Compound Interest and Growth (AA SL)
Money that earns interest on its own interest grows geometrically, not linearly - that's the whole idea behind compound interest. This page covers the annual and non-annual compounding formulas, how to read the compounding frequency out of a question, and the GDC steps that make these questions quick once you know the routine. It's part of the broader Financial Maths topic.
46 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference SL1.4: financial applications of geometric sequences and series, including compound interest calculated yearly, half-yearly, quarterly or monthly. Both formulas below are in the formula booklet.
Annual compounding
\[FV=PV(1+i)^{n}\]
One compound per year - \(i\) is the annual rate as a decimal and \(n\) is the number of years.
✓ In the formula bookletNon-annual compounding
\[FV=PV\left(1+\dfrac{r}{100k}\right)^{kn}\]
For \(k\) compounds per year at nominal annual rate \(r\%\) - divide the rate by \(k\) and multiply the years by \(k\) to get the number of periods.
✓ In the formula bookletNeed the full syllabus wording, GDC finance-solver walkthrough, or the annuity and depreciation formulas too? See Financial Maths.
Worked examples
$3000 is invested at 5% p.a. compounded annually.
Find the value after 6 years.
Worked solution
Compound interest: \(A=P\left(1+\tfrac{r}{100}\right)^n\), with \(P\) the principal, \(r\%\) the annual rate, \(n\) the years. M1
\(P=3000,\ r=5,\ n=6\): \(A=3000(1.05)^6.\) A1
\((1.05)^6=1.340096\), so \(A\approx$4020.29.\) A1
$6000 is invested at 3.6% p.a. compounded monthly.
Find the value after 5 years.
Worked solution
Monthly rate \(i=\tfrac{0.036}{12}=0.003\); periods \(n=5\times12=60.\) M1
\(A=6000(1.003)^{60}\) A1
\(A\approx$7181.41.\) 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) Solve for \(n\).
\(n = \frac{\ln 1.5}{\ln 1.06} = \frac{0.4055}{0.05827} \approx 6.96.\)M1A1
Since the target is only reached after a whole compounding period, round up: \(n = 7\) years. R1A1
Common mistakes
- Confusing simple and compound interest. Simple interest is linear (\(I=Prt\)); compound interest is geometric (\(FV=PV(1+i)^n\)) - mixing the two gives the wrong growth pattern entirely.
- Using the annual rate directly for monthly compounding. A nominal \(6\%\) rate compounded monthly needs a monthly rate of \(0.5\%\) and \(12\) periods per year - not \(6\%\) applied twelve times.
- Rounding the growth factor too early. Rounding \((1.05)^6\) to two decimal places before multiplying by the principal can shift the final answer outside the accepted range - keep full calculator accuracy until the very last step.
Ready to practise properly?
46 compound interest and growth questions, marked instantly like the real exam.
Quick answers
What is the formula for compound interest?
\(FV=PV(1+i)^{n}\) for annual compounding, where \(PV\) is the amount invested, \(i\) is the interest rate as a decimal, and \(n\) is the number of years. For \(k\) compounds per year at nominal annual rate \(r\%\), use \(FV=PV\left(1+\tfrac{r}{100k}\right)^{kn}\).
Do I need to memorise the compound interest formula for the exam?
No - it's in the formula booklet, and most exam questions expect you to solve it on the GDC's finance (TVM) solver rather than substitute into the formula by hand.