N13, R1, R11, G14 - Standard units and compound measures

A numerical answer gains meaning from its unit: a distance, an area and a speed need different conversions. You will convert standard and compound units, read times and timetables, and calculate speed, pay, unit prices, density and pressure using consistent units.

Choosing and converting standard units

Does $72\,\mathrm{km/h}$ mean the same speed as $20\,\mathrm{m/s}$ ? The numbers look very different, but units decide the comparison. Start by changing one unit at a time.

You already know

Multiplication and division undo each other. A smaller unit needs a larger count for the same quantity: one metre contains 100 centimetres.

A measurement gives a number and a unit. Changing the unit changes the number, while the measured quantity stays the same. Use a practical unit: centimetres for a pen, kilometres for a road.

Quantity	Unit equalities to know
Length	$1\,\mathrm{km}=1000\,\mathrm m$ ; $1\,\mathrm m=100\,\mathrm{cm}$ ; $1\,\mathrm{cm}=10\,\mathrm{mm}$
Mass	$1\,\mathrm{kg}=1000\,\mathrm g$ ; 1 tonne $=1000\,\mathrm{kg}$
Capacity	$1\,\mathrm L=1000\,\mathrm{ml}$
Time	1 minute = 60 seconds; 1 hour = 60 minutes; 1 day = 24 hours
Money	£1 = 100 pence

These unit equalities are not on the formula sheet: learn them. Currencies are different units too: you need a stated exchange rate to convert between them.

Worked example 1: smaller units

Question. Convert $2.35\,\mathrm{kg}$ to grams.

Plan. Count 1000 grams for each kilogram.

\begin{aligned} 1\,\mathrm{kg}&=1000\,\mathrm g && \text{write the unit equality}\\ 2.35\,\mathrm{kg}&=2.35\times1000=2350\,\mathrm g && \text{scale the gram count} \end{aligned}

Why this step? Grams are smaller, so more of them fit the same mass.

Answer. $2350\,\mathrm g$ .

Check. $2350\div1000=2.35$ .

Change only the direction of the conversion: start with the gram measurement and recover kilograms.

Worked example 2: larger units

Question. Convert $2350\,\mathrm g$ to kilograms.

Plan. Group the grams into groups of 1000.

\begin{aligned} 1000\,\mathrm g&=1\,\mathrm{kg} && \text{one group is a kilogram}\\ 2350\,\mathrm g&=2350\div1000=2.35\,\mathrm{kg} && \text{count the groups} \end{aligned}

Why this step? The larger unit needs a smaller numerical count.

Answer. $2.35\,\mathrm{kg}$ .

Check. It lies between 2 kg and 3 kg, as 2350 g lies between 2000 g and 3000 g.

Decimals need care in money too: £3.08 is 308 pence, since the $0.08$ pounds is 8 pence. Similarly, $640\,\mathrm{mm}=0.64\,\mathrm m$ .

Watch out

Wrong: $750\,\mathrm g=750\times1000\,\mathrm{kg}$ . Multiplying gives a much larger mass. Correct: $750\div1000=0.75\,\mathrm{kg}$ , since a kilogram contains 1000 grams.

Use the same larger-unit conversion for a different mass.

Clock times, elapsed time and timetables

A clock time tells you when something happens; a duration tells you how long it lasts. Keep them distinct when a journey crosses midnight.

In 12-hour notation, a.m. is before noon and p.m. is after noon: 7:25 p.m. is 19:25. Midnight is 00:00 in 24-hour notation and noon is 12:00. For afternoon and evening times after 12:59 p.m., add 12 to the hour; 12:25 p.m. stays 12:25, while 12:25 a.m. is 00:25. Include the day when it matters.

A decimal duration is a fraction of an hour. $1.4$ hours is 1 hour plus $0.4\times60=24$ minutes, not 1 hour 40 minutes. Conversely, 1 hour 45 minutes is $1+45/60=1.75$ hours. Sixty minutes make an hour, so counting forward through convenient clock times is often safer than subtracting four-digit times.

Worked example 3: a journey across midnight

Question. A train leaves at 23:40 on Monday and arrives at 00:25 on Tuesday. Find the journey time.

Plan. Count to midnight, then from midnight to arrival.

\begin{aligned} 23{:}40\text{ to }00{:}00&=20\text{ minutes} && \text{finish Monday}\\ 00{:}00\text{ to }00{:}25&=25\text{ minutes} && \text{start Tuesday}\\ \text{elapsed time}&=20+25=45\text{ minutes} && \text{combine both intervals} \end{aligned}

Why this step? The arrival is on the next day, so its smaller clock number does not mean a negative duration.

Answer. 45 minutes, or $45/60=0.75$ hours.

Check. Adding 45 minutes to 23:40 gives 00:25 the next day.

Now add an onward connection to this same train journey. Read each service's column down the station rows; read across a row to compare services at that station. A dash means that service does not call there.

Worked example 4: choosing a connection

Question. After that train arrives at Central at 00:25 on Tuesday, a passenger needs at least 10 minutes to change. Which onward service can they catch first, and how long do they wait?

Station	Service A	Service B	Service C
Central, departs Tuesday	00:30	00:42	01:05
Harbour, arrives Tuesday	01:00	01:15	01:40

Plan. Add the minimum change time to arrival, then choose the first departure at or after that time.

\begin{aligned} \text{earliest allowed departure}&=00{:}25+10\text{ min}=00{:}35 && \text{allow time to change}\\ 00{:}30&<00{:}35 && \text{reject service A}\\ \text{waiting time for B}&=42-25=17\text{ minutes} && \text{count from arrival} \end{aligned}

Why this step? The next listed departure is unusable if the connection is too short.

Answer. Service B, leaving at 00:42 and arriving at 01:15; the wait is 17 minutes.

Check. $17\ge10$ , whereas A allows only 5 minutes.

An earlier arrival is not necessarily a shorter journey. When comparing journey durations, use each service's own departure and arrival in the correct station rows.

Watch out

Wrong: 09:10 minus 08:45 is 65 minutes, from $910-845$ . Clock hours contain 60 minutes. Correct: 08:45 to 09:00 is 15 minutes, then 10 more gives 25 minutes. Likewise, $1.25$ hours means 1 hour 15 minutes, rather than 1 hour 25 minutes.

Change only the required connection time and reconsider the departures.

Square units, cubic units and capacity

Area changes in two dimensions and volume in three, so the length conversion factor must be used twice or three times.

A square of side $1\,\mathrm m$ has side $100\,\mathrm{cm}$ , hence area $100\times100=10\,000\,\mathrm{cm}^2$ . A cubic metre contains $100^3=1\,000\,000$ cubic centimetres. Surprisingly, a cube only one metre wide holds a thousand litres: $1\,\mathrm m^3=1000\,\mathrm L$ .

[DIAGRAM: asset_name: Length and area conversion factors; asset_slug: 1ma1_m37_square_conversion; caption: A one-metre side is 100 centimetres, so a one-square-metre area is 10,000 square centimetres.; description: A square has both side dimensions labelled 1 m equals 100 cm, with an area statement 1 square metre equals 100 times 100 square centimetres.]

Capacity is how much a container can hold. Know $1\,\mathrm{ml}=1\,\mathrm{cm}^3$ and $1\,\mathrm L=1000\,\mathrm{cm}^3$ . These conversion relationships are not on the formula sheet: derive square and cubic factors from the length units, and learn the capacity equalities.

Worked example 5: square units

Question. Convert $0.75\,\mathrm m^2$ to square centimetres.

Plan. Square the metre-to-centimetre factor, then apply it to the area.

\begin{aligned} 1\,\mathrm m^2&=100^2=10\,000\,\mathrm{cm}^2 && \text{two length dimensions}\\ 0.75\,\mathrm m^2&=0.75\times10\,000=7500\,\mathrm{cm}^2 && \text{convert the area} \end{aligned}

Why this step? Both length and width contain the factor 100.

Answer. $7500\,\mathrm{cm}^2$ .

Check. Three quarters of $10\,000$ is $7500$ .

Change only the dimension from area to volume; the factor 100 now appears three times.

Worked example 6: cubic units and litres

Question. Convert $0.75\,\mathrm m^3$ to cubic centimetres and litres.

Plan. Cube the length factor, then use 1000 cubic centimetres per litre.

\begin{aligned} 1\,\mathrm m^3&=100^3=1\,000\,000\,\mathrm{cm}^3 && \text{three dimensions}\\ 0.75\,\mathrm m^3&=750\,000\,\mathrm{cm}^3 && \text{multiply by }0.75\\ \text{capacity}&=750\,000\div1000=750\,\mathrm L && \text{convert to litres} \end{aligned}

Why this step? A cubic centimetre is much smaller than a cubic metre, so very many fit inside.

Answer. $750\,000\,\mathrm{cm}^3$ , or $750\,\mathrm L$ .

Check. $0.75\times1000=750$ litres using the cubic-metre equality.

Variables obey the same unit factor: $A\,\mathrm m^2=10\,000A\,\mathrm{cm}^2$ and $V\,\mathrm m^3=1\,000\,000V\,\mathrm{cm}^3$ . If $x$ is a side length in metres, its square's area is $x^2\,\mathrm m^2=10\,000x^2\,\mathrm{cm}^2$ . Do not square a variable that already represents an area. Other useful checks are $1\,\mathrm{cm}^2=100\,\mathrm{mm}^2$ , $1\,\mathrm{cm}^3=1000\,\mathrm{mm}^3$ , and $4500\,\mathrm{cm}^3=4.5\,\mathrm L$ .

Watch out

Wrong: $1\,\mathrm m^2=100\,\mathrm{cm}^2$ . This changes only one dimension. Correct: $1\,\mathrm m^2=100\times100=10\,000\,\mathrm{cm}^2$ . Use three factors for volume.

Change the volume while keeping the same destination units.

Last Accessed

Papers 1, 2 and 3

N13, R1, R11, G14 - Standard units and compound measures

Choosing and converting standard units

Clock times, elapsed time and timetables

Square units, cubic units and capacity

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