Approximation and Error (AI HL)
Every rounded or measured value carries some error against the true value it stands in for, and IB questions expect you to quantify exactly how much. This page covers percentage error, upper and lower bounds, and what happens to error when a measured quantity gets raised to a power - with worked examples and the slip that costs the most marks. It's part of the broader Standard Form & Approximation topic.
23 questions on this sub-topic.
Percentage error and compounding
Covered under IB syllabus reference SL1.6: choosing an appropriate degree of accuracy, upper and lower bounds of rounded numbers, and percentage errors including measurement and rounding errors.
Percentage error
\[\varepsilon = \left|\dfrac{v_A-v_E}{v_E}\right|\times100\%\]
Always divide by the exact (true) value \(v_E\), never the approximate value \(v_A\).
✓ In the formula bookletCompounding error
When a measured quantity is raised to a power in a formula (e.g. \(V\propto r^3\)), the percentage error roughly multiplies by that power - a small error in radius becomes a larger error in volume.
Not in the formula booklet - consequence of the formulaNeed bounds notation and the full syllabus wording? See Standard Form & Approximation.
Worked examples
A quantity has true value \(246\,000.\)
(a) Write it in the form \(a \times 10^{k}\).
(b) It is measured as \(250\,000.\) Find the percentage error, to 3 significant figures.
Worked solution
(a) \(246\,000\) lies between \(10^{5}\) and \(10^{6}\), so \(k=5.\) M1
\(246\,000 = 2.46\times10^{5}.\) A1
(b) \(\varepsilon = \left|\dfrac{v_A-v_E}{v_E}\right|\times100\% = \dfrac{250\,000-246\,000}{246\,000}\times100\%.\) M1
\(\varepsilon = \dfrac{4000}{246\,000}\times100\% \approx 1.63\%.\) A1
A square tile is measured as having side \(20\) cm, but the true side is \(19.6\) cm.
(a) Find the percentage error in the side measurement.
(b) Find the percentage error in the calculated area, to 3 significant figures.
Worked solution
(a) \(\varepsilon = \left|\dfrac{v_A-v_E}{v_E}\right|\times100\% = \dfrac{20-19.6}{19.6}\times100\%.\) M1
\(\varepsilon = \dfrac{0.4}{19.6}\times100\% \approx 2.04\%.\) A1
(b) Measured area \(=20^2=400\) cm\(^2\); true area \(=19.6^2=384.16\) cm\(^2\). M1
\(\varepsilon = \dfrac{400-384.16}{384.16}\times100\%.\) A1
\(\varepsilon = \dfrac{15.84}{384.16}\times100\% \approx 4.12\%.\) A1
Common mistakes
- Dividing by the wrong value in the percentage error formula. The denominator must always be the exact (true) value, never the approximate one - swapping them gives a different, incorrect percentage.
- Forgetting that error compounds through a power. When a linear measurement feeds into an area or volume formula, the resulting percentage error is not the same size as the original - a 2% error in a side length gives roughly double that in the area, not 2% again.
- Halving the tolerance instead of using it directly. A value rounded to the nearest \(0.1\) has bounds \(\pm0.05\), not \(\pm0.1\) - mixing these up shifts every upper/lower bound calculation that follows.
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23 approximation-and-error questions, marked instantly like the real exam.
Quick answers
How do you calculate 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. Always divide by the exact value.
Why does error compound when a measurement is raised to a power?
If a formula raises a measured quantity to a power, such as \(V\propto r^3\), a small percentage error in the measured quantity roughly multiplies by that power in the final answer - a \(2\%\) error in radius becomes roughly a \(6\%\) error in volume.