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Edexcel IGCSE 4MA1/2H, Wednesday 4 June 2025: Worked Solutions, Questions 20 to 26

Sir Faraz Hassan

31 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/2H - Higher Tier - Wednesday 4 June 2025100 marks · 2 hours · Calculator allowed

Back to questions 11 to 19

This is the rest of the paper. Questions 1 to 19, the paper's overview and the frequently asked questions are on the first two pages.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2H (Higher Tier), June 2025 series, sat Wednesday 4 June 2025 –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 20 to 26 of 26

$120$ members of a sports club were asked whether they play cricket $(C)$ or padel $(P)$ or tennis $(T)$

Of these members

$43$ play cricket
$12$ play cricket and padel and tennis
$18$ play cricket and padel
$27$ play cricket and tennis
$32$ play padel and tennis
$29$ do not play cricket or padel or tennis

The number of these members who play only padel is equal to the number of these members who play only tennis.

(a) Complete the Venn diagram to show this information.

[3 marks]

(b) Find $\text{n}(T' \cap C)$ [1 mark]

One of the members who plays cricket is chosen at random.

(b)(c)

[Total 6 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

A sports club with

120

members, each asked whether they play cricket

C

, padel

P

or tennis

T

\text{n}(C) = 43

\text{n}(C \cap P) = 18

\text{n}(C \cap T) = 27

\text{n}(P \cap T) = 32

\text{n}(C \cap P \cap T) = 12

29

members play none of the three, and the number who play only padel equals the number who play only tennis.

Find

(a) The number in each of the eight regions of the Venn diagram. (b)

\text{n}(T' \cap C)

. (c) The probability that a member chosen at random from the cricket players also plays padel.

Plan the Solution

Start in the middle, where all three circles overlap, because that number is given outright.
Take the middle away from each pair total to get the three regions where exactly two sports are played.
Subtract the three known cricket regions from $43$ to leave cricket only.
Everything inside the circles comes to $120 - 29$ , so the two equal regions left over can be found and halved.
Read part (b) off the finished diagram, then write part (c) as a fraction of the $43$ cricket players and not of all $120$ members.

Worked Solution [6 marks]

Rule - Fill a Venn diagram from the centre outwards. Every overlap total already contains the region in the middle, so subtract it:

\text{n}(C \cap P) - \text{n}(C \cap P \cap T)

is the number in both

C

and

P

but not in

T

. When the diagram is complete, all eight regions must add up to

120

Step 1: Write in the middle region

\text{n}(C \cap P \cap T) = 12

(Reason: The question gives the number who play all three sports outright, and every other overlap total contains it, so nothing else can be worked out until it is in place.)

Step 2: Take the middle out of each pair total

18 - 12 = 6

27 - 12 = 15

32 - 12 = 20

(Reason: The

18

who play cricket and padel includes the

12

who play all three, so only

6

play cricket and padel but not tennis. The same subtraction leaves

15

for cricket and tennis only and

20

for padel and tennis only.)

Step 3: Finish the cricket circle

43 - (6 + 12 + 15) = 43 - 33 = 10

(Reason: The cricket circle holds

43

members altogether. Three of its four regions are now known, so taking those off leaves the members who play cricket and nothing else.)

Step 4: Use the total to split the last two regions

120 - 29 = 91

91 - (10 + 6 + 15 + 12 + 20) = 28

\dfrac{28}{2} = 14

(Reason:

29

members play none of the three sports, so

91

are somewhere inside the circles. Taking away the five regions already filled leaves

28

to share between padel only and tennis only, and the question says those two are equal, so each of them is

14

Step 5: Read

\text{n}(T' \cap C)

off the diagram

\text{n}(T' \cap C) = 10 + 6 = 16

(Reason:

T'

is everything outside the tennis circle, so

T' \cap C

is the part of the cricket circle that the tennis circle does not reach. That is cricket only together with cricket and padel only,

10 + 6

Step 6: Write part (c) as a fraction of the cricket players

6 + 12 = 18

\dfrac{18}{43}

(Reason: The member is picked from the cricket players only, so the denominator is

43

and not

120

. Of those

43

, the ones who also play padel are the

6

in cricket and padel only plus the

12

who play all three.)

(a)

C

only

10

P

only

14

T

only

14

, all three

12

C

and

P

only

6

C

and

T

only

15

P

and

T

only

20

, none

29

(b)

16

(c)

\dfrac{18}{43}

Verification

Check 1: Add every region on the finished diagram:

10 + 6 + 14 + 15 + 12 + 20 + 14 + 29

. The eight regions come to

120

, the number of members surveyed, so nobody has been left out or counted twice.

Check 2: Count each circle off the finished diagram -

\text{n}(P) = 6 + 14 + 12 + 20

and

\text{n}(T) = 15 + 12 + 20 + 14

- then put the three circle totals through the inclusion and exclusion rule.

43 + 52 + 61 - 18 - 27 - 32 + 12 = 91

, which is exactly

120 - 29

, so the circle totals agree with the overlap figures the question gives.

Check 3: The padel players inside the cricket circle should come back to the

18

the question states for cricket and padel, and the probability should sit a little under a half.

6 + 12 = 18

, and

\dfrac{18}{43} = 0.4186

to four decimal places, which is just below

0.5

as expected because

18

is a little under half of

43

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The completed Venn diagram	B3	For all numbers in correct regions: $10$ , $6$ , $14$ , $15$ , $12$ , $20$ , $14$ , $29$ . B2 for $5$ , $6$ or $7$ correct numbers. B1 for $3$ or $4$ correct numbers.	✓
(b) $\text{n}(T' \cap C)$	B1	$16$ . Follow through for their $a$ plus their $b$ . Do not follow through if there are no values for $a$ and $b$ .	✓
(c) A correct probability method	M1	For $\dfrac{n}{43}$ where $n < 43$ or $\dfrac{18}{m}$ where $m > 18$ or follow through their $\dfrac{b + c}{p}$ where $p > b + c$ or follow through their $\dfrac{q}{a + b + c + d}$ where $q < a + b + c + d$ . Do not follow through if there are no values for $a$ , $b$ , $c$ and $d$ .	✓
(c) The probability	A1	$\dfrac{18}{43}$ or equivalent, for example $0.41(860\ldots)$ or $41(.860\ldots)\%$ truncated or rounded, or follow through their $\dfrac{b + c}{a + b + c + d}$ as a fraction or a decimal or a percentage.	✓
Note	Note	The printed scheme shades four regions of the cricket circle and names them on its own diagrams: $a$ is $C$ only, $b$ is $C$ and $P$ only, $c$ is the region all three share and $d$ is $C$ and $T$ only. So $a + b + c + d$ is the whole of $C$ and $b + c$ is $C \cap P$ .	✓

Full marks: 6/6

The diagram shows a cuboid.

The edges of the cuboid are $3x$ cm, $2x$ cm and $y$ cm
This cuboid has a volume of $1014$ cm³
Its total surface area is $A$ cm²

Show that $A = 12x^2 + \dfrac{1690}{x}$
You must show every stage of your working. [3 marks]

[Total 3 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A cuboid with edges

3x

cm,

2x

cm and

y

The volume is

1014

cm³

The total surface area is

A

cm²

Find

Show that

A = 12x^2 + \dfrac{1690}{x}

, so the surface area is written in terms of

x

alone

Plan the Solution

Use the volume to link $x$ and $y$ : multiply the three edges together and set the product equal to $1014$ .
Rearrange that equation to make $y$ the subject.
Add the three pairs of faces to get the surface area in terms of $x$ and $y$ .
Replace $y$ by its expression in $x$ , so that only $x$ is left.

Worked Solution [3 marks]

Rule - Cuboid: a cuboid with edges

a

b

and

c

has volume

abc

and total surface area

2ab + 2bc + 2ca

, because the six faces come in three matching pairs.

Step 1: write an equation for the volume

3x \times 2x \times y = 1014

6x^2y = 1014

x^2y = 169

(Reason: (Reason: the volume of a cuboid is the product of its three edges, and

3x \times 2x = 6x^2

. Dividing both sides by

6

gives the tidier form, because

\dfrac{1014}{6} = 169

.))

Step 2: make

y

the subject

y = \dfrac{169}{x^2}

(Reason: (Reason: dividing both sides of

x^2y = 169

x^2

leaves

y

on its own, so from here every

y

can be written in terms of

x

.))

Step 3: write the surface area in terms of

x

and

y

A = 2(3x \times 2x) + 2(3x \times y) + 2(2x \times y)

A = 12x^2 + 6xy + 4xy

A = 12x^2 + 10xy

(Reason: (Reason: the faces come in three pairs. The top and bottom are each

3x

2x

, the front and back are each

3x

y

, and the two ends are each

2x

y

. The two

xy

terms then collect into one.))

Step 4: replace

y

and simplify

A = 12x^2 + 10x \times \dfrac{169}{x^2}

10x \times \dfrac{169}{x^2} = \dfrac{10 \times 169}{x} = \dfrac{1690}{x}

A = 12x^2 + \dfrac{1690}{x}

(Reason: (Reason:

y

has been replaced by

\dfrac{169}{x^2}

. One factor of

x

cancels from the top and the bottom, which turns

\dfrac{x}{x^2}

into

\dfrac{1}{x}

, and that is the expression the question asks for.))

Shown:

A = 12x^2 + \dfrac{1690}{x}

Verification

Check 1: Take

x = 1

. The volume equation gives

y = \dfrac{169}{1^2} = 169

, so the cuboid measures

3

cm by

2

cm by

169

cm. Its volume is

3 \times 2 \times 169 = 1014

cm³, and adding the three pairs of faces gives

2 \times 6 + 2 \times 507 + 2 \times 338 = 1702

cm².

12 \times 1^2 + \dfrac{1690}{1} = 1702

, the same total

Check 2: Take

x = 13

, a value that makes the third edge short. Then

y = \dfrac{169}{13^2} = 1

, so the cuboid measures

39

cm by

26

cm by

1

cm. Its volume is

39 \times 26 \times 1 = 1014

cm³, and adding the three pairs of faces gives

2 \times 1014 + 2 \times 39 + 2 \times 26 = 2158

cm².

12 \times 13^2 + \dfrac{1690}{13} = 2158

, the same total again

Check 3: Read the printed expression backwards. Matching

12x^2 + 10xy

with

12x^2 + \dfrac{1690}{x}

needs

10xy = \dfrac{1690}{x}

, and multiplying both sides by

x

then dividing by

10

gives

x^2y = \dfrac{1690}{10}

x^2y = 169

, which is exactly the volume equation Step 1 produced, so the two ends of the working meet

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Form an equation for the volume in terms of $x$ and $y$	M1	$3x \times 2x \times y = 1014$ oe or $6x^2y = 1014$ oe or $x^2y = 169$ oe. For an equation for volume in terms of $x$ and $y$ .	✓
Write a correct expression for the total surface area	M1 indep	$2 \times 3x \times y + 2 \times 2x \times y + 2 \times 3x \times 2x$ oe or $2 \times (3xy + 2xy + 6x^2)$ oe or $6xy + 4xy + 12x^2$ oe or $10xy + 12x^2$ oe. Independent of the first mark. NB the surface area expression may not be seen explicitly in terms of $y$ , eg $2 \times 3x \times \dfrac{169}{x^2} + 2 \times 2x \times \dfrac{169}{x^2} + 2 \times 3x \times 2x$ oe. It may be fully substituted using $y = \dfrac{169}{x^2}$ oe.	✓
Complete the show that	A1 dep on M2	Using $y = \dfrac{1014}{6x^2}$ $\left(= \dfrac{169}{x^2}\right)$ in the formula for the surface area to obtain a correct expression, eg (SA =) $2 \times 5x \times \dfrac{169}{x^2} + 2 \times 6x^2 = 12x^2 + \dfrac{1690}{x}$ , or equating their surface area equations, eg $10xy + 12x^2 = 12x^2 + \dfrac{1690}{x}$ leading to $x^2y = 169$ oe. Dependent on both method marks. For completing the show that by clearly showing the stages that lead to the given expression for the surface area.	✓
Working required	Note	Working required. Total 3 marks.	✓

Full marks: 3/3

Rectangle $ABCD$ has a square drawn inside it, as shown in the diagram.

The part of the rectangle that lies outside the square is shaded.
The total area of the shaded region is $X \text{ cm}^{2}$ .

$AB = 11.5 \text{ cm}$ correct to the nearest $0.5 \text{ cm}$
$BC = 9.2 \text{ cm}$ correct to $2$ significant figures
side of the square $= 4.1 \text{ cm}$ correct to $2$ significant figures

By considering bounds, work out the value of $X$ to a suitable degree of accuracy.
You must show your working clearly. [4 marks]

X =

[Total 4 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

The length

AB

11.5 \text{ cm}

, correct to the nearest

0.5 \text{ cm}

The length

BC

9.2 \text{ cm}

, correct to

2

significant figures

The side of the square,

s

, is

4.1 \text{ cm}

, correct to

2

significant figures

The shaded region is the rectangle with the square removed, so

X

AB \times BC - s^{2}

Find

The value of

X

to a suitable degree of accuracy: the most accurate value that every possible set of measurements agrees on

Plan the Solution

Write the lower and upper bound of each of the three lengths. A bound sits half a rounding unit either side of the value given.
The shaded area is a difference, so it is largest when the rectangle is largest and the square is smallest, and smallest when the rectangle is smallest and the square is largest.
Work out the upper bound of $X$ and the lower bound of $X$ .
Round both bounds to the same accuracy, starting with the most accurate, and stop at the first degree of accuracy where they agree. That shared value is $X$ .

Worked Solution [4 marks]

Rule - Bounds of a difference: to make

P - Q

as large as possible, take the upper bound of

P

with the lower bound of

Q

. To make it as small as possible, take the lower bound of

P

with the upper bound of

Q

Step 1: Write down the bounds of the three lengths

\dfrac{0.5}{2} = 0.25

\dfrac{0.1}{2} = 0.05

11.25 \leq AB < 11.75

9.15 \leq BC < 9.25

4.05 \leq s < 4.15

(Reason:

AB

is given to the nearest

0.5 \text{ cm}

, so its bounds are

0.25

either side of

11.5

. Giving

9.2

and

4.1

2

significant figures is rounding to the nearest

0.1 \text{ cm}

, so those bounds are

0.05

either side.)

Step 2: The upper bound of

X

11.75 \times 9.25 - 4.05^{2} = 108.6875 - 16.4025 = 92.285

(Reason: The shaded area is as large as it can be when the rectangle is as large as it can be and the square taken out of it is as small as it can be, so this uses

11.75

9.25

and

4.05

Step 3: The lower bound of

X

11.25 \times 9.15 - 4.15^{2} = 102.9375 - 17.2225 = 85.715

(Reason: The shaded area is as small as it can be when the rectangle is as small as it can be and the square taken out of it is as large as it can be, so this uses

11.25

9.15

and

4.15

Step 4: Round both bounds until they agree

92.285 \approx 92 \text{ (2 s.f.)}

85.715 \approx 86 \text{ (2 s.f.)}

92.285 \approx 90 \text{ (1 s.f.)}

85.715 \approx 90 \text{ (1 s.f.)}

(Reason: To

2

significant figures the two bounds give

92

and

86

, which are different, so that is more accuracy than the measurements support. To

1

significant figure both give

90

, so every value the shaded area could take rounds to the same number, and that number is the answer.)

X = 90

(to

1

significant figure)

Verification

Check 1 - round both bounds together: Round the upper bound and the lower bound to the same accuracy and compare them. To

2

significant figures they are

92

and

86

. They disagree at

2

significant figures and agree at

1

, which gives

90

Check 2 - use the measurements as they are stated: Work out the shaded area from the stated lengths themselves:

11.5 \times 9.2 - 4.1^{2}

105.8 - 16.81 = 88.99

, which lies between the two bounds and also rounds to

90

Check 3 - pair the bounds the other way round: Take the largest rectangle with the largest square, and the smallest rectangle with the smallest square:

11.75 \times 9.25 - 4.15^{2} = 91.465

and

11.25 \times 9.15 - 4.05^{2} = 86.535

. Both sit inside

85.715

92.285

, so the pairings used above really are the extreme ones.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$11.25$ , $11.75$ , $9.15$ , $9.25$ , $4.05$ , $4.15$	B1	for a correct bound. Accept $11.74\dot{9}$ or $11.74\overline{9}$ for $11.75$ , $9.24\dot{9}$ or $9.24\overline{9}$ for $9.25$ , $4.14\dot{9}$ or $4.14\overline{9}$ for $4.15$	✓
$11.75 \times 9.25 - 4.05^{2} \; (= 92.285)$	M1	for a correct method to find the upper bound of $X$ , allow $(11.5 < AB \leq 11.75) \times (9.2 < BC \leq 9.25) - \left( (4.05 \leq s < 4.1)^{2} \right)$	✓
$11.25 \times 9.15 - 4.15^{2} \; (= 85.715)$	M1	for a correct method to find the lower bound of $X$ , allow $(11.25 \leq AB < 11.5) \times (9.15 \leq BC < 9.2) - \left( (4.1 < s \leq 4.15)^{2} \right)$	✓
Working required	A1	dep on M2. $90$ , and both the upper bound and the lower bound correct using the correct values $11.25$ , $11.75$ , $9.15$ , $9.25$ , $4.05$ and $4.15$	✓

Full marks: 4/4

$\dfrac{\dfrac{4x^{2} - 4x - 120}{5x^{2} - 180}}{\dfrac{x^{2} + 5x}{10x^{2} + 60x}} = p$ where $p$ is an integer.

Work out the value of $p$
Show clear algebraic working. [4 marks]

p =

[Total 4 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

\dfrac{\dfrac{4x^{2} - 4x - 120}{5x^{2} - 180}}{\dfrac{x^{2} + 5x}{10x^{2} + 60x}} = p

p

is an integer, so every

x

has to cancel out.

Find

The value of

p

Plan the Solution

Factorise all four expressions fully, top and bottom of both fractions.
Dividing by a fraction is multiplying by its reciprocal, so turn the second fraction upside down.
Cancel every bracket that appears on the top and on the bottom, then work out the numbers that are left.

Worked Solution [4 marks]

Rule - Dividing algebraic fractions: factorise every part, turn the second fraction upside down, multiply, then cancel.

Step 1: Factorise the first fraction

4x^{2} - 4x - 120 = 4(x^{2} - x - 30) = 4(x - 6)(x + 5)

5x^{2} - 180 = 5(x^{2} - 36) = 5(x - 6)(x + 6)

(Reason: (Reason: take the common number out first. Then

x^{2} - x - 30

needs two numbers with product

-30

and sum

-1

, which are

-6

and

5

, while

x^{2} - 36

is the difference of two squares.))

Step 2: Factorise the second fraction

x^{2} + 5x = x(x + 5)

10x^{2} + 60x = 10x(x + 6)

(Reason: (Reason: both terms of each expression contain an

x

, and the bottom one also has a factor of

10

, so

10x

comes out. The marks here are for factorising fully, not part way.))

Step 3: Turn the second fraction upside down and multiply

\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}

(Reason: (Reason: dividing by

\dfrac{x(x + 5)}{10x(x + 6)}

is the same as multiplying by its reciprocal

\dfrac{10x(x + 6)}{x(x + 5)}

.))

Step 4: Cancel every factor that appears top and bottom

\dfrac{4(x - 6)(x + 5) \times 10x(x + 6)}{5(x - 6)(x + 6) \times x(x + 5)}

\dfrac{4 \times 10}{5} = \dfrac{40}{5} = 8

(Reason: (Reason:

(x - 6)

(x + 5)

(x + 6)

and

x

each appear on the top and on the bottom, so all four cancel and only the numbers are left.))

p = 8

Verification

Check 1: Put

x = 1

into the original expression. The first fraction becomes

\dfrac{4 - 4 - 120}{5 - 180} = \dfrac{-120}{-175} = \dfrac{24}{35}

and the second becomes

\dfrac{1 + 5}{10 + 60} = \dfrac{6}{70} = \dfrac{3}{35}

, so divide the first by the second.

\dfrac{24}{35} \times \dfrac{35}{3} = 8

Check 2: Try a second value,

x = 2

. Now the first fraction is

\dfrac{16 - 8 - 120}{20 - 180} = \dfrac{-112}{-160} = \dfrac{7}{10}

and the second is

\dfrac{4 + 10}{40 + 120} = \dfrac{14}{160} = \dfrac{7}{80}

. This is a check only: the mark scheme says substituting values of

x

is not an acceptable algebraic method for the marks.

\dfrac{7}{10} \times \dfrac{80}{7} = 8

Check 3: Multiply everything out instead of factorising. One fraction is left,

\dfrac{40x^{4} + 200x^{3} - 1\,440x^{2} - 7\,200x}{5x^{4} + 25x^{3} - 180x^{2} - 900x}

, so each coefficient on the top should be

8

times the one underneath it.

8 \times 5 = 40

8 \times 25 = 200

8 \times (-180) = -1\,440

8 \times (-900) = -7\,200

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)}}{\dfrac{x(x + 5)}{10x(x + 6)}}$ $(= p)$ or $\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}$ $(= p)$ or 2 from $4(x - 6)(x + 5)$ or $5(x - 6)(x + 6)$ or $x(x + 5)$ or $10x(x + 6)$	M1	for factorising 2 or 3 of the quadratics fully - could be implied by 2 factors cancelled correctly. NB factors must be in the form $(ax + b)$ . NB Substitution of values of $x$ into the given equation is not an acceptable algebraic method	✓
$\dfrac{\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)}}{\dfrac{x(x + 5)}{10x(x + 6)}}$ $(= p)$ or $\dfrac{4(x - 6)(x + 5)}{5(x - 6)(x + 6)} \times \dfrac{10x(x + 6)}{x(x + 5)}$ $(= p)$ or $4(x - 6)(x + 5)$ and $5(x - 6)(x + 6)$ and $x(x + 5)$ and $10x(x + 6)$	M1	for factorising all of the quadratics fully - could be implied by 2 factors cancelled correctly. NB factors must be in the form $(ax + b)$	✓
$\dfrac{4x^{2} - 4x - 120}{5x^{2} - 180} \times \dfrac{10x^{2} + 60x}{x^{2} + 5x}$ $(= p)$	M1	for inverting the 2nd fraction (this mark can be awarded at any time and may be awarded with incorrect factorisation if meaning is clear)	✓
Working required	A1	$8$ oe dep on M3	✓
ALT $\dfrac{40x^{4} + 200x^{3} - 1\,440x^{2} - 7\,200x}{5x^{4} + 25x^{3} - 180x^{2} - 900x}$	Note	the printed scheme carries an ALT route: a correct expression with no errors scores M3, and the answer $8$ then scores the A1, oe dep on M3	✓

Full marks: 4/4

The diagram shows a square-based pyramid $ABCDE$ .

$EA = EB = EC = ED$

$M$ is the centre of the horizontal square base $ABCD$
$Q$ is the midpoint of $AB$
Angle $EQM = 80^{\circ}$

$EA : AB = n : 1$

Work out the value of $n$
Give your answer correct to $3$ significant figures. [4 marks]

n =

[Total 4 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

ABCDE

is a square-based pyramid with

EA = EB = EC = ED

M

is the centre of the square base

ABCD

, and that base is horizontal.

Q

is the midpoint of

AB

Angle

EQM = 80^{\circ}

The ratio

EA : AB = n : 1

, so

n

compares a sloping edge with a base edge.

Find

The value of

n

, correct to

3

significant figures.

Plan the Solution

A ratio does not care how big the pyramid is, so give the base the side length $AB = 1$ . Then $n$ is simply the length $EA$ .
All four sloping edges are equal, so $E$ sits directly above the centre $M$ . That makes $EM$ vertical, so angle $EMQ = 90^{\circ}$ and $EQM$ is a right-angled triangle carrying the $80^{\circ}$ .
$M$ is the centre of the square and $Q$ is the midpoint of a side, so $MQ$ is half a side. Cosine in triangle $EQM$ then gives $EQ$ .
Triangle $EAB$ is isosceles because $EA = EB$ , so the line from $E$ to the midpoint $Q$ is perpendicular to $AB$ . Pythagoras in triangle $EQA$ finishes it.

Worked Solution [4 marks]

Rule - Right-angled triangles:

\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}

turns an angle into a side, and Pythagoras

c^2 = a^2 + b^2

turns two sides into the third.

Step 1 - Give the base a side length, then find

MQ

and

AQ

AB = 1

MQ = \dfrac{1}{2} \times 1 = 0.5

AQ = \dfrac{1}{2} \times 1 = 0.5

(Reason: Only the ratio is asked for, so every side length gives the same

n

. The centre of a square is half a side away from the midpoint of any side, which fixes

MQ

; and

Q

is the midpoint of

AB

, which fixes

AQ

Step 2 - Use the right angle at

M

to find

EQ

\cos 80^{\circ} = \dfrac{MQ}{EQ}

EQ = \dfrac{0.5}{\cos 80^{\circ}} = 2.87938\ldots

(Reason: Equal sloping edges put

E

directly above

M

, so

EM

is vertical and angle

EMQ

is a right angle. In triangle

EQM

the side

MQ

lies next to the

80^{\circ}

and

EQ

is the hypotenuse, so cosine is the ratio to use. Keep the unrounded value in the calculator.)

Step 3 - Use the right angle at

Q

to find

EA

EA^2 = EQ^2 + AQ^2

EA^2 = (2.87938\ldots)^2 + 0.5^2 = 8.54085\ldots

EA = \sqrt{8.54085\ldots} = 2.92247\ldots

(Reason:

EA = EB

makes triangle

EAB

isosceles, so the line from

E

to the midpoint

Q

meets

AB

at a right angle. Triangle

EQA

therefore has its right angle at

Q

, with legs

EQ

and

AQ

. Watch which half-length belongs here:

AQ

is half a side, not half a diagonal.)

Step 4 - Write the ratio, then round

EA : AB = 2.92247\ldots : 1

n = 2.92247\ldots

n = 2.92

(Reason: Because

AB

was set to

1

, the ratio

EA : AB

is already in the form

n : 1

, so

n

is the length

EA

. Rounding to

3

significant figures is the last thing done, not the first.)

n = 2.92

Verification

Check 1 - work back to the angle the question gives: Start from the answer. With

EA^2 = 8.54085\ldots

and

AQ = 0.5

, Pythagoras run backwards gives

EQ = \sqrt{8.54085\ldots - 0.25} = 2.87938\ldots

, and then

\cos EQM = \dfrac{0.5}{2.87938\ldots} = 0.17364\ldots

. Angle

EQM = 80^{\circ}

, the angle the question states.

Check 2 - reach the same edge up the axis instead of along the face: The vertical height is

EM = 0.5 \tan 80^{\circ} = 2.83564\ldots

, and

A

is half a diagonal from the centre, so

MA = \sqrt{0.5^2 + 0.5^2} = 0.70710\ldots

. Triangle

EMA

has its right angle at

M

, and it shares nothing with the route above except the given angle.

EA = \sqrt{(2.83564\ldots)^2 + (0.70710\ldots)^2} = 2.92247\ldots

, the same length, so

n = 2.92

again.

Check 3 - is the size sensible? The sloping edge must be longer than the slant height, which must be longer than

MQ

, and the three values do line up:

2.92247\ldots > 2.87938\ldots > 0.5

. An angle of

80^{\circ}

is a steep face, so a sloping edge close to three times the base edge is what to expect. The three lengths are in the right order and the pyramid comes out tall and narrow, which is what

80^{\circ}

demands.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
1	M1	To find $EQ$ : e.g. $MQ = 0.5 \, (= AQ)$ , $\dfrac{0.5}{\cos 80} \, (= 2.87(938\ldots))$ ; e.g. $MQ = x \, (= AQ)$ , $\dfrac{x}{\cos 80} \, (= 5.75\ldots x)$ . NB $\cos 80 = \sin 10$ . Or to find $EM$ : e.g. $MQ = 0.5 \, (= AQ)$ , $0.5 \tan 80 \, (= 2.83(564\ldots))$ ; e.g. $MQ = x \, (= AQ)$ , $x \tan 80 \, (= 5.67\ldots x)$ . NB $\tan 80 = \dfrac{1}{\tan 10}$ . Or to find $EQ$ : e.g. $EM = y$ , $\dfrac{y}{\sin 80} \, (= 1.01\ldots y)$ ; or to find $MQ \, (= AQ)$ : e.g. $EQ = h$ , $h \cos 80 \, (= 0.173\ldots h)$ . Notes: use of a value for the side $AB$ , e.g. $1$ or $2x$ or any value, or let $EM = y$ or any value, or $EQ = h$ .	✓
2	M1	To find $EA^2$ : e.g. $AQ = 0.5 \, (= MQ)$ , $(2.87\ldots)^2 + 0.5^2 \, (= 8.54(08\ldots))$ ; e.g. $AQ = x \, (= MQ)$ , $x^2 + \left(\dfrac{x}{\cos 80}\right)^2$ . Or to find $AM$ : e.g. $MQ = 0.5 \, (= AQ)$ , $\sqrt{0.5^2 + 0.5^2} \, \left(= \dfrac{\sqrt{2}}{2} = 0.707(10\ldots)\right)$ ; e.g. $MQ = x \, (= AQ)$ , $\sqrt{x^2 + x^2} \, (= \sqrt{2x^2} = x\sqrt{2})$ . Or to find $MQ \, (= AQ)$ : e.g. $EM = y$ , $\dfrac{y}{\tan 80} \, (= 0.176\ldots y)$ ; then to find $EA^2$ : e.g. $EQ = h$ , $h^2 + (h \cos 80)^2$ . A value the printed scheme shows in quotation marks may be the candidate's own earlier value.	✓
3	M1	To find $EA$ : e.g. $AQ = 0.5 \, (= MQ)$ , $\sqrt{(2.87\ldots)^2 + 0.5^2} \, \left(= \sqrt{8.54(08\ldots)}\right)$ ; or e.g. $AQ = x \, (= MQ)$ , $\sqrt{x^2 + \left(\dfrac{x}{\cos 80}\right)^2} \, (= 5.84\ldots x)$ . Or e.g. $AQ = 0.5 \, (= MQ)$ , $\sqrt{(2.83\ldots)^2 + (0.707\ldots)^2} \, \left(= \sqrt{8.54(08\ldots)}\right)$ ; e.g. $AQ = x \, (= MQ)$ , $\sqrt{(x \tan 80)^2 + (x\sqrt{2})^2} \, (= 5.84\ldots x)$ . Or e.g. $EM = y$ , $\sqrt{\left(\dfrac{1}{\sin 80}\right)^2 + \left(\dfrac{1}{\tan 80}\right)^2} \, (= 1.03\ldots y)$ , that is $1.03\ldots \times AQ \times \tan 80$ ; e.g. $EQ = h$ , $\sqrt{h^2 + (h \cos 80)^2} \, (= 1.01\ldots h)$ , that is $1.01\ldots \times \dfrac{0.5}{\cos 80}$ .	✓
4	A1	Answer $2.92$ , awrt $2.92$ . Working not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

(a) Express $28 + 24x - 6x^2$ in the form $a - b(x - c)^2$ , where $a$ , $b$ and $c$ are integers. [3 marks]

(b) On the axes below, sketch the curve with equation $y = 28 + 24x - 6x^2$
Show clearly the coordinates of the turning point and the coordinates of the point where the curve meets the $y$ -axis. [3 marks]

[Total 6 marks]

Show solution & mark schemeHide solution & mark scheme

Question 25 - Exam Solution

Understanding the Question

Given

The expression

28 + 24x - 6x^2

The form it must be written in,

a - b(x - c)^2

, with

a

b

and

c

integers

A pair of blank axes for the sketch of

y = 28 + 24x - 6x^2

Find

(a) The completed square form, so the values of

a

b

and

c

(b) The sketch, with the turning point and the point where the curve meets the

y

-axis both labelled with their coordinates

Plan the Solution

Write the expression in descending powers, so the $x^2$ term comes first.
Take the $-6$ out of the $x^2$ and $x$ terms only, and complete the square inside the bracket.
Multiply the $-6$ back in and collect the two constants, which gives $a - b(x - c)^2$ straight away.
For part (b), read the turning point off that completed square, put $x = 0$ in to find where the curve meets the $y$ -axis, and use the sign of the $x^2$ term to decide which way up the curve is.

Worked Solution [6 marks]

Rule - Completing the square with a negative

x^2

term: take that coefficient out of the

x^2

and

x

terms first, complete the square inside the bracket, then multiply back in. In

y = a - b(x - c)^2

with

b

positive, the square is never negative, so the curve has a maximum turning point at

(c,\ a)

Step 1: Put the terms in descending powers and take out the

-6

28 + 24x - 6x^2 = -6x^2 + 24x + 28

-6x^2 + 24x = -6(x^2 - 4x)

28 + 24x - 6x^2 = -6(x^2 - 4x) + 28

(Reason: Dividing

24

-6

gives

-4

, so the bracket is

x^2 - 4x

. The

28

has no

x

in it, so it stays outside the bracket.)

Step 2: Complete the square inside the bracket

(x - 2)^2 = x^2 - 4x + 4

x^2 - 4x = (x - 2)^2 - 4

(Reason: Half of

-4

-2

, so the bracket is

(x - 2)^2

. Squaring that bracket brings in an extra

4

, so

4

is taken away again to keep the value the same.)

Step 3: Multiply the

-6

back in and collect the constants

-6\left[(x - 2)^2 - 4\right] + 28 = -6(x - 2)^2 + 24 + 28

24 + 28 = 52

28 + 24x - 6x^2 = 52 - 6(x - 2)^2

(Reason: Multiplying the

-6

back in changes the sign of the

-4

that sits inside the bracket, and the two constants outside are then collected into one.)

Step 4: Read off the three integers

a - b(x - c)^2 = 52 - 6(x - 2)^2

a = 52

b = 6

c = 2

(Reason: All three are integers, which is what part (a) asks for. Note that

b

6

and not

-6

, because the form already has a minus sign in front of the

b

Step 5: (b) The turning point, straight from the completed square

y = 52 - 6(x - 2)^2

6(x - 2)^2 \geq 0

52 - 6 \times 0 = 52

(2,\ 52)

(Reason: A square is never negative, so

6(x - 2)^2

only takes away from the

52

. The largest value of

y

therefore happens when the bracket is zero, at

x = 2

, and there

y = 52

. Because

y

can never be more than this, the turning point is a maximum.)

Step 6: (b) Where the curve meets the

y

-axis

28 + 24 \times 0 - 6 \times 0^2 = 28

(0,\ 28)

(Reason: Every point on the

y

-axis has

x = 0

, so putting

0

in place of

x

leaves the constant term,

28

Step 7: (b) Draw the sketch

28 + 24 \times 4 - 6 \times 4^2 = 28

(0,\ 28) \quad (2,\ 52) \quad (4,\ 28)

(Reason: The coefficient of

x^2

is negative, so the curve opens downwards. It is symmetrical about the vertical line through the turning point, so the point level with

(0,\ 28)

on the other side is

(4,\ 28)

. Draw one smooth curve through the three points, with the maximum at

(2,\ 52)

, and label the turning point and the

y

-axis crossing with their coordinates.)

(a)

52 - 6(x - 2)^2

(b) a curve opening downwards, turning point

(2,\ 52)

, meeting the

y

-axis at

(0,\ 28)

Verification

Check 1: Multiply the answer out again and compare it with the expression the question prints.

52 - 6(x^2 - 4x + 4) = 52 - 6x^2 + 24x - 24 = 28 + 24x - 6x^2

Check 2: Put

x = 1

into both forms. They are the same expression, so they must give the same value.

28 + 24 \times 1 - 6 \times 1^2 = 46

and

52 - 6 \times (1 - 2)^2 = 46

Check 3: Test the symmetry the sketch depends on. If the turning point really is at

x = 2

, then

x = 0

and

x = 4

are the same distance from it and must give the same

y

28 + 24 \times 0 - 6 \times 0^2 = 28

and

28 + 24 \times 4 - 6 \times 4^2 = 28

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\pm 6\left(x \pm \dfrac{4}{2}\right)^2 \ldots$ or $\pm 6(x \pm 2)^2 \ldots$ or $\pm 6\left[(x \pm 2)^2 \ldots\right]$ or $\pm 6\left[\left(x \pm \dfrac{4}{2}\right)^2 \ldots\right]$ or $\pm 6\left(x \pm \dfrac{24}{2 \times -6}\right)^2 \ldots$	M1	for a start to completing the square, or correct substitution into $a\left(x + \dfrac{b}{2a}\right)^2 + \ldots$ from the formula $a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{(b)^2}{4a} + c$	✓
(a) $-6\left[\left(x - \dfrac{4}{2}\right)^2 - \left(\dfrac{4}{2}\right)^2\right] \ldots$ or $-6\left[(x - 2)^2 - 2^2\right] \ldots$ or $-6\left[\left(x - \dfrac{4}{2}\right)^2 - \left(\dfrac{4}{2}\right)^2 \ldots\right]$ or $-6\left[(x - 2)^2 - 2^2 \ldots\right]$ or $-6\left(x + \dfrac{24}{2 \times -6}\right)^2 - \dfrac{24^2}{4 \times -6} \ldots$	M1	for correctly completing the square, but terms do not need to be simplified and $28$ may or may not be present, or correct simplification of the first two parts of $a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{(b)^2}{4a} (+c)$ . NB: please refer to the ALT mark scheme after (b) for the comparison of coefficients method	✓
(a) $52 - 6(x - 2)^2$ - working not required, so a correct answer scores full marks (unless from obvious incorrect working)	A1	oe eg $-6(x - 2)^2 + 52$	✓
(a) ALT $-bx^2 + 2bcx - bc^2 + a$ and $b = 6$ or $b = -6$	M1	for multiplying out $a - b(x - c)^2$ and $b = 6$ or $b = -6$	✓
(a) ALT $2bc = 24$ or $-bc^2 + a = 28$	M1	for equating coefficients	✓
(a) ALT $52 - 6(x - 2)^2$ - working not required, so a correct answer scores full marks (unless from obvious incorrect working)	A1	oe eg $-6(x - 2)^2 + 52$	✓
(b) a $\cap$ or $\cup$ shaped symmetrical quadratic curve	B1	for drawing a $\cap$ or $\cup$ shaped symmetrical quadratic curve with the turning point in any quadrant	✓
(b) turning point marked as $(2,\ 52)$	B1	for drawing a $\cap$ shaped symmetrical quadratic curve in the correct quadrant with a turning point at $(2,\ 52)$	✓
(b) intersection with the $y$ -axis marked as $(0,\ 28)$ or crossing at $28$ marked	B1	for drawing a $\cap$ shaped symmetrical quadratic curve in the correct quadrant with an intersection on the $y$ -axis marked as $(0,\ 28)$ or marked as $28$ on the $y$ -axis	✓

Full marks: 6/6

Two solid candles, $A$ and $B$ , are mathematically similar.

The height of candle $A$ is $31$ cm
The height of candle $B$ is $18.6$ cm

Given that
$\text{volume of candle } A - \text{volume of candle } B = 735 \text{ cm}^{3}$

work out the volume of candle $A$ [4 marks]

cm³

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 26 - Exam Solution

Understanding the Question

Given

Two solid candles,

A

and

B

, that are mathematically similar.

Height of candle

A

31

cm. Height of candle

B

18.6

cm.

\text{volume of candle } A - \text{volume of candle } B = 735 \text{ cm}^{3}

Neither volume is given on its own, only the gap between them.

Find

The volume of candle

A

, in cubic centimetres.

Plan the Solution

Divide the two heights to get the linear scale factor between the candles.
Cube it. A volume is three-dimensional, so the volume scale factor is the cube of the linear one.
Use it to write the volume of candle $B$ as a fraction of the volume of candle $A$ , so the given difference holds only one unknown.
Solve for the volume of candle $A$ , then rebuild both volumes and check the difference comes back to $735 \text{ cm}^{3}$ .

Worked Solution [4 marks]

Rule for mathematically similar solids: if the linear scale factor is

k

, the area scale factor is

k^{2}

and the volume scale factor is

k^{3}

Step 1: Find the linear scale factor

\dfrac{31}{18.6} = \dfrac{310}{186} = \dfrac{5}{3}

(Reason: Multiply top and bottom by

10

to clear the decimal, then divide both by

62

. Candle

A

\dfrac{5}{3}

times as tall as candle

B

Step 2: Cube the scale factor to get the volume scale factor

\left( \dfrac{5}{3} \right)^{3} = \dfrac{125}{27}

\left( \dfrac{3}{5} \right)^{3} = \dfrac{27}{125}

(Reason: A volume is three-dimensional, so every length being multiplied by

\dfrac{5}{3}

multiplies the volume by that factor three times over. The reversed fraction is the one that takes you from candle

A

down to candle

B

Step 3: Write candle B's volume in terms of candle A's

V_{B} = \dfrac{27}{125} V_{A}

(Reason: Candle

B

is the smaller solid, so its volume is the smaller share. Writing it this way leaves the given difference with only one unknown in it.)

Step 4: Turn the given difference into an equation

V_{A} - \dfrac{27}{125} V_{A} = 735

1 - \dfrac{27}{125} = \dfrac{98}{125}

\dfrac{98}{125} V_{A} = 735

(Reason: The question gives the gap between the volumes, not either volume, so substitute for

V_{B}

and collect the

V_{A}

terms. The gap is

\dfrac{98}{125}

of candle

A

Step 5: Solve for the volume of candle A

V_{A} = 735 \times \dfrac{125}{98}

\dfrac{735}{98} = 7.5

7.5 \times 125 = 937.5

(Reason: Read it as parts: candle

A

125

parts and candle

B

27

of them, so the

98

parts in between are worth

735 \text{ cm}^{3}

. One part is

7.5 \text{ cm}^{3}

, so all

125

parts are

937.5 \text{ cm}^{3}

937.5 \text{ cm}^{3}

Verification

Check 1: Scale the answer back down to candle

B

with the volume scale factor

\dfrac{27}{125}

, then take the difference the question gives.

937.5 \times \dfrac{27}{125} = 202.5

and

937.5 - 202.5 = 735

Check 2: Redo it without ever simplifying the ratio, working straight from the cubed heights

31^{3} = 29791

and

18.6^{3} = 6434.856

\dfrac{29791 \times 735}{29791 - 6434.856} = 937.5

Check 3: Volume divided by height cubed must be the same constant for both candles, since they are mathematically similar.

\dfrac{937.5}{29791} = \dfrac{202.5}{6434.856}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$A : B = 31 : 18.6 \, (= 5 : 3)$ oe or $A^{3} : B^{3} = 31^{3} : 18.6^{3} \, (= 5^{3} : 3^{3})$ oe or $\dfrac{31}{18.6} \left( = \dfrac{5}{3} \right)$ oe or $\dfrac{18.6}{31} \left( = \dfrac{3}{5} \right)$ oe or $\left( \dfrac{18.6}{31} \right)^{3}$ oe or $\dfrac{27}{125}$ oe or $\left( \dfrac{31}{18.6} \right)^{3}$ oe or $\dfrac{125}{27}$ oe	M1	for correct linear SF or volume SF either as a fraction or ratio. Allow $\dfrac{5}{3} = 1.6(6....)$ truncated or rounded	✓
$V_{A} - \left( \dfrac{3}{5} \right)^{3} V_{A} \, (= 735)$ oe or $V_{A} - \dfrac{27}{125} V_{A} \, (= 735)$ oe or $\dfrac{98}{125} V_{A} \, (= 735)$ oe or $\dfrac{V_{A}}{V_{A} - 735} = \dfrac{31^{3}}{18.6^{3}}$ oe or $\left( \dfrac{5}{3} \right)^{3} V_{B} - V_{B} \, (= 735)$ oe or $\dfrac{125}{27} V_{B} - V_{B} \, (= 735)$ oe or $\dfrac{98}{27} V_{B} \, (= 735)$ oe or $\dfrac{V_{B} + 735}{V_{B}} = \dfrac{31^{3}}{18.6^{3}}$ oe or $1 - \left( \dfrac{3}{5} \right)^{3} \left( = \dfrac{98}{125} = 0.784 \right)$ oe or $\left( \dfrac{5}{3} \right)^{3} - 1 \left( = \dfrac{98}{27} = 3.62(962...) \right)$ oe or $5^{3} - 3^{3} \, (= 125 - 27 = 98)$	M1	Note: $735$ is given in the equation. Allow any letter for $V_{A}$ or for $V_{B}$ . $V_{A} - \left( \dfrac{3}{5} \right)^{3} V_{A} \, (= 735)$ can be written as $V_{A} - \dfrac{V_{A}}{\left( \dfrac{5}{3} \right)^{3}} \, (= 735)$ , or $\left( \dfrac{5}{3} \right)^{3} V_{B} - V_{B} \, (= 735)$ can be written as $\dfrac{V_{B}}{\left( \dfrac{3}{5} \right)^{3}} - V_{B} \, (= 735)$	✓
$(V_{A} =) \, 735 \times \dfrac{125}{98}$ oe or $(V_{A} =) \, \dfrac{735}{\dfrac{98}{125}}$ oe or $(V_{A} =) \, \dfrac{31^{3} \times 735}{31^{3} - 18.6^{3}}$ oe or $(V_{B} =) \, 735 \times \dfrac{27}{98} \, (= 202.5)$ oe or $(V_{B} =) \, \dfrac{735}{\dfrac{98}{27}} \, (= 202.5)$ or $(V_{B} =) \, \dfrac{18.6^{3} \times 735}{31^{3} - 18.6^{3}} \, (= 202.5)$ oe or $\dfrac{735}{98} \times 5^{3}$ oe or $7.5 \times 125$ oe or $\dfrac{735}{98} \times 3^{3} \, (= 202.5)$ oe or $7.5 \times 27 \, (= 202.5)$ oe	M1	for a correct method to find $V_{A}$ or $V_{B}$	✓
$937.5$ (working not required, so a correct answer scores full marks unless it comes from obvious incorrect working)	A1	oe allow $938$ from correct working	✓

Full marks: 4/4

Keep revising

That is the whole 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 4MA1/2H, Wednesday 4 June 2025: Worked Solutions, Questions 20 to 26

Worked solutions, questions 20 to 26 of 26

Question 20, Calculator allowed

Question 20 - Exam Solution

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Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

Question 26, Calculator allowed

Question 26 - Exam Solution

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