Module 7 · L2 of 12~30 minMST-12-S2-02⚡ +50 XP available
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.
Today's hook, $5000 is invested at 6% p.a. compounded annually. How much after 2 years? Compare to simple interest at the same rate. Which is better for the investor? Predict before reading on.
0/5QUESTS
Get oriented
Recall what you already know, meet the key ideas and settle the terms.
Worksheets
Practise this lesson
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 on write your gut feeling. We will revisit this at the end of the lesson.
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Key ideas for this lesson
reference
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.
More frequent compounding always produces a higher effective rate than the nominal rate.
Exponential growth
$A = P(1+r)^n$ is an exponential function of $n$. The graph curves upward, growth accelerates as the balance increases.
Divide rate, multiply periods
For monthly compounding at 6% p.a.: rate per period $= 0.06 \div 12 = 0.005$; periods $= n \times 12$.
Effective vs nominal rate
The effective annual rate $(1+r/k)^k - 1$ is always higher than the stated nominal rate when $k > 1$. Use it to compare investment options.
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What you will master
Know
Key facts
Compound interest formula $A = P(1+r)^n$
General formula $A = P(1+r/k)^{kn}$
Compounding frequency values ($k$)
Understand
Concepts
Why compound interest accelerates over time
The effect of compounding frequency on the outcome
Exponential vs linear growth
Can do
Skills
Calculate compound interest for any frequency
Find the effective annual rate
Compare investment options using effective rates
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Key terms
Compound InterestInterest calculated on the principal AND on all previously accumulated interest.
Compounding PeriodThe interval at which interest is calculated and added to the balance.
Compounding Frequency ($k$)The number of compounding periods per year: annually $k=1$, semi-annually $k=2$, quarterly $k=4$, monthly $k=12$, daily $k=365$.
Nominal RateThe stated annual interest rate before compounding effects are applied.
Effective Annual RateThe actual rate earned in one year after accounting for compounding: $(1+r/k)^k - 1$.
Exponential GrowthGrowth where the variable appears in the exponent, the amount added increases each period.
Earn interest on interest
See what changes once interest starts earning interest of its own.
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Interest on interest
core concept
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).
Worked example, step by step: $2000 at 5% p.a. compounded annually for 3 years.
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 $= 2315.25 - 2000 = \$315.25$
Comparison with simple interest: Simple interest on $\$2000$ at 5% for 3 years $= 2000 \times 0.05 \times 3 = \$300$. Compound earns $\$15.25$ more after just 3 years. After 20 years the gap becomes thousands of dollars, that is the power of compounding.
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?
See how often matters
Work out what monthly, quarterly and daily compounding do to the same rate.
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Compounding frequency, how often matters
core concept
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
$r/365$
$365n$
Worked example: $8000 at 4.8% p.a. compounded semi-annually for 4 years.
$k = 2$, 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.
Find the effective annual rate
Turn a nominal rate into the true annual return it actually delivers.
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Effective annual rate, the true rate of return
core concept
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.
Worked example: Nominal rate 6% p.a. compounded monthly ($k = 12$).
$\text{EAR} = (1 + 0.06/12)^{12} - 1 = (1.005)^{12} - 1 \approx 1.0617 - 1 = 0.0617 = 6.17\%$
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.
Worked example, comparison:
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 .
Work the examples and drill
Follow two worked problems, then run the quick-fire practice.
Worked examples · reveal each step
PROBLEM 1 · QUARTERLY COMPOUNDING
$10,000 invested at 8% p.a. compounded quarterly for 5 years. Find the final amount and effective annual rate.
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$P = 10\,000,\quad r = 0.08,\quad k = 4,\quad n = 5$
Effective rate: raise period factor to $k$ then subtract 1
Quick-fire practice · 2 activities
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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).
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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:
Annually
Quarterly
Monthly
Semi-annually
k = 12
k = 2
k = 1
k = 4
Avoid the compounding traps
Fix the kn slip, model both kinds of interest, then name what has changed.
Common errors · the 3 traps that cost marks
Trap 01
Using $kn$ instead of $k$ for the effective rate
The effective annual rate formula uses $k$ (periods per year), not $kn$ (total periods). For 6% compounded monthly: EAR $= (1.005)^{12} - 1$, not $(1.005)^{36} - 1$.
Trap 02
Forgetting to subtract the principal for interest
$A = P(1+r)^n$ gives the total amount, not the interest. To find interest: $I = A - P$. Always re-read the question to see whether it asks for $I$ or $A$.
Trap 03
Not dividing the rate when changing compounding frequency
For quarterly compounding at 8% p.a., the period rate is $0.08 \div 4 = 0.02$. Using $0.08$ directly as the period rate is one of the most common exam errors.
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.
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Modelling both kinds of interest in a spreadsheet
core concept
A spreadsheet is the fastest way to see simple and compound interest side by side, because it computes every year rather than only the one the question asks for. Set up three columns and fill down.
Take $\$5000$ invested at $6\%$ p.a. for $5$ years.
Cell
What goes in it
Fill down?
A2
the year number, $0$
yes, $=$A2+1
B2
simple interest value, $=$5000+5000*0.06*A2
yes
C2
compound value, $=$5000*1.06^A2
yes
Filling down to year $5$ gives:
Year
Simple
Compound
1
$\$5300.00$
$\$5300.00$
2
$\$5600.00$
$\$5618.00$
3
$\$5900.00$
$\$5955.08$
4
$\$6200.00$
$\$6312.38$
5
$\$6500.00$
$\$6691.13$
Select all three columns and insert a line graph. The shapes are the point: simple interest plots as a straight line, because the same $\$300$ is added every year, and compound plots as a curve that pulls away. They are equal at year $1$ and $\$191.13$ apart by year $5$.
The check that catches a broken model. Year $1$ must be identical in both columns. One year of interest cannot have compounded yet, so if your two columns differ at year $1$, a formula is wrong before you look at anything else.
To model interest in a spreadsheet, put the year in one column and fill down. Simple interest is $=P+P*r*n$ and gives a straight line; compound interest is $=P*(1+r)^n$ and gives a curve. Graph both columns against the year to compare them. The two must agree at year 1 and diverge after it.
Pause, copy the two cell formulas and the year-1 check into your book.
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Revisit your thinking
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?
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Practise compound interest
Answer the question bank, then write full short-answer responses.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That 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)?
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Short answer
ApplyBand 42 marks
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)
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ApplyBand 42 marks
SA 2. Which is better: 6.2% p.a. compounded monthly or 6.3% p.a. compounded semi-annually? Show working. (2 marks)
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AnalyseBand 53 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)
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].
