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.