Mathematics Advanced • Year 12 • Module 7 • Lesson 6
Interest Rate Conversions & Multi-Stage Problems
Apply multi-stage compound interest and inflation adjustments to realistic savings, loan and retirement scenarios.
Problem 1 — Bonus rate trap
A bank advertises a "high-interest" online saver: 8% p.a. for the first year, then 2.5% p.a. ongoing. Mei deposits $15,000.
Set up: What are we solving for?
(i) Find the balance after 9 years under the bonus-rate offer. 2 marks
(ii) A competitor offers a flat 4% p.a. for the whole 9 years. Find the balance and state which product wins, and by how much. 2 marks
(iii) Explain in one sentence why "headline rate" advertising can mislead long-term savers, referring to your numbers. 1 mark
Stuck? Revisit lesson § Multi-Stage Compound Interest, Real-World Anchor.Problem 2 — RBA rate cuts
Tom takes out a $300,000 mortgage at 6.4% p.a. compounded annually for the purposes of this question. He plans to pay no principal for 5 years (interest-only equivalent on a single accumulating balance). After 2 years the RBA cuts and his rate drops to 5.4% p.a. for the next 3 years.
Set up: What are we solving for?
(i) Compute the loan balance after 5 years using the multi-stage formula. 2 marks
(ii) Compute what the balance would have been at a flat 6.4% for all 5 years. By how much did the rate cut save Tom in nominal dollars? 2 marks
(iii) Inflation averaged 3.1% p.a. over those 5 years. Convert the saving in (ii) into today's dollars. 2 marks
Problem 3 — Super across two career rates
Ava places $50,000 in a high-growth superannuation fund averaging 8.2% p.a. for 12 years. She then switches to a balanced fund averaging 5.6% p.a. for the next 18 years.
Set up: What are we solving for?
(i) Find the nominal balance after 30 years. 2 marks
(ii) Inflation averages 2.7% p.a. across the 30 years. Find the real value in today's dollars. 2 marks
(iii) Find Ava's annualised real return r_real using r_real = (real value / $50,000)^(1/30) − 1, expressed as a percentage to 2 dp. 2 marks
Stuck on (iii)? The nth root reverses 30 years of compounding to give a single annual rate.Problem 4 — How much does inflation cost a retiree?
Pete retires with $400,000 and leaves it in a term deposit at 5% p.a. for 10 years. Inflation averages 3.5% p.a. for the whole period.
Set up: What are we solving for?
(i) Find Pete's nominal balance after 10 years. 1 mark
(ii) Find the real value (in today's dollars) of that balance. 2 marks
(iii) Compute the real annual return using both the approximation r_real ≈ r_nom − inflation and the exact formula. State the percentage-point gap between them. 2 marks
(iv) Explain in one sentence why the approximation is acceptable here but becomes risky at very high inflation. 1 mark
Problem 5 — Designing a 10-year savings plan
Mira wants $80,000 in nominal terms in 10 years from a starting deposit of $50,000. She plans a 3-stage strategy: 6% p.a. for the first 3 years (high-rate intro term deposit), 4.5% p.a. for the next 4 years, and an unknown rate r for the final 3 years.
Set up: What are we solving for?
(i) Find the balance at the end of stage 2 (after 7 years). 2 marks
(ii) Form an equation involving r and solve to find the rate that delivers exactly $80,000. Give r as a percentage to 2 dp. 3 marks
(iii) If inflation runs at 2.5% p.a., what is the real value of $80,000 in today's dollars? Has Mira's plan actually grown her purchasing power? 2 marks
Stuck on (ii)? Set up: (balance after 7 yrs) × (1 + r)³ = 80,000, then cube-root both sides.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bonus rate trap
Set up. We are comparing total accumulated balances between an introductory-rate product and a flat-rate alternative.
(i) A_bonus = 15,000 × 1.08 × (1.025)⁸ = 15,000 × 1.08 × 1.21840 = $19,738.10.
(ii) A_flat = 15,000 × (1.04)⁹ = 15,000 × 1.42331 = $21,349.67. The flat 4% product wins by about $1,611.57.
(iii) The headline 8% applied for only 1 of the 9 years; the long ongoing rate (2.5%) drags the geometric average well below the steady 4%, so the eye-catching first-year rate did not compensate.
Problem 2 — RBA rate cuts
Set up. We are tracking a debt balance that accrues at two different annual rates across a 5-year horizon, then comparing with a counterfactual.
(i) A = 300,000 × (1.064)² × (1.054)³ = 300,000 × 1.13210 × 1.17085 = $397,612.20.
(ii) Flat counterfactual: 300,000 × (1.064)⁵ = 300,000 × 1.36373 = $409,119. Saving ≈ $11,507 in nominal dollars.
(iii) (1.031)⁵ = 1.16487. Real saving ≈ 11,507 / 1.16487 = $9,878 in today's dollars.
Problem 3 — Super across two career rates
Set up. We are valuing a single lump sum compounding at two different rates over a 30-year horizon, then expressing the result in today's dollars and as an equivalent annual rate.
(i) A = 50,000 × (1.082)¹² × (1.056)¹⁸ = 50,000 × 2.55428 × 2.65780 = $339,395 (rounding to nearest dollar).
(ii) (1.027)³⁰ = 2.22260. Real value = 339,395 / 2.22260 = $152,701 in today's dollars.
(iii) r_real = (152,701 / 50,000)^(1/30) − 1 = (3.05402)^(0.03333) − 1 = 1.03789 − 1 = 0.0379 (≈ 3.79% p.a.).
Problem 4 — Inflation erosion for a retiree
Set up. We are quantifying how much of a nominal return is real purchasing power and how much is eaten by inflation.
(i) Nominal: 400,000 × (1.05)¹⁰ = 400,000 × 1.62889 = $651,557.85.
(ii) Real: 651,557.85 / (1.035)¹⁰ = 651,557.85 / 1.41060 = $461,914.
(iii) Approximation: 5 − 3.5 = 1.5%. Exact: 1.05/1.035 − 1 = 1.449%. Gap ≈ 0.05 percentage points.
(iv) The approximation works because both rates are small; at very high inflation (say 30%) the multiplicative correction (1 + i) in the denominator is no longer ≈ 1, and the gap grows quickly.
Problem 5 — Designing a 10-year savings plan
Set up. We are working forward two stages, then solving algebraically for the final stage rate that delivers a target balance, and finally evaluating the real result against inflation.
(i) Balance after 7 yrs = 50,000 × (1.06)³ × (1.045)⁴ = 50,000 × 1.19102 × 1.19252 = $71,022.18.
(ii) 71,022.18 × (1 + r)³ = 80,000 ⇒ (1 + r)³ = 80,000 / 71,022.18 = 1.12642 ⇒ 1 + r = 1.12642^(1/3) = 1.04055 ⇒ r ≈ 4.06% p.a.
(iii) (1.025)¹⁰ = 1.28008. Real value of $80,000 = 80,000 / 1.28008 = $62,496. Mira's plan grew her starting $50,000 to about $62,496 in today's dollars — a modest real gain of about 25% over 10 years (≈ 2.25% p.a. real).