A part is dimensioned 50 0.2 mm. State the upper and lower limits. [2 marks]

AQA AQA 2018-style practice: A shaft is dimensioned as 20 0.1 mm. State the upper and lower limits and explain what tolerance means.

A good answer states both limits and defines tolerance correctly. The upper limit is 20 + 0.1 = 20.1 mm and the lower limit is 20 - 0.1 = 19.9 mm. Tolerance is the total allowed variation in a dimension, so here the tolerance is 20.1 - 19.9 = 0.2 mm. Any shaft measuring between 19.9 mm and 20.1 mm is acceptable; anything outside those limits is rejected. Markers reward both limits and a clear definition of tolerance as the permitted range, not a single value.

AQA AQA 2021-style practice: A hole is dimensioned with a maximum size of 25.05 mm and a minimum size of 24.95 mm. Calculate the nominal size and the tolerance, and name a tool suitable for checking it.

A good answer works from the limits back to nominal and tolerance, then chooses a capable tool. The nominal (basic) size is the midpoint of the limits: (25.05 + 24.95) / 2 = 25.00 mm. The tolerance is the difference between the limits: 25.05 - 24.95 = 0.10 mm, which can also be written as 25.00 0.05 mm. A tolerance of 0.10 mm is too fine for a steel rule (about 0.5 mm). Vernier callipers (about 0.02 mm) or a micrometer (about 0.01 mm) are capable, or a go/no-go plug gauge gives a fast accept/reject check on the production line. Markers reward the correct nominal, the correct tolerance, and a measuring tool fine enough for a 0.10 mm band.

EnglandEngineeringSyllabus dot point

How do we make sure parts are made accurately and consistently?

Quality control checks, tolerance and how upper and lower limits are stated, and the measuring and gauging equipment used to check parts.

A focused answer to AQA GCSE Engineering on quality control, tolerance with upper and lower limits, and the measuring and gauging equipment (vernier callipers, micrometers, go/no-go gauges) used to check parts.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Quality control
Tolerance
Measuring and checking equipment
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What this dot point is asking

AQA wants you to explain quality control, work out upper and lower limits from a stated tolerance (and back again), and name the right tool to measure or check a part to the needed accuracy. Calculation questions on limits and tolerance are common, so the arithmetic must be quick and reliable.

Quality control

QC includes checking dimensions, surface finish and function, often by sampling parts at set intervals during a production run rather than measuring every single one. Catching a drifting dimension early means scrapping a handful of parts instead of a whole batch, which is why QC is part of keeping cost per item low. Quality control (inspecting the made part) is sometimes contrasted with quality assurance (designing the process so faults are prevented in the first place).

Tolerance

A tighter (smaller) tolerance gives a better, more consistent fit but costs more to make because the process and the measuring must both be more precise. The tolerance chosen must match the job: a decorative bracket can have a loose tolerance, while a bearing housing needs a tight one. Limits can be symmetric (the $\pm$ form) or asymmetric (for example $20^{+0.2}_{-0.0} \text{ mm}$ ), where the upper deviation is added and the lower deviation subtracted.

Measuring and checking equipment

Steel rule: quick measurement to about $0.5 \text{ mm}$ ; suitable only for coarse work.
Vernier callipers: internal, external and depth measurement to about $0.02 \text{ mm}$ .
Micrometer: very accurate external measurement to about $0.01 \text{ mm}$ .
Go/no-go gauge: checks a part is within limits fast; the "go" end is made to the lower limit and should fit, the "no-go" end is made to the upper limit and should not.

Worked example: inspecting a batch of dowel pins against a drawing

A drawing specifies a dowel pin as $\varnothing 8 \pm 0.02 \text{ mm}$ . Plan the quality check.

step Find the limits

Upper limit $= 8 + 0.02 = 8.02 \text{ mm}$ . Lower limit $= 8 - 0.02 = 7.98 \text{ mm}$ . The tolerance is $8.02 - 7.98 = 0.04 \text{ mm}$ .

step Choose a capable tool

A $0.04 \text{ mm}$ band is far too fine for a steel rule. A micrometer (about $0.01 \text{ mm}$ ) is needed to measure the pin reliably within this band.

step Sample and measure

Take a sample from the batch and measure each pin with the micrometer. Suppose three pins read $8.00$ , $8.01$ and $8.03 \text{ mm}$ .

step Accept or reject

The $8.00$ and $8.01 \text{ mm}$ pins are within limits and pass. The $8.03 \text{ mm}$ pin is above the upper limit of $8.02 \text{ mm}$ , so it is rejected. A rising trend toward the upper limit also warns that the machine is drifting and may need resetting before more parts go out of tolerance.

Try this

Q1. A part is dimensioned $50 \pm 0.2 \text{ mm}$ . State the upper and lower limits. [2 marks]

Cue. Upper limit $50.2 \text{ mm}$ , lower limit $49.8 \text{ mm}$ .

Q2. Name a tool that quickly checks whether a part is within tolerance. [1 mark]

Cue. A go/no-go gauge.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20183 marksA shaft is dimensioned as

20 \pm 0.1

mm. State the upper and lower limits and explain what tolerance means.

Show worked answer →

A good answer states both limits and defines tolerance correctly.

The upper limit is $20 + 0.1 = 20.1 \text{ mm}$ and the lower limit is $20 - 0.1 = 19.9 \text{ mm}$ .

Tolerance is the total allowed variation in a dimension, so here the tolerance is $20.1 - 19.9 = 0.2 \text{ mm}$ . Any shaft measuring between $19.9 \text{ mm}$ and $20.1 \text{ mm}$ is acceptable; anything outside those limits is rejected.

Markers reward both limits and a clear definition of tolerance as the permitted range, not a single value.

AQA 20214 marksA hole is dimensioned with a maximum size of

25.05 \text{ mm}

and a minimum size of

24.95 \text{ mm}

. Calculate the nominal size and the tolerance, and name a tool suitable for checking it.

Show worked answer →

A good answer works from the limits back to nominal and tolerance, then chooses a capable tool.

The nominal (basic) size is the midpoint of the limits: $(25.05 + 24.95) / 2 = 25.00 \text{ mm}$ . The tolerance is the difference between the limits: $25.05 - 24.95 = 0.10 \text{ mm}$ , which can also be written as $25.00 \pm 0.05 \text{ mm}$ .

A tolerance of $0.10 \text{ mm}$ is too fine for a steel rule (about $0.5 \text{ mm}$ ). Vernier callipers (about $0.02 \text{ mm}$ ) or a micrometer (about $0.01 \text{ mm}$ ) are capable, or a go/no-go plug gauge gives a fast accept/reject check on the production line.

Markers reward the correct nominal, the correct tolerance, and a measuring tool fine enough for a $0.10 \text{ mm}$ band.

Related dot points

Sources & how we know this

AQA GCSE Engineering (8852) specification — AQA (2017)