What is R3: Quantity as a fraction of another?

To express quantity A as a fraction of quantity B, form A/B and simplify.

How many marks are R3 questions worth?

In verified past-paper appearances, R3 questions on Edexcel GCSE Maths 1MA1 have been worth 1–7 marks.

Which past exam papers has R3 appeared in?

R3 has appeared in 9 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Express one quantity as a fraction of another, where the fraction is less than 1 or greater than 1

Quantity as a fraction of another

Worked answers, methods and verified real exam appearances for R3 on Edexcel GCSE Maths 1MA1.

Explanation

To express quantity $A$ as a fraction of quantity $B$ , form $\dfrac{A}{B}$ and simplify.
The order of the wording fixes the order of the fraction: the quantity before “as a fraction of” is the numerator.
Quantities must first be written in the same units, so that the units cancel.
The result may be less than $1$ , equal to $1$ , or greater than $1$ ; an improper fraction is valid when the first quantity is larger.
Examiners expect a simplified exact fraction unless the question asks for another form.

Worked example

Express $0.84\,\text{m}^2$ as a fraction of $2400\,\text{cm}^2$ . Give the fraction in its simplest form.

1.Convert the area: $0.84\,\text{m}^2=0.84\times10\,000=8400\,\text{cm}^2$ .
2.Form the fraction in the stated order: $\dfrac{8400}{2400}$ .
3.Simplify: $\dfrac{8400}{2400}=\dfrac{84}{24}=\dfrac{7}{2}$ .

Answer: $\dfrac{7}{2}$ .

Common mistakes

Don't reverse the fraction and write the second quantity over the first.
Don't form a fraction before converting the two quantities to the same units.
Don't reject an answer greater than $1$ even though the first quantity is larger.

Exam tip

Underline the quantity named first and place it in the numerator before simplifying.

Worked practice

Tier 1 · Easy

Express

18

as a fraction of

30

. Give the fraction in its simplest form.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\frac{3}{5}$	2	$\frac{18}{30}=\frac{3}{5}$ after dividing the numerator and denominator by $6$ .

Tier 2 · Standard

Express

84

minutes as a fraction of

35

minutes. Simplify your answer.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$\frac{12}{5}$	2	The units already match, so form $\frac{84}{35}$ . Dividing both parts by $7$ gives $\frac{12}{5}$ .

Tier 3 · Hard

A bottle contains

1.2

litres of juice. Sam pours

350\,\text{ml}

of the juice into a jug. Express the amount left in the bottle as a fraction of the amount poured into the jug. Give your answer in its simplest form.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$\dfrac{17}{7}$	3	Convert $1.2$ litres to $1200$ ml, so $1200-350=850$ ml remains. The required fraction is $850/350=17/7$ .

Tier 1 · Easy

Express

45

pence as a fraction of £1.20. Give your answer in its simplest form.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$\dfrac{3}{8}$	2	£1.20 is $120$ pence, so the fraction is $\dfrac{45}{120}=\dfrac{3}{8}$ .

Tier 2 · Standard

A park has

18

oak trees,

12

birch trees and

15

beech trees. What fraction of these trees are birch? Give your answer in its simplest form.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\dfrac{4}{15}$	2	There are $18+12+15=45$ trees in total. The required fraction is $\dfrac{12}{45}=\dfrac{4}{15}$ .

Tier 3 · Hard

Two parcels have masses

1.8\,\text{kg}

and

650\,\text{g}

, then

240\,\text{g}

is removed from the first parcel. Express the first parcel's new mass as a fraction of the new total mass.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\dfrac{12}{17}$	3	The first new mass is $1800-240=1560\,\text{g}$ . The new total is $1560+650=2210\,\text{g}$ , so the fraction is $\dfrac{1560}{2210}=\dfrac{12}{17}$ .

Tier 2 · Standard

A runner travels

1.8\,\text{km}

uphill and then

750\,\text{m}

downhill. Express the total distance as a fraction of the uphill distance. Give your answer in its simplest form.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\dfrac{17}{12}$	3	$1.8\,\text{km}=1800\,\text{m}$ , so the total distance is $1800+750=2550\,\text{m}$ . The required fraction is $\dfrac{2550}{1800}=\dfrac{17}{12}$ .

Tier 3 · Hard

A floor has area

7.2\,\text{m}^2

. Tiles cover

18\,000\,\text{cm}^2

and a mat covers

1.35\,\text{m}^2

. Express the uncovered area as a fraction of the covered area. Give your answer in its simplest form.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$\dfrac{9}{7}$	4	$18\,000\,\text{cm}^2=1.8\,\text{m}^2$ , so the covered area is $1.8+1.35=3.15\,\text{m}^2$ . The uncovered area is $7.2-3.15=4.05\,\text{m}^2$ . The required fraction is $\dfrac{4.05}{3.15}=\dfrac{9}{7}$ .

Tier 3 · Hard

Ribbon A is

2.4\,\text{m}

long and ribbon B is

90\,\text{cm}

long. A

35\,\text{cm}

piece is cut from ribbon A and joined to ribbon B. Express the new length of ribbon B as a fraction of the new length of ribbon A. Give the fraction in its simplest form.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\dfrac{25}{41}$	4	Ribbon A starts at $240\,\text{cm}$ . The new lengths are $240-35=205\,\text{cm}$ and $90+35=125\,\text{cm}$ . The required fraction is $\dfrac{125}{205}=\dfrac{25}{41}$ .

Q10

Tier 3 · Hard

Tank A has capacity

1.8\,\text{m}^3

and is

65\%

full. Tank B contains

420

litres. Express the amount in tank A as a fraction of the total amount in both tanks. Give the fraction in its simplest form.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$\dfrac{39}{53}$	4	$1.8\,\text{m}^3=1800$ litres, so tank A contains $0.65\times1800=1170$ litres. The total is $1170+420=1590$ litres. The fraction is $\dfrac{1170}{1590}=\dfrac{39}{53}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	1F	Q9	7	Non-calculator	Foundation	QP MS
2019-11	2F	Q7	2	Allowed	Foundation	QP MS
2019-06	1H	Q12	3	Non-calculator	Higher	QP MS
2023-11	3F	Q4	1	Allowed	Foundation	QP MS
2022-06	3F	Q12	2	Allowed	Foundation	QP MS
2024-06	1H	Q13	5	Non-calculator	Higher	QP MS
2019-06	3F	Q10	3	Allowed	Foundation	QP MS
2023-06	1F	Q2	1	Non-calculator	Foundation	QP MS
2021-11	2F	Q10	1	Allowed	Foundation	QP MS

Other points in R Ratio, proportion and rates of change

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