What is R8: Ratio, fractions and linear functions?

For parts in ratio a:b, the whole has a+b equal parts, so their fractions of the whole are a/a+b and b/a+b. A constant ratio between variables can also define a linear function.

How many marks are R8 questions worth?

In verified past-paper appearances, R8 questions on Edexcel GCSE Maths 1MA1 have been worth 1–6 marks.

Which past exam papers has R8 appeared in?

R8 has appeared in 8 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Relate ratios to fractions and to linear functions

Ratio, fractions and linear functions

Worked answers, methods and verified real exam appearances for R8 on Edexcel GCSE Maths 1MA1.

Explanation

For parts in ratio $a:b$ , the whole has $a+b$ equal parts, so their fractions of the whole are $\dfrac{a}{a+b}$ and $\dfrac{b}{a+b}$ . A constant ratio between variables can also define a linear function.
If $y:x=m:1$ , then $\dfrac{y}{x}=m$ and $y=mx$ .
Its graph is a straight line through the origin, with gradient $m$ .
A line with a non-zero intercept does not represent a constant ratio.
Examiners may ask you to move between ratio, fraction, equation and graph, so always state the link explicitly.

Worked example

Concentrate and water are mixed in the ratio $2:7$ . Let $c$ be the concentrate volume and $V$ the total volume. Express $V$ as a function of $c$ , then find both volumes when $V=54$ .

1.Concentrate is $\dfrac{2}{2+7}=\dfrac{2}{9}$ of the total, so $c=\dfrac{2}{9}V$ .
2.Rearrange to $V=\dfrac{9}{2}c$ .
3.When $V=54$ , $c=\dfrac{2}{9}\times54=12$ and water $=54-12=42$ .

Answer: $V=\dfrac{9}{2}c$ ; $12$ litres concentrate and $42$ litres water.

Common mistakes

Don't use $\dfrac{a}{b}$ instead of $\dfrac{a}{a+b}$ for a fraction of the whole.
Don't write a constant-ratio graph with a non-zero intercept.
Don't treat the total as one ratio part rather than the sum of the parts.

Exam tip

When linking ratio to a function, verify that the equation gives $y=0$ when $x=0$ .

Worked practice

Tier 1 · Easy

The ratio of cats to dogs at a shelter is

3:5

. What fraction of the animals are cats?

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\frac{3}{8}$	2	There are $3+5=8$ equal parts altogether, of which $3$ are cats. The fraction is $\frac{3}{8}$ .

Tier 2 · Standard

Quantities

y

and

x

are always in the ratio

5:2

. Write

y

as a linear function of

x

, then find

y

when

x=14

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$y=\frac{5}{2}x$ $y=35$	3	$y:x=5:2$ means $y/x=5/2$ , so $y=\frac{5}{2}x$ . At $x=14$ , $y=\frac{5}{2}\times14=35$ .

Tier 3 · Hard

A mixture contains concentrate and water in the ratio

2:7

. Let

c

litres be the concentrate and

V

litres be the total mixture. Express

V

as a linear function of

c

, then find both component volumes when

V=54

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$V=\frac{9}{2}c$ $12$ litres concentrate $42$ litres water	4	The total has $2+7=9$ parts, so $c$ is $2/9$ of $V$ . Hence $V=\frac{9}{2}c$ . If $V=54$ , then $c=\frac{2}{9}\times54=12$ , leaving $54-12=42$ litres of water.

Tier 1 · Easy

The equation

y=4x

describes a constant ratio. Write

y:x

in its simplest form.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$4:1$	1	$y=4x$ means that $y$ is four times $x$ , so $y:x=4:1$ .

Tier 2 · Standard

A rectangle's length and width are in the ratio

5:3

. Let

P

be its perimeter and

w

its width. Express

P

as a linear function of

w

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$P=\dfrac{16}{3}w$	3	Since length:width is $5:3$ , the length is $\dfrac{5}{3}w$ . Therefore $P=2\left(\dfrac{5}{3}w+w\right)=2\times\dfrac{8}{3}w=\dfrac{16}{3}w$ .

Tier 3 · Hard

The pairs

(4,14)

(7,24.5)

and

(10,35)

follow a constant-ratio linear function connecting

x

and

y

. Work out

x

when

y=56

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$x=16$	3	$14\div4=24.5\div7=35\div10=3.5$ , so the constant-ratio function is $y=3.5x$ . When $y=56$ , $3.5x=56$ , giving $x=16$ .

Tier 2 · Standard

In a choir, the numbers of sopranos, altos and tenors are in the ratio

5 : 3 : 2

. Let

t

be the number of tenors and

N

the total number of singers. Express

N

as a linear function of

t

. Then find

N

when

t=18

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$N=5t$ $N=90$	3	The tenors are $2$ of the $10$ ratio parts, so $t=2k$ and $N=10k=5t$ . When $t=18$ , $N=5\times18=90$ .

Tier 3 · Hard

Blue and red beads are used in the ratio

4 : 7

. Each blue bead has mass

3\,\text{g}

and each red bead has mass

5\,\text{g}

. Let

b

be the number of blue beads and

M

grams be the total mass. Express

M

as a linear function of

b

. Work out the number of red beads when

M=282

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$M=\dfrac{47}{4}b$ $42$ red beads	4	There are $\dfrac{7}{4}b$ red beads, so $M=3b+5\left(\dfrac{7}{4}b\right)=\dfrac{47}{4}b$ . When $M=282$ , $b=282\times\dfrac{4}{47}=24$ . The number of red beads is $\dfrac{7}{4}\times24=42$ .

Tier 3 · Hard

Three quantities are in the ratio

2 : 3 : 7

. Let

d

be the difference between the largest quantity and the sum of the other two, and let

T

be the total of all three quantities. Express

T

as a linear function of

d

. Work out

T

when

d=54

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$T=6d$ $T=324$	4	Write the quantities as $2k$ , $3k$ and $7k$ . Then $d=7k-(2k+3k)=2k$ , while $T=12k$ . Therefore $T=6d$ , and when $d=54$ , $T=6\times54=324$ .

Q10

Tier 3 · Hard

A straight line connecting

x

and

y

passes through

(4,11)

and

(10,23)

. Work out the equation of the line. Does the relationship give a constant ratio

y:x

? You must show all your working.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$y=2x+3$ No, the ratio $y:x$ is not constant.	4	The gradient is $\dfrac{23-11}{10-4}=2$ . Using $(4,11)$ gives $11=2\times4+c$ , so $c=3$ and $y=2x+3$ . Also $11:4$ is not equal to $23:10$ , so the ratio $y:x$ is not constant; equivalently, the line does not pass through the origin.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	3F	Q12	2	Allowed	Foundation	QP MS
2019-11	3H	Q14	4	Allowed	Higher	QP MS
2023-06	1F	Q14	5	Non-calculator	Foundation	QP MS
2023-06	2H	Q3	1	Allowed	Higher	QP MS
2021-11	1F	Q12	3	Non-calculator	Foundation	QP MS
2022-06	1H	Q11	3	Non-calculator	Higher	QP MS
2022-11	1H	Q6	6	Non-calculator	Higher	QP MS
2019-06	2H	Q17	4	Allowed	Higher	QP MS

Other points in R Ratio, proportion and rates of change

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