Percentage Error (AI SL)
Percentage error measures how far an approximate or measured value strays from the true value, expressed as a percentage of the true value. This page covers the formula booklet formula, how to keep the sign and denominator correct, and how error behaves when a quantity gets squared or cubed - with worked examples and the mistakes that lose the most marks. It's part of the broader Approximation & Error topic.
21 questions on this sub-topic.
The formula
Covered under IB syllabus reference SL1.6: percentage errors, including finding the maximum percentage error caused by a measurement error - for example, the percentage error in the area of a circle if the radius was measured as \(2.5\) cm to one decimal place.
Percentage error
\[\varepsilon = \left|\dfrac{v_A - v_E}{v_E}\right|\times100\%\]
Always divide by the exact value \(v_E\), and keep the answer positive.
✓ In the formula bookletWhen a quantity is squared or cubed
Percentage error roughly doubles when a linear measurement is squared (e.g. an area), and roughly triples when it's cubed (e.g. a volume).
You still calculate it the same way - find the two actual values first, then apply the formula to those.
Need the full syllabus wording and formula-booklet reference table? See Approximation & Error. For GDC keystrokes on error questions, see the parent topic's GDC guidance.
Worked examples
A length is measured as 12.4 cm; the true value is 12.0 cm.
Find the percentage error.
Worked solution
\(\varepsilon = \left|\dfrac{v_A - v_E}{v_E}\right|\times100\%\), where \(v_A\) is the approximate (measured) value and \(v_E\) the exact value. The denominator is always the exact value, and the absolute value keeps the error non-negative. M1
\(v_A=12.4,\ v_E=12.0\): \(\varepsilon = \left|\frac{12.4-12.0}{12.0}\right|\times100\% = \frac{0.4}{12.0}\times100\%.\) A1
\(\dfrac{0.4}{12.0}=0.0333\ldots\), so \(\varepsilon \approx 3.33\%.\) A1
A square has side measured as 5.0 cm and area calculated from it. The true side is 4.9 cm.
(a) Find the calculated area and the true area.
(b) Find the percentage error in the area.
Worked solution
(a) Step 1 - Area uses the squared side. Calculated \(= 5.0^{2} = 25\) cm²; true \(= 4.9^{2}\) M1 \(= 24.01\) cm². A1
(b) Step 1 - Apply the error formula to the areas (\(v_E = 24.01\)): \(\varepsilon = \left|\frac{25-24.01}{24.01}\right|\times100\%.\) M1
\(\dfrac{0.99}{24.01}\times100\% \approx 4.12\%.\) Note this is about double the \(\approx2\%\) error in the side - squaring roughly doubles the percentage error. A1
Common mistakes
- Dividing by the approximate value instead of the exact value. The formula is \(\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%\) - the denominator must always be the exact (true) value, and the modulus keeps the result positive.
- Rounding intermediate values before the final step. Rounding too early compounds error through a calculation. Keep full calculator precision throughout, and round only the final answer.
- Applying the formula to the original measurement instead of the derived quantity. If a question asks for the error in an area or volume, you must first calculate both the approximate and exact area or volume, then apply the error formula to those - not to the side length itself.
Ready to practise properly?
23 percentage-error questions, marked instantly like the real exam.
Quick answers
What is the formula for percentage error?
\(\varepsilon = \left|\dfrac{v_A - v_E}{v_E}\right|\times100\%\), where \(v_A\) is the approximate value and \(v_E\) is the exact value. The modulus keeps the result positive.
Do you divide by the approximate value or the exact value?
Always the exact (true) value, \(v_E\). Dividing by the approximate value instead is one of the most common mark losses on this topic.