What is R1: Unit conversion and compound measures?

A conversion is multiplication or division by a fixed factor, and the direction is checked by size: 8000 cm^3 is a small volume, so it must become 0.008 m^3, not 8\,000\,000. Lengths use 1 m =100 cm and 1 km =1000 m; areas square the factor and volumes cube it, so 1 m^3=100^3=1\,000\,000 cm^3 and 1 litre =1000 cm^3.

How many marks are R1 questions worth?

In verified past-paper appearances, R1 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has R1 appeared in?

R1 has appeared in 40 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts

Unit conversion and compound measures

Worked answers, methods and verified real exam appearances for R1 on Edexcel GCSE Maths 1MA1.

Definition

Specification R1: change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts.

Explanation

A conversion is multiplication or division by a fixed factor, and the direction is checked by size: $8000$ cm $^3$ is a small volume, so it must become $0.008$ m $^3$ , not $8\,000\,000$ . Lengths use $1$ m $=100$ cm and $1$ km $=1000$ m; areas square the factor and volumes cube it, so $1$ m $^3=100^3=1\,000\,000$ cm $^3$ and $1$ litre $=1000$ cm $^3$ .
Time converts through $60$ : $1$ hour $=3600$ seconds. A compound unit converts one part at a time: $180$ km/h becomes m/s by multiplying by $1000$ and dividing by $3600$ , giving $50$ ; each conversion is a method mark.
Before substituting into a formula, put every quantity into the unit the answer needs, converting millimetres and centimetres to metres for a density in kg/m $^3$ .
In algebraic contexts the same factor multiplies the whole expression.
Write the conversion line down; the mark is for seeing it.

Worked example

A block of wood is

2.4

m long,

35

cm wide and

80

mm high. The mass of the block is

26.88

kg. Work out the density of the wood in kg/m

^3

(4)

(Total for Question 1 is 4 marks)

Working

1.Convert every length to metres: $35$ cm $=0.35$ m and $80$ mm $=0.08$ m.
2.Volume $=2.4\times0.35\times0.08=0.0672$ m $^3$ .
3.Density $=\dfrac{\text{mass}}{\text{volume}}=\dfrac{26.88}{0.0672}=400$ kg/m $^3$ .

Answer: $400$ kg/m $^3$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for converting at least one length to metres, eg $35\div100$ (= 0.35) or $80\div1000$ (= 0.08)	P1
	for a process to find the volume in m $^3$ , eg $2.4\times$ “0.35” $\times$ “0.08” (= 0.0672)	P1	or for the volume in cm $^3$ (= 67 200) then $\div1\,000\,000$
	for a process to find the density, eg $26.88\div$ “0.0672”	P1
	for $400$	A1	cao

Common mistakes

Don't use the length factor $100$ for an area or volume conversion instead of $100^2$ or $100^3$ .
Don't substitute centimetres and metres into the same formula without converting first.
Don't convert only the numerator of a compound unit such as kilometres per hour.

Mark-scheme wording to use

“for 0.008 or 8 × 10^(‒3)” — Pearson 1MA1, November 2022 Paper 3H, Q7(a)
“for conversion from km to m eg 180 × 1000 (= 180 000) … for conversion from hours to seconds eg 180 ÷ (60 × 60) (= 0.05)” — Pearson 1MA1, November 2022 Paper 3H, Q7(b)
“M1 works with volume eg 240000 … M1 uses conversion 1 litre = 1000 cm^3” — Pearson 1MA1, Specimen 1 Paper 1H, Q5

What the examiner wants

Units that match before the formula is used. The June 2024 Paper 3H report describes a density problem in which the volume of flour had to be found from mass and density before anything else, and the November 2024 Paper 1H report warns that a time for $5$ miles at $30$ mph ranging up to $900$ minutes should have been caught by checking the reasonableness of the answer.

Exam tip

For a multi-mark conversion, write the compatible units before the formula so the method mark is unambiguous.

Worked practice

Tier 1 · Easy

A charity walk is

3.6\,\text{km}

long. Change this distance to metres.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$3600\,\text{m}$	1	$3.6\times1000=3600$ , so the distance is $3600\,\text{m}$ .

Tier 2 · Standard

A pump moves water at

540

litres per minute. Change this rate to cubic metres per second.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$0.009\,\text{m}^3/\text{s}$	3	$540$ litres is $0.540\,\text{m}^3$ . Dividing by $60$ converts per minute to per second: $0.540/60=0.009\,\text{m}^3/\text{s}$ .

Tier 3 · Hard

A solid block measures

2.4\,\text{m}

35\,\text{cm}

80\,\text{mm}

and has mass

26.88\,\text{kg}

. Work out its density in

\text{kg}/\text{m}^3

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$400\,\text{kg}/\text{m}^3$	4	Convert the lengths to metres: $35\,\text{cm}=0.35\,\text{m}$ and $80\,\text{mm}=0.08\,\text{m}$ . The volume is $2.4\times0.35\times0.08=0.0672\,\text{m}^3$ , so the density is $26.88/0.0672=400\,\text{kg}/\text{m}^3$ .

Tier 1 · Easy

Change

2.7\,\text{m}^2

to square centimetres.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$27\,000\,\text{cm}^2$	1	$1\,\text{m}^2=10\,000\,\text{cm}^2$ , so $2.7\times10\,000=27\,000\,\text{cm}^2$ .

Tier 2 · Standard

A rectangle is

3x

metres long and

40\,\text{cm}

wide. Express its area in square metres in terms of

x

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$1.2x\,\text{m}^2$	2	$40\,\text{cm}=0.4\,\text{m}$ . The area is $3x\times0.4=1.2x\,\text{m}^2$ .

Tier 3 · Hard

A machine prints

480

labels in

6

minutes, and each label has area

35\,\text{cm}^2

. Work out the area of labels printed in one hour, in square metres.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$16.8\,\text{m}^2$	4	The machine prints $480\div6=80$ labels per minute, so it prints $80\times60=4800$ labels per hour. Their area is $4800\times35=168\,000\,\text{cm}^2=16.8\,\text{m}^2$ .

Tier 2 · Standard

A dairy uses

22.32\,\text{m}^3

of milk in

31

days. Work out the average amount used in litres per hour.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$30$ litres per hour	3	$22.32\,\text{m}^3=22\,320$ litres and $31$ days is $31\times24=744$ hours. The average amount used is $22\,320\div744=30$ litres per hour.

Tier 3 · Hard

A uniform metal sheet has area

0.84\,\text{m}^2

and mass

6.72\,\text{kg}

. Work out the mass, in grams, of a piece with area

350\,\text{cm}^2

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$280\,\text{g}$	4	$0.84\,\text{m}^2=8400\,\text{cm}^2$ , so the piece is $350\div8400=\dfrac{1}{24}$ of the sheet. Its mass is $6.72\div24=0.28\,\text{kg}=280\,\text{g}$ .

Tier 3 · Hard

A roll of material has mass

9.6\,\text{kg}

. Its mass per unit area is

320\,\text{g}/\text{m}^2

and its width is

75\,\text{cm}

. Work out the length of the roll in metres.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$40\,\text{m}$	4	$320\,\text{g}/\text{m}^2=0.32\,\text{kg}/\text{m}^2$ , so the area is $9.6\div0.32=30\,\text{m}^2$ . The width is $0.75\,\text{m}$ , giving length $30\div0.75=40\,\text{m}$ .

Q10

Tier 3 · Hard

A pipe carries water at

420\,\text{cm}^3

per second from 08:35 until 10:20. Calculate the volume carried, in cubic metres.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$2.646\,\text{m}^3$	4	The time is $105\times60=6300$ seconds. The volume is $420\times6300=2\,646\,000\,\text{cm}^3$ . Since $1\,\text{m}^3=1\,000\,000\,\text{cm}^3$ , the volume is $2.646\,\text{m}^3$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	3H	Q6	3	Allowed	Higher	QP MS
2024-11	2F	Q22	3	Allowed	Foundation	QP MS
2019-06	3F	Q3	1	Allowed	Foundation	QP MS
2019-11	1H	Q9	4	Non-calculator	Higher	QP MS
2019-11	3F	Q26	1	Allowed	Foundation	QP MS
2021-11	3H	Q4	4	Allowed	Higher	QP MS
2022-06	2F	Q27	4	Allowed	Foundation	QP MS
2022-06	3H	Q6	4	Allowed	Higher	QP MS
2019-06	3F	Q22	4	Allowed	Foundation	QP MS
2024-06	2F	Q16	3	Allowed	Foundation	QP MS
2022-11	3F	Q14	3	Allowed	Foundation	QP MS
2023-11	2F	Q3	1	Allowed	Foundation	QP MS
2019-11	3H	Q5	1	Allowed	Higher	QP MS
2019-11	2F	Q11	3	Allowed	Foundation	QP MS
2021-11	3F	Q12	4	Allowed	Foundation	QP MS
2022-06	1F	Q1	1	Non-calculator	Foundation	QP MS
2022-11	3F	Q27	4	Allowed	Foundation	QP MS
2022-11	1F	Q14	5	Non-calculator	Foundation	QP MS
2022-06	3F	Q27	4	Allowed	Foundation	QP MS
2022-06	2F	Q19	5	Allowed	Foundation	QP MS
2024-11	3F	Q13	4	Allowed	Foundation	QP MS
2019-11	3F	Q5	1	Allowed	Foundation	QP MS
2019-11	3H	Q6	4	Allowed	Higher	QP MS
2022-06	2F	Q24	5	Allowed	Foundation	QP MS
2024-11	3F	Q11	4	Allowed	Foundation	QP MS
2021-11	2F	Q2	1	Allowed	Foundation	QP MS
2021-11	3F	Q24	4	Allowed	Foundation	QP MS
2022-06	2H	Q4	5	Allowed	Higher	QP MS
2024-06	2F	Q2	1	Allowed	Foundation	QP MS
2022-11	3H	Q7	4	Allowed	Higher	QP MS
2019-06	2H	Q4	4	Allowed	Higher	QP MS
2022-06	2F	Q18	3	Allowed	Foundation	QP MS
2024-11	1F	Q3	1	Non-calculator	Foundation	QP MS
2019-06	2F	Q4	1	Allowed	Foundation	QP MS
2022-06	3F	Q29	2	Allowed	Foundation	QP MS
2023-11	2H	Q7	4	Allowed	Higher	QP MS
2022-06	2H	Q7	4	Allowed	Higher	QP MS
2022-11	2F	Q4	1	Allowed	Foundation	QP MS
2021-11	3F	Q18	1	Allowed	Foundation	QP MS
2023-11	1F	Q6	3	Non-calculator	Foundation	QP MS

Other points in R Ratio, proportion and rates of change

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