). Approximation: 6.5 − 4.2 = 2.30%. The exact value is 0.09 percentage points below the approximation.
Marking notes. 1 mark — correct setup using the exact formula (not the approximation). 1 mark — value rounded to 4 dp. 1 mark — both values stated and the difference correctly identified.
1.3 — Nominal vs real (4 marks)
(a) Sample response. A = 25,000 × (1.052)⁶ = 25,000 × 1.35591 = $33,897.81.
(b) Sample response. (1.025)⁶ = 1.15969. Real value = 33,897.81 / 1.15969 = $29,229.79. Annualised real rate = (29,229.79 / 25,000)^(1/6) − 1 = (1.16919)^(0.16667) − 1 = 1.02634 − 1 = 2.63% p.a.
Marking notes. (a) 1 mark — correct (1.052)⁶ line; 1 mark — correct nominal answer. (b) 1 mark — divides by inflation factor for real value; 1 mark — correct real rate to 2 dp (use nth root, not the approximation).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Alex earns at a flat 5% for 10 years:
A_A = 40,000 × (1.05)¹⁰ = 40,000 × 1.62889 = $65,155.79. [1 mark — Alex.]
Bea uses a two-stage product:
A_B = 40,000 × (1.08)² × (1.04)⁸ = 40,000 × 1.16640 × 1.36857 = 40,000 × 1.59670 = $63,867.92. [1 mark — Bea, with the multi-stage line shown.]
Alex wins in nominal terms by $65,155.79 − $63,867.92 = $1,287.87. [1 mark — explicit comparison and winner.]
Part (b). Inflation factor = (1.03)¹⁰ = 1.34392.
Real_A = 65,155.79 / 1.34392 = $48,481.05.
Real_B = 63,867.92 / 1.34392 = $47,522.45. [1 mark — both real values via division.]
Annualised real rates (10th-root method):
r_real,A = (48,481.05 / 40,000)^(1/10) − 1 = (1.21203)^(0.1) − 1 = 0.0194 (≈ 1.94% p.a.)
r_real,B = (47,522.45 / 40,000)^(1/10) − 1 = (1.18806)^(0.1) − 1 = 0.0174 (≈ 1.74% p.a.) [1 mark — nth-root used; 1 mark — both rates rounded to 4 dp.]
Part (c). Even though Bea's headline 8% is much higher than Alex's 5%, that high rate applies for only 2 of the 10 years; the remaining 8 years compound at 4%, which is lower than Alex's flat 5%. Because compound interest multiplies growth factors (not rates), the geometric average of Bea's product, (1.08²·1.04⁸)^(1/10) − 1 ≈ 4.79%, falls short of Alex's flat 5%. Inflation hits both savers equally, so the nominal gap of $1,288 carries directly into a real gap of about $959 in today's dollars. [1 mark — geometric/factor argument; 1 mark — used own numbers.]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes Alex's flat-rate balance correctly and attempts Bea's two-stage line but with a single-rate error; no inflation adjustment. ≈ 2-3 marks.
Band 4: Both nominal balances correct; identifies winner; attempts inflation adjustment but does not compute the annualised real rates correctly. ≈ 4-5 marks.
Band 5: All four numerical answers correct; provides a partial explanation but does not explicitly link to the geometric-mean idea. ≈ 6-7 marks.
Band 6: All calculations correct; explanation uses both an algebraic reason (rates multiply, not average) and the student's own numerical figures; clean rounding and units throughout. 8/8.