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Edexcel IGCSE 4MA1/2HR, Monday 3 June 2024: Worked Solutions, Questions 11 to 20

Sir Faraz Hassan

13 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/2HR - Higher Tier - Monday 3 June 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 10

This is part two of three. Questions 1 to 10, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2HR (Higher Tier), June 2024 series, sat Monday 3 June 2024 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

Both are PDF files hosted by Pearson: official question paper (PDF) and official mark scheme (PDF).

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

Download printable PDF

All 26 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 11 to 20 of 26

The cumulative frequency table gives information about the time, in hours, that each of $60$ members of a swimming club spent training in one week.

$\begin{array}{|c|c|}\hline \textbf{Time (}t\ \textbf{hours)} & \textbf{Cumulative frequency} \\ \hline 0 < t \leq 5 & 6 \\ \hline 0 < t \leq 10 & 17 \\ \hline 0 < t \leq 15 & 27 \\ \hline 0 < t \leq 20 & 42 \\ \hline 0 < t \leq 25 & 53 \\ \hline 0 < t \leq 30 & 60 \\ \hline \end{array}$

(a) On the grid below, draw a cumulative frequency graph for the information in the table. [2 marks]

(b) Use your graph to find an estimate for the interquartile range of the times. [2 marks]

$25$ members spent more than $W$ hours training.

One of the $60$ members is chosen at random.
This member spent $H$ hours training.

(d) Find the probability that $5 < H \leq 10$ [1 mark]

(b) hours(c) W =(d)

[Total 7 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A cumulative frequency table over

60

members, in classes five hours wide.

The running totals are

6, 17, 27, 42, 53, 60

5, 10, 15, 20, 25

and

30

hours.

25

members trained for more than

W

hours.

Find

(a) the cumulative frequency graph (b) an estimate for the interquartile range of the times (c) the value of

W

(d) the probability that

5 < H \leq 10

Plan the Solution

Plot each running total at the upper end of its class, then join the six points in order.
Read the quartiles off that line at $\dfrac{60}{4} = 15$ and $\dfrac{3 \times 60}{4} = 45$ , then subtract the smaller time from the larger.
For (c), turn $25$ members above $W$ into $60 - 25 = 35$ at or below it, because the graph only ever counts upwards.
For (d), the frequency of a single class is the difference of two neighbouring running totals.

Worked Solution [7 marks]

Rule - Cumulative frequency: plot each running total against the upper boundary of its class. The lower quartile is read at

\dfrac{n}{4}

on the cumulative frequency axis, the upper quartile at

\dfrac{3n}{4}

, and the interquartile range is the upper quartile minus the lower quartile.

Step 1: plot each running total at the top of its class

(5, 6), \; (10, 17), \; (15, 27)

(20, 42), \; (25, 53), \; (30, 60)

(Reason:

27

members trained for

15

hours or less, so the height

27

belongs at the end of that class and not in the middle of it)

Step 2: join the six points in order

6 < 17 < 27 < 42 < 53 < 60

(Reason: a running total never falls, so the six points climb from left to right; join them with a smooth curve or with straight line segments, and every later answer is read off that line)

Step 3: find the two quartile positions on the cumulative frequency axis

\dfrac{60}{4} = 15

\dfrac{3 \times 60}{4} = 45

(Reason: with

60

members, a quarter of the way up the cumulative frequency axis is

15

and three quarters of the way up is

45

)

Step 4: read across to the line, down to the time axis, and subtract

Q_1 \approx 9.1

Q_3 \approx 21.4

21.4 - 9.1 = 12.3

(Reason: the mark scheme allows anything from

8

9.5

for the lower quartile and

21

23

for the upper, so a careful reading of your own graph is enough)

Step 5: turn

25

members above

W

into a cumulative frequency

60 - 25 = 35

W \approx 17.7

(Reason: the graph counts members at or below a time, never above it, so

25

above

W

means

35

at or below

W

, and

35

is the height to read across from)

Step 6: pick one class out of the running totals

17 - 6 = 11

\dfrac{11}{60}

(Reason:

17

members trained for

10

hours or less and

6

of those for

5

hours or less, so the difference is the number of members in the class

5 < H \leq 10

)

(a) the line through

(5, 6)

up to

(30, 60)

(b)

12.3

hours(c)

W = 17.7

(d)

\dfrac{11}{60}

Verification

Check 1 - the column adds back up: Difference the cumulative column into class frequencies and add them:

6 + 11 + 10 + 15 + 11 + 7

6 + 11 + 10 + 15 + 11 + 7 = 60

, the stated number of members

Check 2 - the median falls between the quartiles: Read the median at

\dfrac{60}{2} = 30

, which lands in the class

15 < t \leq 20

the median is

16

hours, which sits between

9.1

and

21.4

as it must

Check 3 - count above W instead of below it: Take the reading used for (c) back off the total:

60 - 35 = 25

25

members above

W

, which is what the question states

Check 4 - the three probabilities cover everybody: Below

5

hours, between

5

and

10

hours, and above

10

hours:

\dfrac{6}{60} + \dfrac{11}{60} + \dfrac{43}{60}

\dfrac{6}{60} + \dfrac{11}{60} + \dfrac{43}{60} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) the cumulative frequency graph	B2	a fully correct graph: the six points at the ends of the intervals, joined with a curve or with line segments	✓
(a) partly correct	(B1)	5 correct points plotted and joined, or 6 correct points plotted but not joined, or 5 or 6 points plotted consistently within each interval rather than at its upper end, at their correct heights and joined, eg plotted at $2.5, 7.5, 12.5, 17.5, 22.5$ and $27.5$	✓
(a) guidance	Note	a bar chart type graph scores zero marks. Ignore any part of the graph drawn before $(5, 6)$	✓
(b) a correct method for the two readings	M1ft	readings taken on the time axis from cumulative frequency $45$ (or $45.75$ ) and from $15$ (or $15.25$ ), or equivalent, shown by lines or by marks on the time axis or just by the correct readings. Follow through from their own graph	✓
(b) the interquartile range	A1ft	a single value in the range $11.5$ to $13.5$ , follow through from their own graph. The two readings themselves are $8$ to $9.5$ and $21$ to $23$	✓
(c) a correct method for W	M1ft	for using or stating $35$ , or for lines or marks showing cumulative frequency $35$ used on the graph, or for an indication on the time axis at the correct point, or just for the correct reading. Follow through from an incorrect graph if the method is shown	✓
(c) the value of W	A1ft	a value in the range $16.5$ to $18.5$ , follow through from their own graph	✓
(d) the probability	B1	$\dfrac{11}{60}$ . Accept $0.18(333\ldots)$ or $18.(333\ldots)\%$ , the bracketed digits being optional, so $0.18$ or better, or $18\%$ or better	✓

Full marks: 7/7

The diagram shows triangle $ACD$ .
$B$ is a point on $AC$ and $E$ is a point on $AD$ , so that $ABC$ and $AED$ are straight lines.
$BE$ is parallel to $CD$

$AE = 10$ cm and $CD = 1.5 \times BE$

(a) Work out the length of $ED$ [2 marks]

$AB = (2x + 5)$ cm and $BC = (3x - 5)$ cm

(b) Work out the value of $x$ [2 marks]

(a) cm(b) x =

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

ABC

and

AED

are straight lines, and

BE

is parallel to

CD

AE = 10

CD = 1.5 \times BE

AB = (2x + 5)

cm and

BC = (3x - 5)

Find

(a) the length of

ED

(b) the value of

x

Plan the Solution

$BE$ is parallel to $CD$ , so triangle $ABE$ and triangle $ACD$ are similar.
The scale factor from the small triangle to the large one is $\dfrac{CD}{BE} = 1.5$ , so every length measured from $A$ is $1.5$ times as long in the large triangle.
Part (a): scale $AE$ up to get $AD$ , then take $AE$ off it, because $E$ sits between $A$ and $D$ .
Part (b): $AC$ can be written two ways, as $AB + BC$ and as $1.5 \times AB$ . Setting them equal gives one equation in $x$ .

Worked Solution [4 marks]

Rule - Similar triangles: a line parallel to one side of a triangle cuts the other two sides in the same ratio, so

AD = k \times AE

and

AC = k \times AB

, where

k = \dfrac{CD}{BE}

Step 1: Find the scale factor, and use it on

AD

k = \dfrac{CD}{BE} = 1.5

AD = 1.5 \times AE = 1.5 \times 10 = 15

(Reason:

BE

is parallel to

CD

, so the angle at

A

is shared and the angles at

B

and

C

are equal corresponding angles. That makes the two triangles similar, and

AD

is the image of

AE

under the same

1.5

Step 2: Take

AE

off

AD

ED = AD - AE

ED = 15 - 10 = 5

(Reason:

E

lies on

AD

between

A

and

D

, so

AD

splits into

AE

and

ED

. The answer is in centimetres because

AE

was.)

Step 3: Write

AC

in two different ways

AC = AB + BC = (2x + 5) + (3x - 5) = 5x

AC = 1.5 \times AB = 1.5(2x + 5) = 3x + 7.5

(Reason:

B

lies on

AC

, so

AC = AB + BC

and the

+5

and the

-5

cancel. The same scale factor

1.5

also takes

AB

AC

, which is where the second expression comes from.)

Step 4: Set the two expressions equal, and solve

5x = 3x + 7.5

2x = 7.5

x = \dfrac{7.5}{2} = 3.75

(Reason: Take

3x

from both sides, then halve.

x

carries no unit here: it is the number that makes the two lengths fit together.)

(a)

ED = 5

cm(b)

x = 3.75

Verification

Check 1: Put

x = 3.75

back into the two expressions:

AB = 2 \times 3.75 + 5 = 12.5

and

BC = 3 \times 3.75 - 5 = 6.25

, so

AC = 12.5 + 6.25 = 18.75

\dfrac{AC}{AB} = \dfrac{18.75}{12.5} = 1.5

, the scale factor the question gives

Check 2: A parallel line cuts both sides in the same ratio, so

ED : AE

must match

BC : AB

. The first pair comes from part (a), the second from part (b), so this ties the two answers together.

\dfrac{5}{10} = \dfrac{6.25}{12.5} = 0.5

Check 3: Solve part (b) a different way.

BC

is half of

AB

, so

\dfrac{3x - 5}{2x + 5} = \dfrac{1}{2}

. Cross-multiplying gives

2(3x - 5) = 2x + 5

6x - 10 = 2x + 5

, so

4x = 15

and

x = \dfrac{15}{4} = 3.75

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $(AD =) 10 \times 1.5$ $(= 15)$ oe	M1	for a complete method to find $AD$ , the scale factor $1.5$ applied to $AE$ . Any equivalent working scores it.	✓
(a) $5$	A1	cao. Working is not required, so a correct answer scores both marks, unless it comes from obviously incorrect working.	✓
(b) $(2x + 5) + (3x - 5) = 1.5(2x + 5)$ oe	M1	for a correct equation in $x$ . Accept any equivalent form, eg $5x = 1.5(2x + 5)$ , or $5x = 3x + 7.5$ , or $\dfrac{3x - 5}{2x + 5} = \dfrac{1}{2}$ .	✓
(b) $3.75$	A1	oe, eg $\dfrac{15}{4}$ or $3\dfrac{3}{4}$ . Working is not required, so a correct answer scores both marks, unless it comes from obviously incorrect working.	✓

Full marks: 4/4

$OAB$ is a sector of a circle. The centre of the circle is $O$ and its radius is $r$ cm.

Angle $AOB = 60^{\circ}$
The perimeter of the sector is $P$ cm.

Work out a formula for $P$ in terms of $r$ .
Write your answer in the form $P = r(c\pi + k)$ , where $c$ and $k$ are numbers to be found. [3 marks]

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

A sector

OAB

of a circle with centre

O

and radius

r

The angle at the centre:

AOB = 60^{\circ}

The perimeter of the sector is

P

Find

A formula for

P

in terms of

r

It must be written as

P = r(c\pi + k)

, so the two numbers

c

and

k

are what the answer really asks for.

Plan the Solution

The perimeter of a sector is the curved arc plus the two straight edges, and both straight edges are radii, so start by writing $P = \text{arc} + 2r$ .
$60^{\circ}$ out of $360^{\circ}$ is one sixth of a full turn, so the arc is one sixth of the whole circumference $2\pi r$ .
Add the two radii, then take a factor of $r$ out of both terms to reach the form the question asks for.

Worked Solution [3 marks]

Rule - Arc length and sector perimeter: an arc that subtends

\theta

at the centre has length

\dfrac{\theta}{360} \times 2\pi r

, and the perimeter of the sector is that arc plus the two radii,

2r

Step 1: Write down what the perimeter is made of

P = \text{arc } AB + OA + OB

OA = OB = r

P = \text{arc } AB + 2r

(Reason: Both straight edges of a sector run from the centre to the circle, so each one is a radius. Together they contribute

2r

, and only the arc is left to find.)

Step 2: Find the length of the arc

\text{arc } AB = \dfrac{60}{360} \times 2 \pi r

\dfrac{60}{360} = \dfrac{1}{6}

\dfrac{1}{6} \times 2 \pi r = \dfrac{\pi r}{3}

(Reason: The arc is the same fraction of the circumference as

60^{\circ}

is of a full turn. Since

60^{\circ}

is one sixth of

360^{\circ}

, the arc is one sixth of

2\pi r

Step 3: Add the two radii

P = \dfrac{\pi r}{3} + 2r

(Reason: The arc and the two radii together make the whole way round the sector, so the two pieces from Step 1 and Step 2 are simply added.)

Step 4: Take out the factor r

P = r\left(\dfrac{1}{3}\pi + 2\right)

c = \dfrac{1}{3} \text{ and } k = 2

(Reason: Both terms contain

r

, and

\dfrac{\pi r}{3} = r \times \dfrac{1}{3}\pi

, so taking

r

outside a bracket gives exactly the form

P = r(c\pi + k)

that was asked for.)

P = r\left(\dfrac{1}{3}\pi + 2\right)

with

c = \dfrac{1}{3}

and

k = 2

Verification

Check 1: Put a number in. With

r = 6

, work the perimeter out directly: the arc is one sixth of

2\pi \times 6

and the two radii add

12

. Then put

r = 6

into the formula and compare.

\dfrac{1}{6} \times 2\pi \times 6 + 12 = 2\pi + 12

and

6\left(\dfrac{1}{3}\pi + 2\right) = 2\pi + 12 \approx 18.28

Check 2: Six of these sectors fit round a point, so six of the arcs must make up the whole circumference. Multiply the arc by

6

and see whether the circumference comes back.

6 \times \dfrac{\pi r}{3} = 2\pi r

, which is the full circumference, so the fraction of the circle used is right.

Check 3: Measure the angle in radians instead, which never uses

360

at all.

60^{\circ} = \dfrac{\pi}{3}

radians, and an arc of angle

\theta

radians has length

r\theta

r \times \dfrac{\pi}{3} + 2r = r\left(\dfrac{1}{3}\pi + 2\right)

, the same formula by a different route.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{60}{360} \times 2 \times \pi \times r$ oe, or $\dfrac{1}{6} \times 2 \times \pi \times r$ oe	M1	for finding the length of the arc	✓
their $\dfrac{60}{360} \times 2 \times \pi \times r + 2r$ oe	M1	dep on M1 for a complete expression from correct working for a method for the perimeter	✓
$P = r\left(\dfrac{1}{3}\pi + 2\right)$	A1	oe, eg $P = r(0.33\ldots\pi + 2)$ or $P = \left(\dfrac{1}{3}\pi + 2\right)r$ or $P = \left(2 + \dfrac{120}{360}\pi\right)r$ or $P = \left(\dfrac{120}{360}\pi + 2\right)r$	✓
Working not required, so a correct answer scores full marks, unless it comes from obviously incorrect working.	Note	guidance printed beside this question in the official mark scheme; it awards nothing on its own	✓

Full marks: 3/3

Camila is going to spin a biased spinner and drop a bent drawing pin.
The spinner has six sections, numbered $1$ to $6$ .
The drawing pin will land either point up or point down.

The probability that the drawing pin will land point up is $0.8$
The probability that the spinner will land on $6$ and the drawing pin will land point up is $0.24$

Work out the probability that the spinner will land on $6$ and the drawing pin will land point down. [3 marks]

[Total 3 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

The drawing pin lands point up with probability

0.8

, so point up and point down are its only two outcomes.

The spinner lands on

6

and the pin lands point up with probability

0.24

The spin and the drop do not affect each other, so the two events are independent.

Find

The probability that the spinner lands on

6

and the drawing pin lands point down.

Plan the Solution

The $0.24$ is already a product of two probabilities, so work backwards from it to the probability of a $6$ .
The pin has only two outcomes, so point down is $1$ minus $0.8$ .
Multiply the two probabilities, because the spinner and the pin are independent.

Worked Solution [3 marks]

Rule - Independent events:

P(A \text{ and } B) = P(A) \times P(B)

, and reading that backwards gives

P(A) = \dfrac{P(A \text{ and } B)}{P(B)}

Step 1: Work back to the probability of a

6

P(6) \times 0.8 = 0.24

P(6) = \dfrac{0.24}{0.8} = 0.3

(Reason: The spin and the drop are independent, so the given

0.24

is the probability of a

6

multiplied by

0.8

. Dividing

0.24

0.8

undoes that multiplication and leaves the probability of a

6

on its own.)

Step 2: Write down the probability that the pin lands point down

P(\text{point down}) = 1 - 0.8 = 0.2

(Reason: The pin lands point up or point down and there is no third outcome, so those two probabilities add to

1

Step 3: Multiply the two probabilities together

P(6 \text{ and point down}) = 0.3 \times 0.2 = 0.06

(Reason: Independent events multiply, so the probability of a

6

and point down is the probability of a

6

times the probability of point down.)

0.06

Verification

Check 1: One spin and one drop have four outcomes in total, so their probabilities must add to

1

. The other two are

P(\text{not } 6) = 1 - 0.3 = 0.7

, then

0.7 \times 0.8 = 0.56

and

0.7 \times 0.2 = 0.14

0.24 + 0.56 + 0.14 + 0.06 = 1

Check 2: Reach the answer without finding the probability of a

6

at all. Point down is

\dfrac{0.2}{0.8} = 0.25

as likely as point up, so scale the given

0.24

by that same ratio.

0.24 \times 0.25 = 0.06

Check 3: Redo the whole question in fractions, where nothing can be lost to a decimal point. The probability of a

6

\dfrac{3}{10}

and point down is

\dfrac{1}{5}

\dfrac{3}{10} \times \dfrac{1}{5} = \dfrac{3}{50} = 0.06

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{0.24}{0.8} = 0.3$	M1	for a correct method to find the probability that the spinner lands on $6$ , or equivalent	✓
$0.3 \times (1 - 0.8)$ or $0.3 \times 0.2$	M1	for a complete method, or equivalent, eg $1 - (0.3 \times 0.8 + 0.7 \times 0.8 + 0.7 \times 0.2)$ or $1 - 0.94$ . The candidate's own value may be used in place of $0.3$	✓
$0.06$	A1	or equivalent, eg $\dfrac{3}{50}$ or $\dfrac{6}{100}$ or $6\%$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working	✓

Full marks: 3/3

The diagram shows three sides of a regular pentagon together with a triangle.

$AB$ , $BC$ and $CD$ are three sides of a regular pentagon and $CDE$ is a triangle.
$BCE$ is a straight line.

$CD = 6.5$ cm
$CE = 3$ cm

Calculate the area of triangle $CDE$ .
Give your answer correct to $3$ significant figures. [3 marks]

cm²

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

AB

BC

and

CD

are three sides of a regular pentagon, so they are equal and the angles at

B

and

C

inside the pentagon are equal.

BCE

is a straight line, so the angle inside the pentagon at

C

and angle

DCE

sit side by side on it.

CD = 6.5

cm and

CE = 3

cm - two sides of triangle

CDE

with the unknown angle between them.

Find

The area of triangle

CDE

, correct to

3

significant figures.

Plan the Solution

Work out one interior angle of a regular pentagon. Its $5$ angles are equal and they add to $(5 - 2) \times 180^\circ$ .
Angle $BCD$ is one of those interior angles, and $BCE$ is straight, so angle $DCE$ is what is left of $180^\circ$ .
Triangle $CDE$ then has two known sides with a known angle between them, which is exactly what $\dfrac{1}{2}ab\sin C$ asks for. No height has to be found.

Worked Solution [3 marks]

Rule - Area from two sides and the included angle:

\text{Area} = \dfrac{1}{2}ab\sin C

, where

a

and

b

are two sides of the triangle and

C

is the angle BETWEEN them. The angle must be the one the two sides enclose, not either of the other two.

Step 1: Work out one interior angle of the regular pentagon

(5 - 2) \times 180^\circ = 540^\circ

\text{interior angle} = \dfrac{540^\circ}{5} = 108^\circ

(Reason: A polygon with

n

sides splits into

n - 2

triangles, so its angles add to

(n - 2) \times 180^\circ

. A pentagon gives

540^\circ

, and in a REGULAR pentagon all

5

angles are equal, so each one takes a fifth of it.)

Step 2: Use the straight line

BCE

to find angle

DCE

\angle DCE = 180^\circ - 108^\circ = 72^\circ

(Reason: Angle

BCD

is the interior angle just found. Angles on a straight line add to

180^\circ

, so angle

DCE

is the rest of that half turn. This is the exterior angle of the pentagon, which is why

\dfrac{360^\circ}{5}

reaches the same

72^\circ

in one step.)

Step 3: Put the two sides and the angle between them into the area formula

\text{Area} = \dfrac{1}{2} \times 6.5 \times 3 \times \sin 72^\circ = 9.2728\ldots

(Reason: The two sides are

CD = 6.5

cm and

CE = 3

cm, and the angle between them at

C

is the

72^\circ

found in Step 2. Make sure the calculator is in degree mode.)

Step 4: Round to

3

significant figures

9.2728\ldots \to 9.27

(Reason: The first three significant figures are

9

2

and

7

. The next digit is

2

, which is below

5

, so the

7

stays as it is. The unit is square centimetres, because both lengths are in centimetres.)

9.27 \text{ cm}^2

Verification

Check 1: Do it without the area formula. Drop a perpendicular from

D

to the straight line

BCE

. Its height is

6.5 \times \sin 72^\circ

, and the base

CE

3

cm, so use half base times height.

h = 6.1818\ldots

and

\dfrac{1}{2} \times 3 \times 6.1818\ldots = 9.2728\ldots

Check 2: A route with no sine in it at all. Find the third side with the cosine rule,

DE^2 = 6.5^2 + 3^2 - 2 \times 6.5 \times 3 \times \cos 72^\circ

, then put all three sides into Heron's formula

\sqrt{s(s - a)(s - b)(s - c)}

DE = 6.2608\ldots

s = 7.8804\ldots

and the formula gives

9.2728\ldots

Check 3: A size check. Two sides of

6.5

cm and

3

cm enclose the largest area they can when the angle between them is

90^\circ

, and that area is

\dfrac{1}{2} \times 6.5 \times 3 = 9.75

cm². The angle here is

72^\circ

, not far below

90^\circ

, so the answer must be a little under

9.75

. It is:

9.27

cm² is just under the

9.75

cm² ceiling, and far above zero, which is where a very small angle would put it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method for an exterior or an interior angle of a regular pentagon: $\dfrac{360}{5} = 72$ oe, or $\dfrac{(5 - 2) \times 180}{5} = 108$ oe, or $\dfrac{540}{5} = 108$ oe	M1	Do not award this mark if $108$ is assigned as an exterior angle, or if $72$ is assigned as an interior angle. Angles written on the diagram other than the exterior or interior angles of the pentagon are ignored, even where they are labelled incorrectly.	✓
Substitute into $\dfrac{1}{2} \times 6.5 \times 3 \times \sin [\text{angle } DCE]$ oe, or find the height first, $h = 6.5 \times \sin [\text{angle } DCE]$ (= 6.18...), and then $\dfrac{1}{2} \times 3 \times h$ oe	M1ft	Follow through on the candidate's own angle $DCE$ when it is substituted in, provided that angle is less than $90^\circ$ .	✓
Working is not required, so a correct answer scores full marks (unless it comes from obviously incorrect working).	A1	For $9.27$ . Accept anything from $9.26$ to $9.28$ , which covers rounding the angle or the area part way through.	✓
Special case: $\dfrac{1}{2} \times 6.5 \times 3 \times \sin 108^\circ = 9.27$ ...	SC B2	The named error is using the pentagon's interior angle as the angle inside triangle $CDE$ , instead of the angle beside it on the straight line. It still produces $9.27$ , because $\sin 108^\circ = \sin 72^\circ$ , so the arithmetic hides the mistake and the work scores $2$ of the $3$ marks rather than all of them. This row carries no mark of its own.	✓

Full marks: 3/3

Six graphs are sketched below, labelled A to F.

For each equation, write down the letter of the graph that could have that equation.

(i) $y = -\dfrac{1}{x}$ [1 mark]

(ii) $y = \sin x^\circ$ [1 mark]

(i)(ii)

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

Six sketch graphs, labelled A to F. The axes carry no numbers, so only the SHAPE of each curve can be used.

Two equations to place:

y = -\dfrac{1}{x}

and

y = \sin x^\circ

Find

One letter for each equation - the sketch that could be its graph.

Plan the Solution

Name the family each equation belongs to. $y = -\dfrac{1}{x}$ is a reciprocal curve; $y = \sin x^\circ$ is a wave that repeats.
Write down the feature that family must show, then look for it. A reciprocal has no value at $x = 0$ , so it is drawn in two separate branches; a sine wave passes through the origin and turns over both above and below the axis.
Two sketches can belong to the same family, so finish each part with one tested value: the sign of $y$ on each side of the $y$ -axis for the reciprocal, and the height at $x = 0$ for the wave.

Worked Solution [2 marks]

Rule - Match a graph by its family first, then by one tested value.

y = \dfrac{k}{x}

has two separate branches and no value at

x = 0

, and the sign of

k

decides which pair of quadrants they lie in.

y = \sin x^\circ

passes through

(0, 0)

, reaches

1

x = 90

, returns to

0

x = 180

, falls to

-1

x = 270

and then repeats every

360^\circ

Step 1: Read off what

y = -\dfrac{1}{x}

must look like

y = -\dfrac{1}{x}, \ x \neq 0

(Reason: Dividing by

0

has no meaning, so the curve has a break at

x = 0

and is drawn in two separate branches, one on each side of the

y

-axis. Only one of the six sketches is drawn in two pieces; the other five are single unbroken curves.)

Step 2: Test the sign of

y

on each side of the

y

-axis

x = -2 \implies y = -\dfrac{1}{-2} = \dfrac{1}{2}

x = 2 \implies y = -\dfrac{1}{2}

(Reason: A negative divided by a negative is positive, so to the left of the

y

-axis the curve lies ABOVE the

x

-axis, and to the right of it the curve lies below. The two branches therefore sit in the second and fourth quadrants: top left and bottom right.)

Step 3: Pick the sketch with those two branches

\text{left branch: } y > 0 \quad \text{right branch: } y < 0

(Reason: Sketch E is the only one drawn in two pieces, and its pieces sit where the signs say: above the axis on the left, below it on the right. Both pieces flatten towards the

x

-axis as

x

grows, which is what

-\dfrac{1}{x}

does once the denominator is large - at

x = 10

the height is only

-0.1

Step 4: Read off what

y = \sin x^\circ

must look like

\sin 0^\circ = 0

\sin 90^\circ = 1

\sin 180^\circ = 0

\sin 270^\circ = -1

\sin 360^\circ = 0

(Reason: The wave starts at the origin, climbs to

1

, comes back down through the axis, drops to

-1

and returns - then does the same again. So the graph must pass THROUGH the origin, and be rising as it does so. Two of the six sketches are waves.)

Step 5: Separate the two wave sketches

\sin 0^\circ = 0

\cos 0^\circ = 1

(Reason: Sketches A and D are both waves. D is at the top of a crest where it meets the

y

-axis, so its graph is at its greatest height when

x = 0

- that is the shape of

y = \cos x^\circ

. Sketch A cuts the axis at the origin and rises, which is what

\sin 0^\circ = 0

requires. So A is the sine curve.)

(i) E(ii) A

Verification

Check 1: Test more points on

y = -\dfrac{1}{x}

, two close to the

y

-axis and one far from it:

x = -0.5

x = 0.5

and

x = 10

y = 2

y = -2

and

y = -0.1

. The left branch climbs steeply near the

y

-axis and the right branch flattens towards the

x

-axis far from it, which is sketch E.

Check 2: Rule the other five sketches out of part (i). B and C are single curves with a value at

x = 0

, A and D are waves that cross the

x

-axis, and F climbs through the origin without a break. None of the five has the break at

x = 0

that

-\dfrac{1}{x}

must have, so E is the only sketch left.

Check 3: Use symmetry for part (ii). Sine is an odd function:

\sin(-30)^\circ = -0.5

while

\sin 30^\circ = 0.5

, so the curve has half-turn symmetry about

O

- whatever it does to the right of the

y

-axis, it does upside down to the left. Sketch A has a crest to the right of

O

and a matching trough the same distance to the left, so it has that symmetry. Sketch D is a mirror image in the

y

-axis instead, which is the symmetry of a cosine curve, so A is the sine.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) E	B1	For E. Working is not required, so the letter alone earns the mark. This is a B mark, given for the answer itself, so there is no method mark on this part and a correct description carrying the wrong letter earns nothing.	✓
(ii) A	B1	For A. Again the letter alone earns the mark. D is the near miss this part is built around: it is the same wave shifted by $90^\circ$ , so it is the graph of $y = \cos x^\circ$ and it scores nothing.	✓

Full marks: 2/2

The functions $\text{f}$ and $\text{g}$ are given by
$\text{f}(x) = \dfrac{x}{2x - 4}$
$\text{g}(x) = 3x + 1$

Given that $\text{fg}(k) = 2$
work out the value of $k$ [3 marks]

k =

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

\text{f}(x) = \dfrac{x}{2x - 4}

and

\text{g}(x) = 3x + 1

, two functions of

x

The composite function satisfies

\text{fg}(k) = 2

Find

The value of

k

, a single number

Plan the Solution

$\text{fg}$ means $\text{g}$ first and $\text{f}$ second, so work out $\text{g}(k) = 3k + 1$ and put the whole of it in place of $x$ in $\text{f}$
Multiply out the bracket in the denominator so the composite becomes one single algebraic fraction.
Set that fraction equal to $2$ , multiply both sides by the denominator to clear the fraction, then solve the linear equation that is left

Worked Solution [3 marks]

Rule - Composite functions:

\text{fg}(x)

means

\text{f}(\text{g}(x))

. The inner function

\text{g}

is applied first, and the whole of its output replaces every

x

\text{f}

Step 1: Work out

\text{g}(k)

and feed it into

\text{f}

\text{g}(k) = 3k + 1

\text{fg}(k) = \text{f}(3k + 1) = \dfrac{3k + 1}{2(3k + 1) - 4}

(Reason:

\text{fg}(k)

means

\text{f}(\text{g}(k))

, so the whole bracket

3k + 1

replaces

x

in the numerator and in the denominator of

\text{f}

at the same time)

Step 2: Simplify the denominator

2(3k + 1) - 4 = 6k + 2 - 4 = 6k - 2

\text{fg}(k) = \dfrac{3k + 1}{6k - 2}

(Reason: The

2

outside the bracket multiplies both terms inside it, and only then are the two numbers collected, so the composite is a single fraction ready to be solved)

Step 3: Set

\text{fg}(k) = 2

and clear the fraction

\dfrac{3k + 1}{6k - 2} = 2

3k + 1 = 2(6k - 2)

3k + 1 = 12k - 4

(Reason: Multiplying both sides by

6k - 2

removes the fraction and leaves an ordinary linear equation, which is the step the mark scheme calls correctly removing the denominator)

Step 4: Solve the linear equation for

k

1 + 4 = 12k - 3k

5 = 9k

k = \dfrac{5}{9}

(Reason: Collecting the

k

terms on the right, where the coefficient is larger, keeps that coefficient positive and avoids a sign slip)

k = \dfrac{5}{9}

Verification

Check 1 - put $k = \dfrac{5}{9}$ back through both functions: Inner function first:

\text{g}\left(\dfrac{5}{9}\right) = 3 \times \dfrac{5}{9} + 1 = \dfrac{5}{3} + 1 = \dfrac{8}{3}

. Feeding that into

\text{f}

gives the denominator

2 \times \dfrac{8}{3} - 4 = \dfrac{4}{3}

\text{fg}\left(\dfrac{5}{9}\right) = \dfrac{8}{3} \times \dfrac{3}{4} = 2

Check 2 - work backwards through the two functions instead: Ask first which input

\text{f}

sends to

2

\dfrac{x}{2x - 4} = 2

gives

x = 2(2x - 4) = 4x - 8

, so

3x = 8

and

x = \dfrac{8}{3}

. That input has to be

\text{g}(k)

3k + 1 = \dfrac{8}{3} \implies 3k = \dfrac{5}{3} \implies k = \dfrac{5}{9}

Check 3 - the value is one the composite is allowed to take: The composite

\dfrac{3k + 1}{6k - 2}

is undefined only where its denominator is zero, that is at

k = \dfrac{1}{3}

. At

k = \dfrac{5}{9}

the denominator is

6 \times \dfrac{5}{9} - 2 = \dfrac{4}{3}

, which is not zero, so the answer is a genuine solution

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Form $\text{fg}(k)$ by substituting $\text{g}(k)$ into $\text{f}$	M1	For a correct expression for $\text{fg}(k)$ or $\text{fg}(x)$ , e.g. $\dfrac{3k + 1}{2(3k + 1) - 4}$ or $\dfrac{3k + 1}{6k - 2}$ , with or without $= 2$ . Or for starting from $\text{f}(x) = 2$ and reaching $x = 2(2x - 4)$ or $x = 4x - 8$ or $x = \dfrac{8}{3}$ . Allow $x$ instead of $k$ for all marks.	✓
Clear the denominator to form a correct equation	M1 dep	Dependent on the first M1, for correctly removing the denominator to form a correct equation, e.g. $3k + 1 = 2(6k - 2)$ or $3k + 1 = 2\left(2(3k + 1) - 4\right)$ or $3k + 1 = 12k - 4$ . Or, by the backwards route, for $\text{g}(k) = \dfrac{8}{3}$ , i.e. $3k + 1 = \dfrac{8}{3}$ .	✓
Solve the linear equation for $k$	A1	For $k = \dfrac{5}{9}$ or any equivalent, e.g. $0.55(555\ldots)$ rounded or truncated, the bracketed digits being optional, so $0.55$ or better, or $0.\dot{5}$ with the recurring dot shown.	✓
A correct answer written down with no working	Note	Working is not required in this question, so a fully correct answer scores all $3$ marks, unless it has come from obviously incorrect working.	✓

Full marks: 3/3

The recurring decimal $0.\dot{3}0\dot{6}$ can be written as the fraction $\dfrac{34}{111}$ .
Use algebra to show that this is correct.
You must show clear algebraic working. [2 marks]

[Total 2 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.\dot{3}0\dot{6}

. The dots mark the first and last digits of the block that repeats, so the digits after the point run

306

over and over without stopping.

The fraction it is claimed to equal,

\dfrac{34}{111}

Find

An algebraic argument that

0.\dot{3}0\dot{6}

is exactly

\dfrac{34}{111}

. The working is the answer here, so a decimal read off a calculator proves nothing on its own.

Plan the Solution

Give the decimal a name: let $x = 0.\dot{3}0\dot{6}$ .
Count the digits in the repeating block. There are three of them, so multiply by $1000$ to move the point past exactly one whole block.
Subtract the original equation from the shifted one. The two never-ending tails are identical, so they cancel and a whole number is left.
Solve that equation, then cancel the fraction by the HCF of its numerator and its denominator.

Worked Solution [2 marks]

Rule - Recurring decimal to fraction: if the repeating block is

n

digits long, multiply by

10^{n}

, subtract the original, and divide by

10^{n} - 1

. The subtraction is what removes the tail that never ends.

Step 1: give the decimal a name

x = 0.\dot{3}0\dot{6} = 0.306306306\ldots

(Reason: Naming the decimal is what makes algebra possible. The dots say the block

306

repeats for ever, so the digits after the point are

306

again and again.)

Step 2: shift the point past one whole block

1000x = 306.306306\ldots

(Reason: The block is three digits long, so multiplying by

1000

moves the point three places. Everything after the point is left exactly as it was, and that is the whole trick.)

Step 3: subtract the original from the shifted copy

1000x - x = 306.306306\ldots - 0.306306\ldots

999x = 306

(Reason: The two tails are identical, so they cancel exactly and a whole number is left behind. Nothing has been rounded, which is what makes this a proof rather than a check.)

Step 4: divide to leave x on its own

x = \dfrac{306}{999}

(Reason: Both sides are divided by the number standing in front of

x

. The fraction that comes out is not yet in its lowest terms.)

Step 5: cancel to lowest terms

\dfrac{306}{999} = \dfrac{34}{111}

(Reason: The HCF of

306

and

999

9

. Dividing top and bottom by that same

9

gives

34

over

111

, which is the fraction the question asked for.)

x = 0.\dot{3}0\dot{6} = \dfrac{306}{999} = \dfrac{34}{111}

Verification

Check 1 - divide the fraction back out: Work out

34

divided by

111

and read the digits after the point.

\dfrac{34}{111} = 0.306306306\ldots

, the block

306

repeating, which is the decimal the question started from.

Check 2 - multiply the fraction back up: Step 3 left

999x = 306

. Putting the fraction back in for

x

must give exactly

306

, with nothing left over.

999 \times \dfrac{34}{111} = \dfrac{33\,966}{111} = 306

Check 3 - the fraction is in its lowest terms: Factorise both parts:

34 = 2 \times 17

and

111 = 3 \times 37

. They share no factor at all, so

\dfrac{34}{111}

cannot be cancelled any further and is the finished fraction.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two correct equations, written ready to subtract, e.g. $1000x = 306.306306\ldots$ and $x = 0.306306\ldots$	M1	M1 for two correct algebraic equations involving the recurring decimal that, when subtracted, give a whole number or a terminating decimal ( $306$ or $306\,000$ ), with the intention to subtract. The larger pair $1\,000\,000x = 306\,306.306306\ldots$ and $1000x = 306.306306\ldots$ earns it just as well.	✓
Subtract, solve and cancel: $999x = 306$ and $\dfrac{306}{999} = \dfrac{34}{111}$	A1	A1 for completion to $\dfrac{34}{111}$ , dep on M1. The larger pair finishes the same way: $999\,000x = 306\,000$ and $\dfrac{306\,000}{999\,000} = \dfrac{34}{111}$ .	✓
Working required	Note	If the recurring dots are not written on both numbers, at least one of them must be shown to at least six significant figures before the M1 can be given. A bare $\dfrac{34}{111}$ with no algebra behind it scores nothing: the question asks for the algebra, and the algebra is what is being marked.	✓

Full marks: 2/2

Noam goes on an electric scooter ride along a coastal path.

For the scooter ride

average speed $= 19$ km/h correct to the nearest whole number
time $= 1.5$ hours correct to one decimal place

Work out the upper bound for the distance Noam travels.
Give your answer correct to $3$ significant figures. [3 marks]

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

An average speed of

19

km/h, correct to the nearest whole number.

A time of

1.5

hours, correct to one decimal place.

Both figures have been rounded, so each one stands for a whole interval of possible values rather than a single exact number.

Find

The upper bound of the distance travelled, correct to

3

significant figures. An upper bound is the value the distance can be pushed as close to as you like without ever quite reaching it, so it is the top end of an interval rather than a distance anyone actually rides.

Plan the Solution

Write $s$ for the speed, $t$ for the time and $d$ for the distance.
Turn each rounded figure back into the interval it came from. A whole number of km/h carries $0.5$ either side; one decimal place carries only $0.05$ either side.
Decide which end of each interval makes the distance as large as it can be.
Multiply those two values, then round the product to $3$ significant figures.

Worked Solution [3 marks]

Rule - Bounds of a product: a figure rounded to a unit

u

lies within

\dfrac{u}{2}

of the figure written down, so its bounds are

\text{figure} - \dfrac{u}{2}

and

\text{figure} + \dfrac{u}{2}

. A product of two positive quantities grows whenever either one of them grows, so

UB_{d} = UB_{s} \times UB_{t}

Step 1: the interval the speed came from

19 - 0.5 = 18.5

19 + 0.5 = 19.5

18.5 \leq s < 19.5

(Reason: The speed was rounded to the nearest whole number, so the rounding unit is

1

km/h and half of it is

0.5

. Every speed from

18.5

up to but not including

19.5

rounds to

19

, which makes

19.5

the upper bound.)

Step 2: the interval the time came from

1.5 - 0.05 = 1.45

1.5 + 0.05 = 1.55

1.45 \leq t < 1.55

(Reason: One decimal place makes the rounding unit

0.1

hours, not

1

hour, so half of it is

0.05

and not

0.5

. A time of

1.55

would itself round up to

1.6

, so the interval stops just short of

1.55

and that value is the upper bound.)

Step 3: multiply the two bounds that make the distance largest

d = s \times t

UB_{d} = UB_{s} \times UB_{t} = 19.5 \times 1.55 = 30.225

(Reason: Distance is speed multiplied by time, and a product of two positive numbers gets bigger whenever either number gets bigger. The largest distance therefore comes from the largest speed together with the largest time.)

Step 4: round to 3 significant figures

30.225 \approx 30.2

(Reason: The first three significant figures of

30.225

are

3

0

and

2

. The digit after them is

2

, which is below

5

, so the tenths digit is left as it stands.)

30.2

Verification

Check 1 - a pair from inside both intervals: Take a speed and a time from just inside each interval,

19.49

km/h and

1.549

hours, and round each one the way the question rounds it.

19.49 \approx 19

to the nearest whole number and

1.549 \approx 1.5

to one decimal place, so this ride fits the question. Its distance is

19.49 \times 1.549 = 30.19001

km, which sits just below the bound and never above it.

Check 2 - divide the product back out: Undo the multiplication: divide the unrounded distance by each upper bound in turn and see whether the other one comes back.

\dfrac{30.225}{1.55} = 19.5

and

\dfrac{30.225}{19.5} = 1.55

, the two upper bounds the product was built from.

Check 3 - every other pairing of the bounds is smaller: The two intervals allow four corner products. Work out the other three:

18.5 \times 1.45

19.5 \times 1.45

and

18.5 \times 1.55

. They come to

26.825

28.275

and

28.675

, all smaller than

30.225

. So the distance lies between

26.825

km and

30.225

km, and pairing an upper bound with a lower one always falls short of the top of that interval.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct bound: $18.5$ or $19.5$ or $1.45$ or $1.55$	B1	B1 for one correct bound. Allow $19.4\dot{9}$ for $19.5$ , and allow $1.54\dot{9}$ for $1.55$ . Those recurring forms are equal to the bounds exactly, not merely close to them.	✓
(distance $=$ ) $19.5 \times 1.55$	M1	M1 for $UB_{s} \times UB_{t}$ , where $19 < UB_{s} \leq 19.5$ and $1.5 < UB_{t} \leq 1.55$ . The upper limits are what let a candidate who writes $19.4\dot{9}$ or $1.549$ keep the method mark.	✓
$30.2$	A1	A1 for $30.2$ . Accept $30.225$ or $30.23$ . The answer must come from the correct figures, $19.5$ and $1.55$ .	✓
Working not required	Note	A correct answer written on the answer line scores all $3$ marks on its own, unless it has obviously come from incorrect working.	✓

Full marks: 3/3

Solve the inequality $6x^2 - 7x - 20 > 0$
You must show clear algebraic working. [4 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The quadratic inequality

6x^2 - 7x - 20 > 0

The coefficient of

x^2

6

, which is positive, so the curve

y = 6x^2 - 7x - 20

is U-shaped.

Find

Every value of

x

that makes

6x^2 - 7x - 20

greater than

0

. A quadratic gives two critical values, so expect an answer in two separate pieces.

Plan the Solution

Factorise $6x^2 - 7x - 20$ by splitting the middle term, using the product $6 \times (-20) = -120$ and the sum $-7$ .
Set each bracket equal to $0$ to get the two critical values.
Test one value from each of the three regions to see where the expression is positive.
Write the answer as two strict inequalities, because a U-shaped curve sits above the $x$ -axis outside its roots.

Worked Solution [4 marks]

Rule - Quadratic inequality: factorise, find the critical values, then read the regions off the shape of the curve. When the coefficient of

x^2

is positive, the expression is positive outside the critical values and negative between them.

Step 1: Factorise

6x^2 - 7x - 20

6 \times (-20) = -120

-15 \times 8 = -120

-15 + 8 = -7

6x^2 - 15x + 8x - 20

3x(2x - 5) + 4(2x - 5)

6x^2 - 7x - 20 = (3x + 4)(2x - 5)

(Reason: Multiply the

6

by the

-20

and look for two numbers with that product whose sum is

-7

. They are

-15

and

8

, so the middle term splits and the four terms pair into two brackets.)

Step 2: Find the two critical values

(3x + 4)(2x - 5) = 0

3x + 4 = 0 \implies x = -\dfrac{4}{3}

2x - 5 = 0 \implies x = \dfrac{5}{2}

(Reason: A product is zero only when one of its factors is zero, so each bracket gives one critical value. These are the two points where the curve meets the

x

-axis.)

Step 3: Test one value in each region

(3 \times (-2) + 4)(2 \times (-2) - 5) = 18

(3 \times 0 + 4)(2 \times 0 - 5) = -20

(3 \times 3 + 4)(2 \times 3 - 5) = 13

(Reason: The two critical values cut the number line into three regions. Testing

x = -2

x = 0

and

x = 3

gives positive, then negative, then positive - which is exactly what a U-shaped curve does.)

Step 4: Write down the solution

x < -\dfrac{4}{3} \quad \text{or} \quad x > \dfrac{5}{2}

(Reason: The two outside regions are the positive ones. The inequality is strict, so the critical values themselves are left out: at

x = -\dfrac{4}{3}

and

x = \dfrac{5}{2}

the expression equals

0

, which is not greater than

0

x < -\dfrac{4}{3} \quad \text{or} \quad x > \dfrac{5}{2}

Verification

Check 1: Expand

(3x + 4)(2x - 5)

again and compare it with the expression the question gives.

6x^2 - 15x + 8x - 20 = 6x^2 - 7x - 20

Check 2: Reach the critical values a second way, with the quadratic formula. The discriminant is

(-7)^2 - 4 \times 6 \times (-20) = 529

, and

\sqrt{529} = 23

x = \dfrac{7 + 23}{12} = \dfrac{5}{2}

and

x = \dfrac{7 - 23}{12} = -\dfrac{4}{3}

Check 3: Substitute a value from each piece of the answer, and one from between the critical values, into

6x^2 - 7x - 20

6(-3)^2 - 7(-3) - 20 = 55

6(4)^2 - 7(4) - 20 = 48

6(0)^2 - 7(0) - 20 = -20

Check 4: Confirm the critical values really are excluded, by putting the upper one back into the expression.

6\left(\dfrac{5}{2}\right)^2 - 7\left(\dfrac{5}{2}\right) - 20 = 0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A first step towards the critical values: factorising to $(3x + 4)(2x - 5)$ , or substituting into $\dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 6 \times (-20)}}{2 \times 6}$ oe, or completing the square to $6\left[\left(x - \dfrac{7}{12}\right)^2 - \left(\dfrac{7}{12}\right)^2\right] - 20$ oe.	M1	If factorising in the form $(ax + b)$ with $a$ and $b$ integers, allow brackets which expand to give $2$ out of $3$ terms correct. If using the formula or completing the square, allow one sign error and some simplification, as far as $\dfrac{7 \pm \sqrt{49 + 480}}{12}$ oe, or $6\left(x - \dfrac{7}{12}\right)^2 - \dfrac{529}{24}$ oe, or $\left(x - \dfrac{7}{12}\right)^2 - \dfrac{529}{144}$ oe.	✓
Both critical values: $x = -\dfrac{4}{3}$ and $x = \dfrac{5}{2}$ oe.	A1	Dependent on the M1, and only for two correct critical values. Accept $-1.3\ldots$ . The candidate may write $<$ , $\leq$ , $>$ or $\geq$ instead of $=$ .	✓
Both regions written in the correct form: $x < a$ and $x > b$ , where $a$ is their lower critical value and $b$ is their upper critical value.	M1ft	Dependent on the M1 and on two critical values having been found. Follow through on the candidate's own values. Also award for $x > \dfrac{5}{2}$ oe alone, or $x < -\dfrac{4}{3}$ oe alone, or $-\dfrac{4}{3} > x > \dfrac{5}{2}$ oe.	✓
$x < -\dfrac{4}{3}$ and $x > \dfrac{5}{2}$ oe.	A1	Dependent on the previous M1. Working is required. Accept $-1.3\ldots$ , or $\left(-\infty, -\dfrac{4}{3}\right), \left(\dfrac{5}{2}, (+)\infty\right)$ , or $\left(-\infty, -\dfrac{4}{3}\right) \cup \left(\dfrac{5}{2}, (+)\infty\right)$ . Do not ignore subsequent working.	✓

Full marks: 4/4

Continue to questions 21 to 26

The remaining 6 questions, with the same full worked solutions and mark schemes

Keep revising

That is part two of three. Read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Edexcel IGCSE 4MA1/2HR, Monday 3 June 2024: Worked Solutions, Questions 11 to 20

Worked solutions, questions 11 to 20 of 26

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Keep revising

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