Mathematics Advanced • Year 12 • Module 7 • Lesson 6
Interest Rate Conversions & Multi-Stage Problems
Build fluency in chaining compound-interest stages and converting between nominal and real (inflation-adjusted) returns.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the multi-stage compound-interest formula:
A = P × (1 + r₁)^____ × (1 + r₂)^____ × … × (1 + r_k)^____
Q1.2 Write the exact formula linking the real and nominal rates:
1 + r_real = _________________________________________
Q1.3 Write the approximate formula (valid for small rates): r_real ≈ ________________________
2. Worked example — $20,000 across two rate stages
Follow every line of working. The reasons are written on the right.
Problem. $20,000 is invested at 6% p.a. for 4 years, then the rate falls to 3.5% p.a. for the next 3 years. Find the final balance.
Step 1 — Identify P, r₁, n₁, r₂, n₂.
P = 20,000; r₁ = 0.06, n₁ = 4; r₂ = 0.035, n₂ = 3.
Reason: each rate period is one "stage" — pair the rate with its time.
Step 2 — Apply the multi-stage formula in one line.
A = 20,000 × (1.06)⁴ × (1.035)³
Reason: chain the growth factors — never average rates.
Step 3 — Evaluate each factor.
(1.06)⁴ = 1.26248; (1.035)³ = 1.10872
A = 20,000 × 1.26248 × 1.10872 = 27,996.12
Conclusion. Final balance ≈ $27,996.12.
3. Faded example — fill in the missing steps
$8,000 is invested at 4.2% p.a. for 5 years, then at 5.6% p.a. for 4 years. Find the real value of the final balance if inflation averages 2.5% p.a. for the whole 9 years. 5 marks
Step 1 — Multi-stage nominal balance.
A = 8,000 × (1.042)^____ × (1.056)^____
(1.042)^____ = ____________ ; (1.056)^____ = ____________
A = 8,000 × ___________ × ___________ = $____________
Step 2 — Inflation factor for n = 9.
(1 + inflation)^n = (1.025)^____ = ____________
Step 3 — Real value = nominal / inflation factor.
Real value = ___________ / ___________ = $____________
Conclusion. The real (today's-dollar) value of the final balance is $____________.
4. Graduated practice
Show every line of substitution. Round money to the nearest cent and rates to two decimal places unless stated.
Foundation — single-step conversions (4 questions)
| Q | Task | Answer |
|---|---|---|
| 4.1 1 | Convert 7.2% p.a. to a monthly rate (as a decimal). | |
| 4.2 1 | Convert 4.8% p.a. to a quarterly rate (as a decimal). | |
| 4.3 1 | Inflation = 3% p.a. Use the approximation r_real ≈ r_nom − inflation to estimate the real rate when r_nom = 5.5%. | |
| 4.4 1 | $1,000 grows for 1 year at 8% p.a. and 1 year at 4% p.a. State the final balance (to the nearest cent). |
Standard — multi-stage compound interest (6 questions)
Show the multi-stage line then the numerical value.
4.5 $5,000 invested at 5% p.a. for 3 years, then 6% p.a. for 2 years. Find A. 2 marks
4.6 $12,000 invested at 4% p.a. for 3 years, then at 5.5% p.a. for 4 years. Find A and the total interest earned. 2 marks
4.7 $15,000 invested at 6% p.a. for 4 years, then 4.5% p.a. for 3 years, then 3% p.a. for 2 years. Find A. 2 marks
4.8 A bank offers 4.5% p.a. for the first 6 months and 2.8% p.a. for the next 6 months. Find the balance after one year on a $20,000 deposit. (Use semi-annual rates: divide each annual rate by 2.) 2 marks
4.9 A nominal return is 7%, inflation is 3.5%. Find the exact real rate r_real (4 dp) and compare with the approximation. 2 marks
4.10 $25,000 grows to $34,242.58 over 8 years (nominal). Inflation averaged 2.8% p.a. for the whole term. Find the real value in today's dollars. 2 marks
Extension — combined concepts (2 questions)
4.11 $30,000 invested at 5.2% p.a. for 5 years, then 4.4% p.a. for 4 years. Inflation averaged 3.1% p.a. over the entire 9 years.
(a) Find the final nominal balance. (b) Find the real value in today's dollars. (c) Find the equivalent annualised real rate (4 dp) using r_real = (real/P)^(1/9) − 1. 4 marks
4.12 Two products are compared on a $10,000 deposit over 6 years. Product X: 5% p.a. throughout. Product Y: 8% p.a. for the first year, then 3% p.a. for 5 years. Show by direct calculation that Product X wins, and explain in one line why averaging the Product Y rates is misleading. 3 marks
5. Self-check the easy 3
Tick once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Multi-stage formula
A = P × (1 + r₁)^n₁ × (1 + r₂)^n₂ × … × (1 + r_k)^n_k.
Q1.2 — Exact real rate
1 + r_real = (1 + r_nominal) / (1 + inflation).
Q1.3 — Approximate real rate
r_real ≈ r_nominal − inflation. (Accurate for small rates; the approximation overstates slightly when rates exceed a few percent.)
Q3 — Faded example: $8,000 over two stages with inflation
Step 1: A = 8,000 × (1.042)⁵ × (1.056)⁴ = 8,000 × 1.22834 × 1.24389 = $12,222.48.
Step 2: (1.025)⁹ = 1.24886.
Step 3: Real value = 12,222.48 / 1.24886 = $9,787.04.
Q4.1 — Monthly rate from 7.2% p.a.
r_month = 0.072 / 12 = 0.006 (i.e. 0.6% per month).
Q4.2 — Quarterly rate from 4.8% p.a.
r_qtr = 0.048 / 4 = 0.012 (i.e. 1.2% per quarter).
Q4.3 — Approximate real rate
r_real ≈ 5.5 − 3 = 2.5%.
Q4.4 — Two single-year stages
A = 1,000 × 1.08 × 1.04 = $1,123.20. (Note: 1.06² = $1,123.60, so averaging the rates overstates by 40 cents.)
Q4.5
A = 5,000 × (1.05)³ × (1.06)² = 5,000 × 1.15763 × 1.1236 = $6,503.05.
Q4.6
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.
Q4.7
A = 15,000 × (1.06)⁴ × (1.045)³ × (1.03)² = 15,000 × 1.26248 × 1.14117 × 1.0609 = $22,931.69 (small rounding may give $22,929–$22,932).
Q4.8
Semi-annual rates: 4.5%/2 = 2.25%, 2.8%/2 = 1.4%. A = 20,000 × 1.0225 × 1.014 = $20,737.62.
Q4.9
Exact: r_real = 1.07/1.035 − 1 = 1.03382 − 1 = 0.0338 (3.38%). Approximation: 7% − 3.5% = 3.5%. The approximation overstates by 0.12 percentage points.
Q4.10
(1.028)⁸ = 1.24718. Real value = 34,242.58 / 1.24718 = $27,455.42 (within a few cents of $27,455.60 depending on rounding).
Q4.11 — Combined multi-stage + inflation
(a) A = 30,000 × (1.052)⁵ × (1.044)⁴ = 30,000 × 1.28859 × 1.18876 = $45,948.74.
(b) (1.031)⁹ = 1.31530. Real value = 45,948.74 / 1.31530 = $34,933.55.
(c) r_real = (34,933.55 / 30,000)^(1/9) − 1 = (1.16445)^0.1111 − 1 = 1.01704 − 1 = 0.0170 (1.70% p.a.).
Q4.12 — Product comparison
Product X: A = 10,000 × (1.05)⁶ = 10,000 × 1.34010 = $13,400.96.
Product Y: A = 10,000 × 1.08 × (1.03)⁵ = 10,000 × 1.08 × 1.15927 = $12,520.13.
Product X wins by about $880. Averaging the Y rates (8% then 3% for 5 years → average ≈ 3.83%) would suggest Y is similar to a fixed 3.83%, but compounding multiplies factors, not rates, so averaging misrepresents the geometric growth.