What is R14: Gradient as rate of change?

The gradient of a straight line is change in y/change in x, so it represents a rate of change with units taken from the axes.

How many marks are R14 questions worth?

In verified past-paper appearances, R14 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has R14 appeared in?

R14 has appeared in 5 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

R14

Interpret the gradient of a straight line graph as a rate of change; recognise and interpret graphs that illustrate direct and inverse proportion

Gradient as rate of change

Worked answers, methods and verified real exam appearances for R14 on Edexcel GCSE Maths 1MA1.

Explanation

The gradient of a straight line is $\dfrac{\text{change in }y}{\text{change in }x}$ , so it represents a rate of change with units taken from the axes.
Choose two well-separated points on the line, not merely nearby grid intersections, and calculate rise over run.
In context, state what the rate means: on a distance-time graph it is speed; on a cost-time graph it is cost per unit time.
A direct-proportion graph is straight through the origin, while an inverse-proportion graph has constant product $xy$ .
Examiners expect both the gradient and its contextual interpretation.

The gradient of a straight line is rise divided by run, with units of vertical-axis units per horizontal-axis unit.

Worked example

A distance-time line passes through $(2,10)$ and $(7,35)$ , with time in seconds and distance in metres. Find and interpret its gradient.

1.Change in distance $=35-10=25\,\text{m}$ .
2.Change in time $=7-2=5\,\text{s}$ .
3.Gradient $=25\div5=5\,\text{m}/\text{s}$ , which is the speed.

Answer: The object travels at $5\,\text{m}/\text{s}$ .

Common mistakes

Don't calculate run divided by rise instead of rise divided by run.
Don't use points that are not both on the straight line.
Don't give a bare gradient without its units or contextual meaning.

Exam tip

For “interpret the gradient”, give a value, compound unit and sentence explaining the rate.

Worked practice

Tier 1 · Easy

A straight distance-time graph passes through

(2,10)

and

(7,35)

, where time is in seconds and distance in metres. Find and interpret its gradient.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Gradient $=5\,\text{m}/\text{s}$ The object travels at $5\,\text{m}/\text{s}$ .	3	Gradient $=(35-10)/(7-2)=25/5=5\,\text{m}/\text{s}$ . On a distance-time graph this is the speed.

Tier 2 · Standard

A phone-call cost graph is modelled by

C=18+0.12m

, where

C

is cost in pounds and

m

is time in minutes. Interpret the gradient and calculate the cost of a

35

-minute call.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	The cost increases by £0.12 per minute. £22.20	3	The coefficient of $m$ is the gradient, so the rate is £0.12 per minute. At $m=35$ , $C=18+0.12\times35=18+4.20=£22.20$ .

Tier 3 · Hard

An inverse-proportion model contains the points

(2,18)

(3,12)

and

(6,6)

. Check that all three coordinates are consistent with the model, write its equation, and find

y

when

x=9

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$xy=36$ for all three points $y=\frac{36}{x}$ $y=4$ when $x=9$	4	The products are $2\times18=36$ , $3\times12=36$ and $6\times6=36$ , so all three points are consistent with the stated inverse-proportion model. Thus $y=36/x$ , and at $x=9$ , $y=36/9=4$ .

Tier 1 · Easy

A graph shows the mass of grain processed, in kilograms, against time in minutes. The straight line passes through the origin and

(6,27)

. Work out its gradient and give the units.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$4.5\,\text{kg}$ per minute	2	Using $(0,0)$ and $(6,27)$ , the gradient is $\dfrac{27-0}{6-0}=4.5\,\text{kg}$ per minute. This is the rate at which grain is processed.

Tier 2 · Standard

Line A passes through

(2,7)

and

(8,25)

, while line B has gradient

2.5

. Work out which line has the greater rate of change.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Line A Line A has gradient $3$ , which is greater than $2.5$ .	3	The gradient of line A is $\dfrac{25-7}{8-2}=\dfrac{18}{6}=3$ . Since $3>2.5$ , line A has the greater rate of change.

Tier 3 · Hard

A straight line passes through

(3,14)

and

(7,30)

. Work out its gradient and give a reason why the line does not represent direct proportion.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Gradient $=4$ It has equation $y=4x+2$ , so it does not pass through the origin.	3	The gradient is $\dfrac{30-14}{7-3}=4$ . Using $(3,14)$ gives $14=4\times3+c$ , so $c=2$ . The line has equation $y=4x+2$ and does not pass through the origin, so it is not direct proportion.

Tier 2 · Standard

A straight-line graph shows the mass of ice left in a cooler. It passes through

(3,870)

and

(11,710)

, where time is in minutes and mass is in grams. Work out the rate at which the ice is melting and the mass of ice at time

0

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$20$ grams per minute $930$ grams at time $0$	3	The gradient is $\dfrac{710-870}{11-3}=\dfrac{-160}{8}=-20$ grams per minute. The negative sign shows the mass is decreasing at $20$ grams per minute. Using $(3,870)$ , the mass at time $0$ is $870+3\times20=930$ grams.

Tier 3 · Hard

A graph shows the volume

V

litres left in a tank after

t

minutes. Two straight-line sections join the points

(0,150)

(5,120)

and

(13,56)

. Work out during which time interval the tank is draining faster, and by how many litres per minute.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	From $5$ to $13$ minutes, by $2$ litres per minute	4	From $0$ to $5$ minutes, the gradient is $\dfrac{120-150}{5-0}=-6$ litres per minute. From $5$ to $13$ minutes, it is $\dfrac{56-120}{13-5}=-8$ litres per minute. The second interval has the greater draining rate, by $8-6=2$ litres per minute.

Tier 3 · Hard

Graph A is straight; it passes through

(0,0)

and through

(5,35)

. Graph B is an inverse-proportion curve through

(4,30)

. Work out the sum of the two

y

-values when

x=10

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$82$	4	For graph A, $y=7x$ , so at $x=10$ , $y=70$ . For graph B, $xy=4\times30=120$ , so at $x=10$ , $y=12$ . The sum is $70+12=82$ .

Q10

Tier 3 · Hard

A straight-line graph shows energy used, in kilowatt-hours, against running time, in hours. The line passes through

(1.5,4.8)

and

(6.5,16.8)

. Work out and interpret the gradient. At this constant rate, work out the running time needed for the energy used to increase by

30

kilowatt-hours.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$2.4$ kilowatt-hours per hour $12.5$ hours	4	The gradient is $\dfrac{16.8-4.8}{6.5-1.5}=\dfrac{12}{5}=2.4$ kilowatt-hours per hour, so energy use increases by $2.4$ kilowatt-hours for each hour of running. An increase of $30$ kilowatt-hours takes $30\div2.4=12.5$ hours.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	3F	Q13	5	Allowed	Foundation	QP MS
2024-11	1F	Q24	1	Non-calculator	Foundation	QP MS
2019-06	3H	Q17	2	Allowed	Higher	QP MS
2024-11	1H	Q7	1	Non-calculator	Higher	QP MS
2022-06	3H	Q8	3	Allowed	Higher	QP MS

Other points in R Ratio, proportion and rates of change

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