Compound Interest
Albert Einstein reportedly called compound interest the eighth wonder of the world. Unlike simple interest, which only pays on the principal, compound interest pays interest on interest. $1000 at 5% compounded annually becomes $1050 after year one, and then 5% is calculated on $1050 in year two, earning $52.50, not just $50. Over decades, this accelerating growth turns small savings into fortunes and modest debts into crushing burdens.
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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
$5000 is invested at 6% p.a. compounded annually. How much after 2 years? Compare this to simple interest at the same rate. Which is better for the investor?
Before reading onwrite your gut feeling. We will revisit this at the end of the lesson.
Compound interest calculates interest on the growing balance rather than just the original principal. Two formulas govern it.
Annual compounding: $A = P(1 + r)^n$, where $P$ is the principal, $r$ is the annual rate as a decimal, and $n$ is the number of years.
Periodic compounding: $A = P\!\left(1 + \dfrac{r}{k}\right)^{kn}$, where $k$ is the number of compounding periods per year.
Key facts
- Compound interest formula $A = P(1+r)^n$
- General formula $A = P(1+r/k)^{kn}$
- Compounding frequency values ($k$)
Concepts
- Why compound interest accelerates over time
- The effect of compounding frequency on the outcome
- Exponential vs linear growth
Skills
- Calculate compound interest for any frequency
- Find the effective annual rate
- Compare investment options using effective rates
Compound interest is calculated on the principal AND on the accumulated interest from all previous periods. Each period the balance grows, so the next period's interest is larger.
$$A = P(1 + r)^n$$Where $P$ is the principal, $r$ is the annual rate as a decimal, and $n$ is the number of years (for annual compounding).
Year 1: $2000 \times 1.05 = \$2100$
Year 2: $2100 \times 1.05 = \$2205$
Year 3: $2205 \times 1.05 = \$2315.25$
Or directly: $A = 2000 \times (1.05)^3 = \$2315.25$
Interest $= 2315.25 - 2000 = \$315.25$
Compound interest: A = P(1 + r)^n. Interest earns interest, each period's interest is added to the balance before the next period's interest is calculated. The compounding effect grows exponentially over time.
Pause, copy A = P(1 + r)^n and write one sentence explaining the exponential growth: each period's interest is added to the balance before the next period's calculation, so the dollar amount of interest increases every period into your book.
Quick check: $4000 is invested at 5% p.a. compounded annually for 6 years. Which expression gives the correct final amount?
Annual compounding uses A = P(1 + r)^n with r as the annual rate and n as years. When a bank compounds monthly, the per-period rate is r/12 and the number of periods is n × 12. The general rule: divide the annual rate by the number of compounding periods per year (k), and multiply years by k to get total periods, more frequent compounding always increases the final balance.
When interest compounds more than once per year, the annual rate is divided by $k$ and the number of periods is multiplied by $k$:
$$A = P\!\left(1 + \frac{r}{k}\right)^{kn}$$| Frequency | $k$ | Rate per period | Periods in $n$ years |
|---|---|---|---|
| Annually | 1 | $r$ | $n$ |
| Semi-annually | 2 | $r/2$ | $2n$ |
| Quarterly | 4 | $r/4$ | $4n$ |
| Monthly | 12 | $r/12$ | $12n$ |
| Daily | 365 | $r/365$ | $365n$ |
$k = 2$, rate per period $= 0.048/2 = 0.024$, periods $= 4 \times 2 = 8$
$A = 8000 \times (1.024)^8 = 8000 \times 1.2060 = \$9648$
Compounding frequency matters: divide the annual rate by the number of periods per year (r_period = r_annual/k) and multiply years by periods per year (n = years × k). More frequent compounding increases the final value.
Pause, copy the two conversion rules for non-annual compounding: per-period rate = annual rate ÷ k, and total periods = years × k, and note that more frequent compounding (larger k) always produces a larger final balance for the same nominal rate into your book.
True or false: For a given annual rate, monthly compounding produces a higher final amount than annual compounding.
Dividing the nominal rate by k and multiplying years by k correctly adjusts the compound interest formula, but how do you compare a 6% p.a. monthly product against a 6.1% p.a. annual product? Convert both to an effective annual rate: EAR = (1 + r_nominal/k)^k − 1 gives the equivalent annual rate accounting for compounding frequency, making any two products directly comparable.
The effective annual rate (EAR) is the actual percentage earned or charged in one year when compounding is taken into account.
$$\text{Effective rate} = \left(1 + \frac{r}{k}\right)^k - 1$$$\text{EAR} = (1 + 0.06/12)^{12} - 1 = (1.005)^{12} - 1 \approx 1.0617 - 1 = 0.0617 = 6.17\%$
The effective rate (6.17%) is higher than the nominal rate (6%) because interest is added 12 times per year rather than once.
Why it matters: When comparing two investment products with different compounding frequencies, always convert both to their effective annual rate. A product offering 6.2% compounded monthly may be better or worse than 6.3% compounded semi-annually, you must calculate both EARs to know.
Option A: 6.2% compounded monthly. EAR $= (1.005167)^{12} - 1 = 6.38\%$
Option B: 6.3% compounded semi-annually. EAR $= (1.0315)^2 - 1 = 6.40\%$
Option B is slightly better despite the lower nominal rate.
Effective annual rate (EAR) = (1 + r_nominal/k)^k − 1. EAR is the equivalent annual rate that produces the same result as more-frequent compounding. Use EAR to compare investments with different compounding frequencies.
Pause, copy EAR = (1 + r_nominal/k)^k − 1 with all variable definitions, and note: always use EAR, not the nominal rate, when comparing two investment products with different compounding frequencies into your book.
Fill the gap: To find the effective annual rate for a nominal rate $r$ compounded $k$ times per year, calculate $(1 + r/k)^k$ then subtract .
Worked examples · reveal each step
$10,000 invested at 8% p.a. compounded quarterly for 5 years. Find the final amount and effective annual rate.
$5000 at 6% p.a. compounded monthly for 3 years. Find $A$ and the effective annual rate.
Quick-fire practice · 2 activities
Calculate compound interest, 4 questions. (a) $4000 at 5% p.a. compounded annually for 6 years. (b) $8000 at 4.8% p.a. compounded semi-annually for 4 years. (c) $6000 at 7.2% p.a. compounded quarterly for 5 years. (d) Find the effective annual rate for (b) and (c).
Compare and analyse. (a) Compare $5000 at 6% simple interest vs 5.5% compounded annually over 10 years, which is larger? (b) $10,000 at 6% compounded annually vs monthly for 20 years, how much more does monthly earn? (c) Why do credit card companies compound interest daily?
Match each compounding frequency to its $k$ value:
Common errors · the 3 traps that cost marks
Top 3 list: Name THREE ways in which compound interest affects everyday Australians, one advantage for savers, one disadvantage for borrowers, and one surprising consequence of compounding over decades.
The answer: Compound $A = 5000 \times (1.06)^2 = \$5618$. Simple $A = 5000(1 + 0.06 \times 2) = \$5600$. Compound earns $18 more after just 2 years.
After 20 years: compound gives $5000 \times (1.06)^{20} = \$16\,035$ vs simple $5000(1 + 0.06 \times 20) = \$11\,000$, a $5,035 difference. Over a working lifetime, this is the difference between a comfortable retirement and struggling. Einstein's "eighth wonder" comment reflects exactly this: the gap starts small but becomes enormous.
What has changed in your understanding? What did you get right? What surprised you?
Pick your answer, then rate your confidencethat tells the system what to drill next.
Q1. $2000 is invested at 5% p.a. compounded annually for 3 years. What is the total amount (to the nearest cent)?
Q2. Which formula correctly gives $A$ for a nominal rate $r$ compounded $k$ times per year over $n$ years?
Q3. The effective annual rate for 6% p.a. compounded monthly is closest to:
Q4. The graph of compound interest amount $A$ against time $n$ is:
Q5. $7500 is invested at 5.4% p.a. compounded quarterly for 5 years. What is the effective annual rate (to 2 decimal places)?
SA 1. $7500 is invested at 5.4% p.a. compounded quarterly for 5 years. (a) Find the final amount. (b) Find the total interest earned. (c) Find the effective annual rate. (2 marks)
SA 2. Which is better: 6.2% p.a. compounded monthly or 6.3% p.a. compounded semi-annually? Show working. (2 marks)
SA 3. (a) Prove that compound interest grows exponentially, not linearly. (b) A credit card charges 18% p.a. compounded daily. Calculate the effective annual rate. (c) Explain why daily compounding is particularly dangerous for borrowers. (3 marks)
Comprehensive answers (click to reveal)
MC 1, C: $A = 2000 \times (1.05)^3 = 2000 \times 1.157625 = \$2315.25$.
MC 2, B: The correct general formula divides the rate by $k$ and raises to $kn$.
MC 3, D: $\text{EAR} = (1.005)^{12} - 1 = 1.0617 - 1 = 6.17\%$.
MC 4, A: $A = P(1+r)^n$ is exponential, the graph curves upward with increasing slope.
MC 5, C: $\text{EAR} = (1 + 0.054/4)^4 - 1 = (1.0135)^4 - 1 = 1.0552 - 1 = 5.52\%$.
SA 1 (2 marks): (a) $A = 7500 \times (1.0135)^{20} \approx \$9802.50$ [0.5]. (b) $I = \$2302.50$ [0.5]. (c) EAR $= 5.52\%$ [1].
SA 2 (2 marks): Monthly EAR $= 6.38\%$; semi-annual EAR $= 6.40\%$. 6.3% semi-annual is slightly better [1 working + 1 conclusion].
SA 3 (3 marks): (a) $A = P(1+r)^n$, $n$ in exponent makes it exponential; $A = P(1+rn)$ is linear [1]. (b) EAR $= (1 + 0.18/365)^{365} - 1 \approx 19.72\%$ [1]. (c) Daily compounding means any unpaid balance accrues interest every single day; the effective rate (19.72%) far exceeds the stated 18%, so debt grows rapidly if only minimum payments are made [1].
Drill 1: (a) $\$5360.38$. (b) $\$9648.42$; EAR $= 4.86\%$. (c) $\$8565.46$; EAR $= 7.39\%$.
Drill 2: (a) Simple $A = \$8000$; compound $A = 5000 \times (1.055)^{10} = \$8546$, compound is larger. (b) Annual: $\$32\,071$; monthly: $\$33\,102$, monthly earns $\$1031$ more. (c) Daily compounding maximises the effective rate charged on any outstanding balance, increasing interest income for the card company.
Five timed questions on compound interest, compounding frequency, and effective annual rates. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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