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Edexcel IGCSE 4MA1/2HR, Monday 3 June 2024: Worked Solutions, Questions 21 to 26

Sir Faraz Hassan

13 Aug 2026

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

Back to questions 11 to 20

This is the rest of the paper. Questions 1 to 20, 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 4MA1/2HR (Higher Tier), June 2024 series, sat Monday 3 June 2024 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

Download printable PDF

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

Worked solutions, questions 21 to 26 of 26

$ABCD$ is a square.

$A$ is the point $(-5, 2)$
$B$ is the point $(3, 5)$

Work out an equation of the line that passes through $B$ and $C$
Give your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers. [4 marks]

[Total 4 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

ABCD

is a square, so all four sides are the same length and each pair of adjacent sides meets at a right angle.

Two adjacent vertices are given:

A(-5, 2)

and

B(3, 5)

, so

AB

is one side of the square.

Find

An equation of the line through

B

and

C

, written in the form

ax + by + c = 0

with

a

b

and

c

integers.

BC

is the side next to

AB

, so the line

BC

is perpendicular to

AB

and it passes through

B

. A gradient and a point are all a line needs, so

C

itself never has to be found.

Plan the Solution

Work out the gradient of $AB$ from the two coordinates the question gives.
Turn that into the gradient of $BC$ with the perpendicular rule.
Substitute $B(3, 5)$ into $y = mx + c$ to find $c$ .
Multiply through by $3$ to clear the fraction, then collect every term on one side so that $a$ , $b$ and $c$ come out as integers.

Worked Solution [4 marks]

Rule - Perpendicular gradients: adjacent sides of a square meet at

90^\circ

, so their gradients multiply to

-1

. One gradient is therefore the negative reciprocal of the other: turn the fraction upside down and change its sign. Gradient from two points: divide the change in

y

by the change in

x

, taking both differences in the same order.

Step 1: Find the gradient of

AB

m_{AB} = \dfrac{5 - 2}{3 - (-5)} = \dfrac{3}{8}

(Reason: Subtract the

y

-coordinates and the

x

-coordinates in the same order,

B

take away

A

both times. The denominator is

3 - (-5) = 8

, not

2

- subtracting a negative adds.)

Step 2: Turn it into the gradient of

BC

\dfrac{3}{8} \times m_{BC} = -1

m_{BC} = -\dfrac{8}{3}

(Reason:

AB

and

BC

are adjacent sides of a square, so they meet at

B

at a right angle and their gradients multiply to

-1

. Invert the fraction and change the sign.)

Step 3: Use

B(3, 5)

to find the intercept

5 = -\dfrac{8}{3} \times 3 + c

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

c = 5 - (-8) = 13

(Reason: The line passes through

B

, so

x = 3

and

y = 5

must fit

y = mx + c

. The

3

cancels the denominator, which is why this question was set with an

x

-coordinate of

3

Step 4: Rearrange into the form the question asks for

y = -\dfrac{8}{3}x + 13

3y = -8x + 39

8x + 3y - 39 = 0

(Reason: Multiplying every term by

3

clears the fraction, and moving the

x

-term and the number across leaves

0

on the right. That gives

a = 8

b = 3

and

c = -39

, which are integers as required.)

8x + 3y - 39 = 0

Verification

Check 1: Put

B(3, 5)

back into the left-hand side. If

B

is on the line it must come out as

0

8 \times 3 + 3 \times 5 - 39 = 24 + 15 - 39 = 0

Check 2: Find

C

for real, as a test of the answer rather than a way to get it. Going from

A

B

8

across and

3

up; turning that through a right angle at

B

puts

C

at either

(6, -3)

(0, 13)

. Both are corners of a genuine square, and both must satisfy the equation - which is why the answer is the same either way.

8 \times 6 + 3 \times (-3) - 39 = 48 - 9 - 39 = 0

and

8 \times 0 + 3 \times 13 - 39 = 39 - 39 = 0

Check 3: Read the gradient back off the answer:

3y = -8x + 39

gives

y = -\dfrac{8}{3}x + 13

. Multiply that gradient by the gradient of

AB

- a right angle must give

-1

-\dfrac{8}{3} \times \dfrac{3}{8} = -1

Check 4: Confirm the shape really is a square and not just a rectangle, by squaring both side lengths with Pythagoras. Going

A

B

8

across and

3

up; going

B

(6, -3)

3

across and

8

down.

8^2 + 3^2 = 73

and

3^2 + 8^2 = 73

, so both sides have length

\sqrt{73}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method for the gradient of $AB$ , eg $\dfrac{5 - 2}{3 - (-5)} = \dfrac{3}{8} = 0.375$ oe, or from $2 = -5m + c$ and $5 = 3m + c$ leading to $(m =) \dfrac{3}{8}$ oe. Or the possible coordinates of $C$ : $(C =) (6, -3)$ or $(0, 13)$ .	M1	For a method to find the gradient of $AB$ , or for finding the possible coordinates of $C$ .	✓
eg $\left[\dfrac{3}{8}\right] \times m = -1$ oe, or their $(m =) -\dfrac{8}{3}$ oe, or $\dfrac{5 - (-3)}{3 - 6} = -\dfrac{8}{3}$ , or $\dfrac{5 - 13}{3 - 0} = -\dfrac{8}{3}$ , each of which is $-2.6(666\ldots)$ as a decimal.	M1ft	(indep) For finding the gradient of $BC$ . Allow the perpendicular gradient to be truncated or rounded to $1$ dp. $\left[\dfrac{3}{8}\right]$ means their gradient of $AB$ .	✓
eg $5 = -\dfrac{8}{3} \times 3 + c$ , or $c = 13$ , or $y = -\dfrac{8}{3}x + 13$ , or $y - 5 = -\dfrac{8}{3}(x - 3)$ oe, or $y - (-3) = -\dfrac{8}{3}(x - 6)$ oe, or $y - 13 = -\dfrac{8}{3}(x - 0)$ oe.	M1ft	(ft dep on the previous M1, on their own perpendicular gradient) For substitution to find $c$ , or for finding an equation for $BC$ . If students find the coordinates of $D$ , which are $(-2, -6)$ or $(-8, 10)$ , then allow this mark for $y - (-6) = -\dfrac{8}{3}(x - (-2))$ oe or $y - 10 = -\dfrac{8}{3}(x - (-8))$ oe.	✓
$8x + 3y - 39 = 0$	A1	oe, with $a$ , $b$ and $c$ integers, eg $16x + 6y - 78 = 0$ or $-8x - 3y + 39 = 0$ or $3y = -8x + 39$ . Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 4/4

Solve the simultaneous equations
$x^2 + y^2 = y + 11$
$y = 3x - 1$
You must show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

x^2 + y^2 = y + 11

, a curve (it is a circle)

y = 3x - 1

, a straight line

Clear algebraic working is asked for, so the values on their own would earn nothing.

Find

Every pair of values

x

and

y

that fits both equations at the same time. A quadratic appears on the way, so expect two solution pairs.

Plan the Solution

The line already gives $y$ in terms of $x$ , so substitute it into the curve equation.
Expand, then collect every term on one side to leave a three term quadratic in $x$ alone.
Solve that quadratic by factorising; the quadratic formula would give the same two values.
Put each value of $x$ back into $y = 3x - 1$ to find its partner value of $y$ .
Keep each pair together, then test both pairs in the original equations.

Worked Solution [5 marks]

Rule, substitution (line into curve): where a line

y = mx + c

meets a curve, replace

y

in the curve equation by

mx + c

, solve the quadratic in

x

that follows, then use the line again to find each

y

Step 1: Substitute the line into the curve

x^2 + (3x - 1)^2 = (3x - 1) + 11

(Reason: The line gives

y

in terms of

x

, so every

y

in the curve equation can be replaced by

3x - 1

. That leaves one equation in

x

only, which is what earns the first method mark.)

Step 2: Expand the squared bracket

(3x - 1)^2 = 9x^2 - 6x + 1

x^2 + 9x^2 - 6x + 1 = 3x + 10

(Reason:

(3x - 1)^2

means

(3x - 1)(3x - 1)

, which gives

9x^2

, two lots of

-3x

and

+1

. Squaring each term separately is the classic slip here. On the right,

-1 + 11 = 10

Step 3: Collect the terms on one side

10x^2 - 6x + 1 - 3x - 10 = 0

10x^2 - 9x - 9 = 0

(Reason: Take everything across so the equation reads

ax^2 + bx + c = 0

. Here

x^2 + 9x^2 = 10x^2

-6x - 3x = -9x

and

1 - 10 = -9

Step 4: Factorise and solve for

x

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

5x + 3 = 0 \implies x = -\dfrac{3}{5} = -0.6

2x - 3 = 0 \implies x = \dfrac{3}{2} = 1.5

(Reason: Two brackets multiply to zero only when one of them is zero, so each bracket gives a value. Checking the factorisation:

5x \times 2x = 10x^2

3 \times (-3) = -9

, and the middle terms are

-15x + 6x = -9x

. The quadratic formula would give

\dfrac{9 \pm \sqrt{81 + 360}}{20} = \dfrac{9 \pm 21}{20}

, the same two values.)

Step 5: Find the matching

y

for each

x

y = 3 \times (-0.6) - 1 = -2.8

y = 3 \times 1.5 - 1 = 3.5

(Reason: Each value of

x

has its own

y

. Substitute into the line rather than the curve: it is linear, so it produces one value and cannot introduce an extra one.)

Step 6: Pair the values correctly

x = -0.6, \; y = -2.8

x = 1.5, \; y = 3.5

(Reason: The pairs must stay together:

x = -0.6

belongs with

y = -2.8

, not with

y = 3.5

. The final mark is for four correct values, correctly labelled, or written as the two coordinate pairs.)

x = -0.6, \; y = -2.8

x = 1.5, \; y = 3.5

Verification

Check 1, the pair $(-0.6, -2.8)$ : Put both values into the curve equation

x^2 + y^2 = y + 11

, working each side out separately, then test the line as well.

0.36 + 7.84 = 8.2

and

-2.8 + 11 = 8.2

, so the two sides agree, and the line gives

3 \times (-0.6) - 1 = -2.8

Check 2, the pair $(1.5, 3.5)$ : The same test on the second pair:

x^2 + y^2

on the left,

y + 11

on the right, then the line.

2.25 + 12.25 = 14.5

and

3.5 + 11 = 14.5

, so this pair fits too, and the line gives

3 \times 1.5 - 1 = 3.5

Check 3, the two roots against the quadratic: A quadratic

ax^2 + bx + c = 0

has roots that add to

-\dfrac{b}{a}

and multiply to

\dfrac{c}{a}

. For

10x^2 - 9x - 9 = 0

those are

\dfrac{9}{10}

and

-\dfrac{9}{10}

-0.6 + 1.5 = 0.9

and

-0.6 \times 1.5 = -0.9

, so the two values are exactly the roots of that quadratic, with nothing missing and nothing spare.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$x^2 + (3x - 1)^2 = 3x - 1 + 11$	M1	For substituting $y = 3x - 1$ into $x^2 + y^2 = y + 11$ to obtain an equation in $x$ only. Substituting $x = \dfrac{\pm y \pm 1}{3}$ to obtain an equation in $y$ only scores the same mark, with either arrangement of the two signs.	✓
$10x^2 - 9x - 9 = 0$	M1 dep	Dependent on the previous M1. For multiplying out and collecting terms to form a three term quadratic in any form of $ax^2 + bx + c = 0$ , with at least two of $a$ , $b$ , $c$ correct. Or equivalent, such as $10x^2 - 9x = 9$ . Working in $y$ instead, the quadratic is $10y^2 - 7y - 98 = 0$ .	✓
$(5x + 3)(2x - 3) = 0$	M1ft	Follow through, dependent on the first M1. For solving their three term quadratic by any correct method: factorising, completing the square, or the formula, allowing one sign error and some simplification, as far as $\dfrac{9 \pm \sqrt{81 + 360}}{20}$ or $\dfrac{7 \pm \sqrt{49 + 3920}}{20}$ . If factorising, brackets which expand to give two of the three terms correct are enough. Correct values for $x$ or for $y$ with no method shown also score it, and the labels may be the wrong way round for this mark only.	✓
$y = 3 \times (-0.6) - 1 = -2.8$ and $y = 3 \times 1.5 - 1 = 3.5$	M1ft	Follow through, dependent on the previous M1. For substituting their two found values of $x$ (or of $y$ ) into either of the two given equations, or for fully correct values of the other variable.	✓
$x = -0.6, \; y = -2.8$ and $x = 1.5, \; y = 3.5$	A1 dep on M2	All four values correct and correctly labelled, or shown correctly as the coordinate pairs $(-0.6, -2.8)$ and $(1.5, 3.5)$ . Or equivalent, so $-\dfrac{3}{5}$ , $-\dfrac{14}{5}$ , $\dfrac{3}{2}$ and $\dfrac{7}{2}$ are accepted.	✓
Working required	Note	The four values on their own score nothing. The question asks for clear algebraic working, so the substitution and the quadratic must be seen.	✓

Full marks: 5/5

The curve $C$ has equation $y = \text{f}(x)$
The minimum point of $C$ has coordinates $(6, -3)$

Write down the coordinates of the minimum point of the curve with equation

(i) $y = \text{f}(x) + 10$ [1 mark]

(ii) $y = \text{f}(3x)$ [1 mark]

(i)(ii)

[Total 2 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

The curve

C

has equation

y = \text{f}(x)

The minimum point of

C

is at

(6, -3)

No formula for

\text{f}

is given, so each answer has to come from the transformation alone.

Find

The minimum point of

y = \text{f}(x) + 10

. The minimum point of

y = \text{f}(3x)

Plan the Solution

Read each new equation as a change to $y = \text{f}(x)$ and decide whether it acts on the output of $\text{f}$ or on the input.
A change outside $\text{f}$ moves the curve vertically, so only the $y$ -coordinate of the minimum point changes.
A change inside $\text{f}$ moves the curve horizontally, so only the $x$ -coordinate of the minimum point changes.
Both parts are write-down marks, so each is one mark for the pair of coordinates and no working is required.

Worked Solution [2 marks]

Rule - Transforming

y = \text{f}(x)

: a constant added outside the function,

\text{f}(x) + a

, translates the curve

a

units up, so a point

(p, q)

becomes

(p, q + a)

. A multiplier applied to

x

inside the function,

\text{f}(ax)

, is a horizontal stretch of scale factor

\dfrac{1}{a}

about the

y

-axis, so

(p, q)

becomes

\left( \dfrac{p}{a}, q \right)

Step 1: (i) the

x

-coordinate does not move

y = \text{f}(x) + 10

(Reason: The

10

is added after

\text{f}

has produced its output, so the input is untouched. The whole curve is translated straight up, and the minimum stays above

x = 6

Step 2: (i) add

10

to the

y

-coordinate

y = -3 + 10 = 7

(6, 7)

(Reason: Every output of

\text{f}

gains

10

, so the least output

-3

becomes

7

. The lowest point of the curve is still the lowest point after it is lifted.)

Step 3: (ii) find the

x

that feeds

6

into

\text{f}

y = \text{f}(3x)

3x = 6

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

(Reason: The new curve does at

x

whatever the old curve does at

3x

. The old curve was at its minimum when its input was

6

, so the new curve is at its minimum where

3x = 6

Step 4: (ii) the

y

-coordinate does not move

y = -3

(2, -3)

(Reason: Nothing has been added to the output of

\text{f}

and nothing multiplies it, so every

y

-value survives unchanged. The curve is squashed sideways towards the

y

-axis, which slides the minimum along but never up or down.)

(i)

(6, 7)

(ii)

(2, -3)

Verification

Check 1: Use a curve that really does have its minimum at

(6, -3)

, say

y = (x - 6)^2 - 3

. Adding

10

gives

y = (x - 6)^2 + 7

, and a square is never negative, so the smallest value is

7

x = 6

. Minimum

(6, 7)

, which is the answer to part (i).

Check 2: Replace

x

with

3x

in the same model curve to get

y = (3x - 6)^2 - 3

. The bracket is zero when

x = 2

and positive everywhere else, so that is where the curve bottoms out, at

y = -3

. Minimum

(2, -3)

, which is the answer to part (ii).

Check 3: Sanity-check the two directions. Adding

10

lifts the curve, so the minimum

y

must rise. Replacing

x

with

3x

squashes the curve towards the

y

-axis by a factor of

3

, so the minimum

x

must end up closer to

0

, not further from it.

7 > -3

and

2 < 6

, so both coordinates moved the way the transformations force them to.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) $(6, 7)$	B1	The coordinates of the minimum point of $y = \text{f}(x) + 10$ , written as a pair. No working is required for this mark.	✓
(ii) $(2, -3)$	B1	The coordinates of the minimum point of $y = \text{f}(3x)$ , written as a pair. No working is required for this mark.	✓

Full marks: 2/2

The diagram shows a solid, $S$ , formed by joining a cone to a hemisphere.

The circular face of the cone and the flat face of the hemisphere have the same centre.

The radius of the circular face of the cone is $x$ cm, and it is equal to the radius of the hemisphere.

The total height of $S$ is $4$ times the radius of the hemisphere.

A different sphere has radius $kx$ cm.
The volume of this sphere is $12.5$ times the volume of $S$

(a) Work out the value of $k$ [4 marks]

A solid, $T$ , is mathematically similar to solid $S$

The volume of $T$ is $512$ times the volume of $S$

The total surface area of $T$ is $d$ times the total surface area of $S$

(b) Find the value of $d$ [1 mark]

(a) k =(b) d =

[Total 5 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

Solid

S

is a cone with a hemisphere on top, joined along one circular face of radius

x

cm.

The total height of

S

4x

cm.

A sphere of radius

kx

cm has volume

12.5

times the volume of

S

Solid

T

is mathematically similar to

S

, with volume

512

times the volume of

S

Find

(a) The value of

k

. (b) The value of

d

, the number the total surface area of

S

is multiplied by to give the total surface area of

T

Plan the Solution

A hemisphere stands its own radius above its flat face, so it takes $x$ of the total height and leaves the rest to the cone.
Write the volume of the cone and the volume of the hemisphere as multiples of $\pi x^3$ , then add them.
Set the sphere's volume equal to $12.5$ times that total. Both $\pi$ and $x^3$ cancel, leaving a number for $k^3$ .
For part (b) no volumes are needed: cube root the volume scale factor to get the length scale factor, then square it for the areas.

Worked Solution [5 marks]

Rule - Volumes, and scale factors between similar solids: a cone is

\dfrac{1}{3}\pi r^2 h

, a sphere is

\dfrac{4}{3}\pi r^3

, and a hemisphere is half of that. If the lengths of two similar solids are in the ratio

n : 1

, their areas are in the ratio

n^2 : 1

and their volumes in the ratio

n^3 : 1

Step 1: Split the total height of

S

h = 4x - x = 3x

(Reason: The hemisphere rises exactly its own radius,

x

cm, above the circular face it shares with the cone. Taking that off the total height

4x

leaves

3x

for the height

h

of the cone.)

Step 2: The volume of the cone

\dfrac{1}{3}\pi x^2 \times 3x = \pi x^3

(Reason: The cone has base radius

x

and height

3x

, and the

3

cancels the

\dfrac{1}{3}

, so the cone is worth exactly

\pi x^3

Step 3: The volume of the hemisphere, and of

S

\dfrac{1}{2} \times \dfrac{4}{3}\pi x^3 = \dfrac{2}{3}\pi x^3

V_{S} = \pi x^3 + \dfrac{2}{3}\pi x^3 = \dfrac{5}{3}\pi x^3

(Reason: A hemisphere is half a sphere of the same radius, so halve

\dfrac{4}{3}\pi x^3

. Adding the two pieces writes the whole of

S

as one multiple of

\pi x^3

Step 4: Set the sphere against

S

\dfrac{4}{3}\pi (kx)^3 = 12.5 \times \dfrac{5}{3}\pi x^3

\dfrac{4}{3}k^3 \pi x^3 = \dfrac{125}{6}\pi x^3

(Reason: The sphere has radius

kx

, so cubing the bracket gives

k^3x^3

. Keep the brackets until they are expanded - dropping them turns

(kx)^3

into

kx^3

, which is a different thing.)

Step 5: Cancel, then cube root

k^3 = \dfrac{125}{6} \times \dfrac{3}{4} = \dfrac{125}{8}

k = \sqrt[3]{\dfrac{125}{8}} = \dfrac{5}{2} = 2.5

(Reason: Every term carries

\pi x^3

, so it divides out and the unknown

x

never has to be found. Both

125

and

8

are cubes, so the cube root is exact.)

Step 6: Part (b), from the volume scale factor

\sqrt[3]{512} = 8

8^2 = 64

(Reason: Volume scales by the cube of the length scale factor, so the lengths of

T

are

8

times those of

S

. Area scales by the square of that same factor, so

d = 64

. Nothing from part (a) is needed here.)

(a)

k = 2.5

(b)

d = 64

Verification

Check 1: Put

k = 2.5

back into the equation. Every volume is a multiple of

\pi x^3

, so only the coefficients need comparing: the sphere gives

\dfrac{4}{3} \times 2.5^3

and the right-hand side gives

12.5 \times \dfrac{5}{3}

\dfrac{4}{3} \times 2.5^3 = \dfrac{125}{6}

and

12.5 \times \dfrac{5}{3} = \dfrac{125}{6}

, so the two sides match.

Check 2: Try a real size instead of a letter. With

x = 2

cm the cone has radius

2

cm and height

6

cm, so

S

has volume

\dfrac{1}{3}\pi \times 4 \times 6 + \dfrac{2}{3}\pi \times 8 = \dfrac{40}{3}\pi

cm³, and the sphere has radius

2.5 \times 2 = 5

cm. The sphere holds

\dfrac{4}{3}\pi \times 125 = \dfrac{500}{3}\pi

cm³, and

12.5 \times \dfrac{40}{3} = \dfrac{500}{3}

, which is the multiple the question asks for.

Check 3: Test part (b) by building

T

. Multiplying every length of

S

8

multiplies its volume by

8^3

, and the question says that multiplier is

512

, so

8

is the right length scale factor.

8^3 = 512

confirms the length scale factor, and

8^2 = 64

is then the area multiplier.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) An expression for the volume of any one of the three solids: the cone $\dfrac{1}{3}\pi x^2 \times 3x$ , the hemisphere $\dfrac{1}{2} \times \dfrac{4}{3}\pi x^3$ , or the sphere $\dfrac{4}{3}\pi (kx)^3$	M1	Any one of the three earns it. Missing brackets around $kx$ are ignored for this mark, and $r$ may be written in place of $x$ for all the method marks.	✓
(a) A correct equation for the volumes, $\dfrac{4}{3}\pi (kx)^3 = 12.5 \times \dfrac{5}{3}\pi x^3$ or equivalent	M1	The equation must be correct, so the volume of $S$ must already be right. If $(kx)^3$ has not been expanded at this stage, the brackets must be seen.	✓
(a) A correct calculation for $k$ or for $k^3$ , such as $k^3 = \dfrac{125}{8}$ or $k = \sqrt[3]{\dfrac{125}{8}}$	M1	A correct equation for $kx$ or for $k^3x^3$ earns this mark just as well, since the $x$ cancels either way.	✓
(a) $k = 2.5$	A1	Or equivalent, for example $\dfrac{5}{2}$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓
(b) $d = 64$	B1	Cube root the volume multiplier to get the length scale factor $8$ , then square it. Answering $512$ or $8$ scores nothing here.	✓

Full marks: 5/5

$OPQR$ is a parallelogram.

$\overrightarrow{OP} = 2\mathbf{a}$ and $\overrightarrow{OR} = 3\mathbf{b}$

The point $M$ lies on $PQ$ so that $PM = \dfrac{1}{4}PQ$

The point $N$ lies on $RQ$ so that $RN = \dfrac{4}{5}RQ$

(a) Work out, in terms of $\mathbf{a}$ and $\mathbf{b}$ , giving each answer in its simplest form
(i) $\overrightarrow{ON}$
[1 mark]
(ii) $\overrightarrow{MR}$ [1 mark]

$MR$ and $ON$ cross at the point $Y$ .
Given that
$OY = k \times ON$

(b) use a vector method to work out the value of $k$ . [4 marks]

(a)(i)(a)(ii)k =

[Total 6 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

OPQR

is a parallelogram, with

\overrightarrow{OP} = 2\mathbf{a}

and

\overrightarrow{OR} = 3\mathbf{b}

M

lies on

PQ

with

PM = \dfrac{1}{4}PQ

, and

N

lies on

RQ

with

RN = \dfrac{4}{5}RQ

MR

and

ON

cross at

Y

, and

OY = k \times ON

Find

(a)

\overrightarrow{ON}

and

\overrightarrow{MR}

, each in terms of

\mathbf{a}

and

\mathbf{b}

in simplest form. (b) the value of

k

, by a vector method.

Plan the Solution

Start with the parallelogram itself: opposite sides are parallel and the same length, so each pair carries the same vector.
Reach every point by travelling along edges. For $\overrightarrow{ON}$ go $O$ to $R$ to $N$ ; for $\overrightarrow{MR}$ go $M$ back to $O$ and on to $R$ .
For part (b) write $\overrightarrow{OY}$ twice - once as a fraction of $\overrightarrow{ON}$ , once as a journey along $\overrightarrow{MR}$ - then match the $\mathbf{a}$ parts and the $\mathbf{b}$ parts to get two equations.
Do not assume $Y$ is the midpoint of $MR$ . It turns out to be, but that is something the working produces, not something the figure gives you.

Worked Solution [6 marks]

Rule - matching coefficients:

\mathbf{a}

and

\mathbf{b}

are not parallel, so if

p\mathbf{a} + q\mathbf{b} = r\mathbf{a} + s\mathbf{b}

then

p = r

and

q = s

. One vector equation therefore gives two ordinary equations.

Step 1: write the sides of the parallelogram as vectors

\overrightarrow{PQ} = \overrightarrow{OR} = 3\mathbf{b}

\overrightarrow{RQ} = \overrightarrow{OP} = 2\mathbf{a}

(Reason: In a parallelogram opposite sides are parallel and equal in length, so they carry exactly the same vector. Everything else in the question is measured along these two sides.)

Step 2: part (a)(i), travel from O to R and then on to N

\overrightarrow{RN} = \dfrac{4}{5}\overrightarrow{RQ} = \dfrac{4}{5} \times 2\mathbf{a} = \dfrac{8}{5}\mathbf{a}

\overrightarrow{ON} = \overrightarrow{OR} + \overrightarrow{RN} = 3\mathbf{b} + \dfrac{8}{5}\mathbf{a}

\overrightarrow{ON} = \dfrac{8}{5}\mathbf{a} + 3\mathbf{b}

(Reason: The fraction is taken along

RQ

, which is

2\mathbf{a}

and not

\mathbf{a}

, so the

\dfrac{4}{5}

multiplies the whole of it. Writing the

\mathbf{a}

term first is what puts the answer in simplest form.)

Step 3: part (a)(ii), travel from M back to O and on to R

\overrightarrow{PM} = \dfrac{1}{4}\overrightarrow{PQ} = \dfrac{1}{4} \times 3\mathbf{b} = \dfrac{3}{4}\mathbf{b}

\overrightarrow{OM} = \overrightarrow{OP} + \overrightarrow{PM} = 2\mathbf{a} + \dfrac{3}{4}\mathbf{b}

\overrightarrow{MR} = \overrightarrow{MO} + \overrightarrow{OR} = -2\mathbf{a} - \dfrac{3}{4}\mathbf{b} + 3\mathbf{b}

\overrightarrow{MR} = \dfrac{9}{4}\mathbf{b} - 2\mathbf{a}

(Reason: Going backwards along a vector reverses its sign, so

\overrightarrow{MO} = -\overrightarrow{OM}

. The two

\mathbf{b}

terms then combine, and

\overrightarrow{OM}

is worth keeping because part (b) needs it again.)

Step 4: part (b), write OY in two different ways

\overrightarrow{OY} = k\overrightarrow{ON} = \dfrac{8}{5}k\mathbf{a} + 3k\mathbf{b}

\overrightarrow{OY} = \overrightarrow{OM} + \lambda\overrightarrow{MR} = 2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \lambda\left(\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}\right)

\overrightarrow{OY} = (2 - 2\lambda)\mathbf{a} + \left(\dfrac{3}{4} + \dfrac{9}{4}\lambda\right)\mathbf{b}

(Reason:

Y

is on both lines, so it can be reached along either. The second journey uses a new letter

\lambda

for the fraction of

\overrightarrow{MR}

travelled, because there is no reason yet to think it equals

k

Step 5: match the a parts and the b parts

\dfrac{8}{5}k = 2 - 2\lambda

3k = \dfrac{3}{4} + \dfrac{9}{4}\lambda

(Reason: The same vector cannot be written two genuinely different ways in terms of

\mathbf{a}

and

\mathbf{b}

, so the coefficients must agree one at a time. Two unknowns, two equations.)

Step 6: solve the pair for k

\lambda = 1 - \dfrac{4}{5}k

3k = \dfrac{3}{4} + \dfrac{9}{4}\left(1 - \dfrac{4}{5}k\right)

3k = \dfrac{3}{4} + \dfrac{9}{4} - \dfrac{9}{5}k

\dfrac{3}{4} + \dfrac{9}{4} = 3

3k + \dfrac{9}{5}k = 3

\dfrac{24}{5}k = 3

k = \dfrac{15}{24} = \dfrac{5}{8}

(Reason: Rearranging the

\mathbf{a}

equation gives

\lambda

in terms of

k

, and substituting it into the

\mathbf{b}

equation leaves one unknown. The fraction then cancels:

\dfrac{15}{24}

has a common factor of

3

(a)(i)

\overrightarrow{ON} = \dfrac{8}{5}\mathbf{a} + 3\mathbf{b}

(a)(ii)

\overrightarrow{MR} = \dfrac{9}{4}\mathbf{b} - 2\mathbf{a}

(b)

k = \dfrac{5}{8}

Verification

Check 1 - put the value back into both journeys: With

k = \dfrac{5}{8}

the first journey gives

\overrightarrow{OY} = \dfrac{5}{8}\left(\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}\right)

, and the

\mathbf{a}

term needs

\dfrac{5}{8} \times \dfrac{8}{5} = 1

. With

\lambda = \dfrac{1}{2}

the second gives

\overrightarrow{OY} = 2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \dfrac{1}{2}\left(\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}\right)

. Both journeys land on

\overrightarrow{OY} = \mathbf{a} + \dfrac{15}{8}\mathbf{b}

Check 2 - test it on numbers: Let

\mathbf{a}

be one step across and

\mathbf{b}

one step up. Then

O

is the origin,

P

(2, 0)

R

(0, 3)

and

Q

(2, 3)

, so

M

(2, 0.75)

and

N

(1.6, 3)

. Crossing the two lines gives

(1, 1.875)

, and

\dfrac{5}{8} \times 1.6 = 1

with

\dfrac{5}{8} \times 3 = 1.875

. The crossing point sits exactly

\dfrac{5}{8}

of the way from

O

N

, which is what

k

means.

Check 3 - what the value of lambda tells you, and what it does not license:

\lambda

came out as

\dfrac{1}{2}

, so

Y

is the midpoint of

MR

. On the numbers of check 2 the midpoint of

M(2, 0.75)

and

R(0, 3)

\dfrac{1}{2} \times (2 + 0) = 1

across and

\dfrac{1}{2} \times (0.75 + 3) = 1.875

up. The same point,

(1, 1.875)

. The midpoint is a result of the working and never a starting point - the mark scheme says as much.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $\overrightarrow{ON} = \overrightarrow{OR} + \dfrac{4}{5}\overrightarrow{RQ} = \dfrac{8}{5}\mathbf{a} + 3\mathbf{b}$	B1	or equivalent, but it must be in simplest form - for example $1.6\mathbf{a} + 3\mathbf{b}$ or $\dfrac{8\mathbf{a} + 15\mathbf{b}}{5}$ .	✓
(a)(ii) $\overrightarrow{MR} = \overrightarrow{MO} + \overrightarrow{OR} = \dfrac{9}{4}\mathbf{b} - 2\mathbf{a}$	B1	or equivalent, but it must be in simplest form - for example $2.25\mathbf{b} - 2\mathbf{a}$ or $\dfrac{9\mathbf{b} - 8\mathbf{a}}{4}$ .	✓
(b) A correct expression for one vector, for example $\overrightarrow{OM} = 2\mathbf{a} + \dfrac{3}{4}\mathbf{b}$ , or $\overrightarrow{OY} = k\left(\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}\right)$ , or $\overrightarrow{YN} = (1 - k)\left(\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}\right)$	M1ft	Follow through the answers the candidate gave in part (a). The reversed forms $\overrightarrow{MO}$ , $\overrightarrow{YO}$ and $\overrightarrow{NY}$ are equally acceptable, and on every method mark any letter may stand for $k$ or for $\lambda$ .	✓
Two independent expressions for the same vector, for example $\overrightarrow{OY} = k\left(\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}\right)$ together with $\overrightarrow{OY} = 2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \lambda\left(\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}\right)$	M1ft	The two expressions may be embedded in one correct equation rather than written out separately.	✓
A correct equation in $k$ alone, for example $3k = \dfrac{3}{4} + \dfrac{9}{4}\left(1 - \dfrac{4}{5}k\right)$ , or the correct value $\lambda = 0.5$	M1	The value of $\lambda$ may not be assumed. A candidate who writes $\lambda = 0.5$ because $Y$ looks like the midpoint of $MR$ has assumed the very thing the vector method is there to establish.	✓
$k = \dfrac{5}{8}$	A1 dep on M2	or equivalent, for example $0.625$ . Dependent on at least two of the method marks above.	✓
The question asks for a vector method, so no mark is available for a value reached by measuring the figure or by assuming a midpoint.	Note	The figure is not drawn accurately, so nothing may be read off it.	✓

Full marks: 6/6

Express $4 - \left[ \dfrac{3x-5}{\dfrac{3x^{2}+x-10}{4x-1}} \right]$ as a single fraction in its simplest form. [4 marks]

[Total 4 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

The expression

4 - \left[ \dfrac{3x-5}{\dfrac{3x^{2}+x-10}{4x-1}} \right]

A division inside the square brackets, and a subtraction outside them. The quadratic

3x^{2}+x-10

is the only part that can be factorised.

Find

One fraction, in its simplest form, with no fraction left inside a fraction.

Plan the Solution

Factorise $3x^{2}+x-10$ first. Nothing can cancel while it is still expanded.
Work inside the square brackets before the subtraction, so the division is dealt with first.
Dividing by a fraction is multiplying by its reciprocal, so turn $\dfrac{3x^{2}+x-10}{4x-1}$ upside down.
Cancel the bracket that appears top and bottom, then write $4$ over the same denominator and subtract.

Worked Solution [4 marks]

Rule - Dividing by a fraction: turn the fraction upside down and multiply.

\dfrac{a}{\left(\dfrac{b}{c}\right)} = a \times \dfrac{c}{b}

Step 1: Factorise the quadratic

3x^{2} + x - 10 = (3x-5)(x+2)

(Reason: Expanding

(3x-5)(x+2)

gives

3x^{2}+6x-5x-10

, and the two middle terms combine to

x

. The bracket

3x-5

is the one the numerator already has, which is why the factorising comes first.)

Step 2: Turn the second fraction upside down and cancel

(3x-5) \times \dfrac{4x-1}{(3x-5)(x+2)} = \dfrac{4x-1}{x+2}

(Reason: Dividing by

\dfrac{(3x-5)(x+2)}{4x-1}

is multiplying by

\dfrac{4x-1}{(3x-5)(x+2)}

. The bracket

3x-5

is now a factor of both the top and the bottom, so it cancels.)

Step 3: Write

4

over the same denominator

4 - \dfrac{4x-1}{x+2} = \dfrac{4(x+2)-(4x-1)}{x+2}

(Reason: Two parts can only be subtracted once they share a denominator, and

4 = \dfrac{4(x+2)}{x+2}

Step 4: Simplify the numerator

\dfrac{4x+8-4x+1}{x+2} = \dfrac{9}{x+2}

(Reason: A whole bracket is being subtracted, so the sign of every term inside it changes. The

x

terms then cancel and a constant is left on top, which is why the answer has no

x

in its numerator.)

\dfrac{9}{x+2}

Verification

Check 1: Put

x = 3

into the original expression. The quadratic is

27+3-10 = 20

, the numerator is

3 \times 3 - 5 = 4

and

4 \times 3 - 1 = 11

, so the bracket is

4 \times \dfrac{11}{20} = \dfrac{11}{5}

4 - \dfrac{11}{5} = \dfrac{9}{5}

, and the answer gives

\dfrac{9}{3+2} = \dfrac{9}{5}

Check 2: Put

x = 0

, where both signs are negative. The quadratic is

-10

and the denominator under it is

-1

, so the fraction inside the brackets is

\dfrac{-10}{-1} = 10

, and

\dfrac{-5}{10} = -\dfrac{1}{2}

4 + \dfrac{1}{2} = \dfrac{9}{2}

, and the answer gives

\dfrac{9}{0+2} = \dfrac{9}{2}

Check 3: Add back the fraction that was subtracted. If the answer is right, the

4

must reappear:

\dfrac{9}{x+2} + \dfrac{4x-1}{x+2} = \dfrac{4x+8}{x+2}

\dfrac{4(x+2)}{x+2} = 4

, which is the

4

the question started from

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Factorise $3x^{2}+x-10$	M1	For $(3x-5)(x+2)$ . It may be seen later on in the working. Alternatively, for combining the two parts into a correct single fraction.	✓
Invert and cancel	M1	For inverting and cancelling to give a correct fraction, $\dfrac{4x-1}{x+2}$ , which implies the first M1. Alternatively, for a correct single fraction whose denominator is factorised.	✓
One fraction over a common denominator	M1	For a correct single fraction, or two correct fractions with a common denominator, $\dfrac{4(x+2)-(4x-1)}{x+2}$ . Alternatively, for a correct fully factorised single fraction, $\dfrac{9(3x-5)}{(3x-5)(x+2)}$ .	✓
Simplify to the final answer	A1	For $\dfrac{9}{x+2}$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect 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 4MA1/2HR, Monday 3 June 2024: Worked Solutions, Questions 21 to 26

Worked solutions, questions 21 to 26 of 26

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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