Compound Interest
Compound interest earns "interest on interest" — creating exponential growth rather than the straight line of simple interest. Master $A = P(1+r)^n$, adjust $r$ and $n$ for any compounding frequency, and understand why more frequent compounding always wins.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
What if your bank not only paid you interest on your original deposit, but also paid interest on last year's interest — and then on that, and so on? After 10 years, the difference between simple and compound interest on the same deposit can be thousands of dollars. Why do you think compound interest grows so much faster than simple interest over time?
Without calculating — write your gut feeling. We'll revisit this at the end of the lesson.
Compound interest in Maths Standard uses one central formula with two possible extensions. Lock the workflow in before the worked examples.
$A = P(1+r)^n$ gives the final amount directly. $I = A - P$ extracts just the interest. The key adjustment step: divide the annual rate by the number of compounding periods per year to get $r$, and multiply years by periods per year to get $n$.
Key facts
- The compound interest formula $A = P(1+r)^n$
- How to find interest earned: $I = A - P$
- How to adjust $r$ and $n$ for quarterly, monthly, semi-annual compounding
- That compound interest gives the total amount $A$, not just $I$
- The Rule of 72 as a reasonableness check
Concepts
- Why compound interest grows exponentially (interest on interest)
- Why more frequent compounding produces a slightly larger final amount
- The difference between the formula giving $A$ vs $I$
- Why both $r$ and $n$ must be adjusted together
Skills
- Apply $A = P(1+r)^n$ for annual, quarterly, monthly, semi-annual compounding
- Calculate interest earned: $I = A - P$
- Compare simple and compound interest on the same investment
- Compare annual and monthly compounding and quantify the difference
Compound interest adds interest to the principal at the end of each period, so the next period's interest is calculated on a larger balance — creating exponential rather than linear growth.
Under simple interest, a $10,000 investment at 5% per annum earns $500 every year. Under compound interest at the same rate:
- Year 1: earns $500 (5% of $10,000) → balance $10,500
- Year 2: earns $525 (5% of $10,500) → balance $11,025
- Year 3: earns $551.25 (5% of $11,025) → balance $11,576.25
The formula $A = P(1 + r)^n$ captures this: each period the balance is multiplied by $(1 + r)$. After $n$ periods, the original principal has been multiplied $n$ times. The difference between compound and simple interest becomes more pronounced as $n$ increases.
Adjusting $r$ and $n$ for compounding frequency — 8% per annum, 3 years:
| Frequency | Periods/year | $r$ per period | $n$ for 3 years |
|---|---|---|---|
| Annually | 1 | 0.08 | 3 |
| Semi-annually | 2 | 0.04 | 6 |
| Quarterly | 4 | 0.02 | 12 |
| Monthly | 12 | 0.08 ÷ 12 | 36 |
Rule: divide annual rate by periods/year for $r$; multiply years by periods/year for $n$.
What to write in your book
- Formula gives total amount $A$: $A = P(1+r)^n$. Interest earned = $I = A - P$.
- Always adjust both $r$ and $n$: $r = \text{annual rate} \div \text{periods/year}$; $n = \text{years} \times \text{periods/year}$.
- "Semi-annual" = twice a year ($n$ doubles, not halves).
- Rule of 72: years to double $\approx 72 \div \text{annual \%}$.
Quick check: $12,000 is invested at 6% per annum compounding quarterly for 2 years. What is the number of compounding periods $n$?
Simple and compound interest produce the same result after one period — the difference emerges from the second period onwards and grows with time.
For short time periods (1–2 years), the difference is small. Over longer periods, compound interest significantly outperforms simple interest for investors. When comparing:
- Calculate the total interest under each method using identical principal, rate, and time.
- Find the difference in interest earned.
- State a conclusion identifying which method produces more interest and by how much.
Key distinction: simple interest is linear ($A = P(1+rn)$); compound interest is exponential ($A = P(1+r)^n$). These are different formulas — do not mix them.
What to write in your book
- Simple: $A = P(1+rn)$ — linear. Compound: $A = P(1+r)^n$ — exponential. Never mix.
- Comparison answer must state which earns more and by how much.
- Show $(1+r)^n$ as a separate line before multiplying by $P$ — earns a method mark.
True or false: The compound interest formula $A = P(1+r)^n$ gives only the interest earned, not the total amount.
Worked examples · 4 problems, reveal step by step
$14,000 is invested at 6% per annum compounding quarterly for 4 years. Calculate: (a) the final amount, and (b) the interest earned.
$r = 0.06 \div 4 = 0.015$ per quarter
$n = 4 \times 4 = 16$ quarters
$A = \$14{,}000 \times (1.015)^{16}$
$(1.015)^{16} = 1.26899\ldots$
$A = \$14{,}000 \times 1.26899 = \$17{,}765.86$
$I = A - P = \$17{,}765.86 - \$14{,}000 = \$3{,}765.86$
$20,000 is invested for 5 years at a nominal rate of 4.5% per annum. Compare the total interest earned under: (a) simple interest, and (b) compound interest compounding annually. Which earns more, and by how much?
$I = Prn = \$20{,}000 \times 0.045 \times 5 = \$4{,}500.00$
$A = \$20{,}000 \times (1.045)^5$
$(1.045)^5 = 1.24618\ldots$
$A = \$24{,}923.56$
$I = \$24{,}923.56 - \$20{,}000 = \$4{,}923.56$
Difference $= \$4{,}923.56 - \$4{,}500.00 = \$423.56$
$\therefore$ Compound interest earns more by $\$423.56$ over 5 years.
Mia takes out a loan of $7,500 at 9% per annum compounding monthly. She makes no repayments. What does she owe after 2 years?
$r = 0.09 \div 12 = 0.0075$ per month
$n = 2 \times 12 = 24$ months
$A = \$7{,}500 \times (1.0075)^{24}$
$(1.0075)^{24} = 1.19641\ldots$
$A = \$7{,}500 \times 1.19641 = \$8{,}973.09$
$10,000 is invested for 3 years at 7.2% per annum. Compare the final amount if interest is compounded annually versus monthly. Which gives more, and by how much?
$A_{\text{annual}} = \$10{,}000 \times (1.072)^3 = \$12{,}319.25$
$r = 0.072 \div 12 = 0.006$; $n = 36$
$A_{\text{monthly}} = \$10{,}000 \times (1.006)^{36} = \$12{,}401.22$
Difference $= \$12{,}401.22 - \$12{,}319.25 = \$81.97$
$\therefore$ Monthly compounding gives more by $\$81.97$.
What to write in your book
- Reliable workflow: write $P$, adjust $r$, adjust $n$, substitute into $A = P(1+r)^n$, show $(1+r)^n$ on its own line, multiply by $P$, subtract $P$ if interest is needed.
- More frequent compounding → slightly larger final amount (interest earns interest sooner).
- For any comparison: use identical $P$, rate, and time; calculate both fully; state conclusion with dollar difference.
Fill the gap: For 7.2% per annum compounding monthly, the rate per period $r$ is per month.
Common errors · the 3 traps that cost marks
What to write in your book
- Both $r$ and $n$ must be adjusted for non-annual compounding.
- Formula gives $A$; interest only is $I = A - P$.
- Semi-annual = 2 periods/year. Quarterly = 4. Monthly = 12.
Match each compounding setup for 8% p.a. over 3 years:
Quick-fire practice · 5 calculations
Calculate the final amount for $9,500 invested at 4.8% per annum compounding annually for 6 years.
A loan of $6,800 compounds monthly at 8.4% per annum for 18 months. What is the amount owed at the end?
$15,000 compounds annually at 5% for 4 years. How much interest is earned?
What is the rate per period if 7.2% per annum compounds monthly?
Compare interest on $18,000 over 4 years at 5.5% p.a. under simple interest and compound interest (annual). Which earns more?
Top 3 list: Name THREE steps you must always do before substituting into $A = P(1+r)^n$ when the compounding is not annual.
Look back at what you wrote in the Think First section. Compound interest earns "interest on interest" — each period's interest is added to the balance before the next period's interest is calculated. This creates exponential growth. The formula $A = P(1+r)^n$ captures this: the balance is multiplied by $(1+r)$ every period, $n$ times.
What has changed? What did you get right? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 1. Calculate the final amount for $9,500 invested at 4.8% per annum compounding annually for 6 years. (2 marks)
SA 2. A loan of $6,800 compounds monthly at 8.4% per annum for 18 months. Write down the adjusted values of $r$ and $n$, then find the amount owed at the end. (3 marks)
SA 3. Compare the interest earned on $18,000 over 4 years at 5.5% per annum under simple interest and compound interest compounding annually. Which earns more, and by how much? (3 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $A = 9500(1.048)^6 = \$12{,}610.09$ · 2: $r = 0.007$, $n = 18$; $A = 6800(1.007)^{18} = \$7{,}707.26$ · 3: $A = 15000(1.05)^4 = 18232.59$; $I = \$3{,}232.59$ · 4: $0.072 \div 12 = 0.006$ · 5: Simple: $I = 18000 \times 0.055 \times 4 = \$3{,}960$; Compound: $A = 18000(1.055)^4 = 22561.83$; $I = \$4{,}561.83$; Compound earns $\$601.83$ more.
SA 1 (2 marks): $A = 9500(1.048)^6$ [1]. $A = \$12{,}610.09$ [1].
SA 2 (3 marks): $r = 0.084 \div 12 = 0.007$ [1], $n = 18$ [1]. $A = 6800(1.007)^{18} = \$7{,}707.26$ [1].
SA 3 (3 marks): Simple: $I = 18000 \times 0.055 \times 4 = \$3{,}960.00$ [1]. Compound: $A = 18000(1.055)^4 = \$22{,}561.83$; $I = \$4{,}561.83$ [1]. Compound earns $\$601.83$ more [1].
Five timed questions on compound interest. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering questions on compound interest. Pool: lessons 1–12.
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