What is R13: Direct and inverse proportion equations?

If X is inversely proportional to Y, then X1/Y and X=k/Y for a constant k; equivalently, XY=k. Foundation questions can require interpreting equations that describe direct and inverse proportion.

How many marks are R13 questions worth?

In verified past-paper appearances, R13 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has R13 appeared in?

R13 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

R13

Understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y; construct and interpret equations that describe direct and inverse proportion

Direct and inverse proportion equations

Worked answers, methods and verified real exam appearances for R13 on Edexcel GCSE Maths 1MA1.

Explanation

If $X$ is inversely proportional to $Y$ , then $X\propto\dfrac{1}{Y}$ and $X=\dfrac{k}{Y}$ for a constant $k$ ; equivalently, $XY=k$ . Foundation questions can require interpreting equations that describe direct and inverse proportion.
Higher tier: construct equations involving a power, such as $y=kx^2$ or $y=\dfrac{k}{x^n}$ .
Substitute a known pair to find $k$ , write the complete equation, then use it to find an unknown.
Check any stated restrictions, such as a positive length, before choosing a root.
Examiners usually award separate method marks for the correct proportional form and for finding the constant.

Worked example

Higher tier: For positive $b$ , $a$ varies inversely as $b^2$ . Given $a=20$ when $b=3$ , find $b$ when $a=7.2$ .

1.Write $a=\dfrac{k}{b^2}$ .
2.Use the first pair: $k=ab^2=20\times3^2=180$ .
3. $7.2=\dfrac{180}{b^2}$ , so $b^2=25$ and the positive value is $b=5$ .

Answer: $b=5$ .

Common mistakes

Don't write $X=kY$ when the relationship is inverse proportion.
Don't use $y=kx$ when the stated proportional quantity is $x^2$ .
Don't find $k$ but never write or use the complete proportional equation.

Exam tip

Translate the wording into a formula containing $k$ before substituting any values.

Worked practice

Tier 1 · Easy

Variables

p

and

q

vary inversely, with recorded values

p=12

and

q=5

. Write their connecting equation.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$p=\frac{60}{q}$	2	Write $p=k/q$ . Using the given pair, $k=pq=12\times5=60$ , so $p=\frac{60}{q}$ .

Tier 2 · Standard

The relationship between

y

and

x^2

is direct proportion. Given

x=3

when

y=45

, form the equation and evaluate

y

x=4

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$y=5x^2$ $y=80$	4	Write $y=kx^2$ . Then $45=k\times3^2$ , so $k=5$ and $y=5x^2$ . At $x=4$ , $y=5\times4^2=80$ .

Tier 3 · Hard

For positive

b

, the variable

a

varies inversely with

b^2

. Given

a=20

b=3

, determine

b

when

a=7.2

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$b=5$	4	Use $a=k/b^2$ . The first pair gives $k=ab^2=20\times9=180$ . Then $7.2=180/b^2$ , so $b^2=25$ . The requested positive value is $b=5$ .

Tier 1 · Easy

xy=14

y=14x

y=x-14

. Write down the letter of the equation that shows

y

is inversely proportional to

x

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	A	1	Inverse proportion has constant product $xy=k$ , so the correct letter is A.

Tier 2 · Standard

The variables

p

and

q

satisfy

p=\dfrac{48}{q}

. When

q

increases from

3

12

, work out the fraction of its original value that

p

becomes.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\dfrac{1}{4}$	3	Originally $p=48\div3=16$ . After the change, $p=48\div12=4$ . The new value as a fraction of the original is $\dfrac{4}{16}=\dfrac{1}{4}$ .

Tier 3 · Hard

A rectangular field has length

x

metres and width

y

metres. The width is inversely proportional to the length. When

x=18

y=14

. Find a formula for

y

in terms of

x

. Then work out the perimeter of the field when

x=21

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$y=\dfrac{252}{x}$ $66\,\text{m}$	4	Write $y=\dfrac{k}{x}$ . Since $k=18\times14=252$ , the equation is $y=\dfrac{252}{x}$ . When $x=21$ , $y=252\div21=12$ , so the perimeter is $2(21+12)=66\,\text{m}$ .

Tier 2 · Standard

The equation

p=\dfrac{72}{q}

connects two inversely proportional variables. State whether

p=9

q=8

satisfies this equation. Work out

q

when

p=12

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Yes, because $9\times8=72$ . $q=6$	3	For the proposed pair, $pq=9\times8=72$ , so it satisfies the equation. When $p=12$ , $12=\dfrac{72}{q}$ , so $12q=72$ and $q=6$ .

Tier 3 · Hard

Higher tier:

u

is directly proportional to

v^3

. When

v=2

u=28

. Write down an equation connecting

u

and

v

. Work out the positive value of

v

when

u=756

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$u=3.5v^3$ $v=6$	4	Write $u=kv^3$ . Since $28=k\times2^3$ , $k=3.5$ and $u=3.5v^3$ . When $u=756$ , $v^3=756\div3.5=216$ . The positive value is $v=6$ .

Tier 3 · Hard

Higher tier: For positive

x

y

is inversely proportional to

x^3

. When

x=2

y=16

. Write an equation connecting

x

and

y

. Work out

x

when

y=2

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$y=\dfrac{128}{x^3}$ $x=4$	4	Write $y=\dfrac{k}{x^3}$ . Using $x=2$ and $y=16$ gives $k=16\times2^3=128$ , so $y=\dfrac{128}{x^3}$ . When $y=2$ , $2=\dfrac{128}{x^3}$ , hence $x^3=64$ and the positive value is $x=4$ .

Q10

Tier 3 · Hard

Higher tier:

A

is directly proportional to

b^2

. When

b=3

A=54

. The value of

b

is decreased by

20\%

. Work out the percentage decrease in

A

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$36\%$ decrease	4	The equation has the form $A=kb^2$ . A $20\%$ decrease multiplies $b$ by $0.8$ , so it multiplies $A$ by $0.8^2=0.64$ . The new value is $64\%$ of the old value, giving a $100\%-64\%=36\%$ decrease. (Using the given pair, $k=54\div9=6$ gives the same result.)

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-06	1H	Q17	4	Non-calculator	Higher	QP MS
2019-06	1H	Q20	4	Non-calculator	Higher	QP MS
2024-11	2H	Q13	5	Allowed	Higher	QP MS
2023-06	1H	Q13	3	Non-calculator	Higher	QP MS
2022-11	1H	Q13	3	Non-calculator	Higher	QP MS
2023-11	2H	Q14	5	Allowed	Higher	QP MS
2021-11	3H	Q8	2	Allowed	Higher	QP MS
2024-06	2F	Q22	2	Allowed	Foundation	QP MS
2024-06	2H	Q12	4	Allowed	Higher	QP MS
2019-11	3H	Q16	3	Allowed	Higher	QP MS

Other points in R Ratio, proportion and rates of change

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