Australian Curriculum v9 / ACiQ Year 10 Mathematics - Unit 3 - Measurement error and result accuracy

Measurement error and result accuracy

Use absolute and percentage error, bounds and calculator-supported working to judge the accuracy of measurement results.

Updated 2026-07-26 - 12 min read

Measurement error and result accuracy is taught here as a connected set of decisions, not a list of facts. Work through the prerequisite recall, explicit models, carefully faded examples, misconception repairs and transfer task before using the target in Check, Practice, Review or Rapid Revision.

This note is designed to work with the guided lessons, curated practice, flashcards, Tutor context, Review and Rapid Revision for the same canonical target. The same three evidence checks are used throughout, so feedback can route a learner back to the precise idea that needs repair.

Calculate absolute and percentage error

Absolute error is |measured − accepted|; percentage error divides absolute error by the accepted or reference value and multiplies by 100%. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by calculate absolute and percentage error and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: The denominator is the accepted/reference quantity unless the task defines a different comparison.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 1.1

Measured 49.2, accepted 50. Find absolute error.

Step 1 - identify the governing idea: Absolute error is |measured − accepted|; percentage error divides absolute error by the accepted or reference value and multiplies by 100%.

Step 2 - apply it to this evidence: |49.2 − 50| = 0.8.

Result: 0.8

The relationship is visible in the working: |49.2 − 50| = 0.8. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • −0.8 — It does not agree with the required relationship: |49.2 − 50| = 0.8.
  • 99.2 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 1.6 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.2

Measured 49.2, accepted 50. Find percentage error.

Step 1 - identify the governing idea: Absolute error is |measured − accepted|; percentage error divides absolute error by the accepted or reference value and multiplies by 100%.

Step 2 - apply it to this evidence: 0.8/50 × 100% = 1.6%.

Result: 1.6%

The relationship is visible in the working: 0.8/50 × 100% = 1.6%. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 0.8% — It does not agree with the required relationship: 0.8/50 × 100% = 1.6%.
  • 1.63% — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 98.4% — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.3

A 2.40 m length is reported as 2.37 m. Find absolute error.

Step 1 - identify the governing idea: Absolute error is |measured − accepted|; percentage error divides absolute error by the accepted or reference value and multiplies by 100%.

Step 2 - apply it to this evidence: |2.37 − 2.40| = 0.03 m.

Result: 0.03 m

The relationship is visible in the working: |2.37 − 2.40| = 0.03 m. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 0.03% — It does not agree with the required relationship: |2.37 − 2.40| = 0.03 m.
  • 4.77 m — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • −0.03 m — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Use measurement bounds

A value rounded to the nearest unit lies from half a unit below inclusive to half a unit above exclusive. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by use measurement bounds and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Rounding represents an interval of possible original values.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 2.1

A length is 12 cm to the nearest centimetre. Which interval applies?

Step 1 - identify the governing idea: A value rounded to the nearest unit lies from half a unit below inclusive to half a unit above exclusive.

Step 2 - apply it to this evidence: Half a centimetre lies on each side of 12.

Result: 11.5 cm ≤ L < 12.5 cm

The relationship is visible in the working: Half a centimetre lies on each side of 12. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 12 ≤ L < 13 — It does not agree with the required relationship: Half a centimetre lies on each side of 12.
  • 11 ≤ L ≤ 12 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 11.9 ≤ L ≤ 12.1 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.2

A mass is 3.4 kg to the nearest 0.1 kg. Find the upper bound.

Step 1 - identify the governing idea: A value rounded to the nearest unit lies from half a unit below inclusive to half a unit above exclusive.

Step 2 - apply it to this evidence: Half of 0.1 kg is 0.05 kg.

Result: 3.45 kg

The relationship is visible in the working: Half of 0.1 kg is 0.05 kg. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 3.5 kg — It does not agree with the required relationship: Half of 0.1 kg is 0.05 kg.
  • 3.41 kg — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 3.35 kg — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.3

Why is the upper rounding bound usually excluded?

Step 1 - identify the governing idea: A value rounded to the nearest unit lies from half a unit below inclusive to half a unit above exclusive.

Step 2 - apply it to this evidence: At the halfway point the chosen convention assigns it to the adjacent interval.

Result: It would round to the next stated value

The relationship is visible in the working: At the halfway point the chosen convention assigns it to the adjacent interval. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Upper values are impossible — It does not agree with the required relationship: At the halfway point the chosen convention assigns it to the adjacent interval.
  • The measurement has no maximum — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Intervals cannot include endpoints — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Report calculator-supported results

Show the model and substitution, retain guard digits and give a final answer with units, justified accuracy and context. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by report calculator-supported results and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Technology performs operations but does not identify the relationship, assumptions, units or interpretation.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 3.1

Which line best communicates a calculated area?

Step 1 - identify the governing idea: Show the model and substitution, retain guard digits and give a final answer with units, justified accuracy and context.

Step 2 - apply it to this evidence: It shows formula, substitution, result, unit and accuracy.

Result: A = π(4.2)² = 55.4 m² to 1 d.p.

The relationship is visible in the working: It shows formula, substitution, result, unit and accuracy. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 55.4176944 — It does not agree with the required relationship: It shows formula, substitution, result, unit and accuracy.
  • Area = calculator — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 4.2π — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.2

Inputs are to 2 significant figures. Which final form is safest for 18.746?

Step 1 - identify the governing idea: Show the model and substitution, retain guard digits and give a final answer with units, justified accuracy and context.

Step 2 - apply it to this evidence: The result should not imply unsupported precision.

Result: 19 to 2 significant figures

The relationship is visible in the working: The result should not imply unsupported precision. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 18.746000 — It does not agree with the required relationship: The result should not imply unsupported precision.
  • 18 exactly — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 20.00000 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.3

A model yields a negative number of items. What should happen?

Step 1 - identify the governing idea: Show the model and substitution, retain guard digits and give a final answer with units, justified accuracy and context.

Step 2 - apply it to this evidence: Counts cannot be negative, so algebra alone is insufficient.

Result: Reject or reinterpret it using the contextual domain

The relationship is visible in the working: Counts cannot be negative, so algebra alone is insufficient. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Report the negative count — It does not agree with the required relationship: Counts cannot be negative, so algebra alone is insufficient.
  • Take absolute value silently — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Add units and accept it — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Retrieval check

Try these without looking back at the examples.

  1. Measured 49.2, accepted 50. Find percentage error.
  2. A mass is 3.4 kg to the nearest 0.1 kg. Find the upper bound.
  3. Inputs are to 2 significant figures. Which final form is safest for 18.746?

Answers

  1. 1.6% — 0.8/50 × 100% = 1.6%.
  2. 3.45 kg — Half of 0.1 kg is 0.05 kg.
  3. 19 to 2 significant figures — The result should not imply unsupported precision.

Transfer task

Find an unfamiliar example from school, daily life, a credible news source or another subject. Explain which of the three evidence checks applies. Complete the task, then audit your own response: identify the evidence used, the relationship applied, one plausible misconception and the final reasonableness check. If a peer could not reproduce your reasoning, add the missing step.

Sources