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Edexcel IGCSE 4MA1/1HR, Thursday 15 May 2025: Worked Solutions, Questions 20 to 26

Sir Faraz Hassan

31 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1HR - Higher Tier - Thursday 15 May 2025100 marks · 2 hours · Calculator allowed

Back to questions 12 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 4MA1/1HR (Higher Tier), June 2025 series, sat Thursday 15 May 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

The diagram shows a cuboid $ABCDEFGH$ .

$AB = 9$ cm
$AF = 7$ cm
$FC = 18$ cm

Work out the length of $BC$ .
Give your answer correct to $3$ significant figures. [3 marks]

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

A cuboid

ABCDEFGH

, with the rectangle

ABGF

as the end face nearest the reader.

AB = 9

cm and

AF = 7

cm, the two edges of that end face.

FC = 18

cm, the diagonal running from

F

through the inside of the solid to

C

Find

The length of

BC

, the depth of the cuboid, correct to

3

significant figures.

Plan the Solution

Pythagoras only works inside one right-angled triangle, and no single triangle here holds $AB$ , $AF$ and $FC$ together. So use it twice.
First across the end face. $F$ and $B$ are opposite corners of the rectangle $ABGF$ , so $FB$ is its diagonal.
Then in triangle $FBC$ . The edge $BC$ leaves the end face at right angles, which puts the right angle at $B$ and makes $FC$ the hypotenuse.
Carry $FB$ forward as a squared value. Taking the root and squaring it again only undoes the work, and rounding halfway through would move the third significant figure.

Worked Solution [3 marks]

Rule - Pythagoras, used twice: in any right-angled triangle

h^2 = a^2 + b^2

, where

h

is the hypotenuse. Across a face it gives the face diagonal; in the triangle standing on that diagonal it gives the edge that leaves the face.

Step 1: The diagonal

FB

across the end face

FB^2 = AF^2 + AB^2

7^2 + 9^2 = 49 + 81 = 130

(Reason: Angle

FAB

is a right angle, because

AF

is a vertical edge of the cuboid and

AB

is a horizontal one. Stop at

130

: the next step wants

FB^2

, not

FB

Step 2: Triangle

FBC

, right-angled at

B

FC^2 = FB^2 + BC^2

18^2 - 130 = 324 - 130 = 194

(Reason: The edge

BC

is perpendicular to the whole end face

ABGF

, so it is perpendicular to

FB

as well. That places the right angle at

B

and makes

FC

the hypotenuse, so it is the two shorter squares that add to give the longest one.)

Step 3: Square root, then round

BC = \sqrt{194} = 13.9284\ldots

BC = 13.9

(Reason: Take the positive root: a length cannot be negative. The fourth significant figure is a

2

, so the third one stays as it is and

13.9284\ldots

rounds to

13.9

BC = 13.9

Verification

Check 1: Put the answer back into three-dimensional Pythagoras,

FC^2 = AB^2 + AF^2 + BC^2

. The three edge squares must rebuild

18^2

81 + 49 + 194 = 324

, and

18^2 = 324

Check 2: Test the rounding rather than the arithmetic. Anything that rounds to

13.9

3

significant figures lies between

13.85

and

13.95

, so

194

has to sit between their squares.

13.85^2 = 191.8225

and

13.95^2 = 194.6025

, and

194

lies between the two

Check 3: Reach the same place along a different pair of triangles, through the base diagonal

AC

. Triangle

FAC

is right-angled at

A

, so

AC^2 = FC^2 - AF^2

; then triangle

ABC

is right-angled at

B

, so

BC^2 = AC^2 - AB^2

324 - 49 = 275

, then

275 - 81 = 194

, the same square as before

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(AC^2 =) \; 18^2 - 7^2 \; (= 275)$ or $(AC =) \; \sqrt{18^2 - 7^2} \; (= \sqrt{275} \text{ or } 5\sqrt{11} \text{ or } 16.5(831\ldots))$ or $(FB^2 =) \; 9^2 + 7^2 \; (= 130)$ or $(FB =) \; \sqrt{9^2 + 7^2} \; (= \sqrt{130} \text{ or } 11.4(017\ldots))$ or $(GC^2 =) \; 18^2 - 9^2 \; (= 243)$ or $(GC =) \; \sqrt{18^2 - 9^2} \; (= \sqrt{243} \text{ or } 9\sqrt{3} \text{ or } 15.5(884\ldots))$ or $18^2 = (BC)^2 + 7^2 + 9^2$ oe	M1	for method to find $AC^2$ or $AC$ or $FB^2$ or $FB$ or $GC^2$ or $GC$ or for a correct equation using $BC^2$ and $18$ and $7$ and $9$ Other longer ways to find $AC$ , $FB$ , $GC$ may be used but must be a complete method, eg $\angle FCA = \sin^{-1}\left(\dfrac{7}{18}\right) \; (= 22.88\ldots)$ and $AC = \dfrac{7}{\tan 22.88\ldots}$ , where the printed scheme's quotation marks round $22.88\ldots$ mean the candidate's own angle may be used.	✓
eg $275 - 9^2 \; (= 194)$ or $16.5\ldots^2 - 9^2 \; (= 194)$ or $18^2 - 130 \; (= 194)$ or $18^2 - 11.4\ldots^2 \; (= 194)$ or $243 - 7^2 \; (= 194)$ or $15.5\ldots^2 - 7^2 \; (= 194)$ or $18^2 - 7^2 - 9^2 \; (= 194)$ or $\angle FCB = \sin^{-1}\left(\dfrac{11.4\ldots}{18}\right) \; (= 39.3(036\ldots))$ and $\cos 39.3\ldots = \dfrac{BC}{18}$ or $\tan 39.3\ldots = \dfrac{11.4\ldots}{BC}$ oe Every value the printed scheme sets in quotation marks - $275$ , $130$ , $243$ , $16.5\ldots$ , $11.4\ldots$ , $15.5\ldots$ and $39.3\ldots$ - may be the candidate's own value carried through from the first mark.	M1	for a complete method to find $BC^2$ Other longer ways to find $BC$ may be used but must be a complete method, leading to a trigonometric equation in $BC$ .	✓
$13.9$ A correct answer scores full marks, unless it comes from obviously incorrect working.	A1	accept $13.8$ to $14$	✓

Full marks: 3/3

A curve has equation $y = f(x)$ and it has exactly one turning point.
The coordinates of that turning point are $(-6, 9)$

(a) Write down the coordinates of the turning point of the curve with equation $y = f(3x)$ [1 mark]

The curve $C$ , with equation $y = g(x)$ , is transformed to give the curve $S$ , with equation $y = g(x + a) + b$
Under this transformation the point $(4, -5)$ on $C$ is mapped to the point $(1, -16)$ on $S$

(b) Write down the value of $a$ and the value of $b$ [2 marks]

(a)a =b =

[Total 3 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

y = f(x)

has exactly one turning point, and it is at

(-6, 9)

The curve

C

y = g(x)

, is transformed into the curve

S

y = g(x + a) + b

The point

(4, -5)

C

is mapped to the point

(1, -16)

S

Find

(a) the coordinates of the turning point on

y = f(3x)

(b) the value of

a

and the value of

b

Plan the Solution

Read $f(3x)$ as a change to the INPUT of $f$ . The curve turns when that input is $-6$ , so solve $3x = -6$ .
A change inside the bracket moves the curve sideways only, so the $y$ -coordinate of the turning point is untouched.
For part (b), write the image of $(x, y)$ under $y = g(x + a) + b$ as $(x - a, y + b)$ , then match it against the two given points one coordinate at a time.
Both parts say "write down", so no method is demanded - but knowing which way each transformation goes is what keeps the signs right.

Worked Solution [3 marks]

Rule - Function transformations:

y = f(ax)

divides every

x

-coordinate by

a

and leaves the

y

-coordinate unchanged;

y = g(x + a) + b

moves every point

a

to the left and

b

up.

Step 1: (a) Find the new x-coordinate

3x = -6

x = \dfrac{-6}{3} = -2

(Reason: The curve

y = f(3x)

takes at

x

the value that

y = f(x)

takes at

3x

, so it turns where the input to

f

-6

. Divide by

3

Step 2: (a) The y-coordinate does not move

y = 9

(-2, 9)

(Reason: Replacing

x

3x

changes only the input to

f

, never its output, so the height of the turning point stays at

9

. This is a horizontal stretch of scale factor

\dfrac{1}{3}

Step 3: (b) Say what the transformation does to one point

x_{S} = x_{C} - a

y_{S} = y_{C} + b

(Reason:

g(x + a)

slides the curve

a

units in the negative

x

-direction, and the

+ b

slides it

b

units up. So a point at

(x, y)

C

lands at

(x - a, y + b)

S

Step 4: (b) Compare the x-coordinates

4 - a = 1

a = 4 - 1 = 3

(Reason: The point

(4, -5)

C

is mapped to

(1, -16)

S

, so the

4

has to become a

1

Step 5: (b) Compare the y-coordinates

-5 + b = -16

b = -16 + 5 = -11

(Reason: The same point drops from

-5

-16

, so

b

comes out negative: the curve has moved

11

down as well as

3

to the left.)

(a)

(-2, 9)

(b)

a = 3

and

b = -11

Verification

Check 1: Put

x = -2

into

f(3x)

. The input to

f

becomes

3 \times (-2) = -6

, which is exactly where

f

turns.

f(3 \times (-2)) = f(-6) = 9

, so the turning point is

(-2, 9)

Check 2: Reach the same point a different way: read

y = f(3x)

as a horizontal stretch of scale factor

\dfrac{1}{3}

and stretch the

x

-coordinate directly. A stretch of

\dfrac{1}{3}

must pull the point towards the

y

-axis, so the answer has to be closer to

0

than

-6

is.

-6 \times \dfrac{1}{3} = -2

, and the

y

-coordinate is unchanged at

9

Check 3: Put

a = 3

and

b = -11

back into

y = g(x + a) + b

and test it at

x = 1

. The point

(4, -5)

lying on

C

tells us

g(4) = -5

g(1 + 3) - 11 = -5 - 11 = -16

, which is the

y

-coordinate on

S

Check 4: Read the answer as a translation -

3

to the left and

11

down - and move the point on the grid instead of using algebra.

(4 - 3, -5 - 11) = (1, -16)

, the point given on

S

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$(-2, 9)$	✓
(b)	B1	$a = 3$	✓
(b)	B1	$b = -11$	✓

Full marks: 3/3

Solve these simultaneous equations.

$x^2 + y^2 = 41$
$2x + y = 3$

You must show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

The circle

x^2 + y^2 = 41

The straight line

2x + y = 3

One equation is quadratic and the other is linear, so this is a straight line meeting a circle.

Find

The pairs of values of

x

and

y

that satisfy both equations at the same time. Substituting the line into the circle leaves a quadratic, so expect

2

solution pairs.

Plan the Solution

Rearrange the linear equation to make $y$ the subject, because $y$ already has a coefficient of $1$ .
Substitute that expression for $y$ into $x^2 + y^2 = 41$ so that only $x$ is left.
Expand the bracket, collect everything on one side, and solve the three term quadratic.
Put each value of $x$ back into the linear equation to get its partner value of $y$ .

Worked Solution [5 marks]

Rule - Substitution (line into curve): make one letter the subject of the linear equation, substitute it into the quadratic equation, solve the quadratic that results, then find the partner value for each root.

Step 1: make

y

the subject of the linear equation

2x + y = 3

y = 3 - 2x

(Reason:

y

has a coefficient of

1

, so rearranging costs no fractions and keeps the substitution clean.)

Step 2: substitute into

x^2 + y^2 = 41

and expand

x^2 + (3 - 2x)^2 = 41

x^2 + 9 - 12x + 4x^2 = 41

(Reason: Squaring a bracket needs every term.

(3 - 2x)^2

has a middle term of

-12x

, and that is the term most often lost.)

Step 3: collect the terms into a three term quadratic

5x^2 - 12x + 9 = 41

5x^2 - 12x - 32 = 0

(Reason: A quadratic can only be factorised, or put through the formula, once every term is on one side and the other side is

0

Step 4: solve the quadratic

(5x + 8)(x - 4) = 0

x = -\dfrac{8}{5} \text{ or } x = 4

(Reason: The brackets multiply back out to

5x^2 - 12x - 32

, and a product is zero only when one of its factors is zero.)

Step 5: find the value of

y

that goes with

x = -\dfrac{8}{5}

y = 3 - 2 \times \left(-\dfrac{8}{5}\right) = 3 + \dfrac{16}{5} = \dfrac{31}{5}

(Reason: Substituting into the linear equation is quicker than substituting into the circle, and it gives one value of

y

rather than two.)

Step 6: find the value of

y

that goes with

x = 4

y = 3 - 2 \times 4 = 3 - 8 = -5

(Reason: Each root of the quadratic carries its own partner value, so the two roots give two separate solution pairs.)

x = -\dfrac{8}{5}

y = \dfrac{31}{5}

x = 4

y = -5

Verification

Check 1 - the first pair: Put

x = -\dfrac{8}{5}

and

y = \dfrac{31}{5}

into both of the original equations.

\dfrac{64}{25} + \dfrac{961}{25} = \dfrac{1025}{25} = 41

and

-\dfrac{16}{5} + \dfrac{31}{5} = \dfrac{15}{5} = 3

Check 2 - the second pair: Put

x = 4

and

y = -5

into both of the original equations.

4^2 + (-5)^2 = 16 + 25 = 41

and

2 \times 4 + (-5) = 8 - 5 = 3

Check 3 - the roots of the quadratic: For

5x^2 - 12x - 32 = 0

the two roots must add to

\dfrac{12}{5}

and multiply to

-\dfrac{32}{5}

-\dfrac{8}{5} + 4 = \dfrac{12}{5}

and

-\dfrac{8}{5} \times 4 = -\dfrac{32}{5}

Check 4 - eliminate the other letter instead: Substitute

x = \dfrac{3 - y}{2}

instead. The quadratic in

y

must give the same two values of

y

5y^2 - 6y - 155 = 0

factorises as

(5y - 31)(y + 5) = 0

, giving

y = \dfrac{31}{5}

and

y = -5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$x^2 + (-2x + 3)^2 = 41$ or $\left(\dfrac{-y + 3}{2}\right)^2 + y^2 = 41$	M1	Substitution of $y = \pm 2x \pm 3$ (or $x = \dfrac{\pm y \pm 3}{2}$ ) into $x^2 + y^2 = 41$ to obtain an equation in $x$ only (or $y$ only).	✓
eg $5x^2 - 12x - 32 \, (= 0)$ oe or $5x^2 - 12x = 32$ ; or eg $5y^2 - 6y - 155 \, (= 0)$ or $5y^2 - 6y = 155$	M1ft	Dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of $ax^2 + bx + c \, (= 0)$ where at least $2$ coefficients ( $a$ or $b$ or $c$ ) are correct.	✓
$(5x + 8)(x - 4) \, (= 0)$ or $(x =) \dfrac{12 \pm \sqrt{(-12)^2 - 4 \times 5 \times (-32)}}{2 \times 5}$ or $5\left[\left(x - \dfrac{6}{5}\right)^2 - \left(\dfrac{6}{5}\right)^2\right] - 32 \, (= 0)$ , which should give $(x =) -\dfrac{8}{5}, \, 4$ ; or the same three methods on the $y$ quadratic, eg $(5y - 31)(y + 5) \, (= 0)$ , which should give $(y =) \dfrac{31}{5}, \, -5$	M1ft	Dep on M1 method to solve their three term quadratic using any correct method (allow one sign error and some simplification - allow as far as eg $\dfrac{12 \pm \sqrt{144 + 640}}{10}$ or $\dfrac{6 \pm \sqrt{36 + 3100}}{10}$ , or if factorising allow brackets which expanded give $2$ out of $3$ terms correct) or correct values for $x$ or correct values for $y$ .	✓
eg $2 \times 4 + y = 3$ and $2 \times -\dfrac{8}{5} + y = 3$ ; or eg $2x + \dfrac{31}{5} = 3$ and $2x - 5 = 3$	M1ft	Dep on previous M1 for substituting their $2$ found values of $x$ or $y$ into one of the two given equations or their rearranged equation used in the substitution, or for one correct pair of values.	✓
$x = -\dfrac{8}{5}$ , $y = \dfrac{31}{5}$ , $x = 4$ , $y = -5$ . Working required: if the correct answers come from incorrectly using $y = 2x - 3$ oe, award M4A0.	A1	oe, dep on M2, for all $4$ values (allow coordinates).	✓
If they find the values of $y$ but think they are the values of $x$ , then the maximum mark is $3$ .	Note	A guidance row printed with the scheme. It carries no mark of its own.	✓

Full marks: 5/5

The chords $PTR$ and $QTS$ of a circle cross at the point $T$ .

$PT = (3y - 5)$ cm, $QT = (y + 3)$ cm, $RT = 4x$ cm and $ST = (x + 2)$ cm.

Work out an expression for $y$ in terms of $x$ . [5 marks]

y =

[Total 5 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

Two chords of the same circle,

PTR

and

QTS

, crossing at

T

inside the circle.

PT = (3y - 5)

cm and

RT = 4x

cm are the two parts of

PTR

QT = (y + 3)

cm and

ST = (x + 2)

cm are the two parts of

QTS

Find

An expression for

y

in terms of

x

. Expect an algebraic fraction:

y

sits inside two of the four lengths, so it can only be freed by factorising and then dividing.

Plan the Solution

Turn the four lengths into one equation using the intersecting chords rule.
Expand both sides so that every term is written out on its own.
Gather every term containing $y$ on one side and every term without it on the other.
Factorise so that $y$ appears exactly once, then divide by the bracket beside it.

Worked Solution [5 marks]

Rule - Intersecting chords: when two chords of a circle cross at a point inside it, the two parts of one chord have the same product as the two parts of the other, so

PT \times RT = QT \times ST

Step 1: use the intersecting chords rule

PT \times RT = QT \times ST

(3y - 5) \times 4x = (y + 3)(x + 2)

(Reason: The two chords cross at

T

, so the two parts of

PTR

multiply to give the same value as the two parts of

QTS

. That single equation is what carries both unknowns.)

Step 2: expand both sides

12xy - 20x = xy + 2y + 3x + 6

(Reason: Multiply

4x

through the first bracket, and expand the pair of brackets on the right in the usual way. Nothing is collected yet - every term is simply written out.)

Step 3: collect the y terms on one side

12xy - xy - 2y = 3x + 20x + 6

11xy - 2y = 23x + 6

(Reason: Every term carrying a

y

goes to the left; every term without one goes to the right. Note that

12xy

and

xy

can be subtracted because they are like terms.)

Step 4: factorise so that y appears once

y(11x - 2) = 23x + 6

(Reason:

y

is a common factor of

11xy

and

-2y

, so it comes outside a bracket. This is the step the whole question is built around: until

y

appears once, it cannot be made the subject.)

Step 5: divide by the bracket

y = \dfrac{23x + 6}{11x - 2}

(Reason: Dividing both sides by

(11x - 2)

leaves

y

on its own, and the answer is a single algebraic fraction in

x

y = \dfrac{23x + 6}{11x - 2}

Verification

Check 1: Put

x = 2

into the answer, which gives

\dfrac{46 + 6}{22 - 2} = \dfrac{13}{5}

, then work out the four lengths and compare the two products.

PT \times RT = \dfrac{14}{5} \times 8 = \dfrac{112}{5}

and

QT \times ST = \dfrac{28}{5} \times 4 = \dfrac{112}{5}

Check 2: Work backwards. Multiply the answer by its own denominator and put every term back where it came from; the expansion in Step 2 should reappear.

y(11x - 2) = 23x + 6

opens out to

11xy - 2y = 23x + 6

, and moving the terms back gives

12xy - 20x = xy + 2y + 3x + 6

, which is exactly

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

expanded.

Check 3: Test a second value, in case

x = 2

was lucky. At

x = 1

the answer gives

y = \dfrac{29}{9}

, and all four lengths come out positive, as lengths must.

\dfrac{14}{3} \times 4 = \dfrac{56}{3}

and

\dfrac{56}{9} \times 3 = \dfrac{56}{3}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(y + 3)(x + 2) = 4x(3y - 5)$ or $\dfrac{3y - 5}{x + 2} = \dfrac{y + 3}{4x}$ or $\dfrac{x + 2}{3y - 5} = \dfrac{4x}{y + 3}$ or $\dfrac{3y - 5}{y + 3} = \dfrac{x + 2}{4x}$ or $\dfrac{4x}{x + 2} = \dfrac{y + 3}{3y - 5}$	M1	for correct use of intersecting chords theorem to form an equation	✓
eg $xy + 2y + 3x + 6 = 12xy - 20x$ oe	M1	for expanding the brackets or for removing the fractions and expanding the brackets, allow one error in one term. We can ft $4x(y + 3) = (x + 2)(3y - 5)$ oe for the 2nd, 3rd and 4th method marks	✓
eg $20x + 3x + 6 = 12xy - xy - 2y$ oe or $23x + 6 = 11xy - 2y$	M1ft	dep on previous M1 for correctly collecting all the $y$ terms on one side and non- $y$ terms on the other side	✓
eg $23x + 6 = y(11x - 2)$ or $20x + 3x + 6 = y(12x - x - 2)$	M1ft	dep on 2nd M mark for factorising, for $y$ , an equation in the form $ax + b = cxy + dy$ (may not be simplified). The factorisation must be correct	✓
$\dfrac{23x + 6}{11x - 2}$ . Correct answer scores full marks (unless from obvious incorrect working)	A1	oe	✓

Full marks: 5/5

The opening $3$ terms of an arithmetic series are

$(2x + 5) \qquad (3y - 4) \qquad (4x - 2)$

in which $x$ and $y$ are constants.

The first $9$ terms of the series have a sum of $216$

Work out the value of $x$ and the value of $y$ .
You must show clear algebraic working. [6 marks]

x =y =

[Total 6 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

The first

3

terms of an arithmetic series are

(2x + 5)

(3y - 4)

and

(4x - 2)

The series is arithmetic, so consecutive terms differ by the same amount

d

The first

9

terms have a sum of

216

Find

The value of

x

. The value of

y

Plan the Solution

Two unknowns need two equations. The first comes from the definition of an arithmetic series: the gap from term $1$ to term $2$ equals the gap from term $2$ to term $3$ .
Terms $1$ and $3$ are two common differences apart, and both are written in $x$ only. So $d$ can be found with no $y$ in it at all, which is what keeps the second equation short.
Put that $a$ and that $d$ into the sum formula with $n = 9$ . Everything in it is now an $x$ , so it solves on its own.
Finish by putting $x$ back into the first equation to get $y$ .

Worked Solution [6 marks]

Rule - Arithmetic series: consecutive terms differ by a constant

d

, and the sum of the first

n

terms is

S_n = \dfrac{n}{2}\left[2a + (n - 1)d\right]

, where

a

is the first term.

Step 1: Use the common difference to link the two letters

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

3y - 2x - 9 = 4x - 3y + 2

6y - 6x = 11

(Reason: (Reason: in an arithmetic series every term is the one before it plus the same

d

, so term

2

minus term

1

must equal term

3

minus term

2

. Collecting the letters on the left gives the first of the two equations.))

Step 2: Write

d

in terms of

x

alone

2d = (4x - 2) - (2x + 5)

2d = 2x - 7

d = x - 3.5

(Reason: (Reason: term

3

is two steps along from term

1

, so the difference between them is

2d

and not

d

. Both of those terms are written in

x

, so the

y

never enters and

d

comes out in

x

on its own.))

Step 3: Build an equation from the sum of the first

9

terms

S_n = \dfrac{n}{2}\left[2a + (n - 1)d\right]

216 = \dfrac{9}{2}\left[2(2x + 5) + 8(x - 3.5)\right]

(Reason: (Reason: here

a = 2x + 5

is the first term,

d

is the common difference found in Step 2, and

n = 9

. The bracket now holds nothing but

x

, so this is one equation in one unknown.))

Step 4: Solve for

x

216 = \dfrac{9}{2}\left[4x + 10 + 8x - 28\right]

216 = \dfrac{9}{2}(12x - 18)

48 = 12x - 18

12x = 66

x = \dfrac{66}{12} = \dfrac{11}{2}

(Reason: (Reason: multiply out both brackets, tidy the bracket to

12x - 18

, then undo the

\dfrac{9}{2}

by multiplying both sides by

\dfrac{2}{9}

. The answer is left as a fraction because

\dfrac{11}{2}

is exact.))

Step 5: Substitute back to find

y

6y - 6\left(\dfrac{11}{2}\right) = 11

6y - 33 = 11

6y = 44

y = \dfrac{44}{6} = \dfrac{22}{3}

(Reason: (Reason: the equation from Step 1 is the only one still carrying a

y

, so it is the one to substitute into.

\dfrac{22}{3}

does not terminate as a decimal, which is why the fraction is the answer to give.))

x = \dfrac{11}{2}

y = \dfrac{22}{3}

Verification

Check 1: Put

x = \dfrac{11}{2}

and

y = \dfrac{22}{3}

back into the three printed terms and look at the gaps between them. The terms are

16

18

and

20

, so the common difference is

2

both times, and the series really is arithmetic.

Check 2: Add the

9

terms without the formula, by pairing the first with the last. The ninth term is

16 + 8 \times 2 = 32

, so the average term is

\dfrac{16 + 32}{2} = 24

9 \times 24 = 216

, which is the total the question states.

Check 3: An arithmetic series with an odd number of terms has its middle term equal to the mean, so the fifth term must be

\dfrac{216}{9} = 24

. Counting up from the first term, the fifth term is

16 + 4 \times 2 = 24

, so the two agree.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(d =) (3y - 4) - (2x + 5) \; (= 3y - 2x - 9)$ or $(2x + 5) + d = 3y - 4$ oe or $(d =) (4x - 2) - (3y - 4) \; (= 4x - 3y + 2)$ or $(3y - 4) + d = (4x - 2)$ or $(2d =) (4x - 2) - (2x + 5) \; (= 2x - 7)$ or $(2x + 5) + 2d = (4x - 2)$ oe	M1	For a correct expression or equation using the common difference. It may be in terms of $d$ for this mark, and an expression for $-d$ or $-2d$ is allowed.	✓
eg $216 = \dfrac{9}{2}\left[2(2x + 5) + (9 - 1)d\right]$ or the same equation with the candidate's own $3y - 2x - 9$ , $4x - 3y + 2$ or $x - 3.5$ written in place of $d$ or $216 = \dfrac{9}{2}\left[2(2(d + 3.5) + 5) + 8d\right]$	M1	For a correct equation for the sum of $9$ terms in $x$ and $d$ , or in terms of $x$ and $y$ , or in terms of $x$ , or in terms of $d$ . For $3y - 2x - 9$ allow $(3y - 4) - (2x + 5)$ , or allow the candidate's own incorrect simplification of it where that simplification is shown. For $4x - 3y + 2$ allow $(4x - 2) - (3y - 4)$ , or allow the candidate's own incorrect simplification of it where that simplification is shown. Similarly for the candidate's own $x - 3.5$ and the candidate's own $d + 3.5$ .	✓
Left-hand column: eg $6x - 6y = -11$ oe and $12y - 6x = 55$ oe or $6x - 6y = -11$ oe and $18x - 12y = 11$ oe or $2x - 2d = 7$ oe and $8d + 4x = 38$ oe Right-hand column: eg $d = x - 3.5$ oe and $48 = 12x - 18$ oe or $x = d + 3.5$ oe and $48 = 12d + 24$ oe or $d = 4.75 - 0.5x$ oe and $3x - 2 = 14.50$ oe One equation must come from the differences and one equation must come from the sum of the terms.	M2	Left-hand column: $2$ correct equations in terms of $x$ and $y$ in the form $px + qy = r$ oe, or $2$ correct equations in terms of $x$ and $d$ in the form $px + qd = r$ oe. Right-hand column: $2$ correct equations in terms of $x$ and $y$ where one is substituted into the other to get a correct equation in the form $px + q = r$ or $py + q = r$ , or $2$ correct equations in terms of $x$ and $d$ where one is substituted into the other to get a correct equation in the form $px + q = r$ or $pd + q = r$ . If not M2 then M1 for one correct equation in any of the required forms from the left-hand or the right-hand column.	✓
$x = \dfrac{11}{2}$	A1 dep on M2	Working is required. Or equivalent, for example $5.5$ . The printed scheme states this as A2 for both values together, with A1 for either value on its own, so this row and the next carry one of those two marks each.	✓
$y = \dfrac{22}{3}$	A1 dep on M2	Working is required. Or equivalent; allow $7.3(33\ldots)$ . The printed scheme states this as A2 for both values together, with A1 for either value on its own, so this row and the one above carry one of those two marks each.	✓
A correct pair of values with no algebra shown scores nothing here.	Note	The question demands clear algebraic working, and the printed scheme marks the final row "Working required". Trial and improvement that lands on $x = 5.5$ and $y = \dfrac{22}{3}$ earns no accuracy mark without the two equations behind it.	✓

Full marks: 6/6

The diagram shows two sectors of circles that meet at $C$ .

$BAC$ is a sector of a circle, centre $A$

$BCD$ is a sector of a circle, centre $C$

Angle $BAC = 40^\circ$
Angle $BCD = 130^\circ$
Area of the shaded segment $= 28 \text{ cm}^2$

Work out the length of the arc $BD$ .
Give your answer correct to $3$ significant figures. [6 marks]

[Total 6 marks]

Show solution & mark schemeHide solution & mark scheme

Question 25 - Exam Solution

Understanding the Question

Given

BAC

is a sector of a circle, centre

A

, with angle

BAC = 40^\circ

BCD

is a sector of a circle, centre

C

, with angle

BCD = 130^\circ

The shaded segment, between the chord

BC

and the arc

BC

, has area

28 \text{ cm}^2

Find

The length of the arc

BD

, correct to

3

significant figures

Plan the Solution

The shaded piece is a segment: it is what is left when triangle $ABC$ is cut out of sector $BAC$ .
Call the radius of the first circle $r$ . The sector and the triangle are both multiples of $r^2$ , so the area $28$ gives one equation in $r^2$ .
The second sector has centre $C$ , so its radius is $CB$ and not $r$ . Get $CB$ out of triangle $ABC$ .
The arc is then $\dfrac{130}{360}$ of the circumference of a circle of radius $CB$ .

Worked Solution [6 marks]

Rule - Sector, segment and arc: a sector of radius

r

with angle

\theta

has arc length

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

and area

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

. A triangle with two sides

r

and the angle

\theta

between them has area

\dfrac{1}{2} r^2 \sin\theta

. A segment is the sector minus that triangle.

Step 1: Write the shaded segment in terms of the radius

\dfrac{40}{360} \times \pi \times r^2 - \dfrac{1}{2} \times r^2 \times \sin 40^\circ = 28

(Reason: Both

AB

and

AC

are radii of the circle with centre

A

, so write each of them as

r

. Cutting triangle

ABC

away from sector

BAC

leaves exactly the shaded segment, so sector minus triangle is

28

Step 2: Take

r^2

out as a factor and solve

r^2 \left( \dfrac{40}{360} \times \pi - \dfrac{1}{2} \times \sin 40^\circ \right) = 28

\dfrac{40}{360} \times \pi - \dfrac{1}{2} \times \sin 40^\circ = 0.0276720\ldots

r^2 = \dfrac{28}{0.0276720\ldots} = 1011.85\ldots

r = \sqrt{1011.85\ldots} = 31.8096\ldots

(Reason: Every term carries

r^2

, so the bracket is a plain number the calculator works out once. Keep the unrounded value on the calculator - rounding

r

here would drift into the final answer.)

Step 3: Find

BC

, the radius of the second sector

BC^2 = 2 \times 1011.85\ldots - 2 \times 1011.85\ldots \times \cos 40^\circ

BC^2 = 473.4565\ldots

BC = \sqrt{473.4565\ldots} = 21.7590\ldots

(Reason: The second sector has centre

C

, so its radius is

CB

and not

r

. In triangle

ABC

the two sides from

A

are both

r

and the angle between them is

40^\circ

, so the cosine rule gives

BC

. Splitting the isosceles triangle in half gives the same value,

BC = 2r \sin 20^\circ

Step 4: Take

\dfrac{130}{360}

of the circumference

\text{arc } BD = \dfrac{130}{360} \times 2 \times \pi \times 21.759 = 49.3696\ldots

(Reason: The arc

BD

\dfrac{130}{360}

of the whole circumference of the circle with centre

C

, and that circle has radius

CB

Step 5: Round to

3

significant figures

\text{arc } BD = 49.4 \text{ cm (3 s.f.)}

(Reason: The first three significant figures are

4

9

and

3

, and the digit after them is

6

, so the

3

rounds up.)

49.4

Verification

Check 1: Put the unrounded

r^2 = 1011.85\ldots

back into the segment: work out the sector and the triangle separately, then subtract.

\dfrac{40}{360} \times \pi \times r^2 = 353.203

\dfrac{1}{2} \times r^2 \times \sin 40^\circ = 325.203

, and

353.203 - 325.203 = 28

, the area the question gives.

Check 2: Find

BC

a completely different way. Triangle

ABC

is isosceles, so its base angles are

\dfrac{180 - 40}{2} = 70

degrees each and the sine rule applies.

BC = \dfrac{r \times \sin 40^\circ}{\sin 70^\circ} = 21.7590\ldots

, the same value the cosine rule gave.

Check 3: Check the size a third way. A full circle of radius

21.7590\ldots

has circumference

136.72

, and

130^\circ

is a little over a third of a full turn.

\dfrac{130}{360} \times 136.72 = 49.37

, which agrees with the answer to the accuracy shown.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
e.g. $\dfrac{40}{360}\pi r^2 - \dfrac{1}{2} r^2 \sin 40 (= 28)$ oe or $\dfrac{40}{360}\pi r^2 = 28 + \dfrac{1}{2} r^2 \sin 40$ oe	M1	For a correct expression for the area of the shaded region. Allow $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$ . $\sin 40 = 0.64\ldots$	✓
$(r^2 =) 992$ to $1024$ , $(r =) 31.8(096\ldots)$	A1	Allow answers in the range $31.5$ to $32.0$	✓
e.g. $(BC^2 =) 2 \times 31.8^2 - 2 \times 31.8^2 \cos 40 (= 473.4\ldots)$ or $\dfrac{0.5BC}{31.8} = \sin 20$ or $\dfrac{BC}{\sin 40} = \dfrac{31.8}{\sin (70)}$	M1	For a correct first step to find $BC$ using their clearly identified radius, e.g. $r = \ldots$ or seen on the diagram. The printed scheme puts that $31.8$ in quotation marks, so the candidate may use their own radius here. NB $\dfrac{180 - 40}{2} = 70$ . $\sin 20 = 0.34\ldots$ , $\sin 70 = 0.93\ldots$ or $0.94$	✓
e.g. $(BC =) \sqrt{2 \times 31.8^2 - 2 \times 31.8^2 \cos 40} (= 21.7\ldots)$ or $(BC =) 2 \times 31.8 \sin 20 (= 21.7\ldots)$ or $BC = \dfrac{31.8 \sin 40}{\sin (70)} (= 21.7\ldots)$	M1 dep	Dep on the previous M1. For a complete method to find $BC$ . $\cos 40 = 0.76\ldots$ or $0.77$	✓
e.g. $\dfrac{130}{360} \times 2 \times \pi \times 21.7$	M1 dep	Dep on the previous M1. For a complete method to find the length of the arc $BD$ . The printed scheme puts that $21.7$ in quotation marks, so the candidate may use their own $BC$ .	✓
$49.4$	A1	Accept $48.9$ to $49.7$ . A correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 6/6

$R$ , $S$ and $T$ are three mathematically similar storage jars.

The volume of jar $S$ is $72.8\%$ more than the volume of jar $R$

The height of jar $R$ is $h \text{ cm}$
The height of jar $T$ is $6h \text{ cm}$

The surface area of jar $S$ is $A \text{ cm}^2$
The surface area of jar $T$ is $kA \text{ cm}^2$

Work out the value of $k$ [4 marks]

k =

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 26 - Exam Solution

Understanding the Question

Given

Three mathematically similar jars

R

S

and

T

The volume of

S

72.8\%

more than the volume of

R

Height of

R

h \text{ cm}

, height of

T

6h \text{ cm}

Surface area of

S

A \text{ cm}^2

, surface area of

T

kA \text{ cm}^2

Find

The value of

k

. Since

kA

compares with

A

, the value of

k

is the area scale factor from

S

T

- not the one from

R

T

Plan the Solution

Turn $72.8\%$ more into a volume scale factor from $R$ to $S$ .
Cube root it, because lengths are the cube root of volumes for similar solids.
Read the length scale factor from $R$ to $T$ straight off the heights.
Chain the two to get $S$ to $T$ , then square it, because areas scale by the square.

Worked Solution [4 marks]

Rule - similar solids: if the lengths are in the ratio

1 : n

, then the areas are in the ratio

1 : n^2

and the volumes are in the ratio

1 : n^3

Step 1: the volume scale factor from

R

S

1 + \dfrac{72.8}{100} = 1.728

(Reason: (Reason: a volume that is

72.8\%

more is

172.8\%

of the original, so every volume is multiplied by

1.728

.))

Step 2: cube root it for the length scale factor from

R

S

\sqrt[3]{1.728} = 1.2

(Reason: (Reason: volumes scale by the cube of the length scale factor, so the length scale factor is the cube root of

1.728

. The heights of

R

and

S

are therefore in the ratio

1 : 1.2

.))

Step 3: the length scale factor from

S

T

\dfrac{6}{1.2} = 5

(Reason: (Reason: the heights put

R

S

and

T

in the ratio

1 : 1.2 : 6

, so going from

S

T

multiplies every length by

5

.))

Step 4: square it, because

k

compares areas

5^2 = 25

(Reason: (Reason: areas scale by the square of the length scale factor, so the surface area of

T

25

times the surface area of

S

, giving

kA = 25A

.))

k = 25

Verification

Check 1 - cube the length factor back: If the lengths of

R

and

S

really are in the ratio

1 : 1.2

, then cubing

1.2

must return the volume factor the percentage gave.

1.2^3 = 1.728

Check 2 - do it entirely in volumes: The heights put

R

and

T

in the ratio

1 : 6

, so their volumes are in the ratio

1 : 216

. Dividing by the volume factor from

R

S

leaves the volume factor from

S

T

, whose cube root is the length factor.

\dfrac{216}{1.728} = 125

and

5^3 = 125

Check 3 - go through the areas instead: Areas from

R

S

scale by

1.2^2

, and areas from

R

T

scale by

6^2

. Dividing the second by the first jumps straight from

S

T

without ever naming a length factor.

\dfrac{36}{1.44} = 25

Mark Scheme Breakdown

Step	Mark	Description	Got it?
e.g. $\sqrt[3]{1.728}$ giving $1.2$ oe or lengths $R : S$ as $1 : 1.2$ oe, e.g. $5 : 6$ or lengths $R : S : T$ as $1 : 1.2 : 6$ oe, e.g. $5 : 6 : 30$ or volumes $S : T$ as $1.728 : 216$ , or $\dfrac{216}{1.728}$ giving $125$	M1	for method to find the scale factor between the heights of $R$ and $S$ or for a correct ratio for the lengths $R : S$ or for a correct ratio for the lengths $R : S : T$ or for a correct ratio or scale factor for the volumes $S : T$	✓
e.g. $\dfrac{6}{1.2}$ giving $5$ , or $\sqrt[3]{\dfrac{216}{1.728}}$ giving $5$ or lengths $S : T$ as $1 : 5$ , or as $\sqrt[3]{1.728} : \sqrt[3]{216}$ or areas $S : T$ as $1.2^2 : 6^2$ oe, or as $6^2 : 30^2$ oe or areas $R : S : T$ as $1 : 1.2^2 : 6^2$ oe, or as $5^2 : 6^2 : 30^2$ oe	M1	for method to find the scale factor between the heights of $S$ and $T$ or for a correct ratio of the heights of $S$ and $T$ in the form $1 : n$ or for a correct method to find the ratio for the areas $S : T$ follow through their ratio of lengths or for a correct method to find the ratio for the areas $R : S : T$ follow through their ratio of lengths (maybe seen as the two separate ratios of $R : S$ and $R : T$ )	✓
e.g. $5^2$ or areas $S : T$ as $1^2 : 5^2$ , or as $1 : 25$ or $k$ from $\dfrac{36}{1.44}$ oe	M1	for squaring the scale factor of the heights of $S$ and $T$ or for a correct ratio in the form $1 : n$ for the areas $S : T$ follow through their ratio of areas, may be implied by their final answer or for a correct calculation using the area ratio of $S : T$ to find the value of $k$	✓
Correct answer scores full marks (unless from obvious incorrect working) $k = 25$	A1	cao	✓

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 4MA1/1HR, Thursday 15 May 2025: Worked Solutions, Questions 20 to 26

Worked solutions, questions 20 to 26 of 26

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

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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