Standard and Compound Units for University Admission

Q: What is the difference between capacity and volume?

Capacity (measured in $ml$ or $l$) usually refers to the amount of liquid a container can hold: volume (measured in $cm^3$ or $m^3$) refers to the physical space occupied by an object. Numerically, $1\text{ ml}$ is exactly equal to $1\text{ cm}^3$.

Q: Is a tonne the same as a thousand kilograms?

Yes, in the metric system, $1$ tonne ($t$) is equal to $1000\text{ kg}$.

Updated August 2025

This guide covers the essential principles of unit conversion required for university mathematics admission tests. It explains how to move between standard units of mass, length, area, volume, and time, while also mastering compound units like density and pressure. A particular focus is placed on multi-dimensional conversions and algebraic problem solving.

Core concept

Unit conversion involves multiplying or dividing by specific factors to translate a measurement between different scales. For area and volume, the conversion factor is the linear scale factor squared or cubed respectively: for compound units, each component part must be converted individually.

Standard Units

To succeed in mathematics admission tests, you must be comfortable using and converting between standard units. The following are the standard categories of measurement and their common units:

Mass: Milligrams ( $mg$ ), grams ( $g$ ), kilograms ( $kg$ ), and tonnes ( $t$ ).
Force: Newtons ( $N$ ).
Length: Millimetres ( $mm$ ), centimetres ( $cm$ ), metres ( $m$ ), and kilometres ( $km$ ).
Area: Square millimetres ( $mm^2$ ), square centimetres ( $cm^2$ ), square metres ( $m^2$ ), and square kilometres ( $km^2$ ).
Capacity and Volume: Millilitres ( $ml$ ), litres ( $l$ ), cubic millimetres ( $mm^3$ ), cubic centimetres ( $cm^3$ ), and cubic metres ( $m^3$ ).

There are specific relationships between capacity and volume that you must memorise:

$1\text{ ml} = 1\text{ cm}^3$
$1\text{ l} = 1\text{ dm}^3 = 1000\text{ cm}^3$
$1000\text{ l} = 1\text{ m}^3$

Generally, small quantities of liquid are measured in $ml$ or $l$ , while larger volumes, such as those in swimming pools or reservoirs, are measured in $m^3$ .

Time Measures

Time follows a non-decimal system. Key intervals include:

Year: $12$ months, or $365$ days ( $366$ in leap years). Leap years occur nearly every $4$ years.
Century: $100$ years.
Millennium: $1000$ years.
Standard blocks: $60$ seconds = $1$ minute: $60$ minutes = $1$ hour: $24$ hours = $1$ day: $7$ days = $1$ week.

Compound Units

Compound units are formed by combining two or more standard measurements. For example, average speed is calculated by dividing the distance travelled by the time taken. If distance is in $km$ and time is in hours, the units are kilometres per hour ( $km/h$ or $km\ h^{-1}$ ).

Other common compound units include:

Density: Mass divided by volume ( $g/cm^3$ or $kg/m^3$ ).
Rates of Pay: Pay received divided by time worked ( $£/h$ ).
Unit Pricing: Total cost divided by the number of items ( $£/\text{item}$ ).
Pressure: Force divided by area ( $N/m^2$ ).

Worked Example: Unit Cost

Problem: $50$ boxes of sweets cost $£215$ . What is the unit cost per box?

Solution: The unit cost is the cost of $1$ box. Divide the total cost by the quantity: $£215 \div 50 = £4.30$ per box.

Changing Between Standard Units

When converting units, ensure you apply the factor correctly to the dimension of the measurement.

Measurement	Conversion Factors
Length	$1\text{ km} = 1000\text{ m}$ : $1\text{ m} = 100\text{ cm} = 1000\text{ mm}$ : $1\text{ cm} = 10\text{ mm}$
Area	$1\text{ km}^2 = 1000^2\text{ m}^2 = 1,000,000\text{ m}^2$ : $1\text{ m}^2 = 100^2\text{ cm}^2 = 10,000\text{ cm}^2$ : $1\text{ cm}^2 = 10^2\text{ mm}^2 = 100\text{ mm}^2$
Volume	$1\text{ m}^3 = 100^3\text{ cm}^3 = 1,000,000\text{ cm}^3$ : $1\text{ cm}^3 = 10^3\text{ mm}^3 = 1000\text{ mm}^3$
Mass	$1\text{ kg} = 1000\text{ g}$ : $1\text{ g} = 1000\text{ mg}$

Worked Example: Area Conversion

Problem: How many $cm^2$ are in $2.65\text{ m}^2$ ?

Solution: Since $1\text{ m} = 100\text{ cm}$ , then $1\text{ m}^2 = (100)^2\text{ cm}^2 = 10,000\text{ cm}^2$ . Therefore, $2.65\text{ m}^2 = 2.65 \times 10,000 = 26,500\text{ cm}^2$ .

Changing Between Compound Units

To change compound units, you must convert each component unit separately.

Worked Example: Density Conversion

Problem: The density of a metal is $20\text{ g cm}^{-3}$ . Convert this to $kg\text{ m}^{-3}$ .

Solution:

Think of $20\text{ g cm}^{-3}$ as $\frac{20\text{ g}}{1\text{ cm}^3}$ .
Convert grams to kilograms: $20\text{ g} = \frac{20}{1000}\text{ kg} = 0.02\text{ kg}$ .
Convert cubic centimetres to cubic metres: $1\text{ cm}^3 = \frac{1}{1,000,000}\text{ m}^3$ .
Perform the division: $\frac{0.02\text{ kg}}{1/1,000,000\text{ m}^3} = 0.02 \times 1,000,000 = 20,000\text{ kg m}^{-3}$ .

Worked Example: Speed Problem Solving

Problem: A car travels $35,000\text{ m}$ in $30$ minutes. What is its average speed in $km/h$ ?

Solution:

Convert the distance to kilometres: $35,000\text{ m} = \frac{35,000}{1000} = 35\text{ km}$ .
Convert the time to hours: $30\text{ min} = \frac{30}{60} = 0.5$ hours.
Calculate speed: $\text{Average speed} = \frac{35\text{ km}}{0.5\text{ h}} = 70\text{ km h}^{-1}$ .

Key takeaways

Area conversion factors are the square of linear factors: volume factors are the cube of linear factors.
The standard relationship $1\text{ ml} = 1\text{ cm}^3$ is the fundamental bridge between capacity and volume.
Compound units like $kg\ m^{-3}$ can be written using negative indices such as $kg\ m^{-3}$ .
When converting compound units, process the numerator and denominator units independently before simplifying.
In leap years, there are $366$ days: this occurs nearly every $4$ years.

Tips

In exam conditions, always double-check if your answer's magnitude makes sense. For example, if you convert a density from $g/cm^3$ to $kg/m^3$ , the number should increase significantly because a cubic metre is much larger than a cubic centimetre.

Cautions

The most common error is forgetting to square or cube the conversion factor for area and volume. Forgetting that $1\text{ m}^3 = 1,000,000\text{ cm}^3$ (not $100$ or $1000$ ) will lead to large scale errors in physical science problems.

Insight

Compound units are essentially fractions. Treating them algebraically (e.g. $Speed = Distance \cdot Time^{-1}$ ) allows you to use the laws of indices to simplify complex units and ensure that your final derived unit matches the required dimensions of the physical quantity.

Worked Examples

Example 1

A piece of metal of mass 50 g is at thermal equilibrium in a hot liquid at temperature

T

.

The metal is removed from the liquid and immediately placed in 100 g of water that is at 20 °C.

The water is stirred and reaches a final temperature of 26 °C.

| material | specific heat capacity / J kg

^{-1}

°C

^{-1}

|
|---|---|
| hot liquid | 2000 |
| metal | 350 |
| water | 4200 |

What is the temperature

T

of the hot liquid?

(Assume that heat transfers to or from the surroundings are negligible.)

A:38 °C

B:51 °C

C:150 °C

D:170 °C

E:480 °C

Practice Questions

Practice Question 1

A light, metal wire of length 2.5 m and cross-sectional area

1.8 \times 10^{-6} m^2

is suspended vertically. A mass of 7.2 kg is attached to the lower end of the wire. The wire extends by 0.50 mm.

What is the Young modulus of the metal and how much energy is stored in the extended wire?

(gravitational field strength =

10 N kg^{-1}

; assume that the wire obeys Hooke's law and that changes in the cross-sectional area are negligible)

| | Young modulus / Pa | energy stored / J |
| :--- | :--- | :--- |
| A |

5.0 \times 10^{-12}

| 0.018 |
| B |

5.0 \times 10^{-12}

| 0.036 |
| C |

2.0 \times 10^{11}

| 0.018 |
| D |

2.0 \times 10^{11}

| 0.036 |
| E |

2.0 \times 10^{14}

| 18 |
| F |

2.0 \times 10^{14}

| 36 |

A:Young modulus / Pa:

5.0 \times 10^{-12}

, energy stored / J: 0.018

B:Young modulus / Pa:

5.0 \times 10^{-12}

, energy stored / J: 0.036

C:Young modulus / Pa:

2.0 \times 10^{11}

, energy stored / J: 0.018

D:Young modulus / Pa:

2.0 \times 10^{11}

, energy stored / J: 0.036

E:Young modulus / Pa:

2.0 \times 10^{14}

, energy stored / J: 18

F:Young modulus / Pa:

2.0 \times 10^{14}

, energy stored / J: 36

Frequently asked questions

Why do I multiply by 10,000 to convert m squared to cm squared instead of 100?

Because area is two-dimensional. A square metre is $100\text{ cm}$ wide and $100\text{ cm}$ high: therefore, its area is $100 \times 100 = 10,000\text{ cm}^2$ .

What is the difference between capacity and volume?

Capacity (measured in $ml$ or $l$ ) usually refers to the amount of liquid a container can hold: volume (measured in $cm^3$ or $m^3$ ) refers to the physical space occupied by an object. Numerically, $1\text{ ml}$ is exactly equal to $1\text{ cm}^3$ .

How do I handle a rate like miles per gallon if I need to convert to kilometres per litre?

Convert the miles to kilometres in the numerator and the gallons to litres in the denominator. Then divide the new numerator by the new denominator to find the value in the new compound unit.

Is a tonne the same as a thousand kilograms?

Yes, in the metric system, $1$ tonne ( $t$ ) is equal to $1000\text{ kg}$ .

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