What is A15: Gradients and areas under graphs?

Higher tier only. Gradient is change in the vertical coordinate divided by change in the horizontal coordinate, with units formed from the axis units.

How many marks are A15 questions worth?

In verified past-paper appearances, A15 questions on Edexcel GCSE Maths 1MA1 have been worth 3–7 marks.

Which past exam papers has A15 appeared in?

A15 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.

A15

Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Higher only

Gradients and areas under graphs

Worked answers, methods and verified real exam appearances for A15 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. Gradient is change in the vertical coordinate divided by change in the horizontal coordinate, with units formed from the axis units.
For a curve, draw a tangent at the required point and use two well-separated points on that tangent, not two points on the curve.
Estimate an area under a non-linear graph by dividing it into strips and applying the trapezium rule.
On a distance-time graph gradient represents speed; on a velocity-time graph gradient represents acceleration and area represents displacement.
The examiner expects a numerical result, correct units and an interpretation of its sign where relevant.

Worked example

A distance-time graph is a straight line from $(12,150)$ to $(32,510)$ , with time in seconds and distance in metres. Calculate and interpret its gradient.

1.Use change in distance divided by change in time: $\dfrac{510-150}{32-12}$ .
2.Evaluate $\dfrac{360}{20}=18$ .
3.Attach the units metres per second and interpret the constant straight-line gradient.

Answer: $18$ m/s; the object travels at a constant speed of $18$ m/s.

Common mistakes

Don't use two points on the curve instead of two points on the tangent when estimating a gradient.
Don't add trapezium heights without multiplying by half the strip width.
Don't report a velocity-time area in m/s instead of metres.

Exam tip

For an estimate, leave the tangent or trapezia visible because the method marks depend on the construction.

Worked practice

Tier 1 · Easy

A distance-time graph is a straight line from

(12,150)

(32,510)

, where time is in seconds and distance is in metres. Calculate 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	$18$ m/s The object travels at a constant speed of $18$ m/s.	3	The gradient is $(510-150)/(32-12)=360/20=18$ . Distance divided by time has units m/s, so this is the object's constant speed.

Tier 2 · Standard

A velocity-time graph joins the points

(0,0)

(6,15)

(14,15)

and

(20,3)

with straight lines. Work out the distance travelled in the first

20

seconds.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$219$ m	4	The distance is the area under the graph. From $0$ to $6$ seconds the triangle has area $\frac12\times6\times15=45$ . From $6$ to $14$ seconds the rectangle has area $8\times15=120$ . From $14$ to $20$ seconds the trapezium has area $\frac12(15+3)\times6=54$ . The total is $45+120+54=219$ m.

Tier 3 · Hard

A curved velocity-time graph passes through the values

v=18,25,29,24,10

m/s at

t=0,3,6,9,12

seconds. Use four trapezia to estimate the distance travelled. A tangent at

t=6

passes through

(3,38)

and

(11,14)

; estimate the acceleration then.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Estimated distance $=276$ m Estimated acceleration $=-3$ m/s $^2$	6	With strip width $3$ , the trapezium estimate is $\frac{3}{2}[18+2(25)+2(29)+2(24)+10]=276$ m. The tangent gradient is $(14-38)/(11-3)=-24/8=-3$ . On a velocity-time graph this gradient is acceleration, so the estimate is $-3$ m/s $^2$ .

Tier 1 · Easy

A straight-line graph of a machine's value in £ against its age in years passes through

(2,1160)

and

(7,860)

. Work out the gradient. Explain what the gradient represents.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$-60$ The machine loses £60 in value each year.	3	The gradient is $\dfrac{860-1160}{7-2}=\dfrac{-300}{5}=-60$ . The vertical coordinate is value in £ and the horizontal coordinate is age in years, so the negative gradient means the machine loses £60 in value each year.

Tier 2 · Standard

A distance-time curve passes through

A(2,18)

B(7,83)

and

C(12,123)

, with time in seconds and distance in metres. Work out the average speed from

A

B

and from

B

C

. Which interval has the greater average speed?

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	From $A$ to $B$ : $13$ m/s From $B$ to $C$ : $8$ m/s The interval from $A$ to $B$ has the greater average speed.	3	From $A$ to $B$ , the average speed is $\dfrac{83-18}{7-2}=\dfrac{65}{5}=13$ m/s. From $B$ to $C$ , it is $\dfrac{123-83}{12-7}=\dfrac{40}{5}=8$ m/s. Therefore the average speed is greater from $A$ to $B$ .

Tier 3 · Hard

The speeds of a cyclist at

t=0,2,4,6,8

seconds are

3,7,10,8,4

m/s. The speed-time curve lies above the straight line segment in each strip. Use four trapezia to estimate the distance travelled during the

8

seconds. State, with a reason, whether your estimate is an underestimate or an overestimate.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$57$ m Underestimate, because the curve lies above the top of each trapezium, so the trapezia miss some of the area.	4	Each strip has width $2$ , so the trapezium estimate is $\dfrac{2}{2}[3+2(7)+2(10)+2(8)+4]=57$ m. The curve lies above the straight line segments used as the tops of the trapezia, so the trapezia omit some area and the result is an under-estimate.

Tier 2 · Standard

A tangent to a curve showing fuel volume

V

litres against time

t

minutes passes through

(8,46)

and

(17,19)

. Estimate the gradient of the curve at the point of contact. Explain what this gradient represents.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$-3$ litres per minute The fuel volume is decreasing at $3$ litres per minute at that time.	3	The tangent gradient is $\dfrac{19-46}{17-8}=\dfrac{-27}{9}=-3$ litres per minute. The negative sign shows that the volume is decreasing, at a rate of $3$ litres per minute at the point of contact.

Tier 3 · Hard

The flow rate into a tank is measured at

5

-minute intervals. At times

0,5,10,15,20

minutes, the flow rates are

12,18,25,21,14

litres per minute respectively. Use four trapezia to estimate the volume added during the

20

minutes. Hence estimate the mean flow rate during this time.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Estimated volume $=385$ litres Estimated mean flow rate $=19.25$ litres per minute	4	The strip width is $5$ , so the trapezium estimate is $\dfrac{5}{2}[12+2(18)+2(25)+2(21)+14]=385$ litres. The mean flow rate is estimated volume divided by time: $385/20=19.25$ litres per minute.

Tier 3 · Hard

A boat's speed is recorded at uneven time intervals. At times

0,2,5,9,12

seconds, its speeds are

4,10,16,13,7

m/s respectively. Use four trapezia to estimate the distance travelled, then estimate the mean speed during the

12

seconds.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Estimated distance $=141$ m Estimated mean speed $=11.75$ m/s	5	The strip widths are $2$ , $3$ , $4$ and $3$ seconds, so calculate each trapezium separately. Their areas are $\frac12(4+10)(2)=14$ , $\frac12(10+16)(3)=39$ , $\frac12(16+13)(4)=58$ and $\frac12(13+7)(3)=30$ . The estimated distance is $14+39+58+30=141$ m, giving an estimated mean speed of $141/12=11.75$ m/s.

Q10

Tier 3 · Hard

A curved financial graph shows the value

V

pounds of a machine against its age

t

years. At

t=7

, the graph has value £28 500. A tangent there passes through

(4,31200)

and

(10,25800)

. Estimate the annual change in value and express the magnitude of this change as a percentage of the value at

t=7

, giving the percentage to one decimal place.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	The value changes by -£900 per year, so the machine is losing £900 in value per year $3.2\%$	4	The tangent gradient is $(25800-31200)/(10-4)=-5400/6=-900$ pounds per year. Its magnitude as a percentage of the value at the contact point is $900/28500\times100=3.157894\ldots\%$ , which rounds to $3.2\%$ . This value is $0.079$ final-digit units from the nearest one-decimal-place rounding boundary.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2021-11	3H	Q15	3	Allowed	Higher	QP MS
2019-11	2H	Q21	5	Allowed	Higher	QP MS
2022-06	1H	Q14	4	Non-calculator	Higher	QP MS
2022-11	2H	Q21	4	Allowed	Higher	QP MS
2024-11	3H	Q22	3	Allowed	Higher	QP MS
2019-11	3H	Q19	3	Allowed	Higher	QP MS
2021-11	3H	Q21	4	Allowed	Higher	QP MS
2024-06	2H	Q14	6	Allowed	Higher	QP MS
2019-06	2H	Q14	7	Allowed	Higher	QP MS

Other points in A Algebra · graphs

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