Compound Interest and Growth (AI SL)
Compound interest is what happens when interest itself starts earning interest - each period's growth is calculated on the current balance, not just the original amount, so a savings account or investment grows faster and faster over time. This page covers the annual and non-annual versions of the model and sits within the wider Financial Maths topic.
51 questions on this sub-topic.
Simple vs. compound growth
Covered under IB syllabus reference SL1.4 - financial applications of geometric sequences and series, including compound interest, annual depreciation, and calculating the real value of an investment given an interest rate and an inflation rate. Compound interest may be calculated yearly, half-yearly, quarterly, or monthly.
Simple interest
\[I = Prt\]
Interest is earned only on the original principal \(P\) - it stays constant each period, since there's no interest-on-interest.
✓ In the formula bookletAnnual compounding
\[FV = PV(1+\tfrac{r}{100})^n\]
Interest is added once per year, so \(n\) is simply the number of years.
✓ In the formula bookletNon-annual compounding
\[FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}\]
Divide the annual rate by \(k\) periods, and multiply the years by \(k\) - e.g. monthly means \(k=12\).
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table? See Financial Maths.
Worked examples
$2000 is invested at 4% p.a. compounded annually.
Find the value after 5 years.
Worked solution
Annual compound interest: \(A = P\left(1 + \tfrac{r}{100}\right)^n\), where \(P\) is the amount invested, \(r\%\) the annual rate, and \(n\) the number of years. M1
\(P = 2000,\ r = 4,\ n = 5\): \(A = 2000(1.04)^5 = 2000(1.21665).\) A1
, so \(A \approx $2433.31.\) A1
$4000 is invested at 6% p.a. compounded annually.
(a) Write an equation for the value to reach $6000.
(b) Find the least whole number of years required.
Worked solution
(a) Equation. Starting at $4000 growing at 6% annually, after \(n\) years: \(4000(1.06)^n = 6000 \;\Rightarrow\; (1.06)^n = 1.5.\)A1
(b) \(n = \frac{\ln 1.5}{\ln 1.06}.\)M1
\(n = \frac{0.4055}{0.05827} \approx 6.96.\)A1
Since the target is only reached after a whole compounding period, round up.R1
\(n = 7\) years. A1
Common mistakes
- Using \(I=Prt\) for compound growth. Simple interest formulas only apply when there's genuinely no interest-on-interest - anything described as "compounded" needs \(FV=PV(1+\tfrac{r}{100})^n\) instead.
- Forgetting to convert the rate and time for non-annual compounding. Monthly compounding needs the annual rate divided by 12 and the number of years multiplied by 12 - using the annual rate directly overstates the growth.
- Rounding \(n\) the wrong way when solving for time. Logs give a decimal number of years or periods, but interest is only credited at the end of a whole period - "when does it first exceed..." questions need rounding up, not to the nearest integer.
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Quick answers
What is the formula for compound interest on the IB syllabus?
\(FV = PV\left(1+\tfrac{r}{100}\right)^n\) for annual compounding, or \(FV = PV\left(1+\tfrac{r}{100k}\right)^{kn}\) when interest is compounded \(k\) times per year (e.g. \(k=12\) for monthly).
How is compound interest different from simple interest?
Simple interest (\(I = Prt\)) earns the same amount every period, based only on the original principal. Compound interest earns interest on the interest already added, so the growth accelerates over time.