Convert 4700 to kilo-ohms. [1 mark]

AQA AQA 2018-style practice: A rectangular steel plate measures 200 mm by 150 mm. Calculate its area in mm^2 and convert your answer to cm^2.

A good answer shows the area calculation and the correct unit conversion. Area = length × width = 200 × 150 = 30000 mm^2. To convert to cm^2, remember 1 cm = 10 mm, so 1 cm^2 = 100 mm^2. Divide by 100: 30000 / 100 = 300 cm^2. Markers reward the correct area and the correct area-unit conversion (dividing by 100, not 10).

AQA AQA 2021-style practice: A solid steel cylinder has a radius of 20 mm and a length of 50 mm. Steel has a density of 7800 kg/m^3. Calculate the mass of the cylinder in kilograms. Use π = 3.14.

A good answer finds the volume, converts to consistent units, then uses mass = density × volume. Volume of a cylinder = π r^2 × length = 3.14 × 20^2 × 50 = 3.14 × 400 × 50 = 62800 mm^3. Convert to cubic metres: 1 m^3 = 10^9 mm^3, so 62800 mm^3 = 62800 × 10^-9 = 6.28 × 10^-5 m^3. Mass = density × volume = 7800 × 6.28 × 10^-5 = 0.49 kg (to 2 significant figures). Markers reward the correct volume, the conversion to m^3, and mass = density × volume giving about 0.49 kg.

EnglandEngineeringSyllabus dot point

What core maths do engineers use for measurement and quantities?

Units and prefixes, rearranging formulae, ratio and proportion, areas and volumes, and using standard form for engineering calculations.

A focused answer to AQA GCSE Engineering on the core maths engineers use, including SI units and prefixes, rearranging formulae, ratio and proportion, area and volume, and standard form.

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
Units and prefixes
Rearranging formulae
Ratio, proportion, area and volume
Standard form
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What this dot point is asking

AQA wants you to handle units and prefixes, rearrange formulae, use ratio and proportion, calculate areas and volumes, and write and read numbers in standard form, all in engineering contexts. Calculation marks are won by showing the method, keeping units consistent, and stating the unit with the final answer.

Units and prefixes

The biggest source of error is mixing units within one calculation, for example using millimetres in one term and metres in another. Convert everything to a single consistent set of units before substituting numbers, and the arithmetic looks after itself.

Rearranging formulae

The rule is to undo operations in reverse order and to keep both sides balanced. A quick check is to substitute simple numbers back into the rearranged formula and confirm the original still holds, which catches most slips. The same approach works for any engineering formula in the paper, for example rearranging $P = V \times I$ to find current as $I = \dfrac{P}{V}$ , or rearranging density $= \dfrac{\text{mass}}{\text{volume}}$ to find mass as mass $=$ density $\times$ volume. Treat the unknown as the thing you want alone, then move every other term to the far side one operation at a time.

Ratio, proportion, area and volume

Ratio and proportion: used for gear ratios, scale drawings and mixing. A scale of $1:50$ means $1$ unit on paper is $50$ in real life.
Area: for a rectangle, $\text{area} = \text{length} \times \text{width}$ ; for a circle, $\text{area} = \pi r^2$ . Remember $1 \text{ cm}^2 = 100 \text{ mm}^2$ .
Volume: for a cuboid, $\text{volume} = \text{length} \times \text{width} \times \text{height}$ ; for a cylinder, $\text{volume} = \pi r^2 \times \text{length}$ . Remember $1 \text{ cm}^3 = 1000 \text{ mm}^3$ .

Standard form

Standard form writes a number as a value between $1$ and $10$ times a power of ten, for example $47000 = 4.7 \times 10^{4}$ and $0.0033 = 3.3 \times 10^{-3}$ . It keeps large and small engineering numbers manageable and makes multiplying and dividing them quick by adding or subtracting the powers.

Worked example: rearranging and substituting to find a missing length

A rectangular plate must have an area of $24000 \text{ mm}^2$ and a width of $120 \text{ mm}$ . Find its length, then give the area in $\text{cm}^2$ .

step Choose and rearrange the formula

Area $= \text{length} \times \text{width}$ . Rearrange to make length the subject: length $= \dfrac{\text{area}}{\text{width}}$ .

step Substitute the values

length $= \dfrac{24000}{120} = 200 \text{ mm}$ . Keep the unit (mm) with the answer.

step Convert the area to cm squared

Because $1 \text{ cm}^2 = 100 \text{ mm}^2$ , divide by $100$ : $24000 / 100 = 240 \text{ cm}^2$ . Note you divide by $100$ for an area, not by $10$ .

step Sanity check

A $200 \text{ mm}$ by $120 \text{ mm}$ plate has area $200 \times 120 = 24000 \text{ mm}^2$ , matching the requirement, so the rearrangement and substitution are correct.

Try this

Q1. Convert $4700 \text{ }\Omega$ to kilo-ohms. [1 mark]

Cue. $4700 / 1000 = 4.7 \text{ k}\Omega$ .

Q2. Write $56000$ in standard form. [1 mark]

Cue. $5.6 \times 10^{4}$ .

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 rectangular steel plate measures

200 \text{ mm}

150 \text{ mm}

. Calculate its area in

\text{mm}^2

and convert your answer to

\text{cm}^2

Show worked answer →

A good answer shows the area calculation and the correct unit conversion.

Area $= \text{length} \times \text{width} = 200 \times 150 = 30000 \text{ mm}^2$ .

To convert to $\text{cm}^2$ , remember $1 \text{ cm} = 10 \text{ mm}$ , so $1 \text{ cm}^2 = 100 \text{ mm}^2$ . Divide by $100$ : $30000 / 100 = 300 \text{ cm}^2$ .

Markers reward the correct area and the correct area-unit conversion (dividing by $100$ , not $10$ ).

AQA 20214 marksA solid steel cylinder has a radius of

20 \text{ mm}

and a length of

50 \text{ mm}

. Steel has a density of

7800 \text{ kg/m}^3

. Calculate the mass of the cylinder in kilograms. Use

\pi = 3.14

Show worked answer →

A good answer finds the volume, converts to consistent units, then uses mass $=$ density $\times$ volume.

Volume of a cylinder $= \pi r^2 \times \text{length} = 3.14 \times 20^2 \times 50 = 3.14 \times 400 \times 50 = 62800 \text{ mm}^3$ .

Convert to cubic metres: $1 \text{ m}^3 = 10^9 \text{ mm}^3$ , so $62800 \text{ mm}^3 = 62800 \times 10^{-9} = 6.28 \times 10^{-5} \text{ m}^3$ .

Mass $=$ density $\times$ volume $= 7800 \times 6.28 \times 10^{-5} = 0.49 \text{ kg}$ (to 2 significant figures).

Markers reward the correct volume, the conversion to $\text{m}^3$ , and mass $=$ density $\times$ volume giving about $0.49 \text{ kg}$ .

Related dot points

Sources & how we know this

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