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Edexcel IGCSE 4MA1 Paper 2H, November 2024: Worked Solutions, Questions 16 to 25

Sir Faraz Hassan

30 Jul 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 2H - Higher Tier - November 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 15

This is the rest of the paper. Questions 1 to 15, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 2H (Higher Tier), November 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 25 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 16 to 25 of 25

The points $A$ , $B$ , $C$ and $D$ all lie on a circle whose centre is $O$

(a) (i) Work out the size of angle $AOD$
[1 mark]
(ii) Give a reason for your answer to part (a)(i). [1 mark]

(b) Work out the size of angle $CAO$ [1 mark]

(a)(i) °(a)(ii)(b) °(c) °

[Total 5 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

A

B

C

and

D

lie on a circle with centre

O

, in that order round the circle.

The angle marked at

C

, between the chords

CA

and

CD

, is

\angle ACD = 54^{\circ}

The angle marked at

D

, between the chord

DC

and the radius

DO

, is

\angle ODC = 28^{\circ}

OA

OC

and

OD

are all radii of the same circle, so the triangles

OAD

OCD

and

OAC

are every one of them isosceles.

Find

The size of

\angle AOD

The circle theorem that justifies that answer, written in words The size of

\angle CAO

The size of

\angle ABC

Plan the Solution

Start at the marked $54^{\circ}$ . It stands on the arc $AD$ , and so does the angle at the centre, so doubling gives $\angle AOD$ straight away.
Use the isosceles triangle $OCD$ to turn the marked $28^{\circ}$ into the second angle at the centre, $\angle COD$ .
Two of the three angles at $O$ are then known, so the third, $\angle AOC$ , is what is left of the full turn.
Triangle $OAC$ is isosceles as well, so halving what is left of $180^{\circ}$ gives part (b).
For part (c), split $\angle ADC$ at the radius $DO$ , then use the fact that $ABCD$ is a cyclic quadrilateral.

Worked Solution [5 marks]

Rule - Angles in a circle: the angle at the centre is twice the angle at the circumference when both stand on the same arc, so

\angle AOD = 2 \times \angle ACD

. Angles round the point

O

add to

360^{\circ}

, the angles of a triangle add to

180^{\circ}

, and opposite angles of a cyclic quadrilateral add to

180^{\circ}

Step 1: double the marked angle, for part (a)(i)

\angle AOD = 2 \times \angle ACD

\angle AOD = 2 \times 54^{\circ} = 108^{\circ}

(Reason:

\angle ACD

has its vertex on the circle and

\angle AOD

has its vertex at the centre, and both of them stand on the same arc

AD

, so the one at the centre is double the one at the circumference.)

Step 2: say which theorem that was, for part (a)(ii)

\angle ACD = \dfrac{1}{2} \times \angle AOD

(Reason: The mark here is for the words, not for the arithmetic a second time. Either direction of the same theorem earns it: the angle at the centre is twice the angle at the circumference, or the angle at the circumference is half the angle at the centre. Whichever wording is used, it must be clear that

\angle ACD

and

\angle AOD

stand on the same arc.)

Step 3: turn the marked 28 degrees into an angle at the centre

\angle OCD = \angle ODC = 28^{\circ}

\angle COD = 180^{\circ} - 28^{\circ} - 28^{\circ} = 124^{\circ}

(Reason:

OC

and

OD

are both radii, so triangle

OCD

is isosceles and its two base angles are equal. The three angles of that triangle then add to

180^{\circ}

Step 4: the third angle at the centre

\angle AOC = 360^{\circ} - 108^{\circ} - 124^{\circ} = 128^{\circ}

(Reason:

\angle AOD

\angle COD

and

\angle AOC

fill the whole turn at

O

with nothing left over, so the three of them add to

360^{\circ}

Step 5: halve what is left in triangle OAC, for part (b)

\angle CAO = \angle ACO = \dfrac{180^{\circ} - 128^{\circ}}{2} = 26^{\circ}

(Reason:

OA

and

OC

are radii, so triangle

OAC

is isosceles with its apex at

O

. The apex angle is

128^{\circ}

, and the two equal base angles share what is left of

180^{\circ}

Step 6: the base angles of triangle OAD

\angle ODA = \angle OAD = \dfrac{180^{\circ} - 108^{\circ}}{2} = 36^{\circ}

(Reason:

OA

and

OD

are radii as well, so triangle

OAD

is isosceles too, with its apex angle the

108^{\circ}

found in step 1. Taking that apex angle off

180^{\circ}

leaves an amount that the two equal base angles share between them.)

Step 7: put the two pieces of the angle at D together

\angle ADC = \angle ODA + \angle ODC

\angle ADC = 36^{\circ} + 28^{\circ} = 64^{\circ}

(Reason:

O

lies inside the angle at

D

, so the radius

DO

cuts

\angle ADC

into exactly the two pieces already found, and they add.)

Step 8: use the cyclic quadrilateral, for part (c)

\angle ABC + \angle ADC = 180^{\circ}

\angle ABC = 180^{\circ} - 64^{\circ} = 116^{\circ}

(Reason:

A

B

C

and

D

lie on the circle in that order, so

ABCD

is a cyclic quadrilateral.

\angle ABC

and

\angle ADC

are a pair of opposite angles in it, and opposite angles of a cyclic quadrilateral add to

180^{\circ}

(a)(i)

\angle AOD = 108^{\circ}

(a)(ii) The angle at the centre is twice the angle at the circumference, both standing on the arc

AD

(b)

\angle CAO = 26^{\circ}

(c)

\angle ABC = 116^{\circ}

Verification

Check 1: Add the three angles at the centre. If

\angle AOD

\angle COD

and

\angle AOC

are all right, they must make exactly one full turn.

108^{\circ} + 124^{\circ} + 128^{\circ} = 360^{\circ}

Check 2: Reach part (c) the second way the mark scheme allows, with no cyclic quadrilateral at all. From

B

the arc through

D

is covered by the two angles at the centre

108^{\circ}

and

124^{\circ}

, so the angle at

B

is half of their total.

\dfrac{108^{\circ} + 124^{\circ}}{2} = \dfrac{232^{\circ}}{2} = 116^{\circ}

Check 3: Add up triangle

ACD

. The radius

AO

cuts

\angle CAD

into the

26^{\circ}

of part (b) and the

36^{\circ}

of step 6, giving

\angle CAD = 62^{\circ}

. The other two angles of that triangle are the marked

54^{\circ}

and the

64^{\circ}

of step 7.

62^{\circ} + 54^{\circ} + 64^{\circ} = 180^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $108$	B1	Accept $252$ as well, which is the reflex angle $AOD$ at $O$ . No working is required for this mark.	✓
(a)(ii) A correct reason	B1	Dependent on part (a)(i) being correct. Accept: the angle at the centre (midpoint, origin, middle) is twice the angle at the circumference (side, edge, arc), or any equivalent, for instance that the inscribed angle is half of the central angle. The angle symbol is accepted in place of the word angle, and $2 \times$ or the word double is accepted in place of twice.	✓
(b) $26$	B1	No working is required for this mark.	✓
(c) $\angle ABC = 180^{\circ} - 64^{\circ}$ or $\dfrac{108^{\circ} + 124^{\circ}}{2}$	M1ft	Follow through their $\dfrac{180^{\circ} - 108^{\circ}}{2}$ used in $28^{\circ} + 36^{\circ} = 64^{\circ}$ . Only their $108$ is followed through.	✓
(c) $116$	A1	Correct answer only.	✓
Note - a correct answer with no working	-	A correct answer scores full marks, unless it plainly follows from obviously incorrect working.	✓

Full marks: 5/5

Two vectors are given below.

$\overrightarrow{FG} = \begin{pmatrix} -5 \\ 2 \end{pmatrix} \qquad \overrightarrow{HG} = \begin{pmatrix} 4 \\ 14 \end{pmatrix}$

Work out the magnitude of the vector $\overrightarrow{HF}$ [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

\overrightarrow{FG} = \begin{pmatrix} -5 \\ 2 \end{pmatrix}

\overrightarrow{HG} = \begin{pmatrix} 4 \\ 14 \end{pmatrix}

Two column vectors joining the three points

F

G

and

H

. Neither of them runs from

H

F

, so that vector has to be built first.

Find

The magnitude

\left| \overrightarrow{HF} \right|

- a length, so expect one positive number, and no unit is given here.

Plan the Solution

Travel from $H$ to $F$ by way of $G$ , which gives $\overrightarrow{HF} = \overrightarrow{HG} + \overrightarrow{GF}$ .
Only $\overrightarrow{FG}$ is given, so reverse it: $\overrightarrow{GF} = -\overrightarrow{FG}$ .
Add the two column vectors component by component to get $\overrightarrow{HF}$ .
Treat those two components as the legs of a right-angled triangle and apply Pythagoras. The magnitude is a length, so take the positive root.

Worked Solution [3 marks]

Rule - Magnitude of a column vector:

\left| \begin{pmatrix} a \\ b \end{pmatrix} \right| = \sqrt{a^2 + b^2}

. Reversing a vector reverses the sign of both of its components, so

\overrightarrow{GF} = -\overrightarrow{FG}

Step 1: build a route from

H

F

\overrightarrow{HF} = \overrightarrow{HG} + \overrightarrow{GF}

\overrightarrow{GF} = -\overrightarrow{FG} = -\begin{pmatrix} -5 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}

(Reason: Going from

H

G

and then from

G

F

finishes at

F

, so the two journeys add to give

\overrightarrow{HF}

. Walking a vector backwards changes the sign of both components.)

Step 2: work out the components of

\overrightarrow{HF}

\overrightarrow{HF} = \begin{pmatrix} 4 \\ 14 \end{pmatrix} + \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}

(Reason: Add the top components together and the bottom components together. This is the same as

\overrightarrow{HG} - \overrightarrow{FG}

, which is the form the mark scheme states.)

Step 3: apply Pythagoras to the two components

\left| \overrightarrow{HF} \right| = \sqrt{9^2 + 12^2}

\left| \overrightarrow{HF} \right| = \sqrt{81 + 144} = \sqrt{225}

(Reason: The components

9

and

12

are the two legs of a right-angled triangle, and the magnitude is its hypotenuse.)

Step 4: take the positive square root

\left| \overrightarrow{HF} \right| = \sqrt{225} = 15

(Reason:

225

is a square number, so the magnitude is exact and no rounding is needed. A magnitude is a length, so the negative root is discarded.)

\left| \overrightarrow{HF} \right| = 15

Verification

Check 1 - travel the other way: Build the reverse journey instead:

\overrightarrow{FH} = \overrightarrow{FG} + \overrightarrow{GH} = \begin{pmatrix} -5 \\ 2 \end{pmatrix} + \begin{pmatrix} -4 \\ -14 \end{pmatrix} = \begin{pmatrix} -9 \\ -12 \end{pmatrix}

. A vector and its reverse must have the same length.

\sqrt{(-9)^2 + (-12)^2} = \sqrt{225} = 15

, the same magnitude.

Check 2 - an enlarged 3, 4, 5 triangle: The components

9

and

12

are three times

3

and three times

4

, so this is a

3, 4, 5

triangle enlarged by scale factor

3

3 \times 5 = 15

, which agrees with the square root, and no calculator is needed to see it.

Check 3 - put the points on a grid: Pin

H

at the origin. Then

G

is at

(4, 14)

, and since

\overrightarrow{FG}

runs from

F

G

, the point

F

is at

(4 - (-5),\ 14 - 2) = (9, 12)

. Now use the distance formula.

\sqrt{(9 - 0)^2 + (12 - 0)^2} = \sqrt{225} = 15

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct calculation for $\overrightarrow{HF}$ or $\overrightarrow{FH}$ , for example $\begin{pmatrix} 4 \\ 14 \end{pmatrix} - \begin{pmatrix} -5 \\ 2 \end{pmatrix} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}$ or equivalent.	M1	For this mark, allow the vector written as coordinates. Also allow, for example, $9\mathbf{i} + 12\mathbf{j}$ . The reverse vector $\begin{pmatrix} -9 \\ -12 \end{pmatrix}$ scores this mark too.	✓
$\sqrt{9^2 + 12^2}$ or $\sqrt{(-9)^2 + (-12)^2}$	M1 indep	Awarded independently of the first mark. Allow a complete method using their own $\overrightarrow{HF}$ or $\overrightarrow{FH}$ , provided it came from $\begin{pmatrix} \pm 4 \\ \pm 14 \end{pmatrix} - \begin{pmatrix} \pm 5 \\ \pm 2 \end{pmatrix}$ , allowing any sign error. If $(-9)^2$ and $(-12)^2$ are used, condone missing brackets if they are recovered.	✓
$15$	A1	From fully correct figures. Use of $\begin{pmatrix} 9 \\ -12 \end{pmatrix}$ would give the correct answer of $15$ , because the squares are the same, but it would not gain this accuracy mark.	✓
Note - no mark of its own	-	A correct answer scores full marks unless it comes from obviously incorrect working. Watch out for a correct answer from wrong working, for example $-5 + 2 + 4 + 14 = 15$ : on these particular vectors the four components happen to total the right number, and that earns nothing.	✓

Full marks: 3/3

The straight line $L$ is perpendicular to the line with equation $2x + y = 9$
The line $L$ passes through the point with coordinates $(8, 11)$

Find an equation for $L$
Give your answer in the form $y = mx + c$ [4 marks]

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The line

L

is perpendicular to the line

2x + y = 9

L

passes through the point

(8, 11)

Two facts, and each one fixes a different part of the answer.

Find

An equation for

L

, written in the form

y = mx + c

. So one gradient

m

and one intercept

c

are needed.

Plan the Solution

Rearrange $2x + y = 9$ into $y = mx + c$ form, so its gradient can be read off.
Turn that gradient into the perpendicular gradient using the negative reciprocal.
Substitute $x = 8$ and $y = 11$ to find $c$ .
Write the finished equation in the form the question asks for.

Worked Solution [4 marks]

Rule - Perpendicular gradients: if two lines are perpendicular then

m_{1} \times m_{2} = -1

, so

m_{2} = \dfrac{-1}{m_{1}}

- the negative reciprocal. The gradient of a line is only visible once its equation is in

y = mx + c

form.

Step 1: Rearrange the given line into

y = mx + c

form

2x + y = 9

y = -2x + 9

(Reason: taking

2x

from both sides leaves

y

on its own, and the number in front of

x

is then the gradient, so the given line has gradient

-2

)

Step 2: Turn that into the gradient of

L

m_{L} \times (-2) = -1

m_{L} = \dfrac{-1}{-2} = \dfrac{1}{2}

(Reason: perpendicular gradients multiply to give

-1

, so divide

-1

by the gradient of the given line - the negative reciprocal)

Step 3: Use the point

(8, 11)

to find

c

11 = \dfrac{1}{2} \times 8 + c

11 = 4 + c

c = 7

(Reason: the point lies on

L

, so putting

x = 8

and

y = 11

into

y = mx + c

leaves

c

as the only unknown)

Step 4: Write the equation in the form asked for

y = \dfrac{1}{2}x + 7

(Reason: the question asks for

y = mx + c

, so the gradient

\dfrac{1}{2}

and the intercept

7

go straight into that form)

y = \dfrac{1}{2}x + 7

Verification

Check 1: Put

x = 8

into the answer and see whether

y

comes back as

11

\dfrac{1}{2} \times 8 + 7 = 4 + 7 = 11

, so

L

does pass through

(8, 11)

Check 2: Multiply the two gradients together. Perpendicular lines must give

-1

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

, so the two lines really are at right angles

Check 3: Compare directions instead of gradients. At

x = 0

the answer gives

y = 7

, and the given line steps

1

across for every

2

down. From

(0, 7)

(8, 11)

8

across and

4

up, and

8 \times 1 + 4 \times (-2) = 0

, which is what perpendicular directions do

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$y = -2x + 9$ or gradient of the given line $= -2$	M1	a rearrangement with the correct $x$ term, or a statement that the gradient of the given line is $-2$	✓
gradient of the perpendicular $= \dfrac{1}{2}$ oe, eg $\dfrac{-1}{-2}$	M1ft	for a statement that the gradient of the perpendicular line is $\dfrac{1}{2}$ , or for it being implied by an equation of a line with gradient $\dfrac{1}{2}$ . A student who goes straight to this stage is awarded M2. The mark is also available for the perpendicular gradient of whatever they indicate the gradient of the original line to be	✓
eg $11 = \dfrac{1}{2} \times 8 + c$ or $y - 11 = \dfrac{1}{2}(x - 8)$ oe, or $c = 7$	M1dep	dep on the previous M1 being awarded; a correct method to find the equation of the perpendicular line, using their gradient of the perpendicular line and $(8, 11)$	✓
$y = \dfrac{1}{2}x + 7$	A1	a correct equation for the line in the form $y = mx + c$ , as requested. oe	✓
If no other marks are awarded: $y = -\dfrac{1}{2}x + 15$	SC B1	the named error behind this special case is taking the reciprocal of $-2$ without changing its sign, which gives gradient $-\dfrac{1}{2}$ and then $c = 15$ . This row carries no mark of its own; it records what that one wrong answer is worth	✓
Note	note	A correct answer scores full marks, unless it comes from obviously incorrect working. This row earns nothing on its own.	✓

Full marks: 4/4

The curve $C$ has equation $y = f(x)$ .
There is exactly one minimum point on $C$ , and its coordinates are $(5,\ 4)$ .

For each equation below, write down the coordinates of the minimum point on that curve.

(i) $y = f(x + 5)$
[1 mark]
(ii) $y = 3f(x)$
[1 mark]
(iii) $y = f(x) - 7$ [1 mark]

(i)(ii)(iii)

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

A curve

C

with equation

y = f(x)

. The function

f

itself is never stated.

C has exactly one minimum point, at

(5,\ 4)

Find

(i) the coordinates of the minimum point on

y = f(x + 5)

(ii) the coordinates of the minimum point on

y = 3f(x)

(iii) the coordinates of the minimum point on

y = f(x) - 7

Plan the Solution

For each equation, ask one question first: is the change inside the bracket of $f(\ldots)$ , or outside it?
Inside the bracket acts on $x$ and does the opposite of what it looks like, so $f(x + 5)$ moves the curve $5$ to the left.
Outside the bracket acts on $y$ and does exactly what it looks like, so $3f(x)$ multiplies every $y$ -coordinate by $3$ , and $f(x) - 7$ lowers every $y$ -coordinate by $7$ .
Then move the single point $(5,\ 4)$ three times, once per equation. Nothing else about $f$ is needed, and none of the three parts depends on the others.

Worked Solution [3 marks]

Rule - transforming

y = f(x)

: a change inside the bracket is horizontal and reversed, and leaves

y

alone; a change outside the bracket is vertical and direct, and leaves

x

alone.

Step 1: (i)

y = f(x + 5)

- the change is inside the bracket

x + 5 = 5 \implies x = 0

y = 4

(Reason: The minimum of

f

happens when the number fed into

f

5

. Here the number fed in is

x + 5

, so

x + 5 = 5

and therefore

x = 0

- the curve has moved

5

to the left. Nothing has been done to the output of

f

, so the height stays at

4

Step 2: (ii)

y = 3f(x)

- the change is outside the bracket

x = 5

y = 3 \times 4 = 12

(Reason: The number fed into

f

is untouched, so the lowest point is still reached at

x = 5

. What has changed is the output: it is multiplied by

3

, so the

y

-coordinate is stretched away from the

x

-axis by scale factor

3

Step 3: (iii)

y = f(x) - 7

- the change is outside the bracket

x = 5

y = 4 - 7 = -3

(Reason: Again the number fed into

f

is untouched, so

x = 5

. Subtracting

7

after

f

has acted lowers every point by

7

, so the height

4

becomes

-3

. The minimum is now below the

x

-axis, which is allowed.)

(i)

(0,\ 4)

(ii)

(5,\ 12)

(iii)

(5,\ -3)

Verification

Check 1: Feed each answer back into its own equation. At

x = 0

, the first curve gives

f(0 + 5) = f(5) = 4

. At

x = 5

, the second gives

3f(5) = 3 \times 4 = 12

and the third gives

f(5) - 7 = 4 - 7 = -3

. The three heights come out as

4

12

and

-3

, which are the three answers.

Check 2: Test a curve that really does have its only minimum at

(5,\ 4)

, for instance

f(x) = (x - 5)^2 + 4

. Transforming it gives

f(x + 5) = x^2 + 4

3f(x) = 3(x - 5)^2 + 12

and

f(x) - 7 = (x - 5)^2 - 3

, and the minimum of each can be read straight off the completed square. The vertices are

(0,\ 4)

(5,\ 12)

and

(5,\ -3)

, matching the answers. Repeating the test with

f(x) = |x - 5| + 4

gives the same three points, so the answers do not depend on which curve is chosen.

Check 3: Count which coordinate each transformation is even allowed to change. A change inside the bracket can move

x

only; a change outside it can move

y

only. Part (i) keeps

y = 4

, and parts (ii) and (iii) keep

x = 5

. Each answer changes exactly one coordinate, as it must.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) $(0,\ 4)$	B1	Both coordinates correct, in the right order. Accept $x = 0$ and $y = 4$ written separately on the answer line. One coordinate alone scores nothing.	✓
(ii) $(5,\ 12)$	B1	Both coordinates correct, in the right order. $(5,\ 7)$ scores nothing: it comes from reading the multiplier as a translation up $3$ .	✓
(iii) $(5,\ -3)$	B1	Both coordinates correct, in the right order. $(5,\ 11)$ scores nothing: it comes from adding $7$ instead of subtracting it. Each part is marked independently of the other two, so a wrong answer earlier costs nothing here.	✓

Full marks: 3/3

Find the set of values of $x$ for which
$10x^2 + 11x - 21 < 0$
You must show clear algebraic working. [3 marks]

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The quadratic inequality

10x^2 + 11x - 21 < 0

, to be solved with algebraic working shown.

The coefficient of

x^2

10

, which is positive, so

y = 10x^2 + 11x - 21

is a parabola that opens upwards.

Find

Every value of

x

that makes the expression negative. Expect an interval between two critical values, not a single number.

Plan the Solution

Solve the matching equation $10x^2 + 11x - 21 = 0$ to get the two critical values. Factorising is quickest here, and the formula or completing the square give the same pair.
Decide which region satisfies the inequality. An upwards parabola sits below the $x$ -axis only between its two roots, and a test value settles it.
Write the answer as one double inequality. The inequality is strict, so both critical values are excluded.

Worked Solution [3 marks]

Rule, quadratic inequality: solve

ax^2 + bx + c = 0

for the critical values, then when

a > 0

and the inequality reads

< 0

, the solution is the region between them.

Step 1: factorise the quadratic

10x^2 + 11x - 21 = (10x + 21)(x - 1)

(10x + 21)(x - 1) = 10x^2 - 10x + 21x - 21 = 10x^2 + 11x - 21

(Reason: The brackets have to multiply to

-21

and leave a middle term of

+11x

. Expanding straight back is the quickest way to be certain the pair chosen is the right one.)

Step 2: set each factor to zero to get the critical values

10x + 21 = 0 \implies x = -\dfrac{21}{10} = -2.1

x - 1 = 0 \implies x = 1

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

x

-axis.)

Step 3: choose the region where the expression is negative

10(0)^2 + 11(0) - 21 = -21

-2.1 < x < 1

(Reason: The parabola opens upwards, so it dips below the axis only between the two critical values. Testing

x = 0

, which lies between them, gives

-21

, and that is negative, so the middle region is the one wanted. Because the inequality is strict, neither endpoint belongs to the answer.)

-2.1 < x < 1

Verification

Check 1: Reach the critical values a second way, with the quadratic formula on

10x^2 + 11x - 21 = 0

. The discriminant is

11^2 - 4 \times 10 \times (-21) = 121 + 840 = 961

, and

\sqrt{961} = 31

x = \dfrac{-11 + 31}{20} = 1

and

x = \dfrac{-11 - 31}{20} = -2.1

, the same two critical values that factorising gave.

Check 2: Test one value taken from each of the three regions:

x = -2.5

x = 0

and

x = 1.5

. The three values of the expression are

14

-21

and

18

. Only the middle one is negative, so only

-2.1 < x < 1

satisfies the inequality.

Check 3: Complete the square instead of factorising:

10(x + 0.55)^2 - 3.025 - 21 < 0

(x + 0.55)^2 < 2.4025

, so

-1.55 < x + 0.55 < 1.55

, which gives

-2.1 < x < 1

once more.

Check 4: Substitute the two critical values themselves, to confirm the endpoints are excluded rather than included. Each one gives exactly zero:

10(-2.1)^2 + 11(-2.1) - 21 = 44.1 - 23.1 - 21 = 0

and

10 + 11 - 21 = 0

. Zero is not less than zero, so

x = -2.1

and

x = 1

are correctly left out.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to solve the quadratic: the correct factors $(10x + 21)(x - 1)$ , or correct substitution into the formula, or correctly completing the square	M1	Substitution into the formula may be simplified only as far as $\dfrac{-11 \pm \sqrt{121 + 840}}{20}$ and still earn this mark. Writing $(10)(x + 2.1)(x - 1)$ is not a correct factorisation: it is working backwards from calculator answers.	✓
The correct critical values $x = 1$ and $x = -2.1$	A1	Dependent on the method mark. Critical values that did not come from a correct method earn nothing here.	✓
$-2.1 < x < 1$ , with working shown	A1	Dependent on the method mark, and working is required. Accept any equivalent form: $x > -2.1$ and $x < 1$ (do not penalise 'or'), or $-\dfrac{21}{10} < x < 1$ , or the open interval written $(-2.1, 1)$ .	✓

Full marks: 3/3

The Venn diagram below records how many items lie in each region of set $A$ , set $B$ and set $C$ .
In the diagram $x$ is an integer.

Given that $\mathrm{n}(A \cup B)' = 26$
work out $\mathrm{n}(A' \cap C)$ [4 marks]

n(A' ∩ C) =

[Total 4 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

One set only:

3x

items in

A

alone,

4x

B

alone,

5x

C

alone

Overlaps:

x + 2

A

and

B

only,

x + 7

A

and

C

only,

2x

B

and

C

only,

x

in all three

3x + 2

items lie inside the rectangle but outside all three circles, and

x

is an integer

\mathrm{n}(A \cup B)' = 26

Find

\mathrm{n}(A' \cap C)

, the number of items in

C

that are not in

A

Plan the Solution

Read $(A \cup B)'$ off the diagram: the two regions that touch neither $A$ nor $B$
Set their total equal to $26$ and solve for $x$
Read $A' \cap C$ off the diagram the same way, as a multiple of $x$
Substitute the value of $x$ into that multiple

Worked Solution [4 marks]

Rule - Complements and intersections are read as REGIONS:

\mathrm{n}(S')

counts everything in the universal set that is not in

S

, and

A' \cap C

is the part of

C

that lies outside

A

. Name the regions first, then add only those regions.

Step 1: pick out the regions that lie outside

A \cup B

\mathrm{n}(A \cup B)' = 5x + (3x + 2)

5x + 3x + 2 = 26

(Reason:

A \cup B

is everything inside circle

A

or circle

B

, so its complement is what is left of the universal set: the part of

C

outside both circles,

5x

, and the

3x + 2

outside all three)

Step 2: solve the equation for

x

8x + 2 = 26

8x = 24

x = \dfrac{24}{8} = 3

(Reason: Collect the

x

terms, take

2

from each side, then divide by

8

. The value comes out a whole number, as the question promised)

Step 3: pick out the regions that make up

A' \cap C

\mathrm{n}(A' \cap C) = 5x + 2x = 7x

(Reason:

A' \cap C

is the part of

C

outside

A

: the

5x

region and the

2x

region that

C

shares with

B

. The

x + 7

and

x

regions sit inside

A

, so they are left out)

Step 4: substitute

x = 3

7 \times 3 = 21

(Reason: Multiply the coefficient found in step

3

by the value of

x

found in step

2

)

\mathrm{n}(A' \cap C) = 21

Verification

Check 1: Put

x = 3

back into the two regions outside

A \cup B

: the

C

only region holds

5 \times 3 = 15

and the region outside every circle holds

3 \times 3 + 2 = 11

15 + 11 = 26

, the given complement

Check 2: Count

A' \cap C

a different way: take the whole of

C

, which is

10 + 3 + 6 + 15 = 34

, then remove the part of

C

inside

A

, which is

10 + 3 = 13

34 - 13 = 21

Check 3: Add every region:

9 + 5 + 12 + 3 + 10 + 6 + 15 + 11 = 71

items in the universal set, while the six regions inside

A \cup B

hold

9 + 5 + 12 + 3 + 10 + 6 = 45

71 - 45 = 26

, so the diagram has been read consistently

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$5x + 3x + 2 = 26$ oe	M1	a correct equation for $x$	✓
$x = 3$	A1	the correct value of $x$ . A correct value of $x$ on its own scores M1A1, unless it comes from obviously incorrect working	✓
$7 \times 3$ , or $15 + 6$ oe	M1ft	use of their positive value of $x$ in $7x$ , i.e. use of the correct regions of the Venn diagram for the set required ( $15$ being $5 \times 3$ and $6$ being $2 \times 3$ )	✓
$21$	A1	cao. A correct answer scores full marks, unless it comes from obviously incorrect working	✓

Full marks: 4/4

Find all the solutions of the simultaneous equations

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

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 = 3

, an equation with a squared term in each letter

x + 2 = y

, a linear equation

One curve and one straight line, so the pairs being looked for are where they meet

Find

Every pair of values of

x

and

y

that satisfies both equations at the same time. A quadratic appears, so expect two solution pairs Clear algebraic working is asked for, so the answer alone earns nothing

Plan the Solution

Rearrange the linear equation so that $y$ is written in terms of $x$
Replace every $y$ in the quadratic equation by $x + 2$ , which leaves one letter only
Expand, collect like terms and write the result as $ax^2 + bx + c = 0$
Factorise to get the two values of $x$
Put each value of $x$ back into $y = x + 2$ so that every $x$ leaves with its own $y$
Test both pairs in both of the original equations

Worked Solution [5 marks]

Rule - Substitution (line into curve): make one letter the subject of the LINEAR equation, substitute it into the quadratic equation, and solve the three-term quadratic that comes out. Each root is one solution PAIR, so every value found is sent back through the linear equation to collect its partner.

Step 1: make

y

the subject of the linear equation

x + 2 = y

y = x + 2

(Reason: The linear equation is the easy one to rearrange, and turning it round costs nothing: it now reads as an instruction, telling us what to write in place of every

y

in the other equation)

Step 2: substitute into the quadratic equation

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

(Reason:

y

appears twice in the quadratic equation, once squared and once on its own, so BOTH are replaced by

x + 2

. Each replacement goes inside brackets, or the squaring and the signs come out wrong)

Step 3: expand the squared bracket

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

x^2 + (x^2 + 4x + 4) + (x + 2) = 3

(Reason:

(x + 2)^2

means

(x + 2)(x + 2)

, which gives

x^2

, then two lots of

2x

, then

4

. Writing

x^2 + 4

and losing the middle term is the commonest slip in this question)

Step 4: collect the like terms and make one side zero

2x^2 + 5x + 6 = 3

2x^2 + 5x + 3 = 0

(Reason: The two squared terms add to

2x^2

. The

x

term from the expanded bracket and the single

x

that arrived with the curve's own

y

add to

5x

, and the two constants add to

6

. Taking

3

from each side leaves the standard three-term form with zero on the right)

Step 5: factorise, then solve each bracket for

x

(2x + 3)(x + 1) = 0

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

x + 1 = 0 \implies x = -1

(Reason: For

2x^2 + 5x + 3

the two numbers needed multiply to give

2 \times 3 = 6

and add to give

5

, which are

2

and

3

; splitting the middle term with them gives these brackets. A product is zero only when one factor is zero, so each bracket is set to zero in turn. The quadratic formula or completing the square would reach the same two values)

Step 6: pair each

x

with its own

y

x = -\dfrac{3}{2} \implies y = -\dfrac{3}{2} + 2 = \dfrac{1}{2}

x = -1 \implies y = -1 + 2 = 1

(Reason: Each value of

x

goes back into

y = x + 2

, the simpler equation, so no new quadratic is needed. The values must stay in their pairs:

-\dfrac{3}{2}

belongs with

\dfrac{1}{2}

, and swapping partners would give a point lying on neither graph)

x = -\dfrac{3}{2}, y = \dfrac{1}{2}

x = -1, y = 1

Verification

Check 1: Test the first pair in the quadratic equation. Squaring

-\dfrac{3}{2}

gives

\dfrac{9}{4}

, squaring

\dfrac{1}{2}

gives

\dfrac{1}{4}

, and the separate

y

adds

\dfrac{1}{2}

\dfrac{9}{4} + \dfrac{1}{4} + \dfrac{1}{2} = 3

, and the line gives

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

, so both equations hold

Check 2: Test the second pair the same way, in both original equations rather than in any line of the working

1 + 1 + 1 = 3

, and

-1 + 2 = 1

, so both equations hold

Check 3: Eliminate the other letter instead. Substituting

x = y - 2

into the quadratic equation gives

(y - 2)^2 + y^2 + y = 3

, which collects to

2y^2 - 3y + 1 = 0

and factorises as

(2y - 1)(y - 1) = 0

The

y

values come out as

\dfrac{1}{2}

and

1

without using the quadratic in

x

at all

Check 4: Use the sum and product of the roots of

2x^2 + 5x + 3 = 0

: they must be

-\dfrac{5}{2}

and

\dfrac{3}{2}

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

and

-\dfrac{3}{2} \times (-1) = \dfrac{3}{2}

, so the two roots are the right pair for that quadratic

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$x^2 + (x + 2)^2 + x + 2 = 3$ , or $(y - 2)^2 + y^2 + y = 3$	M1	substitution of the linear equation into the quadratic equation. Allow one sign error in the substituted expression	✓
$2x^2 + 5x + 3 [= 0]$ oe (any form with three terms), or $2y^2 - 3y + 1 [= 0]$ oe	M1	dep on the first M1. Simplified to a three-term quadratic with 2 or 3 of the 3 terms correct	✓
$(2x + 3)(x + 1) [= 0]$ , or $x = \dfrac{-5 + \sqrt{5^2 - 4 \times 2 \times 3}}{2 \times 2}$ and $x = \dfrac{-5 - \sqrt{5^2 - 4 \times 2 \times 3}}{2 \times 2}$ , or $2[(x + \dfrac{5}{4})^2 - \dfrac{25}{16}] + 3 = 0$ , leading to $x = -\dfrac{3}{2}$ and $x = -1$	M1ft	dep on the first M1, for solving their three-term quadratic by any correct method. If factorising, brackets which expand to give 2 of the 3 terms correct are enough; if using the formula, allow one sign error and only partial simplification; if completing the square, as far as the form shown. Leads to the correct values of $x$ OR the correct values of $y$ (the other letter may be used throughout)	✓
( $y =$ ) $-\dfrac{3}{2} + 2$ and $-1 + 2$ oe, or a correct pair of values	M1	dep on the previous M1, for a correct method to find both of the other two values, or for a correct pair of values	✓
$x = -\dfrac{3}{2}, y = \dfrac{1}{2}$ and $x = -1, y = 1$	A1	oe dep on M2, for all 4 values. Working is required, so a correct answer with no algebraic working scores nothing	✓

Full marks: 5/5

Show that
$\dfrac{\dfrac{16x^2 - 36}{x - 7}}{\dfrac{2x^2 + 7x + 6}{x^2 - 5x - 14}} - (7 + 8x) = n$
where $n$ is an integer to be found.
You must show clear algebraic working. [4 marks]

[Total 4 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

\dfrac{\dfrac{16x^2 - 36}{x - 7}}{\dfrac{2x^2 + 7x + 6}{x^2 - 5x - 14}} - (7 + 8x) = n

Three quadratics, and every one of them factorises.

n

is an integer, so every

x

must disappear by the end.

Find

The value of

n

Plan the Solution

Factorise all three quadratics. $16x^2 - 36$ has a common factor of $4$ and then a difference of two squares.
Dividing by a fraction is the same as multiplying by its reciprocal, so turn the second fraction upside down.
Cancel every bracket that appears top and bottom. What is left is linear.
Subtract $(7 + 8x)$ and collect like terms. The $x$ terms cancel and a constant is all that survives.

Worked Solution [4 marks]

Rule - Dividing algebraic fractions: factorise everything first, then

\dfrac{\dfrac{A}{B}}{\dfrac{C}{D}} = \dfrac{A}{B} \times \dfrac{D}{C}

, and cancel any bracket that appears in both a numerator and a denominator.

Step 1: Factorise

16x^2 - 36

16x^2 - 36 = 4(4x^2 - 9)

4(4x^2 - 9) = 4(2x - 3)(2x + 3)

(Reason: Take the common factor of

4

out first. What is left,

4x^2 - 9

, is a difference of two squares, because

4x^2 = (2x)^2

and

9 = 3^2

Step 2: Factorise the other two quadratics

2x^2 + 7x + 6 = (2x + 3)(x + 2)

x^2 - 5x - 14 = (x - 7)(x + 2)

(Reason: For

2x^2 + 7x + 6

split the middle term:

4

and

3

multiply to

12

and add to

7

. For

x^2 - 5x - 14

look for two numbers multiplying to

-14

and adding to

-5

, which are

-7

and

2

. Notice that

(x + 2)

and

(2x + 3)

have both appeared twice - that is the hint that everything will cancel.)

Step 3: Multiply by the reciprocal, then cancel

\dfrac{\dfrac{4(2x - 3)(2x + 3)}{x - 7}}{\dfrac{(2x + 3)(x + 2)}{(x - 7)(x + 2)}} = \dfrac{4(2x - 3)(2x + 3)}{x - 7} \times \dfrac{(x - 7)(x + 2)}{(2x + 3)(x + 2)}

= 4(2x - 3) = 8x - 12

(Reason: Turning the divisor upside down puts

(x - 7)

on top and

(2x + 3)

underneath, so

(x - 7)

(x + 2)

and

(2x + 3)

each cancel. Only the

4

and the bracket

(2x - 3)

survive.)

Step 4: Subtract

(7 + 8x)

and collect like terms

8x - 12 - (7 + 8x) = 8x - 12 - 7 - 8x

= (8x - 8x) + (-12 - 7) = -19

(Reason: Watch the bracket: subtracting

(7 + 8x)

changes the sign of both terms inside it. The

8x

terms then cancel and only the constant

-19

is left, which is exactly why the question can promise that

n

is an integer.)

n = -19

Verification

Check 1: Put

x = 0

into the original expression. The first fraction is

\dfrac{-36}{-7} = \dfrac{36}{7}

and the second is

\dfrac{6}{-14} = -\dfrac{3}{7}

, so the division gives

\dfrac{36}{7} \times \left(-\dfrac{7}{3}\right) = -12

-12 - (7 + 0) = -19

Check 2: Put

x = 1

in instead, so a second, unrelated value tests the same identity. The first fraction is

\dfrac{-20}{-6} = \dfrac{10}{3}

and the second is

\dfrac{15}{-18} = -\dfrac{5}{6}

, so the division gives

\dfrac{10}{3} \times \left(-\dfrac{6}{5}\right) = -4

-4 - (7 + 8) = -19

Check 3: Do it without cancelling at all. Keep

2x + 3

underneath and combine the whole expression over it:

16x^2 - 36 - (7 + 8x)(2x + 3) = -38x - 57

. If the answer really is a constant, that numerator must be that constant times

(2x + 3)

\dfrac{-38x - 57}{2x + 3} = \dfrac{-19(2x + 3)}{2x + 3} = -19

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$16x^2 - 36 = 4(2x - 3)(2x + 3)$ or $(4x - 6)(4x + 6)$ or $(8x - 12)(2x + 3)$	M1	Factorise the first numerator, in any correct factorised form.	✓
$2x^2 + 7x + 6 = (2x + 3)(x + 2)$ and $x^2 - 5x - 14 = (x - 7)(x + 2)$	M1	Factorise both of the other quadratics. Allow $2x^2 + 7x + 6 = (4x + 6)(0.5x + 1)$ , since that also cancels with $(4x + 6)$ .	✓
The two fractions, divided and cancelled, give $4(2x - 3)$ or $2(4x - 6)$ or $8x - 12$ .	Note	Any one of these on its own gains the first two M marks. Award $M2$ for any fraction with a completely simplified non-linear numerator and non-linear denominator that will cancel to $-19$ .	✓
$4(2x - 3) - (7 + 8x)\ (= n)$	M1	A linear expression that should give the correct value for $n$ . Allow invisible brackets, ie $8x - 12 - 7 + 8x$ . Alternatively, an expression clearly showing the numerator is $-19$ times the denominator, eg $\dfrac{-38x - 57}{2x + 3}\ (= n)$ . This mark implies the previous M marks, as not all of the factorising is necessary.	✓
Working required: $n = -19$	A1	The integer $-19$ , with the algebraic working shown. An answer with no working scores no marks.	✓

Full marks: 4/4

$A$ , $B$ and $C$ are three control posts on level ground at an orienteering course.

$AB = 8.4$ metres, $BC = 9.2$ metres

$B$ is on a bearing of $067^\circ$ from $A$
$C$ is on a bearing of $129^\circ$ from $B$

Calculate the bearing of $A$ from $C$ .
Give your answer correct to the nearest degree. [6 marks]

[Total 6 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

AB = 8.4

BC = 9.2

Bearing of

B

from

A

067^\circ

Bearing of

C

from

B

129^\circ

Level ground, and the paper supplies no figure, so the sketch is yours to draw

Find

The bearing of

A

from

C

, correct to the nearest degree

Plan the Solution

Sketch it first: a north line at each post, then the two given bearings. Nothing can be worked out until the angle INSIDE the triangle at $B$ is known.
The north lines are parallel, so that angle comes straight from the two bearings: $67^\circ + 51^\circ$ .
Two sides with the angle between them is the cosine rule, which gives $AC$ . Then the sine rule gives the angle at $C$ .
An angle is not a bearing. Reverse the $129^\circ$ to get the bearing of $B$ from $C$ , then turn through the angle at $C$ to reach $A$ .

Worked Solution [6 marks]

Cosine rule:

b^2 = a^2 + c^2 - 2ac\cos B

. Sine rule:

\dfrac{\sin A}{a} = \dfrac{\sin B}{b}

. A back bearing is the forward bearing

\pm 180^\circ

Find the angle inside the triangle at

B

180 - 129 = 51

51 + 67 = 118

(Reason: The north lines at

A

and

B

are parallel, so the

67^\circ

A

reappears at

B

between

BA

and the southward direction (alternate angles). North round to south at

B

is a straight

180^\circ

, so

BC

lies

51^\circ

from that same southward direction. The two pieces together make angle

ABC

Use the cosine rule to find

AC^2

AC^2 = 8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos 118^\circ

70.56 + 84.64 + 72.5615 = 227.7615

(Reason: Angle

ABC

sits between the two known sides, so the cosine rule applies with no extra work.

\cos 118^\circ

is negative, which is why the third term ADDS

72.5615

instead of taking it away.)

Square root to get the length

AC

AC = \sqrt{227.7615} = 15.0918 \text{ m}

(Reason: The mark scheme accepts anything from

15

15.1

here, but leave the unrounded value in the calculator: the marks that follow depend on figures that come from correct working.)

Set up the sine rule for angle

ACB

\dfrac{\sin ACB}{8.4} = \dfrac{\sin 118^\circ}{15.0918}

(Reason: Angle

ACB

faces the side

AB = 8.4

, and the known angle

ABC = 118^\circ

faces the side

AC

just found. A known angle opposite a known side is exactly what the sine rule needs.)

Solve for angle

ACB

\sin ACB = \dfrac{8.4 \times \sin 118^\circ}{15.0918} = 0.4914

ACB = \sin^{-1} 0.4914 = 29.4^\circ

(Reason: Angle

ABC

is obtuse, so the other two angles are both acute and the inverse sine gives the one wanted with no ambiguity.)

Turn the angle at

C

into a bearing

129 + 180 = 309

309 - 29.4 = 279.6

(Reason: Reversing the

129^\circ

journey gives the bearing of

B

from

C

309^\circ

. On the sketch, turning from

CB

round to

CA

is an ANTICLOCKWISE turn of

29.4^\circ

, so the angle is taken away from

309^\circ

. Bearings are given to three figures, and to the nearest degree this is

280^\circ

The bearing of

A

from

C

280^\circ

Verification

Check 1: Drop the triangle rules and use coordinates instead. Put

A

at the origin:

B

8.4\sin 67^\circ = 7.73

east and

8.4\cos 67^\circ = 3.28

north of it, and

C

is a further

9.2\sin 129^\circ = 7.15

east and

9.2\cos 129^\circ = -5.79

north of

B

. So

A

lies

14.88

west and

2.51

north of

C

, which puts the bearing in the final quarter turn.

360^\circ - \tan^{-1}\dfrac{14.88}{2.51} = 360^\circ - 80.4^\circ = 279.6^\circ

Check 2: Go round the other way, through

A

instead of

C

. The cosine rule on the third angle gives

\cos BAC = \dfrac{8.4^2 + 227.7615 - 9.2^2}{2 \times 8.4 \times 15.0918} = 0.8428

, so

BAC = 32.6^\circ

. The bearing of

C

from

A

is therefore

67^\circ + 32.6^\circ

, and the answer is its back bearing.

(67^\circ + 32.6^\circ) + 180^\circ = 99.6^\circ + 180^\circ = 279.6^\circ

Check 3: A cheap sanity check on both angles at once: the three angles of the triangle must close to

180^\circ

. If either of the two calculated angles were wrong, this sum would miss.

118^\circ + 29.4^\circ + 32.6^\circ = 180^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$67 + 51 = 118$ , or the angle at $B$ split into $67$ and $51$	M1	A diagram showing $118$ , or $118$ used in a further calculation.	✓
$AC^2 = 8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos 118^\circ \; (= 227.7615\ldots)$	M1	A correct method to find $AC^2$ . Accept $227$ to $228$ .	✓
$AC = \sqrt{8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos 118^\circ} \; (= 15.09\ldots)$	M1	A correct method to find the length $AC$ . Accept $15$ to $15.1$ .	✓
$\dfrac{\sin ACB}{8.4} = \dfrac{\sin 118^\circ}{15.09}$ or $\cos ACB = \dfrac{9.2^2 + 15.09^2 - 8.4^2}{2 \times 9.2 \times 15.09}$	M1	Dependent on the previous method marks. A correct statement of the sine rule, or of the cosine rule, to find angle $ACB$ or angle $BAC$ . The figures used must come from correct working.	✓
$ACB = \sin^{-1}\left(\dfrac{8.4 \sin 118^\circ}{15.09}\right) \; (= 29.4\ldots)$	M1	A completely correct statement for angle $ACB$ , accept $29$ to $30$ , or for angle $BAC$ , accept $32$ to $33$ .	✓
$309 - 29.4 = 279.6$ , so the bearing is $280$	A1	Allow $279$ to $280$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Note - a scale drawing	note	A correct answer read off an accurate scale drawing gains full marks, but a slightly inaccurate one gains $0$ marks. Look for angles and lengths written on the candidate's own diagram.	✓

Full marks: 6/6

The diagram shows an equilateral triangle $ABC$ and a circle, centre $O$

Each of $AB$ , $BC$ and $CA$ is a tangent to the circle.
The circle has radius $x$ cm
The three regions shown shaded have a total area of $nx^{2}$ $\text{cm}^{2}$

Work out the value of $n$
Give your answer correct to $3$ significant figures. [5 marks]

n =

[Total 5 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

An equilateral triangle

ABC

with a circle, centre

O

, inside it.

AB

BC

and

CA

are all tangents, so the circle touches each side exactly once.

The radius of the circle is

x

cm.

The three shaded corner regions have total area

nx^{2}

\text{cm}^{2}

Find

The value of

n

, correct to

3

significant figures. Note that

x

never has to be given a value: it cancels.

Plan the Solution

A radius drawn to a point of contact is perpendicular to the tangent, so the distance from $O$ to each side is $x$ . That makes $O$ the centre of the inscribed circle.
Join $O$ to $A$ , $B$ and $C$ . Each $60^\circ$ vertex angle is bisected, so a right-angled triangle with a $30^\circ$ angle sits at every corner.
Use one of those right-angled triangles to get half a side, then the whole side, then the height.
Shaded area = area of triangle minus area of circle. Divide by $x^{2}$ and what is left is $n$ .

Worked Solution [5 marks]

Rule - Tangent and inscribed circle: the radius to a point of contact is perpendicular to the tangent, and the line from a vertex to the centre bisects that vertex angle. Shaded area = area of

ABC

minus area of the circle.

Step 1: Work out half a side from the tangent

\angle OBM = 30^\circ

\tan 30^\circ = \dfrac{OM}{MB} = \dfrac{x}{MB}

MB = \dfrac{x}{\tan 30^\circ} = x\tan 60^\circ = \sqrt{3}\,x

(Reason:

M

is the point where the circle touches

AB

. The radius to a point of contact is perpendicular to the tangent, so

OM = x

and

\angle OMB = 90^\circ

OB

bisects the

60^\circ

angle at

B

, leaving

30^\circ

Step 2: Write down the side and the perpendicular height

AB = 2 \times \sqrt{3}\,x = 2\sqrt{3}\,x

CM = MB \times \tan 60^\circ = \sqrt{3}\,x \times \sqrt{3} = 3x

(Reason: By symmetry

M

is the midpoint of

AB

, so

AB = 2MB

. In triangle

CMB

the angle at

B

60^\circ

, which gives a height of

3x

- three radii, which is what the picture suggests.)

Step 3: Area of triangle

ABC

\text{Area of }ABC = \dfrac{1}{2} \times AB \times CM

= \dfrac{1}{2} \times 2\sqrt{3}\,x \times 3x = 3\sqrt{3}\,x^{2}

\dfrac{1}{2} \times 3.46410 \times 3 = 5.19615

(Reason: In decimals

2\sqrt{3} = 3.46410

5

decimal places, so the area of

ABC

5.19615\,x^{2}

. Keeping the surd form

3\sqrt{3}\,x^{2}

until the very end avoids rounding twice.)

Step 4: Area of the circle

\text{Area of circle} = \pi x^{2}

\pi x^{2} = 3.14159\,x^{2}

(Reason: The radius is

x

, so the circle's area is

\pi x^{2}

. Both areas now carry a factor of

x^{2}

, which is exactly why the answer does not depend on the radius.)

Step 5: Subtract, then read off

n

\text{Shaded area} = 3\sqrt{3}\,x^{2} - \pi x^{2} = (3\sqrt{3} - \pi)x^{2}

5.19615 - 3.14159 = 2.05456

n = 2.05 \text{ (3 s.f.)}

(Reason: Dividing by

x^{2}

leaves

n = 3\sqrt{3} - \pi

, a pure number: the shaded area is

2.05456

times

x^{2}

whatever the radius happens to be. Rounding to

3

significant figures gives

2.05

n = 2.05

correct to

3

significant figures

Verification

Check 1: Put a real radius in and work in numbers only. With

x = 100

, the side is

200\tan 60^\circ = 346.410

and the area of

ABC

\dfrac{1}{2} \times 346.410^{2} \times \sin 60^\circ = 51961.5

. The circle has area

\pi \times 100^{2} = 31415.9

\dfrac{51961.5 - 31415.9}{100^{2}} = 2.05456

Check 2: Compare the two areas instead of subtracting them. The circle fills

\dfrac{\pi}{3\sqrt{3}} = 0.60460

of the triangle, so the shaded part is

0.39540

of it.

0.39540 \times 5.19615 = 2.05456

Check 3: Build the triangle from the six congruent right-angled triangles that

OA

OB

and

OC

create. Each has legs

x

and

\sqrt{3}\,x

, so each has area

\dfrac{\sqrt{3}}{2}x^{2}

6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3}

, the same coefficient as Step 3.

Check 4: Is the size sensible? The circle must fill most of the triangle but not all of it, so

n

must be a good deal smaller than

5.19615

and bigger than

0

0 < 2.05 < 5.19615

, and the shaded fraction

0.39540

looks right for three corners.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct tangent ratio linking the radius to half a side, e.g. $\tan 30^\circ = \dfrac{x}{0.5AB}$ or $\tan 60^\circ = \dfrac{0.5AB}{x}$ oe	M1	Also allow $0.5AB = \dfrac{x}{\tan 30^\circ}$ , $0.5AB = x\tan 60^\circ$ or $\dfrac{1}{2}AB = \sqrt{3}\,x$ . The scheme also allows the off-spec route from a known height of $3x$ : $\tan 60^\circ = \dfrac{3x}{0.5AB}$ or $\sin 60^\circ = \dfrac{3x}{BC}$ .	✓
An expression for a side, e.g. $AB = 2x\tan 60^\circ$ , $\dfrac{2x}{\tan 30^\circ}$ , $2\sqrt{3}\,x$ or $3.46\ldots x$	M1	Expression for a side of the triangle ( $AB$ or $BC$ or $AC$ ), OR the area of one or more of the six triangles, e.g. $\dfrac{1}{2} \times x\tan 60^\circ \times x$ , $\dfrac{1}{2} \times \sqrt{3}\,x \times x$ or $\dfrac{\sqrt{3}}{2}x^{2}$ .	✓
A correct expression for the area of $ABC$ , e.g. $6 \times \dfrac{1}{2} \times x\tan 60^\circ \times x$ , $\dfrac{1}{2}(2x\tan 60^\circ)^{2}\sin 60^\circ$ , $0.5 \times 2\sqrt{3}\,x \times 3x$ , $3\sqrt{3}\,x^{2}$ or $5.19\ldots x^{2}$	M1	Any equivalent correct form scores, including $\dfrac{1}{2}\left(\dfrac{2x}{\tan 30^\circ}\right)^{2}\sin 60^\circ$ and the Pythagoras route $\dfrac{1}{2} \times 2\sqrt{3}\,x \times \sqrt{(2\sqrt{3}x)^{2} - (\sqrt{3}x)^{2}}$ .	✓
A correct expression or equation for the shaded area, e.g. $3\sqrt{3}\,x^{2} - \pi x^{2}$ , $6 \times \dfrac{\sqrt{3}}{2}x^{2} - \pi x^{2}$ or $5.19\ldots x^{2} - \pi x^{2}$	M1	Triangle minus circle, correctly formed. An equation set equal to $nx^{2}$ scores here too, e.g. $3\sqrt{3}\,x^{2} - \pi x^{2} = nx^{2}$ .	✓
The value of $n$	A1	Accept $2.05$ to $2.06$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Guidance for the whole question	Note	Any value may be used for the radius provided it is used consistently: a solution that works throughout with $x = 100$ earns every mark, reaching $51961.52 - \pi \times 100^{2} = 100^{2}n$ . Examiners are also told to look for values written on the diagram.	✓

Full marks: 5/5

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 Paper 2H, November 2024: Worked Solutions, Questions 16 to 25

Worked solutions, questions 16 to 25 of 25

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