What formulas do I need for density and pressure in GCSE Maths?

Use $\rho=\dfrac{m}{V}$ for density and $P=\dfrac{F}{A}$ for pressure. Their rearranged forms are $m=\rho V$, $V=\dfrac{m}{\rho}$, $F=PA$ and $A=\dfrac{F}{P}$.

What units are used for density?

Common density units are $\text{g/cm}^3$ and $\text{kg/m}^3$. The mass and volume values must be compatible with the required density unit before you substitute them into the formula.

What units are used for pressure?

When force is measured in newtons and area in square metres, pressure is measured in $\text{N/m}^2$. This is equivalent to pascals, where $1\text{ Pa}=1\text{ N/m}^2$.

How do I convert square centimetres to square metres?

Divide by $10\,000$ because $1\text{ m}^2=10\,000\text{ cm}^2$. For example, $600\text{ cm}^2=0.06\text{ m}^2$.

How do I rearrange the density formula to find volume?

Start with $\rho=\dfrac{m}{V}$. Multiply both sides by $V$, then divide by $\rho$, giving $V=\dfrac{m}{\rho}$.

Are density and pressure on Foundation or Higher GCSE Maths papers?

Density and pressure can be assessed on Foundation and Higher GCSE Maths papers as compound units. The precise presentation varies by exam board, while Higher questions may use more demanding algebra, conversions or multi-step reasoning.

Density and Pressure GCSE Maths Revision Guide

Density and pressure GCSE Maths questions test your ability to use compound measures. You need to select the correct formula, make the units consistent, substitute accurately and rearrange the formula when the required quantity is not already the subject.

Both topics are part of ratio, proportion and rates of change. They can appear on Foundation and Higher papers, although Higher questions may involve more demanding unit conversions, algebraic rearrangement or several connected steps. This guide explains the methods step by step and includes density and pressure worked examples followed by exam-style questions.

What are density and pressure?

Density and pressure are compound measures. A compound measure combines two or more measurements into a new quantity.

Density compares mass with volume. It describes how much mass is packed into each unit of volume.
Pressure compares force with area. It describes how much force acts on each unit of area.

The word per indicates division. For example, grams per cubic centimetre means grams divided by cubic centimetres.

Measure	Formula	Common units
Density	$density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}}$	$g/cm3\text{g/cm}^3$ , $kg/m3\text{kg/m}^3$
Pressure	$pressure=forcearea\text{pressure}=\dfrac{\text{force}}{\text{area}}$	$N/m2\text{N/m}^2$ , pascals ( $Pa\text{Pa}$ )

The formulas look similar because both involve dividing an amount by the space over which that amount is distributed. However, mass and force are different quantities, so the formulas cannot be used interchangeably.

The density formula

Using $ρ\rho$ for density, $m$ for mass and $V$ for volume, the density formula is

ρ=mV.\rho=\frac{m}{V}.

You must be able to use all three forms:

ρ=mV,m=ρV,V=mρ.\rho=\frac{m}{V}, \qquad m=\rho V, \qquad V=\frac{m}{\rho}.

A formula triangle may help you remember these relationships, but understanding the rearrangement is safer. Starting with $ρ=m/V\rho=m/V$ , multiply both sides by $V$ to obtain $ρV=m\rho V=m$ . To find $V$ , divide both sides by $ρ\rho$ .

Density worked example 1: finding density

A metal block has a mass of $g540\text{ g}$ and a volume of $cm360\text{ cm}^3$ . Find its density.

Step 1: Choose the formula.

density=massvolume\text{density}=\frac{\text{mass}}{\text{volume}}

Step 2: Substitute the values.

density=54060=9\text{density}=\frac{540}{60}=9

Step 3: Give the appropriate unit.

g/cm3\boxed{9\text{ g/cm}^3}

The unit follows from dividing grams by cubic centimetres.

Density worked example 2: finding volume

A liquid has a mass of $g680\text{ g}$ and a density of $g/cm30.8\text{ g/cm}^3$ . Find its volume.

Rearrange the formula so that volume is the subject:

V=mρ.V=\frac{m}{\rho}.

Substitute and calculate:

V=6800.8=850.V=\frac{680}{0.8}=850.

Therefore,

cm3.\boxed{V=850\text{ cm}^3}.

A density of $g/cm30.8\text{ g/cm}^3$ means each cubic centimetre has a mass of $g0.8\text{ g}$ . Dividing the total mass by this amount gives the number of cubic centimetres.

Units and conversions in density questions

Before substituting values, check that the mass and volume units match the density unit. If density is given in $g/cm3\text{g/cm}^3$ , use mass in grams and volume in cubic centimetres.

Useful conversions include:

g1\text{ kg}=1000\text{ g}

cm1\text{ m}=100\text{ cm}

For volume, the length conversion must be cubed:

cm3.1\text{ m}^3=(100\text{ cm})^3=1\,000\,000\text{ cm}^3.

This is a frequent source of errors. It is incorrect to say that $cm31\text{ m}^3=100\text{ cm}^3$ .

Density worked example 3: converting units

A substance has a mass of $kg1.8\text{ kg}$ and a volume of $cm3600\text{ cm}^3$ . Find its density in $g/cm3\text{g/cm}^3$ .

Convert kilograms to grams first:

g.1.8\text{ kg}=1800\text{ g}.

Now use the density formula:

ρ=1800600=3.\rho=\frac{1800}{600}=3.

Therefore,

g/cm3.\boxed{3\text{ g/cm}^3}.

The pressure formula

In GCSE Maths, pressure is normally calculated using

P=FA,P=\frac{F}{A},

where $P$ is pressure, $F$ is force and $A$ is area.

The rearranged forms are

P=FA,F=PA,A=FP.P=\frac{F}{A}, \qquad F=PA, \qquad A=\frac{F}{P}.

If force is measured in newtons and area in square metres, pressure is measured in newtons per square metre:

N/m2.\text{N/m}^2.

One newton per square metre is also called one pascal:

N/m2.1\text{ Pa}=1\text{ N/m}^2.

Pressure worked example 1: finding pressure

A force of $N480\text{ N}$ acts over an area of $m20.6\text{ m}^2$ . Calculate the pressure.

Use

P=FA.P=\frac{F}{A}.

Substitute the values:

P=4800.6=800.P=\frac{480}{0.6}=800.

Therefore,

N/m2\boxed{800\text{ N/m}^2}

or, equivalently, $Pa\boxed{800\text{ Pa}}$ .

Pressure worked example 2: finding force

The pressure on a surface is $N/m22500\text{ N/m}^2$ . The contact area is $m20.12\text{ m}^2$ . Find the force.

Rearrange the formula:

F = P A .

Then

F=2500×0.12=300.F=2500\times0.12=300.

The force is

N.\boxed{300\text{ N}}.

Pressure worked example 3: finding contact area

A box exerts a force of $N72\text{ N}$ and creates a pressure of $N/m230\text{ N/m}^2$ . Find its contact area.

Use

A=FP.A=\frac{F}{P}.

Therefore,

A=7230=2.4.A=\frac{72}{30}=2.4.

The area is

m2.\boxed{2.4\text{ m}^2}.

Area conversions in pressure questions

Pressure problems may provide area in square centimetres but require pressure in $N/m2\text{N/m}^2$ . Since

cm,1\text{ m}=100\text{ cm},

it follows that

cm2.1\text{ m}^2=10\,000\text{ cm}^2.

To convert from $cm2\text{cm}^2$ to $m2\text{m}^2$ , divide by $00010\,000$ . To convert from $m2\text{m}^2$ to $cm2\text{cm}^2$ , multiply by $00010\,000$ .

For example,

m2.350\text{ cm}^2=\frac{350}{10\,000}=0.035\text{ m}^2.

Do not divide by $100$ . Area is two-dimensional, so the linear conversion factor must be squared.

A reliable method for density and pressure exam questions

Use the following method whenever you meet a compound-measure problem:

Identify the required quantity. Decide whether you need density, mass, volume, pressure, force or area.
Write the relevant formula. This can earn a method mark and reduces the risk of using the numbers in the wrong order.
Rearrange before substituting, if necessary.
Check the units. Convert measurements so that they are compatible.
Substitute with units visible in your working.
Calculate without rounding too early. Keep the full calculator value until the final line.
State the final unit. A numerical answer without its compound unit may be incomplete.
Check whether the answer is reasonable. Density should be mass divided by volume, while increasing area for the same force should reduce pressure.

For board-specific topic placement, use MathsGenie's GCSE Maths topics by exam board and tier.

Common mistakes and how to avoid them

Dividing in the wrong order

Density is mass divided by volume, not volume divided by mass. Pressure is force divided by area, not area divided by force. Write the formula before entering anything into your calculator.

Multiplying when division is required

To find mass, use $m=ρVm=\rho V$ , so multiplication is correct. To find volume, use $V=m/ρV=m/\rho$ , so division is required. Rearrange algebraically rather than guessing the operation.

Ignoring squared and cubed units

Length, area and volume conversions use different scale factors. If $cm1\text{ m}=100\text{ cm}$ , then $cm21\text{ m}^2=10\,000\text{ cm}^2$ and $cm31\text{ m}^3=1\,000\,000\text{ cm}^3$ .

Mixing incompatible units

Using kilograms with a density in $g/cm3\text{g/cm}^3$ produces an incorrect answer unless you convert first. Choose the unit required by the question, then make every measurement compatible with it.

Confusing mass with force

Mass may be measured in grams or kilograms, whereas force is measured in newtons. In a GCSE Maths pressure question, use the force stated in the question. Do not treat a mass in kilograms as though it were automatically the same numerical value in newtons.

Omitting the unit

Compound units communicate what the number means. Typical answers include $g/cm37.8\text{ g/cm}^3$ , $kg/m31200\text{ kg/m}^3$ and $N/m2450\text{ N/m}^2$ .

Embedded exam-style question

A solid cuboid has dimensions $cm20\text{ cm}$ by $cm8\text{ cm}$ by $cm5\text{ cm}$ . Its mass is $kg2.4\text{ kg}$ . Calculate its density in $g/cm3\text{g/cm}^3$ .

First find the cuboid's volume:

cm3.V=20\times8\times5=800\text{ cm}^3.

Convert the mass:

g.2.4\text{ kg}=2400\text{ g}.

Then calculate density:

g/cm3.\rho=\frac{2400}{800}=3\text{ g/cm}^3.

g/cm3\boxed{3\text{ g/cm}^3}

Try 4 more density and pressure practice questions.

Density and pressure exam questions

Attempt each question before checking the answers. Show the formula, substitutions and units as you would in an examination.

Question 1

A material has a mass of $g975\text{ g}$ and a volume of $cm3125\text{ cm}^3$ . Calculate its density.

Answer:

g/cm3.\rho=\frac{975}{125}=\boxed{7.8\text{ g/cm}^3}.

Question 2

A liquid has a density of $g/cm31.2\text{ g/cm}^3$ and a volume of $cm3450\text{ cm}^3$ . Calculate its mass.

Answer:

g.m=\rho V=1.2\times450=\boxed{540\text{ g}}.

Question 3

A force of $N900\text{ N}$ acts on an area of $m20.15\text{ m}^2$ . Calculate the pressure.

Answer:

N/m2.P=\frac{900}{0.15}=\boxed{6000\text{ N/m}^2}.

Question 4

A force of $N240\text{ N}$ acts over an area of $cm2800\text{ cm}^2$ . Calculate the pressure in $N/m2\text{N/m}^2$ .

Answer: Convert the area first.

m2800\text{ cm}^2=0.08\text{ m}^2

N/m2.P=\frac{240}{0.08}=\boxed{3000\text{ N/m}^2}.

For further exam practice, MathsGenie provides Edexcel Speed and Density questions, additional question-by-question practice and Eduqas Speed and Density questions. These resources include worked solutions or marking support so that you can compare your method with the mark scheme.

Conclusion

For density, remember $ρ=m/V\rho=m/V$ . For pressure, remember $P = F / A$ . Rearrange the relevant formula carefully, make every unit compatible and include the correct compound unit in your answer.

Unit conversion is often the decisive part of a multi-step question, particularly when square or cubic units are involved. After learning the formulas, use MathsGenie's revision materials and exam-style practice to apply them under exam conditions and check your work against the provided solutions and mark schemes.

Sources and referenced URLs

Density and Pressure GCSE Maths Revision Guide

What are density and pressure?

Density and pressure are compound measures. A compound measure combines two or more measurements into a new quantity.

Density compares mass with volume. It describes how much mass is packed into each unit of volume.
Pressure compares force with area. It describes how much force acts on each unit of area.

The word per indicates division. For example, grams per cubic centimetre means grams divided by cubic centimetres.

Measure	Formula	Common units
Density	$density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}}$	$g/cm3\text{g/cm}^3$ , $kg/m3\text{kg/m}^3$
Pressure	$pressure=forcearea\text{pressure}=\dfrac{\text{force}}{\text{area}}$	$N/m2\text{N/m}^2$ , pascals ( $Pa\text{Pa}$ )

The density formula

Using $ρ\rho$ for density, $m$ for mass and $V$ for volume, the density formula is

ρ=mV.\rho=\frac{m}{V}.

You must be able to use all three forms:

ρ=mV,m=ρV,V=mρ.\rho=\frac{m}{V}, \qquad m=\rho V, \qquad V=\frac{m}{\rho}.

Density worked example 1: finding density

A metal block has a mass of $g540\text{ g}$ and a volume of $cm360\text{ cm}^3$ . Find its density.

Step 1: Choose the formula.

density=massvolume\text{density}=\frac{\text{mass}}{\text{volume}}

Step 2: Substitute the values.

density=54060=9\text{density}=\frac{540}{60}=9

Step 3: Give the appropriate unit.

g/cm3\boxed{9\text{ g/cm}^3}

The unit follows from dividing grams by cubic centimetres.