Mathematics Advanced • Year 12 • Module 7 • Lesson 2
Compound Interest in Depth
Apply the transposed compound-interest formulas (with logarithms) to planning, investing and debt scenarios.
Problem 1 — Plan a retirement deposit (find P)
A graduate wants to have $250,000 saved at age 60. She is currently 35 and will make a single lump-sum deposit today. The account pays 6.0% p.a. compounded annually.
Set up: What are we solving for?
(i) Identify P, r, n and A in this scenario. Which is unknown? 1 mark
(ii) Write the transposed formula and find the required lump sum, to the nearest cent. 3 marks
(iii) If she delays the deposit by 5 years (so n = 20), how much more must she deposit to reach the same goal? 2 marks
Stuck? Revisit lesson § Finding the Principal.Problem 2 — Infer growth rate from a house sale (find r)
A house was purchased in 2018 for $850,000 and sold in 2026 for $1,200,000. Assume compound annual growth.
Set up: What are we solving for?
(i) Find the implied annual growth rate r to two decimal places. 2 marks
(ii) Using the rate from (i), project the house value at the end of 2030 (12 years from 2018). 2 marks
(iii) Use logarithms to find the year in which the house first exceeds $1,500,000 at the same rate (give answer as "year X from 2018"). 3 marks
Problem 3 — Doubling time (find n with logs and with the Rule of 72)
Four common savings rates: 3%, 5%, 8%, 12% per annum compound (annual compounding).
Set up: What are we solving for?
(i) For each rate, compute the exact doubling time using logarithms. Record results in the table.
| Rate | Rule-of-72 estimate (yrs) | Exact doubling time (yrs) | % error of Rule of 72 |
|---|---|---|---|
| 3% | |||
| 5% | |||
| 8% | |||
| 12% |
4 marks
(ii) Around which rate does the Rule of 72 give zero error, and why does this happen mathematically? 2 marks
(iii) Berkshire Hathaway has compounded at approximately 20% p.a. for 58 years. Using the Rule of 72, how many times has $1,000 invested in 1965 doubled? Give the approximate final dollar amount. 2 marks
Stuck? Revisit lesson § Rule of 72 — Real-World Anchor.Problem 4 — Credit-card debt grows fast (find n in days)
A credit card charges 18% p.a. compounded daily on unpaid balances. A user leaves $2,000 unpaid.
Set up: What are we solving for?
(i) State the daily rate and explain why we cannot just use r = 0.18 with n in years here. 2 marks
(ii) How many days will it take the balance to reach $2,500? Give the answer to the nearest whole day. 2 marks
(iii) Convert that answer to years and months, and write one line of advice to the user about minimum-payment behaviour on credit-card debt. 2 marks
Problem 5 — Present value of a future payment (find P, multi-rate)
A government bond pays a single lump sum of $50,000 in 8 years. Three buyers use different discount rates because they have different alternative investment opportunities.
Buyer X: discount rate 3% p.a. compounded annually.
Buyer Y: discount rate 5% p.a. compounded annually.
Buyer Z: discount rate 8% p.a. compounded annually.
Set up: What are we solving for?
(i) Compute the fair price (present value) for each buyer. 3 marks
(ii) Which buyer would offer the highest price for the bond, and why does using a lower discount rate raise the present value? 2 marks
(iii) If the seller wants at least $34,000 for the bond, which buyers can transact with her, and at what minimum discount rate (to 2 dp) would the deal still close? 2 marks
Stuck? Use P = A / (1 + r)ⁿ and rearrange for r.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Retirement deposit
Set up. We are finding the present value (today's deposit) that grows to a target future value at a fixed rate.
(i) A = 250,000; r = 0.06; n = 25; P unknown.
(ii) P = A / (1 + r)ⁿ = 250,000 / (1.06)²⁵ = 250,000 / 4.29187 = $58,251.20.
(iii) Delayed (n = 20): P = 250,000 / (1.06)²⁰ = 250,000 / 3.20714 = $77,949.55. Extra required = 77,949.55 − 58,251.20 = $19,698.35 — a 5-year delay costs nearly $20,000 of additional deposit.
Problem 2 — House growth rate
Set up. We are inferring the annual compound growth rate from two values and using it to project and to invert with logarithms.
(i) r = (1,200,000/850,000)1/8 − 1 = (1.4118)0.125 − 1 = 1.04393 − 1 = 0.0439 ≈ 4.39% p.a.
(ii) A12 = 850,000(1.0439)¹² = 850,000 × 1.67395 = $1,422,860.91.
(iii) Solve 850,000(1.0439)ⁿ ≥ 1,500,000 ⇒ (1.0439)ⁿ ≥ 1.7647 ⇒ n ≥ ln 1.7647 / ln 1.0439 = 0.5681 / 0.04298 = 13.22 years. So the house first exceeds $1.5 m at year 14 from 2018 (i.e., 2032).
Problem 3 — Doubling time table
Set up. We are computing exact doubling time at each rate and quantifying the Rule-of-72 error.
(i) Table values (exact = ln 2 / ln(1 + r)):
- 3%: estimate 24.00 yr, exact 23.45 yr, error 2.36%
- 5%: estimate 14.40 yr, exact 14.21 yr, error 1.36%
- 8%: estimate 9.00 yr, exact 9.01 yr, error 0.07%
- 12%: estimate 6.00 yr, exact 6.12 yr, error 1.95%
(ii) Rule of 72 gives ≈ zero error near r ≈ 8%. Mathematically this is because the two approximations cancel: replacing 69.31 (= 100 · ln 2) with 72 over-states, while approximating ln(1 + r) by r under-states; the two errors balance near 8%.
(iii) At 20%, doublings = 72 ÷ 20 = 3.6 years per double; 58 ÷ 3.6 ≈ 16 doublings. $1,000 × 2¹⁶ = $1,000 × 65,536 ≈ $65 million (textbook quotes "over $36 million" using the exact 17.6%-effective figure; either is acceptable as an order-of-magnitude estimate).
Problem 4 — Credit-card debt
Set up. We are solving for n in days using daily compounding because the rate is per-year but interest is added each day.
(i) Daily rate r = 0.18 / 365 ≈ 0.000493. Using r = 0.18 with n in years would assume only one compounding event per year, badly under-stating the true growth (which compounds 365 times per year).
(ii) n = ln(2,500/2,000) / ln(1.000493) = 0.22314 / 0.0004929 ≈ 452.5 days, i.e. 453 days to the nearest whole day.
(iii) 453 days ≈ 1 year 88 days (about 14.5 months). Advice: minimum payments barely dent the principal because daily compounding adds interest faster than tiny repayments subtract; always pay more than the minimum to break the exponential.
Problem 5 — Bond present values
Set up. We are computing the present value of a single $50,000 payment in 8 years at three discount rates, then inverting.
(i) X (3%): P = 50,000/(1.03)⁸ = 50,000/1.26677 = $39,470.46. Y (5%): P = 50,000/(1.05)⁸ = 50,000/1.47746 = $33,841.92. Z (8%): P = 50,000/(1.08)⁸ = 50,000/1.85093 = $27,013.40.
(ii) Buyer X (3% discount rate) offers the highest price. A lower discount rate means the investor's alternative opportunities earn less, so $50,000 in 8 years is worth relatively more today — there is less "opportunity cost" to wait.
(iii) X and Y can transact (their offers exceed $34,000); Z cannot. To find the break-even rate: 34,000 = 50,000 / (1 + r)⁸ ⇒ (1 + r)⁸ = 1.47059 ⇒ 1 + r = 1.470591/8 = 1.04937 ⇒ r ≈ 4.94% p.a. Any buyer with a discount rate below 4.94% will offer at least $34,000.