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Edexcel IGCSE 4MA1/2HR, Monday 3 June 2024: Worked Solutions and Mark Schemes

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

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.

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Every question with a full worked solution and mark scheme - free PDF

Worked solutions, questions 1 to 10 of 26

Express $1\,400$ as a product of powers of its prime factors.
You must show your working clearly. [3 marks]

[Total 3 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The number

1\,400

Working has to be shown, so the marks are for the factorising and not only for the last line.

Find

1\,400

written as a product of powers of its prime factors. The answer must be in index form, so equal primes are collected into one power each.

Plan the Solution

Divide by the smallest prime that goes in exactly, then divide the quotient again, and keep going until the quotient reaches one.
Start with $2$ , because $1\,400$ is even. Move up to the next prime only when the one being used stops dividing exactly.
Write the primes out as a full list first, then collect the equal ones into powers.
Multiply the powers back together, and factorise from a different starting split, to check.

Worked Solution [3 marks]

Rule - Prime factorisation: divide by the smallest prime that goes in exactly, keep dividing each new quotient until it reaches

1

, then collect equal primes into powers.

Step 1: take out the twos

1\,400 = 2 \times 700

700 = 2 \times 350

350 = 2 \times 175

(Reason: Each of these is even, so

2

divides it exactly. The quotient

175

is odd, so this is where the twos stop.)

Step 2: move up to the next prime that divides

175 = 5 \times 35

35 = 5 \times 7

(Reason:

175

is odd, and its digits add to

13

, so neither

2

nor

3

divides it. It ends in

5

, so

5

does.)

Step 3: stop, and write the full list

1\,400 = 2 \times 2 \times 2 \times 5 \times 5 \times 7

(Reason: The last quotient is

7

, which is prime, so the dividing is finished. The list holds the six primes that were divided out, in the order they came out.)

Step 4: collect equal primes into powers

2 \times 2 \times 2 = 2^{3}

5 \times 5 = 5^{2}

1\,400 = 2^{3} \times 5^{2} \times 7

(Reason: There are three

2

s and two

5

s, so each becomes a power. A prime that appears once, here the

7

, is written with no index.)

1\,400 = 2^{3} \times 5^{2} \times 7

Verification

Check 1: Multiply the powers back together:

2^{3} = 8

and

5^{2} = 25

8 \times 25 = 200

, and

200 \times 7 = 1\,400

, the number the question started from.

Check 2: Start from a completely different split,

1\,400 = 14 \times 100

, and factorise each piece on its own.

14 = 2 \times 7

and

100 = 2^{2} \times 5^{2}

, which between them give three

2

s, two

5

s and one

7

again.

Check 3: Count what the indices claim. They should add up to the number of primes in the list, and every base should itself be prime.

3 + 2 + 1 = 6

, which is the length of the list, and

2

5

and

7

are all prime, with no factor of

1

left in.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two correct stages of the factorisation, for example $1\,400 = 2 \times 700$ and then $700 = 2 \times 350$ .	M1	Award for $2$ correct stages with no incorrect stage, or for at least $3$ stages with no more than one incorrect stage. Each stage gives two factors and may be a list, a factor tree or a ladder, so $2$ , $2$ , $350$ earns it. Other correct openings include $2 \times 7 \times 100$ , $2 \times 5 \times 140$ , $5 \times 7 \times 40$ and $5 \times 5 \times 56$ .	✓
The complete list of prime factors, $2 \times 2 \times 2 \times 5 \times 5 \times 7$ .	M1 dep	Dependent on the first M1. Award for $2 \times 2 \times 2 \times 5 \times 5 \times 7$ , or for the list $2^{3}$ , $5^{2}$ , $7$ , or for $2^{3} + 5^{2} + 7$ . Ignore any $1$ s. It may be seen inside a fully correct factor tree or ladder rather than written out on the answer line.	✓
The answer in index form, $2^{3} \times 5^{2} \times 7$ .	A1	Dependent on both method marks. Any order is accepted, and $2^{3} \cdot 5^{2} \cdot 7$ is accepted, but the answer must be in index form because that is what the question asks for. Do not accept a $1$ in the final answer.	✓

Full marks: 3/3

The grid shows shape $A$ and shape $B$ .

(a) Describe fully the single transformation that maps shape $A$ onto shape $B$ . [2 marks]

(b) On the grid above, rotate shape $A$ through $180^\circ$ about $(-1, 0)$ .
Label your shape $C$ . [2 marks]

(a)

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Shape

A

has vertices

(2, 1)

(2, 2)

(3, 2)

and

(4, 1)

Shape

B

has vertices

(5, -4)

(5, -3)

(6, -3)

and

(7, -4)

Both shapes are drawn on the same grid, and part (b) gives the centre

(-1, 0)

and the angle

180^\circ

Find

(a) the single transformation that maps

A

onto

B

, described fully (b) the image of shape

A

after that rotation, drawn on the grid and labelled

C

Plan the Solution

Part (a): line the vertices up in pairs. Same size, same way round and same way up rules out rotation, reflection and enlargement, so it is a translation; subtracting a pair of coordinates gives the vector.
Describe fully means the name of the transformation AND everything that pins it down. For a translation that is the vector, and nothing else.
Part (b): a rotation of $180^\circ$ about $(a, b)$ sends $(x, y)$ to $(2a - x, \; 2b - y)$ . Apply it to each vertex in turn, plot the four images and join them in the same order.

Worked Solution [4 marks]

Rule - Translation: a translation is described by one column vector, the top number the movement in

x

and the bottom number the movement in

y

. Rule - Half turn: a rotation of

180^\circ

about

(a, b)

maps

(x, y)

(2a - x, \; 2b - y)

, so the centre is the midpoint of every point and its image.

Step 1: Pair the vertices of

A

with the vertices of

B

A: (2, 1), \; (2, 2), \; (3, 2), \; (4, 1)

B: (5, -4), \; (5, -3), \; (6, -3), \; (7, -4)

(Reason: Both shapes are trapeziums of the same size, the same way round and the same way up, so nothing has been turned, flipped or resized. A translation is the only single transformation left.)

Step 2: Count the move on one pair of matching vertices

5 - 2 = 3

-4 - 1 = -5

(Reason: The vertex

(2, 1)

of shape

A

goes to

(5, -4)

of shape

B

. The

x

-coordinate goes up by

3

, which is

3

squares right; the

y

-coordinate goes down by

5

, which is

5

squares down.)

Step 3: Check the same move on a second vertex

4 + 3 = 7

1 - 5 = -4

(Reason: The vertex

(4, 1)

lands on

(7, -4)

, which is the matching vertex of shape

B

. Every vertex moves by the same amount, and that is what makes it one single translation.)

Step 4: Write the translation as a column vector

\begin{pmatrix} 3 \\ -5 \end{pmatrix}

(Reason: The top number is the movement in the

x

-direction, the bottom number the movement in the

y

-direction, and a negative bottom number means downwards. The word and the vector are worth a mark each, so an answer with only one of them scores

1

out of

2

Step 5: Write down what a rotation of

180^\circ

about

(-1, 0)

does

(x, y) \to (2 \times (-1) - x, \; 2 \times 0 - y)

(x, y) \to (-2 - x, \; -y)

(Reason: A half turn sends every point to the point the same distance the other side of the centre, so the centre sits exactly halfway between a point and its image. Doubling the centre and subtracting the point is that sentence written as arithmetic, one coordinate at a time. Note that the

y

-coordinate is negated as well as the

x

-coordinate.)

Step 6: Turn each vertex of shape

A

(2, 1) \to (-4, -1)

(2, 2) \to (-4, -2)

(3, 2) \to (-5, -2)

(4, 1) \to (-6, -1)

(Reason: Take

(2, 1)

as the pattern:

-2 - 2 = -4

for the

x

-coordinate and

-1

for the

y

-coordinate. Work through the vertices in the order they are joined, so the images can be joined in that order too.)

Step 7: Plot the four images, join them and label the shape

C

C: (-6, -1), \; (-4, -1), \; (-4, -2), \; (-5, -2)

(Reason: Joining the images in the same order as the vertices of shape

A

puts the sloping side of

C

on the far side from the sloping side of

A

, which is what a half turn does. The label is part of the instruction, although the mark scheme condones it being missing.)

(a) Translation by the vector

\begin{pmatrix} 3 \\ -5 \end{pmatrix}

(b) Shape

C

drawn at

(-6, -1), \; (-4, -1), \; (-4, -2), \; (-5, -2)

Verification

Check 1: Add the vector to every vertex of shape

A

(2, 1) \to (5, -4)

(2, 2) \to (5, -3)

(3, 2) \to (6, -3)

(4, 1) \to (7, -4)

. the four vertices of shape

B

, so the vector maps the whole shape and not just one corner

Check 2: The centre of a rotation is the midpoint of a point and its image. Halfway between

(2, 2)

and

(-4, -2)

\left( \dfrac{2 + (-4)}{2}, \; \dfrac{2 + (-2)}{2} \right)

(-1, 0)

, the centre the question gives

Check 3: Turn shape

C

through

180^\circ

about

(-1, 0)

as well. A half turn undoes a half turn, so the images should come back to where they started, and the four sides should still measure

1

1

\sqrt{2}

and

2

. shape

A

again, side for side

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Translation	B1	for translation, with none of reflection, rotation, enlargement, mirrored, turned or flipped also stated. Allow translated, translating, translate, and a misspelling such as translat. Move with translation is acceptable.	✓
(a) (vector =) $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$	B1	for the vector. Both marks are needed for a full description: the word on its own, or the vector on its own, scores $1$ .	✓
(b) Shape drawn at $(-6, -1), \; (-4, -1), \; (-4, -2), \; (-5, -2)$	B2	condone a missing label $C$ .	✓
(b) Partial credit	B1	if not B2, then B1 for a correct trapezium drawn with the correct orientation in the wrong position, or for $3$ points plotted correctly. The wrong position usually comes from turning about the origin instead of about $(-1, 0)$ .	✓

Full marks: 4/4

These four numbers are written in order of size, smallest first.

$x \qquad x \qquad y \qquad 15$

Here $x$ and $y$ are integers.

For these four numbers,
the median is $12.5$
the range is $4$

Work out the value of $x$ and the value of $y$ . [2 marks]

x =y =

[Total 2 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

The four numbers

x

x

y

15

, already written in ascending order of size.

Both

x

and

y

are integers.

The median is

12.5

and the range is

4

Find

The value of

x

and the value of

y

. Two numbers are wanted, and the paper prints a separate answer line for each of them.

Plan the Solution

The list is already in order, so the smallest number is $x$ and the largest is $15$ . The range is built from those two alone, so it gives $x$ by itself. Start there.
A list of four numbers has no single middle number, so the median is the mean of the middle two, $x$ and $y$ . Use that second, once $x$ is known.
Finish by writing the four numbers out in full and reading the median and the range back off them.

Worked Solution [2 marks]

Rule - Median and range of an ordered list: for four numbers in order the median is the mean of the middle two, and the range is the largest minus the smallest.

Step 1: read what the order already tells you

x \leq x \leq y \leq 15

(Reason: The four numbers are written in ascending order of size, so the smallest is

x

and the largest is

15

. The middle two, the ones the median is built from, are

x

and

y

Step 2: use the range to find x

15 - x = 4

x = 15 - 4 = 11

(Reason: The range is the largest minus the smallest, which here is

15 - x

. That equation holds only

x

, so it can be solved on its own, before anything is known about

y

Step 3: use the median to find y

\dfrac{x + y}{2} = 12.5

x + y = 2 \times 12.5 = 25

y = 25 - 11 = 14

(Reason: With four numbers the median is the mean of the middle two, so multiplying the median by

2

gives the total of that middle pair. Taking the known

x

away from

25

leaves

y

Step 4: write the list out and check it is a legal list

11 \quad 11 \quad 14 \quad 15

(Reason: The question sets two conditions that are easy to forget: the numbers are integers, and they rise from left to right. Both hold here, so this list is a legitimate answer to the question as it is set.)

x = 11

y = 14

Verification

Check 1: Read the range straight off the finished list

11

11

14

15

: largest minus smallest.

15 - 11 = 4

, which is the range the question gives.

Check 2: Read the median off the same list. The two middle numbers are

11

and

14

\dfrac{11 + 14}{2} = \dfrac{25}{2} = 12.5

, which is the median the question gives.

Check 3: Test the pair the other way round,

x = 14

and

y = 11

, to confirm that only one pair fits. The list would then read

14 \quad 14 \quad 11 \quad 15

, which is not in ascending order of size, so it is rejected. The ordering is the condition that separates the two pairs.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both values correct: $x = 11$ and $y = 14$ .	B2	Award both marks for the pair $x = 11$ and $y = 14$ , however they are reached. The question does not ask for working, so a correct pair with nothing written down still scores $2$ . The values may appear on the answer lines in either order of writing, as long as each is against its own letter.	✓
Only one of the two values correct.	B1	If the answer does not score B2, award one mark for $x = 11$ or for $y = 14$ , with the other value wrong or missing.	✓
Special case: the two values written the wrong way round, $x = 14$ and $y = 11$ .	SC B1	One mark, and no more, for the pair reversed. This is worth a mark because the two numbers themselves are right: the middle pair still totals $25$ and the smallest is still $4$ below $15$ . What fails is the ordering, since $14$ , $14$ , $11$ , $15$ does not rise from left to right, and the question states that the list is in ascending order of size.	✓

Full marks: 2/2

$\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
$A = \{\text{factors of } 6\}$
$B = \{\text{prime numbers}\}$

(a) List all the members of the set
(i) $A \cup B$
[1 mark]
(ii) $A'$ [1 mark]

Ravinder claims that $A \cap B = \varnothing$
Ravinder is wrong.
(b) Explain why. [1 mark]

$C$ is a set with $4$ members, and
$A \cap C$ has $2$ members
$B \cap C$ has $2$ members
No member belongs to both $A \cap C$ and $B \cap C$ .
(c) List all $4$ members of set $C$ [2 marks]

(a)(i)(a)(ii)(b)(c)

[Total 5 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

- the universal set, so every member of every set here comes from this list

A = \{\text{factors of } 6\}

and

B = \{\text{prime numbers}\}

, both taken from

\mathcal{E}

C

has

4

members, with

n(A \cap C) = 2

and

n(B \cap C) = 2

, and those two sets share no member

Find

(a)(i) the members of

A \cup B

(a)(ii) the members of

A'

(b) why the claim

A \cap B = \varnothing

is wrong (c) the

4

members of set

C

Plan the Solution

Write $A$ and $B$ out as lists first. Every part of the question is then read straight off those two lists.
A factor of $6$ divides $6$ exactly, and a prime number has exactly two factors - so $1$ is a factor of $6$ but is not prime.
For (c), ask what a member of $A \cap C$ is not allowed to be. If it were prime it would sit in $B \cap C$ too, and the question says those two sets share nothing.

Worked Solution [5 marks]

Rule -

A \cup B

is every member that is in

A

or in

B

, each written once;

A'

is every member of

\mathcal{E}

that is not in

A

; and

A \cap B

is what the two sets share. The empty set

\varnothing

has no members at all.

Step 1: List the factors of

6

6 = 1 \times 6

6 = 2 \times 3

A = \{1, 2, 3, 6\}

(Reason: The factor pairs give every whole number that divides

6

exactly, and all four of them are in

\mathcal{E}

Step 2: List the prime numbers in

\mathcal{E}

B = \{2, 3, 5, 7\}

(Reason: A prime has exactly two factors. So

1

is not prime (it has only one factor), and

9 = 3 \times 3

is not prime either.)

Step 3: Part (a)(i) - take the union

A \cup B = \{1, 2, 3, 5, 6, 7\}

(Reason: The union collects everything that is in

A

or in

B

2

and

3

belong to both sets, but each member is written only once.)

Step 4: Part (a)(ii) - take the complement

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

A = \{1, 2, 3, 6\}

A' = \{4, 5, 7, 8, 9, 10\}

(Reason: The complement is everything left in

\mathcal{E}

once the members of

A

are crossed out. Work along

\mathcal{E}

in order, so that nothing at the end of the list is missed.)

Step 5: Part (b) - find what

A

and

B

A \cap B = \{2, 3\}

(Reason:

2

and

3

are factors of

6

and are also prime, so the intersection has two members. The empty set has none, so

A \cap B = \varnothing

cannot be right.)

Step 6: Part (c) - which members can be in

A \cap C

A = \{1, 2, 3, 6\}

A \cap C = \{1, 6\}

(Reason: A member of

A \cap C

must not be prime: if it were, it would be in

B \cap C

as well, and those two sets share nothing. That rules out

2

and

3

, leaving exactly the

2

members needed.)

Step 7: Part (c) - which members can be in

B \cap C

B = \{2, 3, 5, 7\}

B \cap C = \{5, 7\}

(Reason: The same argument the other way round.

2

and

3

are factors of

6

, so they would fall in

A \cap C

too, which leaves

5

and

7

Step 8: Part (c) - put set

C

together

C = \{1, 6\} \cup \{5, 7\} = \{1, 5, 6, 7\}

(Reason: The two pairs have no member in common, so together they give the

4

members

C

is allowed. Any other choice breaks one of the three conditions.)

(a)(i)

A \cup B = \{1, 2, 3, 5, 6, 7\}

(a)(ii)

A' = \{4, 5, 7, 8, 9, 10\}

(b)

2

and

3

are in both sets, so

A \cap B = \{2, 3\}

, which is not empty(c)

C = \{1, 5, 6, 7\}

Verification

Check 1: Count the union a second way, with

n(A \cup B) = n(A) + n(B) - n(A \cap B)

4 + 4 - 2 = 6

, and the listed union has

6

members

Check 2: A set and its complement must fill

\mathcal{E}

between them with no overlap, so their sizes add to

10

4 + 6 = 10

, and no number appears in both

A

and

A'

Check 3: Test

C = \{1, 5, 6, 7\}

against all three conditions in the question.

A \cap C = \{1, 6\}

has

2

members,

B \cap C = \{5, 7\}

has

2

members, and the two share none

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $A \cup B = \{1, 2, 3, 5, 6, 7\}$	B1	In any order, with no repeats.	✓
(a)(ii) $A' = \{4, 5, 7, 8, 9, 10\}$	B1	In any order, with no repeats.	✓
(b) $2$ (or $3$ , or both) is a member of $A$ and of $B$	B1	For naming the shared element with an explanation that shows the meaning of intersection and of the empty set. $A \cap B = \{2, 3\}$ on its own also scores. If a number is named as common, that number must be correct. The word sector is allowed in place of set. This is not an exhaustive list of acceptable answers.	✓
(c) $C = \{1, 5, 6, 7\}$	B2	For all four correct members, in any order.	✓
(c) partial credit, if B2 is not earned	(B1)	For three correct values with no more than one incorrect, or for four correct values with no more than one incorrect.	✓

Full marks: 5/5

A craft studio makes hanging ornaments from copper wire.
The diagram shows the design for one ornament.
The design is a circle inside a square $ABCD$

The circle touches the square at the points $E$ , $F$ , $G$ and $H$

The area of the square is $81$ cm²

Work out the total length of copper wire needed to make the square and the circle.
Give your answer correct to $3$ significant figures. [4 marks]

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A circle sits inside the square

ABCD

and touches all four sides, at

E

F

G

and

H

The area of the square is

81

cm²

The wire has to make both outlines - the square and the circle - so the two lengths are added

Find

The total length of wire, in cm, correct to

3

significant figures

Plan the Solution

The area of a square is side times side, so the side is the square root of $81$ .
Four equal sides give the perimeter, which is the wire that makes the square.
The circle touches all four sides, so it fits exactly across the square: its diameter is the side of the square. That single sentence is what the whole question turns on.
Add the perimeter and the circumference, and round only at the very end - rounding part way through moves the answer.

Worked Solution [4 marks]

Rule - A square of side

s

has area

s^2

and perimeter

4s

. A circle of diameter

d

has circumference

\pi d

, which is the same as

2\pi r

. A circle that touches all four sides of a square has

d

equal to the side of that square.

Step 1: Work out the side of the square

s^2 = 81

s = \sqrt{81} = 9

(Reason: The area of a square is side times side, so the side is the square root of the area. A length is positive, so the negative root is not used here. The side is

9

cm.)

Step 2: Work out the perimeter of the square

4 \times 9 = 36

(Reason: All four sides of a square are equal, so the perimeter is

4

lots of

9

cm. That is

36

cm of wire for the square on its own.)

Step 3: Find the diameter of the circle

d = s = 9

(Reason: The circle touches the square at

E

F

G

and

H

, so

HF

runs right across the circle through its centre and lies along the full width of the square. The diameter is therefore the side of the square,

9

cm, and the radius is half of that,

4.5

cm.)

Step 4: Work out the circumference of the circle

C = \pi d

C = \pi \times 9 = 28.2743\ldots

(Reason: The circumference formula multiplies

\pi

by the DIAMETER - the full distance across the circle, not the distance from the centre to the edge. Keep the decimal running on the calculator; rounding it here would move the final answer.)

Step 5: Add the two lengths of wire

36 + 28.2743\ldots = 64.2743\ldots

(Reason: The wire makes the square and the circle, so the total is the perimeter plus the circumference. The two shapes touch, but no wire is shared between them - each outline is drawn completely.)

Step 6: Round to

3

significant figures

64.2743\ldots \to 64.3

(Reason: The first three significant figures are

6

4

and

2

. The next digit is

7

, which is

5

or more, so the

2

rounds up to

3

. The unit is centimetres, because every length in the question is in centimetres.)

64.3

Verification

Check 1: Square the side to get back to the area the question gives.

9 \times 9 = 81

, the area printed in the question, so the side is right

Check 2: Work backwards. Take the perimeter off the total, then divide what is left by the diameter - it should give

\pi

64.2743 - 36 = 28.2743

and

\dfrac{28.2743}{9} = 3.1416

, which is

\pi

4

decimal places

Check 3: A rough estimate, done without a calculator:

\pi

is a little over

3

, so the circumference is a little over

3

diameters.

36 + 3 \times 9 = 63

, so the answer should be a little above

63

cm - and

64.3

cm is

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\sqrt{81} = 9$ or $9$ or $9 \times 9 = 81$	M1	For a method to find the length of the side of the square. A bare $9$ with no working also earns it. It may be seen on the diagram rather than in the working.	✓
$4 \times 9 = 36$	M1	For the perimeter of the square, or any equivalent method. The scheme writes the $9$ in quotation marks, so the candidate's own value for the side follows through. The first M mark can be implied by $36$ .	✓
$\pi \times 9 = 28.2743\ldots$ or $9\pi$	M1	For a correct expression for the circumference, from $2\pi r$ or $\pi D$ . The scheme writes the $9$ in quotation marks here too, so the candidate's own side value follows through. The first M mark can be implied by $28.2(743\ldots)$ rounded or truncated to $1$ decimal place, so by $28.2$ or $28.3$ , or by $9\pi$ .	✓
$64.3$	A1	Accept any value from $64.26$ to $64.3$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓

Full marks: 4/4

(a) Solve the equation $\dfrac{2f}{3} = 4f - 17$
You must show clear algebraic working. [3 marks]

(b) Simplify $(e + 12)^0$ , given that $e > 0$ [1 mark]

(d) Factorise fully $20x^5y + 12x^3y^4$ [2 marks]

(a) f =(b)(c)(d)

[Total 8 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Part (a): the equation

\dfrac{2f}{3} = 4f - 17

, with the algebraic working itself asked for

Part (b): the expression

(e + 12)^0

, together with the condition

e > 0

Part (c): the fraction

\dfrac{12a^4h^6}{4ah^2}

, to be simplified fully

Part (d): the expression

20x^5y + 12x^3y^4

, to be factorised fully

Find

(a) The value of

f

that makes the two sides equal (b) The value of

(e + 12)^0

(c) The fraction written as one term, with each letter appearing once (d) The expression written as a product, with the highest common factor outside the bracket

Plan the Solution

Part (a): a fraction in an equation is cleared by multiplying every term on BOTH sides by the denominator. Then gather the letter terms on one side and the numbers on the other, and divide by whatever is multiplying $f$ .
Part (b): the condition $e > 0$ is not decoration. It guarantees the bracket is not zero, and only a number that is not $0$ can be raised to the power $0$ .
Part (c): deal with the numbers and with each letter separately. The numbers divide; powers of the same letter subtract their indices.
Part (d): the highest common factor is the highest common factor of the numbers, together with the LOWEST power of each letter that appears in both terms. Divide each term by it to find what goes inside the bracket.

Worked Solution [8 marks]

Rule - Balance: multiplying both sides of an equation by the same number that is not

0

keeps it true, and every term must be multiplied. Indices with the same base:

\dfrac{x^m}{x^n} = x^{m-n}

and

x^0 = 1

for every

x

that is not

0

. Factorising fully: outside the bracket goes the highest common factor of the numbers together with the lowest power of each letter that appears in every term.

Step 1: Part (a) - clear the fraction

\dfrac{2f}{3} = 4f - 17

3 \times \dfrac{2f}{3} = 3 \times (4f - 17)

2f = 12f - 51

(Reason: The only denominator is

3

, so multiplying both sides by

3

clears it. Multiply every term on the right, not just the first one:

3 \times 4f = 12f

and

3 \times 17 = 51

. Leaving the

17

untouched is exactly the slip the mark scheme carries a follow-through for.)

Step 2: Part (a) - collect the letter terms

2f - 12f = -51

-10f = -51

(Reason: Subtract

12f

from both sides, so the

f

terms sit together on the left and the number sits on the right. Since

2 - 12 = -10

, the left-hand side becomes

-10f

. This is the two-term equation the second method mark is for.)

Step 3: Part (a) - divide by the coefficient

f = \dfrac{-51}{-10} = \dfrac{51}{10}

\dfrac{51}{10} = 5.1

(Reason: Divide both sides by

-10

. A negative divided by a negative is positive, so the two minus signs cancel and

f

comes out positive. The mark scheme gives

\dfrac{51}{10}

and allows any equivalent form, so the decimal

5.1

scores the same mark.)

Step 4: Part (b) - use the zero index

e > 0 \implies e + 12 > 12

(e + 12)^0 = 1

(Reason: Any number except

0

raised to the power

0

1

. The condition

e > 0

makes the bracket larger than

12

, so it certainly is not

0

and the rule applies. The size of the bracket makes no difference whatsoever: the answer is

1

for every allowed value of

e

Step 5: Part (c) - divide the numbers, subtract the indices

\dfrac{12a^4h^6}{4ah^2} = \dfrac{12}{4} \times \dfrac{a^4}{a^1} \times \dfrac{h^6}{h^2}

\dfrac{12}{4} = 3

\dfrac{a^4}{a^1} = a^{4-1} = a^3

\dfrac{h^6}{h^2} = h^{6-2} = h^4

\dfrac{12a^4h^6}{4ah^2} = 3a^3h^4

(Reason: Split the fraction into a number part and one part for each letter. The numbers are divided; the letters follow the index law, so their indices are subtracted. Remember that

a

written on its own means

a^1

, so its index is

1

and the

a

does not vanish from the answer.)

Step 6: Part (d) - find the highest common factor

20x^5y = 4x^3y \times 5x^2

12x^3y^4 = 4x^3y \times 3y^3

(Reason: The highest common factor of

20

and

12

4

. The lowest power of

x

in the two terms is

x^3

, and the lowest power of

y

y

itself, so the highest common factor of the whole expression is

4x^3y

. Taking out something smaller,

2x^3y

say, does factorise the expression but not fully, and fully is what is asked for.)

Step 7: Part (d) - write it as a product

20x^5y + 12x^3y^4 = 4x^3y(5x^2 + 3y^3)

(Reason: The highest common factor goes outside the bracket and what is left of each term goes inside. Then check the bracket:

5x^2

and

3y^3

share no number factor and no letter, so nothing further can come out and the factorisation is complete.)

(a)

f = \dfrac{51}{10}

(that is,

f = 5.1

)(b)

1

(c)

3a^3h^4

(d)

4x^3y(5x^2 + 3y^3)

Verification

Check 1: Part (a): put

f = 5.1

back into the equation the question printed, and work the two sides out separately.

\dfrac{2 \times 5.1}{3} = \dfrac{10.2}{3} = 3.4

and

4 \times 5.1 - 17 = 20.4 - 17 = 3.4

, so the two sides agree

Check 2: Part (b): try an allowed value. Take

e = 4

, which satisfies

e > 0

, and work the bracket out before applying the index.

(4 + 12)^0 = 16^0 = 1

, and every other positive value of

e

gives

1

in exactly the same way

Check 3: Part (c): multiply the answer back by the denominator. It must return the numerator the question printed.

3a^3h^4 \times 4ah^2 = 12a^4h^6

, which is the numerator in the question

Check 4: Part (d): expand the brackets again and compare, term by term, with the expression the question printed.

4x^3y \times 5x^2 = 20x^5y

and

4x^3y \times 3y^3 = 12x^3y^4

, which add to give the original expression

Check 5: Parts (c) and (d) with numbers in place of the letters: take

a = 2

and

h = 1

in part (c), and

x = 2

and

y = 1

in part (d).

\dfrac{12 \times 16}{4 \times 2} = 24

and

3 \times 8 = 24

; then

20 \times 32 + 12 \times 8 = 736

and

32 \times 23 = 736

, so both answers survive a numerical test

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$2f = 12f - 51$ or $2f = -51 + 12f$	M1	For a correct first step: multiplying both sides by $3$ correctly and expanding. Writing the right-hand side as two terms each over $2$ also earns it, for example $\dfrac{f}{3} = 2f - \dfrac{17}{2}$ or $0.3f = 2f - 8.5$ . Decimals are allowed to $1$ decimal place or better, rounded or truncated.	✓
$-10f = -51$ or $10f = 51$	M1	For a correct two-term equation in the form $af = b$ , for example $\dfrac{5f}{3} = \dfrac{17}{2}$ or $17 = \dfrac{10f}{3}$ or $3.3f = 17$ . Follow through is allowed from these three equations only: $2f = 12f - 17$ , $2f = 4f - 51$ and $6f = 12f - 51$ , or any equivalent of them. Decimals are allowed to $1$ decimal place or better, rounded or truncated.	✓
$f = \dfrac{51}{10}$	A1	Or any equivalent, for example $5.1$ . Dependent on at least one of the method marks. Working is required on this part, so the answer alone scores nothing.	✓
$1$	B1	The single mark is for the answer, and no working is expected for it.	✓
$3a^3h^4$	B2	Or any equivalent. B1 for a product in the form $ka^p h^q$ in which two of $k$ , $p$ and $q$ are correct, with multiplication signs allowed, for example $5a^3h^4$ or $\dfrac{12a^3h^4}{4}$ . Also allow $3a^3$ or $a^3h^4$ or $3h^4$ for B1, as long as it is not added to any other term.	✓
$4x^3y(5x^2 + 3y^3)$	B2	B1 for any correct factorisation whose factor outside the bracket has at least two parts to it, for example $2x^3y(10x^2 + 6y^3)$ or $x^3y(20x^2 + 12y^3)$ or $2x(10x^4y + 6x^2y^4)$ or $4y(5x^5 + 3x^3y^3)$ or $4x^3(5x^2y + 3y^4)$ . B1 is also given for the correct highest common factor with a two-term bracket in which at most one term is wrong, for example $4x^3y(5x^2 + \ldots)$ or $4x^3y(\ldots + 3y^3)$ .	✓

Full marks: 8/8

$\dfrac{3^{-2} \times 3^{5}}{3^{10}} = 3^{n}$
Work out the value of $n$ [2 marks]

n =

[Total 2 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

The equation

\dfrac{3^{-2} \times 3^{5}}{3^{10}} = 3^{n}

Every power on both sides has the same base,

3

, which is what makes the index laws usable here

Find

The value of

n

: the single index the whole left-hand side collapses to

Plan the Solution

Take the top line first. Multiplying two powers of the same base ADDS their indices, so $3^{-2} \times 3^{5}$ becomes one power of $3$ .
Then divide. Dividing powers of the same base SUBTRACTS the indices, and the one that comes off is the index downstairs.
Both sides are then a single power of $3$ . Two powers of the same base are equal only when their indices are equal, so $n$ can be read straight off.

Worked Solution [2 marks]

Rule - Index laws with a common base:

a^{m} \times a^{n} = a^{m+n}

and

\dfrac{a^{m}}{a^{n}} = a^{m-n}

. A negative index is not a mistake to be tidied away:

a^{-k} = \dfrac{1}{a^{k}}

, so an index below

0

is a perfectly good answer.

Step 1: Add the indices on the top line

3^{-2} \times 3^{5} = 3^{-2 + 5} = 3^{3}

(Reason: Multiplying two powers of the same base adds their indices, so the

-2

and the

5

are added rather than the numbers

\dfrac{1}{9}

and

243

being multiplied out. A negative index behaves as a subtraction inside that sum, which is why

-2 + 5

comes to

3

and not to

7

. This single line is what the mark scheme's method mark is for.)

Step 2: Take away the index on the bottom

\dfrac{3^{3}}{3^{10}} = 3^{3 - 10} = 3^{-7}

(Reason: Dividing powers of the same base subtracts the indices, and the index that comes off is the one downstairs. Both powers here have base

3

, so the whole fraction collapses into a single power of

3

. The index that results must be SMALLER than the one it started from, because the bottom of the fraction is the bigger power.)

Step 3: Match the indices on the two sides

3^{n} = 3^{-7}

n = -7

(Reason: The left-hand side is now one power of

3

and the right-hand side always was one. Two powers of the same base are equal only when their indices are equal, so the index on the left IS

n

. Nothing is calculated on this line; it is a comparison.)

n = -7

Verification

Check 1: Count all three indices in one sum instead of in two steps. Multiplying puts an index in and dividing takes one out, so the

-2

, the

5

and the

10

combine on a single line.

-2 + 5 - 10 = -7

, the same index the two-step working reached

Check 2: Put the index laws aside and work each side out as an ordinary fraction. If the answer is right, the two sides must be the same number.

\dfrac{3^{-2} \times 3^{5}}{3^{10}} = \dfrac{27}{59049} = \dfrac{1}{2187}

and

3^{-7} = \dfrac{1}{2187}

, so the two sides agree exactly

Check 3: Test the SIGN on its own, without finding the index at all. The bottom of the fraction,

3^{10}

, is far bigger than the top, so the left-hand side is less than

1

- and only a negative power of

3

is less than

1

\dfrac{1}{2187}

is less than

1

, so

n

is negative, which rules out

7

and leaves

-7

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$3^{-2} \times 3^{5} = 3^{3}$ or $-2 + 5 - 10$	M1	For a correct application of an index rule as a first step, or a correct calculation for $n$ . Any equivalent earns it, whichever pair of powers is combined first: $3^{3}$ or $(3^{-2}) \times 3^{-5}$ or $\dfrac{3^{3}}{3^{10}}$ or $\dfrac{3^{5}}{3^{12}}$ or $\dfrac{3^{-2}}{3^{5}}$ or $3^{-12}$ with the $3^{5}$ still to be multiplied in. The index sums score it just as well: $-2 + 5 - 10$ or $-12 + 5$ or $3 - 10$ .	✓
$n = -7$	A1	For $-7$ . The scheme also allows the answer written as the power itself, $3^{-7}$ .	✓
Working is not required on this question	Note	A correct answer on its own scores both marks, unless it has clearly come from incorrect working.	✓

Full marks: 2/2

In a clearance sale at an electrical goods shop, all normal prices are reduced by $17\%$
The sale price of a table fan is $6\,225$ rupees.
Work out the normal price of the table fan. [3 marks]

rupees

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

In the sale, every normal price is reduced by

17\%

Sale price of the table fan:

6\,225

rupees

Find

The normal price of the table fan, in rupees.

Plan the Solution

A reduction of $17\%$ leaves $83\%$ of the normal price, so the sale price is $0.83$ of it.
The normal price has been multiplied by $0.83$ , so divide by $0.83$ to undo that. This is a reverse percentage: never add $17\%$ back on to the sale price.
Finish by taking $17\%$ off the answer and looking for the sale price again.

Worked Solution [3 marks]

Rule - Reverse percentage:

\text{normal price} = \dfrac{\text{sale price}}{\text{multiplier}}

, and the multiplier for a reduction of

17\%

0.83

Step 1: Turn the reduction into a multiplier

1 - 0.17 = 0.83

(Reason: Start from the whole normal price, which is

1

, and account for the

17\%

the sale takes off it. What is left is the fraction of the normal price a shopper actually pays.)

Step 2: Divide the sale price by the multiplier

\dfrac{6\,225}{0.83}

(Reason: The normal price was multiplied by

0.83

to give

6\,225

rupees, so dividing

6\,225

0.83

undoes the multiplication and gives the normal price back.)

Step 3: Work out the division

\dfrac{6\,225}{0.83} = 7\,500

(Reason: The division is exact, so the normal price is

7\,500

rupees, which is larger than the sale price of

6\,225

rupees, as a reduction requires.)

7\,500

rupees

Verification

Check 1: Take

17\%

of the answer and subtract it. The sale price should come back.

0.17 \times 7\,500 = 1\,275

and

7\,500 - 1\,275 = 6\,225

Check 2: Reach the answer a different way. The sale price is

83

per cent of the normal price, so divide by

83

to get one per cent, then multiply by

100

\dfrac{6\,225}{83} \times 100 = 7\,500

Check 3: Test the trap. Adding

17\%

on to the sale price is a different calculation, so it must give a different number.

1.17 \times 6\,225 = 7\,283.25

, which is not

7\,500

, so a reduction is undone by dividing, never by adding the same percentage on.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Find the multiplier for the reduction, or one per cent of the normal price	M1	for $1 - 0.17$ or $0.83$ or $\dfrac{83}{100}$ , or $100 - 17 = 83$ per cent, or $\dfrac{6\,225}{83} = 75$ or equivalent	✓
Divide the sale price by the multiplier	M1	for $\dfrac{6\,225}{0.83}$ or $\dfrac{6\,225}{83} \times 100$ or $6\,225 \times \dfrac{100}{83}$ or $75 \times 100$ or equivalent. The multiplier here may be the candidate's own value from the first mark.	✓
State the normal price	A1	for $7\,500$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Express $6.04 \times 10^{5}$ as an ordinary number.
[1 mark]
(b) Express $0.000\,07$ in standard form.
[1 mark]
(c) Work out $\dfrac{7.6 \times 10^{10}}{4 \times 10^{5} - 2 \times 10^{4}}$
Give your answer in standard form. [2 marks]

(a)(b)(c)

[Total 4 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Part (a):

6.04 \times 10^{5}

, already written in standard form.

Part (b):

0.000\,07

, written as an ordinary number.

Part (c):

\dfrac{7.6 \times 10^{10}}{4 \times 10^{5} - 2 \times 10^{4}}

, with a subtraction in the denominator.

Find

Part (a): the same number written out in full. Part (b): the same number written in standard form. Part (c): the value of the fraction, in standard form.

Plan the Solution

Standard form is $a \times 10^{n}$ , where $a$ is at least $1$ and less than $10$ , and $n$ is a whole number. All three parts are that one definition, read in three different directions.
In part (a) the index is positive, so the ordinary number is the bigger of the two: the decimal point moves $5$ places to the right.
In part (b) the number is smaller than $1$ , so the index is negative. Count the places from the decimal point to the first significant figure, not the zeros.
In part (c) the denominator has to be worked out before anything is divided. $4 \times 10^{5}$ and $2 \times 10^{4}$ carry different powers of ten, so their front numbers cannot be subtracted while they stand as they are.
Then divide: front number by front number, and the indices subtracted. Finish by checking the front number still lies between $1$ and $10$ .

Worked Solution [4 marks]

Rule - Standard form:

a \times 10^{n}

with

1 \leq a < 10

. To divide,

\dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m-n}

Step 1: Part (a) - move the decimal point five places to the right

10^{5} = 100\,000

6.04 \times 10^{5} = 6.04 \times 100\,000 = 604\,000

(Reason: Multiplying by

10^{5}

is multiplying by

100\,000

, so every digit moves five columns to the left and the decimal point ends up five places further right. The

6

lands in the hundred-thousands column, and the columns left empty behind it are filled with zeros.)

Step 2: Part (b) - count the places to the first significant figure

0.000\,07 = \dfrac{7}{100\,000}

\dfrac{7}{100\,000} = \dfrac{7}{10^{5}} = 7 \times 10^{-5}

(Reason: The only non-zero digit is the

7

, so the front number is

7

. It sits five places after the decimal point, which makes the index

-5

, and the index is negative because the number is smaller than

1

. There are only four zeros after the point, so counting zeros instead of places gives

10^{-4}

and an answer ten times too big.)

Step 3: Part (c) - write both terms of the denominator out in full

4 \times 10^{5} = 400\,000

2 \times 10^{4} = 20\,000

(Reason: The two terms carry different powers of ten, so their front numbers cannot simply be subtracted. Writing each one out in full lines the digits up in the right columns, and the subtraction then becomes ordinary arithmetic.)

Step 4: Part (c) - subtract, then put the denominator back into standard form

400\,000 - 20\,000 = 380\,000

380\,000 = 3.8 \times 10^{5}

(Reason: Taking twenty thousand away from four hundred thousand leaves three hundred and eighty thousand. This is the value the mark scheme wants for its method mark, in whichever of its forms it is written.)

Step 5: Part (c) - divide the front numbers, and subtract the indices

\dfrac{7.6}{3.8} = 2

\dfrac{10^{10}}{10^{5}} = 10^{10-5} = 10^{5}

(Reason: A division in standard form splits into two easy divisions: the front numbers divide to give

2

, and the powers of ten are handled by subtracting the indices. Dividing the indices instead of subtracting them is the slip to avoid here.)

Step 6: Part (c) - put the two halves back together

\dfrac{7.6 \times 10^{10}}{3.8 \times 10^{5}} = 2 \times 10^{5}

(Reason: The front number

2

lies between

1

and

10

, so the answer is already in standard form and nothing more needs doing to it. Writing

200\,000

instead is the same number in the wrong form, and the question asked for standard form.)

(a)

604\,000

(b)

7 \times 10^{-5}

(c)

2 \times 10^{5}

Verification

Check 1: Reverse part (a). Divide the answer by

100\,000

and the number the question printed should come back.

\dfrac{604\,000}{100\,000} = 6.04

Check 2: Reverse part (b). Turn the answer back into an ordinary number and compare it with the number the question printed.

7 \times 10^{-5} = \dfrac{7}{100\,000}

, and

\dfrac{7}{100\,000}

written out is

0.000\,07

Check 3: Reverse part (c). Multiplying the answer by the denominator must give the numerator back.

2 \times 10^{5} \times 3.8 \times 10^{5} = 7.6 \times 10^{10}

Check 4: Reach part (c) a second way, in ordinary numbers only, so that nothing depends on the index laws.

\dfrac{76\,000\,000\,000}{380\,000} = 200\,000

, which is

2 \times 10^{5}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Part (a): the number written out in full	B1	for $604\,000$	✓
Part (b): the number written in standard form	B1	for $7 \times 10^{-5}$	✓
Part (c): simplify the denominator	M1	for $380\,000$ or $3.8 \times 10^{5}$ or $38 \times 10^{4}$ or equivalent	✓
Part (c): the answer, in standard form	A1	for $2 \times 10^{5}$ . Accept $2.0 \times 10^{5}$ or $2.00 \times 10^{5}$ and so on, and accept a dot or a comma in place of the multiplication sign. Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Part (c): special case - the right digits, in the wrong form	SC B1	for $200\,000$ or $20 \times 10^{4}$ or $0.2 \times 10^{6}$ or equivalent, or for $2 \times 10^{n}$ with $n$ other than $5$ , when it is given as the final answer. The first three are the right number written in a form that is not standard form. Not for incorrect simplification of the denominator.	✓

Full marks: 4/4

$ABCDEF$ is a hexagon.
The measurements of the hexagon are shown on the diagram.

The lines $AB$ , $FC$ and $ED$ are parallel, and angle $BCF$ is $30^\circ$ .

Work out the area of $ABCDEF$ .
You must show all your working. [5 marks]

cm²

[Total 5 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A hexagon

ABCDEF

with

AB = 11

cm,

BC = 5

cm,

ED = 23

cm and

EF = 4.7

Angle

BCF = 30^\circ

, and

AB

FC

and

ED

are parallel

Right angles at

E

and at

D

, marked on the diagram

Find

The area of the hexagon

ABCDEF

, in

\text{cm}^2

Plan the Solution

The dashed line $FC$ cuts the hexagon into two shapes you already have formulas for: the rectangle $FCDE$ underneath and the trapezium $ABCF$ on top.
The right angles at $E$ and $D$ are what make $FCDE$ a rectangle, so $FC$ is the same length as $ED$ .
The trapezium's height is not $BC$ itself. Drop a perpendicular from $B$ to $FC$ to make a right-angled triangle with $BC$ as its hypotenuse, and use the $30^\circ$ angle.
Work out the two areas and add them.

Worked Solution [5 marks]

Rule - Rectangle:

\text{Area} = \text{length} \times \text{width}

. Trapezium:

\text{Area} = \dfrac{1}{2}(a + b)h

, where

a

and

b

are the two parallel sides and

h

is the perpendicular distance between them.

Step 1: Cut the hexagon along

FC

FC = ED = 23 \text{ cm}

(Reason:

FC

is parallel to

ED

, and the right angles at

E

and

D

make

FCDE

a rectangle, so its opposite sides are equal)

Step 2: Area of the rectangle

FCDE

23 \times 4.7 = 108.1 \text{ cm}^2

(Reason: The rectangle is

23

cm long and

4.7

cm high, because

EF

is one of its sides)

Step 3: Height of the trapezium

ABCF

\sin 30^\circ = \dfrac{h}{5}

h = 5 \sin 30^\circ = 2.5 \text{ cm}

(Reason: Dropping a perpendicular from

B

FC

gives a right-angled triangle in which

BC = 5

cm is the hypotenuse and

h

is the side opposite the

30^\circ

angle)

Step 4: Area of the trapezium

ABCF

\dfrac{1}{2} \times (11 + 23) \times 2.5 = 42.5 \text{ cm}^2

(Reason: The parallel sides are

AB = 11

cm and

FC = 23

cm, and the height is the perpendicular distance between them from Step 3)

Step 5: Add the two areas

108.1 + 42.5 = 150.6 \text{ cm}^2

(Reason: The rectangle and the trapezium meet along

FC

and do not overlap, so together they make up the whole hexagon)

150.6 \text{ cm}^2

Verification

Check 1: Build the area a different way. The hexagon sits inside a rectangle

23

cm by

7.2

cm, with one triangle cut off above

FA

and another above

BC

. The two triangle bases add up to

23 - 11 = 12

cm and both triangles have height

2.5

cm.

23 \times 7.2 - \dfrac{1}{2} \times 12 \times 2.5 = 150.6

Check 2: A trapezium's area is also its mean width times its height. The mean of the parallel sides is

\dfrac{11 + 23}{2} = 17

cm, which should give the same trapezium area as Step 4.

17 \times 2.5 = 42.5

Check 3: Check the size is sensible. The hexagon contains the rectangle

FCDE

and fits inside the

23

cm by

7.2

cm rectangle, so its area must lie between those two.

108.1 < 150.6 < 165.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$23 \times 4.7 = 108.1$	B1	The area of the rectangle $FCDE$ , or equivalent. Awarded whatever else is done, and it may be embedded in $23 \times (4.7 + 2.5) = 165.6$ .	✓
$\sin 30^\circ = \dfrac{x}{5}$	M1	A correct trigonometric statement for the height $x$ of the trapezium, or equivalent, for example $\dfrac{x}{\sin 30^\circ} = \dfrac{5}{\sin 90^\circ}$ . The mark is also earned without trigonometry for the height itself: $5 \cos 30^\circ$ $(= 4.33\ldots)$ for the horizontal run, then $x^2 = 5^2 - (5 \cos 30^\circ)^2$ $(= 6.25)$ .	✓
$x = 5 \sin 30^\circ = 2.5$	M1	The height of the trapezium worked out correctly. Accept $x = \sqrt{5^2 - (5 \cos 30^\circ)^2}$ reaching $2.5$ .	✓
$\dfrac{1}{2} \times (11 + 23) \times 2.5 = 42.5$	M1	A correct method for the area of the trapezium $ABCF$ , or for the whole shape, using their own height.	✓
$108.1 + 42.5 = 150.6$	A1	The area of the hexagon. Accept anything that rounds to $150.6$ , allow $151$ , and accept $\dfrac{753}{5}$ .	✓

Full marks: 5/5

Continue to questions 11 to 20

The remaining 10 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 26 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Higher tier targets grades 4 to 9, so the lower grades 1 to 3 are only reachable on the tier below. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 2FR and Paper 2HR, so the lowest grades on this Higher paper are the ones the two tiers share.

Yes. The paper states in its own instructions that without sufficient working, correct answers may be awarded no marks. Several questions ask you to show your working clearly or to show clear algebraic working, and on those a bare answer scores nothing. That is why every solution here sets out the method mark by mark.

Yes, a Higher tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume and curved surface area of a cylinder, the volume and curved surface area of a cone, the volume and surface area of a sphere, the area of a triangle from two sides and the included angle, the sine rule, the cosine rule, the sum of an arithmetic series and the quadratic formula. Other results, such as Pythagoras theorem and the trigonometric ratios for right-angled triangles, still have to be recalled. Nothing may be written on the formulae page.

Both are published by Pearson Edexcel and are linked directly from this page as PDF files. The solutions here are original: every question has been reworded, but all the numbers match the original paper, so the answers agree with the official mark scheme. This resource reproduces neither the exam paper nor the official mark scheme.

Keep revising

Once you have worked through this paper, 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 and Mark Schemes

Worked solutions, questions 1 to 10 of 26

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Frequently asked questions

Keep revising

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