What is R10: Direct and inverse proportion?

Write the relationship as an equation with a constant before doing anything numerical: y=kx, y=k/x, or at Higher tier y=kx^2, y=k√(x), y=k/x^2. Substitute the given pair to find k, write the formula out, then substitute the new value.

How many marks are R10 questions worth?

In verified past-paper appearances, R10 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has R10 appeared in?

R10 has appeared in 28 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

R10

Solve problems involving direct and inverse proportion, including graphical and algebraic representations

Direct and inverse proportion

Worked answers, methods and verified real exam appearances for R10 on Edexcel GCSE Maths 1MA1.

Definition

Specification R10: solve problems involving direct and inverse proportion, including graphical and algebraic representations. If $y$ is directly proportional to $x$ then $y=kx$ ; if $y$ is inversely proportional to $x$ then $y=\dfrac{k}{x}$ , where $k$ is the constant of proportionality.

Explanation

Write the relationship as an equation with a constant before doing anything numerical: $y=kx$ , $y=\dfrac{k}{x}$ , or at Higher tier $y=kx^2$ , $y=k\sqrt{x}$ , $y=\dfrac{k}{x^2}$ . Substitute the given pair to find $k$ , write the formula out, then substitute the new value.
For inverse proportion the product $xy$ is constant, so the same problem can be solved by $xy=k$ ; for direct proportion the ratio $\dfrac{y}{x}$ is constant and doubling $x$ doubles $y$ , while for $y=kx^3$ doubling $x$ multiplies $y$ by $8$ , not $6$ .
The graph of direct proportion is a straight line through the origin; inverse proportion is a decreasing curve approaching both axes.
In a practical problem such as seed for a lawn, scale by the ratio of areas and compare with what is available, and finish with a supported yes or no.
Writing $k$ on the wrong side, $y=kx^2$ for $y=\dfrac{k}{x^2}$ , loses every mark.

Direct proportion gives a straight line through the origin; inverse proportion gives a decreasing reciprocal curve for positive values.

Worked example

y

is inversely proportional to

x

. When

x=6

y=12

. Work out the value of

y

when

x

is increased by

25\%

(3)

(Total for Question 1 is 3 marks)

Working

1.Write the relationship: $y=\dfrac{k}{x}$ , so $xy=k$ .
2.Find $k$ from the given pair: $k=6\times12=72$ , giving $y=\dfrac{72}{x}$ .
3.The new value of $x$ is $6\times1.25=7.5$ .
4.Substitute: $y=\dfrac{72}{7.5}=9.6$ .

Answer: $y=9.6$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for establishing a method linked to proportion, eg $y=\dfrac{k}{x}$ or $k=6\times12$ (= 72) or $12=k\div6$	M1
	(dep) for substitution with the new value, eg $y=72\div(6\times1.25)$ or $72\div7.5$	M1	oe
	for $9.6$	A1	cao

Common mistakes

Don't write $y=kx$ for an inverse-proportion relationship.
Don't draw a direct-proportion line that does not pass through the origin.
Don't find the constant correctly and fail to use it in a complete equation.

Mark-scheme wording to use

“Establishing method linked to proportion eg d=k÷c or 25=k÷280 M1 (dep) substitution eg d = 7000 ÷ 350 or 25 × 280 ÷ 350 oe A1 cao” — Pearson 1MA1, Sample Paper 2H, Q13
“M1 Finds constant 36 × 1.5 (=54) or 6/1.5 =4 M1 54 ÷ 6 or 36 ÷ 4 A1 9 cao” — Pearson 1MA1, Specimen 2 Paper 1H, Q10
“for process using ratio of areas, eg 140 / 45 (= 3.1...) … for Yes supported by correct figures” — Pearson 1MA1, June 2020 Paper 3H, Q4(a)

What the examiner wants

The equation with $k$ in the right place. The June 2024 Paper 2H report says that, as is often the case with inverse proportion, a significant number lost all marks by working with $f=kd^2$ rather than $f=\dfrac{k}{d^2}$ , and that the most common wrong sketch for a linear relationship was a direct proportion graph.

Exam tip

Write $y=kx$ or $y=\dfrac{k}{x}$ before substituting values; this is the key method step.

Worked practice

Tier 1 · Easy

A direct variation links

x

and

y

. The pair

x=6

y=18

is known. Determine

y

x=10

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$y=30$	3	$y=kx$ . Using $18=6k$ gives $k=3$ , so when $x=10$ , $y=3\times10=30$ .

Tier 2 · Standard

Variables

x

and

y

vary inversely. One recorded pair is

x=3

y=14

. Determine

y

x=7

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$y=6$	3	For inverse proportion, $xy=k$ . Here $k=3\times14=42$ , so at $x=7$ , $y=42/7=6$ .

Tier 3 · Hard

x

and

y

are inversely proportional. Initially

x=6

and

y=12

. The value of

x

is increased by

25\%

. Find the new value of

y

and the percentage decrease in

y

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	New $y=9.6$ $20\%$ decrease	5	The constant product is $xy=6\times12=72$ . Increasing $x$ by $25\%$ gives $x=6\times1.25=7.5$ , so $y=72/7.5=9.6$ . The decrease is $12-9.6=2.4$ , and $2.4/12\times100=20\%$ .

Tier 1 · Easy

y=11x

y=\dfrac{11}{x}

y=x+11

. Write down the letter of the equation that shows

y

is directly 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	A direct-proportion equation has the form $y=kx$ , so the correct letter is A.

Tier 2 · Standard

A direct-proportion graph has gradient

2.8

. Work out the

x

-coordinate of the point on the graph where

y=35

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=12.5$	2	The graph has equation $y=2.8x$ . Therefore $x=35\div2.8=12.5$ .

Tier 3 · Hard

The pairs

(2,30)

(3,20)

(5,12)

and

(8,8)

are meant to show inverse proportion. Work out the corrected value of

y

when

x=8

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$y=7.5$	3	For the first three pairs, $xy=2\times30=3\times20=5\times12=60$ . The constant product is $60$ , so when $x=8$ , $y=60\div8=7.5$ .

Tier 2 · Standard

A fixed amount of feed lasts

18

goats for

28

days. Each goat eats the same amount each day. Work out how many days the feed would last

24

goats.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$21$ days	3	The number of goats and the number of days are inversely proportional, so the constant product is $18\times28=504$ . For $24$ goats, the feed lasts $504\div24=21$ days.

Tier 3 · Hard

The mass of a cable is directly proportional to its length. A

3.6\,\text{m}

length has mass

1.26\,\text{kg}

. A drum holding some of this cable has total mass

8.4\,\text{kg}

, and the empty drum has mass

1.4\,\text{kg}

. Work out the length of cable on the drum.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$20\,\text{m}$	4	The mass per metre is $1.26\div3.6=0.35\,\text{kg}/\text{m}$ . The cable on the drum has mass $8.4-1.4=7\,\text{kg}$ . Its length is $7\div0.35=20\,\text{m}$ .

Tier 3 · Hard

Six identical scanners can process one batch of documents in

14

minutes. After all six scanners have worked for

4

minutes, two scanners stop. The remaining scanners continue at the same rate. Work out the total time taken to process the batch.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$19$ minutes	5	The complete batch requires $6\times14=84$ scanner-minutes. In the first $4$ minutes, $6\times4=24$ scanner-minutes of work are completed, leaving $60$ scanner-minutes. Four scanners need $60\div4=15$ more minutes, so the total time is $4+15=19$ minutes.

Q10

Tier 3 · Hard

The volume

V

litres delivered by a pump is directly proportional to the running time

t

minutes. The pump delivers

270

litres in

7.5

minutes. Write an equation connecting

V

and

t

. A tank of capacity

750

litres already contains

138

litres. Work out how long the pump must run to fill the tank.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$V=36t$ $17$ minutes	4	Write $V=kt$ . Then $k=270\div7.5=36$ , so $V=36t$ . The tank needs $750-138=612$ litres. Therefore $36t=612$ , giving $t=17$ minutes.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-06	1F	Q18	2	Non-calculator	Foundation	QP MS
2023-11	3F	Q25	2	Allowed	Foundation	QP MS
2022-06	1H	Q17	4	Non-calculator	Higher	QP MS
2019-06	1H	Q20	4	Non-calculator	Higher	QP MS
2022-06	2H	Q11	3	Allowed	Higher	QP MS
2024-11	2H	Q13	5	Allowed	Higher	QP MS
2023-06	3F	Q23	2	Allowed	Foundation	QP MS
2023-11	1F	Q10	3	Non-calculator	Foundation	QP MS
2023-06	1H	Q13	3	Non-calculator	Higher	QP MS
2024-06	2H	Q3	2	Allowed	Higher	QP MS
2022-11	2F	Q16	3	Allowed	Foundation	QP MS
2019-06	1F	Q23	5	Non-calculator	Foundation	QP MS
2019-11	2F	Q28	3	Allowed	Foundation	QP MS
2023-11	3H	Q6	2	Allowed	Higher	QP MS
2023-06	3H	Q4	2	Allowed	Higher	QP MS
2019-06	2F	Q17	4	Allowed	Foundation	QP MS
2022-11	1H	Q13	3	Non-calculator	Higher	QP MS
2019-06	3H	Q9	3	Allowed	Higher	QP MS
2019-11	2F	Q19	3	Allowed	Foundation	QP MS
2023-11	2H	Q14	5	Allowed	Higher	QP MS
2021-11	3H	Q8	2	Allowed	Higher	QP MS
2019-06	2F	Q18	4	Allowed	Foundation	QP MS
2019-06	2F	Q22	3	Allowed	Foundation	QP MS
2023-06	1F	Q16	2	Non-calculator	Foundation	QP MS
2021-11	1F	Q17	5	Non-calculator	Foundation	QP MS
2019-11	2H	Q8	3	Allowed	Higher	QP MS
2021-11	3H	Q17	4	Allowed	Higher	QP MS
2021-11	3H	Q9	3	Allowed	Higher	QP MS

Other points in R Ratio, proportion and rates of change

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