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Edexcel IGCSE 4MA1/1H, Thursday 16 May 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

6 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1H - Higher Tier - Thursday 16 May 2024100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1H (Higher Tier), June 2024 series, sat Thursday 16 May 2024 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

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

Worked solutions, questions 1 to 14 of 25

The first four terms of an arithmetic sequence are shown below.

$1 \qquad 4 \qquad 7 \qquad 10$

(a) Work out an expression, in terms of $n$ , for the $n$ th term of this sequence. [2 marks]

A different arithmetic sequence has $n$ th term $5n + 17$

(b) Work out the $12$ th term of this sequence. [1 mark]

(a)(b)

[Total 3 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

An arithmetic sequence whose first four terms are

1

4

7

10

A second, different arithmetic sequence, given by its

n

th term

5n + 17

Find

(a) An expression in terms of

n

for the

n

th term of the first sequence. (b) The

12

th term of the second sequence.

Plan the Solution

Subtract each term from the next to find the common difference $d$ of the first sequence.
Put $a$ and $d$ into the standard formula and simplify it to the form $dn + c$ .
Part (b) already gives a formula, so nothing has to be built there: substitute the position number into it.

Worked Solution [3 marks]

Rule -

n

th term of an arithmetic sequence:

a + (n - 1)d

, where

a

is the first term and

d

is the common difference. Expanded, this is always

dn + (a - d)

Step 1: find the common difference

4 - 1 = 3

7 - 4 = 3

10 - 7 = 3

(Reason: (Reason: the same gap of

3

appears every time, so the sequence really is arithmetic with

d = 3

.))

Step 2: put

a = 1

and

d = 3

into the formula

a + (n - 1)d = 1 + (n - 1) \times 3

= 1 + 3n - 3

= 3n - 2

(Reason: (Reason: the

3n

comes from the common difference, and the

-2

is the adjustment that pulls

3n

back onto the first term.))

Step 3 (part b): substitute

n = 12

into

5n + 17

5 \times 12 + 17 = 60 + 17 = 77

(Reason: (Reason: the

12

th term is the term in position

12

, so the position number is what goes in place of

n

in the formula the question hands you.))

(a)

3n - 2

(b)

77

Verification

Check 1: Put

n = 1

and

n = 4

into

3n - 2

and compare with the first and fourth terms printed in the question.

3 \times 1 - 2 = 1

and

3 \times 4 - 2 = 10

, which are exactly the first and fourth terms printed.

Check 2: Reach part (b) a different way. The first term of that sequence is

5 \times 1 + 17

, and it goes up in

5

s, so the

12

th term is

11

steps of

5

after it.

22 + 11 \times 5 = 22 + 55 = 77

Check 3: Test the shape of the part (a) answer rather than its values: consecutive terms of

3n - 2

must differ by the common difference.

(3(n + 1) - 2) - (3n - 2) = 3

, the common difference found in step 1.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correct method for the $n$ th term: $3n + k$ where $k \neq -2$ , or $3 \times n + k$ , or $n \times 3 + k$ .	M1	The multiplier of $n$ must be $3$ ; the constant $k$ may be zero or absent.	✓
(a) $3n - 2$	A1	oe, eg $1 + (n - 1)3$ or $3 \times n - 2$ or $n \times 3 - 2$ or $3x - 2$ . Working is not required, so a correct answer scores full marks unless it clearly comes from incorrect working. Allow $T_n$ , $U_n$ or $a_n$ for the $n$ th term, but only M1 for $n = 3n - 2$ oe or $x = 3x - 2$ .	✓
(b) $77$	B1	cao - correct answer only. The formula is given, so the single mark is for the value $77$ and no method is required.	✓

Full marks: 3/3

$450$ workers at a shipping depot were asked how they travelled to work on Tuesday.
Each worker walked or travelled by tram or travelled by taxi or travelled by bicycle.
Each worker used just one method of travel.

One of these workers is chosen at random.
The table shows information about the probability of each method of travel.

$\begin{array}{|c|c|c|c|c|}\hline \textbf{Method of travel} & \text{walk} & \text{tram} & \text{taxi} & \text{bicycle} \\ \hline \textbf{Probability} & 0.20 & x & 2x & 0.26 \\ \hline \end{array}$

Work out how many of the $450$ workers travelled by taxi. [4 marks]

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

450

workers, each using exactly one of four methods of travel.

The probabilities are walk

0.20

, tram

x

, taxi

2x

, bicycle

0.26

The taxi probability is exactly twice the tram probability.

Find

How many of the

450

workers travelled by taxi.

Plan the Solution

Add the four probabilities and set the total equal to $1$ .
Subtract the two known probabilities to see what $x + 2x$ is worth.
Divide by $3$ to get $x$ , then double it for the taxi probability.
Multiply the taxi probability by $450$ .

Worked Solution [4 marks]

Rule - Probability: for one set of outcomes that covers every case with no overlap, the probabilities add to

1

, and the expected number in a group is

\text{probability} \times \text{total}

Step 1: Set the four probabilities equal to 1

0.20 + x + 2x + 0.26 = 1

(Reason: Every worker used just one of the four methods, so the four outcomes cover all

450

workers exactly once and no worker twice. Probabilities like that always add to

1

Step 2: Take out the two probabilities that are known

0.20 + 0.26 = 0.46

3x = 1 - 0.46 = 0.54

(Reason: Walking and cycling account for

0.46

of the workers, so the tram and taxi groups together must account for the remaining

0.54

Step 3: Find

x

, then the taxi probability

\dfrac{0.54}{3} = 0.18

2 \times 0.18 = 0.36

(Reason: The two unknown cells are

x

and

2x

, which is

3x

altogether, so

x = 0.18

. The taxi cell is twice that, so the taxi probability is

0.36

Step 4: Turn the probability into a number of workers

0.36 \times 450 = 162

(Reason: Multiplying a probability by the total number of workers gives how many of the

450

fall into that group.)

162

workers travelled by taxi

Verification

Check 1: Put

x = 0.18

back into the table and add all four probabilities.

0.20 + 0.18 + 0.36 + 0.26 = 1

Check 2: Work in people instead of probabilities. Walking gives

0.20 \times 450 = 90

and cycling gives

0.26 \times 450 = 117

, so take those away and split what is left in the ratio

1 : 2

450 - 207 = 243

, and

\dfrac{2}{3} \times 243 = 162

Check 3: Add the four group sizes and see whether they come back to

450

90 + 81 + 162 + 117 = 450

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 - (0.20 + 0.26)$ or $0.54$ oe, or $0.20 + x + 2x + 0.26 = 1$ oe, or $x + 2x = 0.54$ oe	M1	Showing clear understanding that the total of the probabilities is $1$ . If the probabilities are given as percentages then the per cent sign must be seen.	✓
$\dfrac{0.54}{3} = 0.18$ or $\dfrac{2}{3} \times 0.54 = 0.36$ or $0.54 \times 450 = 243$	M1	For a correct method to find $x$ or $2x$ .	✓
$(2 \times) \, 0.18 \times 450$ oe, or $81$ , or $0.36 \times 450$ oe	M1	Or for $\dfrac{81}{450}$ or $\dfrac{162}{450}$ .	✓
$162$	A1	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Alternative method, carrying the same four marks	Note	The same marks are available for working in people throughout: $(0.2 \times 450) + (0.26 \times 450) = 207$ for the first mark, $450 - 207 = 243$ for the second, and $\dfrac{2}{3} \times 243 = 162$ for the third and the answer.	✓

Full marks: 4/4

Work out the highest common factor (HCF) of $72$ and $108$
You must show your working clearly. [2 marks]

[Total 2 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

The two numbers

72

and

108

Working must be shown, so the method itself carries a mark.

Find

The highest common factor (HCF) of

72

and

108

, the largest number that divides into both of them exactly

Plan the Solution

Break $72$ down into a product of prime factors.
Do the same for $108$ .
Compare the two products and keep only the primes that appear in both, each taken to the lower power.
Multiply those shared prime factors together to give the HCF.

Worked Solution [2 marks]

Rule - HCF from prime factors: write each number as a product of primes, then multiply the primes the two numbers share, each one taken to the LOWER power it appears with. Taking the higher power would give the LCM instead.

Step 1: Write

72

as a product of prime factors

72 = 8 \times 9

72 = 2 \times 2 \times 2 \times 3 \times 3

72 = 2^3 \times 3^2

(Reason: halving repeatedly strips out every factor of

2

, and what is left,

9

, is a power of

3

)

Step 2: Write

108

as a product of prime factors

108 = 4 \times 27

108 = 2 \times 2 \times 3 \times 3 \times 3

108 = 2^2 \times 3^3

(Reason: the same method on the second number, which stops at two twos because

27

is odd)

Step 3: Take the lower power of each shared prime

2^2 = 4

3^2 = 9

(Reason: the HCF has to divide BOTH numbers, so each prime may be used only as often as it appears in the number that has fewer of it, and here the twos are limited by

108

while the threes are limited by

72

)

Step 4: Multiply the shared prime factors together

2^2 \times 3^2 = 4 \times 9 = 36

(Reason: every prime factor in this product sits inside both numbers, so the product divides both, and nothing larger can)

36

Verification

Check 1: Divide each of the original numbers by

36

and confirm that both divisions come out exactly.

\dfrac{72}{36} = 2

and

\dfrac{108}{36} = 3

, both whole numbers, so

36

really is a common factor.

Check 2: List every factor of each number and compare the two lists. This is the mark scheme's alternative method and it uses no prime factors at all. The factors shared by both numbers are

1, 2, 3, 4, 6, 9, 12, 18

and

36

, and the largest of them is

36

Check 3: For any two numbers, the HCF multiplied by the LCM equals the product of the numbers. The LCM here takes the HIGHER power of each prime,

2^3 \times 3^3 = 216

36 \times 216 = 7776

and

72 \times 108 = 7776

, so the two sides agree.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Any correct valid method with no errors, eg $72 = 2 \times 2 \times 2 \times 3 \times 3$ and $108 = 2 \times 2 \times 3 \times 3 \times 3$	M1	Also earned by starting to list at least four different factors of each number with no errors, by a factor tree or a ladder diagram, by a fully correct Venn diagram, or by any other clear method such as a table. Ignore $1$ in a list of prime factors.	✓
$36$	A1	Dependent on the method mark. Accept $2^2 \times 3^2$ or equivalent. Working is required, so an answer written down with no method earns nothing.	✓

Full marks: 2/2

Rosa records the number of kilometres her delivery van travels each month.

In April, the van travelled $943$ kilometres.
This is $15\%$ more than the number of kilometres the van travelled in March.

Work out the number of kilometres the van travelled in March. [3 marks]

kilometres

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

April's total is

943

kilometres.

April is

15\%

more than March, so the

15\%

is an increase ON THE MARCH FIGURE, not on April's.

Find

The number of kilometres the van travelled in March.

Plan the Solution

This is a reverse percentage: the $943$ is the answer to an increase, not the starting point.
Turn the $15\%$ increase into a single multiplier.
Write March as an unknown, multiply it by that multiplier to make $943$ , then divide to undo the increase.
Finish by increasing the answer by $15\%$ again - it must come back to $943$ .

Worked Solution [3 marks]

Rule - Reverse percentage: an increase multiplies, so undoing it divides. If

\text{new} = \text{original} \times \text{multiplier}

, then

\text{original} = \dfrac{\text{new}}{\text{multiplier}}

Step 1: Turn the

15\%

increase into a multiplier

100\% + 15\% = 115\%

115\% = \dfrac{115}{100} = 1.15

(Reason: An increase keeps the original

100\%

and adds

15\%

on top, so the whole of April is

115\%

of March.)

Step 2: Write the equation for April

1.15m = 943

(Reason: Let

m

be the number of kilometres in March. Multiplying

m

1.15

is what produced April's

943

kilometres.)

Step 3: Divide by the multiplier to undo the increase

m = \dfrac{943}{1.15} = 820

(Reason: Dividing by

1.15

reverses the increase and takes April back to March. Taking

15\%

off April is a different calculation, because that

15\%

would be measured against the wrong month.)

820

kilometres

Verification

Check 1: Put the answer back into the question: increase

820

15\%

and see whether April's figure returns.

0.15 \times 820 = 123

and

820 + 123 = 943

Check 2: Reach the same answer without the multiplier. If

115\%

943

kilometres, find

1\%

and then

100\%

. This is the mark scheme's second route.

\dfrac{943}{115} = 8.2

8.2 \times 100 = 820

Check 3: Test the direction. March must be the smaller month, and the gap between the two months must be

15\%

of the MARCH figure.

943 - 820 = 123

and

\dfrac{123}{820} = 0.15 = 15\%

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct first step: $1 + 0.15 = 1.15$ or $x + 0.15x = 943$ or $100\% + 15\% = 115\%$ or $\dfrac{943}{115} = 8.2$ oe	M1	Any one of these earns the first mark: the multiplier, an equation in the March value, the percentage total, or one per cent of April. Recognising that April is $115\%$ of March is the whole of this mark.	✓
A complete method to reverse the increase: $\dfrac{943}{1.15}$ or $\dfrac{943}{115} \times 100$ or $943 \times \dfrac{100}{115}$ oe or $8.2 \times 100$	M1	Awarded on the candidate's own multiplier or own value of $1\%$ from the first row, so the figures in quotation marks on the official scheme need not be correct. This mark is dependent on the first.	✓
$820$	A1	Working not required, so the correct answer scores full marks (unless it comes from obvious incorrect working). An answer of $801.55$ earns nothing here: it comes from taking $15\%$ off April instead of dividing by $1.15$ .	✓

Full marks: 3/3

The diagram shows a regular pentagon $ABCDE$ .
A straight line is drawn from the vertex $E$ to the point $F$ , as shown.

Angle $AEF = 96^\circ$

Work out the size of the obtuse angle $FED$ .
Show your working clearly. [4 marks]

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A regular pentagon

ABCDE

A straight line

EF

drawn from the vertex

E

, with

F

outside the pentagon.

Angle

AEF = 96^\circ

Find

The size of the obtuse angle

FED

Plan the Solution

Three angles meet at the point $E$ : the given angle $AEF$ , the pentagon's own angle $AED$ , and the angle $FED$ that is wanted.
So the only missing piece is the interior angle of a regular pentagon. Get it from the angle sum $(n - 2) \times 180^\circ$ .
Then take both known angles away from the full turn of $360^\circ$ .

Worked Solution [4 marks]

Angles at a point add up to

360^\circ

. Each interior angle of a regular

n

-sided polygon is

\dfrac{(n - 2) \times 180^\circ}{n}

Step 1: Find the angle sum of the pentagon

(5 - 2) \times 180 = 540

(Reason: A pentagon splits into

3

triangles from one vertex, and each triangle holds

180^\circ

, so its five interior angles add up to

540^\circ

Step 2: Find one interior angle of the regular pentagon

\dfrac{540}{5} = 108

\angle AED = 108^\circ

(Reason: Regular means all five interior angles are equal, so share the

540^\circ

between the

5

vertices. The pentagon's own angle at

E

is the angle

AED

, because

A

and

D

are the vertices either side of

E

Step 3: Use the angles at the point E

\angle AEF + \angle AED + \angle FED = 360^\circ

360 - 96 - 108 = 156

\angle FED = 156^\circ

(Reason: The three angles at

E

make one complete turn, so subtract the given

96^\circ

and the pentagon's

108^\circ

from

360^\circ

\angle FED = 156^\circ

Verification

Check 1: Work with exterior angles instead, so the interior angle is never used. Extend

AE

beyond

E

: the angle between that extension and

ED

is the exterior angle

\dfrac{360}{5} = 72

, and the angle between the extension and

EF

180 - 96 = 84

72 + 84 = 156

Check 2: Add the three angles that meet at

E

and see whether they close one full turn.

96 + 108 + 156 = 360

Check 3: The question asks for the obtuse angle, so a correct answer has to sit between

90^\circ

and

180^\circ

. Going round the point the other way gives

360 - 156 = 204

, which is reflex, not obtuse.

156^\circ

is obtuse, so it is the angle the question wants

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(5 - 2) \times 180 = 540$ or $\dfrac{360}{5} = 72$	M1	A correct first step towards an angle of the regular pentagon: either the angle sum or the exterior angle. If angles are written on the diagram they must come from correct working and be correctly assigned.	✓
$\dfrac{540}{5} = 108$ or $180 - 72 = 108$ or $180 - 96 = 84$	M1	Any one of these: the interior angle of the pentagon, or the angle between $EF$ and $AE$ extended. Follow through on the candidate's own value from the first mark.	✓
$72 + 84$ or $360 - (96 + 108)$ or $180 - (108 - 84)$	M1	A complete method that reaches the required angle, again on the candidate's own earlier values.	✓
$156$	A1	Working is required on this question, so an answer given with no working scores nothing.	✓

Full marks: 4/4

(a) Multiply out the brackets and simplify
$(m + 5)(m - 8)$ [2 marks]

(b) Solve the equation
$3n - 4 = \dfrac{5n + 6}{3}$
You must show clear algebraic working. [3 marks]

(a)(b) n =

[Total 5 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Part (a): the product

(m + 5)(m - 8)

, two brackets to be multiplied out.

Part (b): the equation

3n - 4 = \dfrac{5n + 6}{3}

, which has a fraction on one side only.

Find

Part (a): the expansion, written as a simplified quadratic in

m

. Part (b): the value of

n

that makes both sides equal. The equation is linear, so expect exactly one value.

Plan the Solution

Part (a): multiply each term in the first bracket by each term in the second. That gives $4$ terms, and only the two $m$ terms can be combined.
Part (b): the fraction is the obstacle, so multiply every term on both sides by $3$ to clear it.
Then collect the $n$ terms on one side and the number terms on the other, and divide by the coefficient of $n$ .

Worked Solution [5 marks]

Rule - Expanding a pair of brackets:

(x + a)(x + b) = x^2 + (a + b)x + ab

. Rule - Clearing a fraction from an equation: multiply every term on both sides by the denominator, so the equation stays balanced.

Part (a), Step 1: multiply every term in the first bracket by every term in the second

(m + 5)(m - 8) = m \times m + m \times (-8) + 5 \times m + 5 \times (-8)

= m^2 - 8m + 5m - 40

(Reason: Each of the two terms in

(m + 5)

meets each of the two terms in

(m - 8)

, so a correct expansion always has

4

terms before anything is simplified. Keeping the

-8

with its sign is what stops the middle term coming out wrong.)

Part (a), Step 2: collect the like terms

m^2 - 8m + 5m - 40 = m^2 - 3m - 40

(Reason: The two middle terms are like terms, because both are a number of

m

, so they combine into a single

m

term. The

m^2

term and the number term have nothing to pair with, so they are unchanged.)

Part (b), Step 1: clear the fraction

3 \times (3n - 4) = 3 \times \dfrac{5n + 6}{3}

9n - 12 = 5n + 6

(Reason: Multiplying by

3

cancels the denominator on the right. On the left the bracket must be multiplied out, so both

3n

and

-4

are tripled, not just the first term.)

Part (b), Step 2: collect the n terms on one side and the numbers on the other

9n - 5n = 6 + 12

4n = 18

(Reason: Subtracting

5n

from both sides and adding

12

to both sides keeps the equation balanced, and leaves one

n

term against one number.)

Part (b), Step 3: divide by the coefficient of n

n = \dfrac{18}{4} = \dfrac{9}{2}

(Reason: Dividing both sides by

4

leaves

n

on its own.

\dfrac{18}{4}

cancels by

2

\dfrac{9}{2}

, which is

4.5

. A fraction is a perfectly acceptable final form here.)

(a)

m^2 - 3m - 40

(b)

n = \dfrac{9}{2}

, that is

4.5

Verification

Check 1: Part (a): put

m = 2

into the original product and into the simplified answer. They must give the same number.

(2 + 5)(2 - 8) = 7 \times (-6) = -42

and

2^2 - 3 \times 2 - 40 = 4 - 6 - 40 = -42

Check 2: Part (a): a second value,

m = -1

, in case

m = 2

agreed by luck. A sign slip in the middle term shows up here.

(-1 + 5)(-1 - 8) = 4 \times (-9) = -36

and

(-1)^2 - 3 \times (-1) - 40 = 1 + 3 - 40 = -36

Check 3: Part (b): substitute

n = \dfrac{9}{2}

back into both sides of the original equation. Left side

3 \times 4.5 - 4 = 9.5

and right side

\dfrac{5 \times 4.5 + 6}{3} = \dfrac{28.5}{3} = 9.5

Check 4: Part (b): solve a second way, by splitting the fraction instead of clearing it, so the two methods are independent.

3n - 4 = \dfrac{5}{3}n + 2

gives

\dfrac{4}{3}n = 6

, so

n = 6 \times \dfrac{3}{4} = \dfrac{9}{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $m^2 - 8m + 5m - 40$	M1	For any $3$ correct terms out of the $4$ , or for all $4$ terms correct ignoring signs, or for $m^2 - 3m$ , or for $-3m - 40$ .	✓
(a) $m^2 - 3m - 40$	A1	Working is not required in this part, so a correct answer scores full marks, unless it comes from obviously incorrect working.	✓
(b) $9n - 12 = 5n + 6$	M1	For removal of the fraction and multiplying out the left-hand side. Or for separating the fraction on the right, as in $3n - 4 = \dfrac{5}{3}n + \dfrac{6}{3}$ or equivalent.	✓
(b) $9n - 5n = 6 + 12$ or $4n = 18$	M1ft	Follow through, dependent on a $4$ term equation, for correctly rearranging their $4$ term equation so that the $n$ terms are on one side and the number terms on the other. Equivalents such as $-12 - 6 = 5n - 9n$ or $n = \dfrac{-18}{-4}$ also score.	✓
(b) $n = \dfrac{9}{2}$	A1	Dependent on M2. Or equivalent, for example $\dfrac{18}{4}$ or $4.5$ or $4\dfrac{1}{2}$ . Working is required in this part.	✓

Full marks: 5/5

$\mathcal{E} = \{23,\ 24,\ 25,\ 26,\ 27,\ 28,\ 29,\ 30,\ 31,\ 32,\ 33,\ 34\}$
$A = \{\text{even numbers}\}$
$B = \{23,\ 29,\ 31\}$
$C = \{\text{multiples of }3\}$

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

(b) Is it true that $B \cap C = \varnothing$ ?
Tick one box below.
Then write down a reason for your answer. [1 mark]

The set $D$ has $4$ members, and $D \cap (A \cup C) = \varnothing$

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

[Total 5 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

The universal set

\mathcal{E} = \{23,\ 24,\ 25,\ 26,\ 27,\ 28,\ 29,\ 30,\ 31,\ 32,\ 33,\ 34\}

A = \{\text{even numbers}\}

B = \{23,\ 29,\ 31\}

and

C = \{\text{multiples of }3\}

, each taken from inside

\mathcal{E}

For part (c): the set

D

has

4

members and

D \cap (A \cup C) = \varnothing

Find

(a)(i) the members of

B \cup C

(a)(ii) the members of

A' \cap C

(b) whether

B \cap C = \varnothing

is true, with a reason (c) the members of

D

Plan the Solution

Write $A$ and $C$ out in full first. Both are described in words, and only members of $\mathcal{E}$ count.
Then read each symbol as an instruction: $\cup$ collects everything in either set, $\cap$ keeps only what is in both, and $A'$ is everything in $\mathcal{E}$ that is not in $A$ .
For (c), a set that meets $A \cup C$ in nothing at all can only be made from what is left over, so list $A \cup C$ and take the rest of $\mathcal{E}$ .

Worked Solution [5 marks]

Rule - Set notation:

\cup

means in one set or the other,

\cap

means in both sets at once,

A'

means inside

\mathcal{E}

but outside

A

, and

\varnothing

is the set with no members.

Step 1: write

A

and

C

out as lists

A = \{24,\ 26,\ 28,\ 30,\ 32,\ 34\}

C = \{24,\ 27,\ 30,\ 33\}

(Reason: (Reason: both sets are described in words, so they must be turned into lists before any symbol can be used. Only the numbers

23

34

are available, so the multiples of

3

stop at

33

.))

Step 2: part (a)(i), collect everything in

B

or in

C

B \cup C = \{23,\ 29,\ 31\} \cup \{24,\ 27,\ 30,\ 33\}

B \cup C = \{23,\ 24,\ 27,\ 29,\ 30,\ 31,\ 33\}

(Reason: (Reason: a union keeps every member of either set, written once each and in order. Nothing appears in both lists here, so all

7

members survive.))

Step 3: part (a)(ii), write down

A'

and then intersect it with

C

A' = \{23,\ 25,\ 27,\ 29,\ 31,\ 33\}

A' \cap C = \{23,\ 25,\ 27,\ 29,\ 31,\ 33\} \cap \{24,\ 27,\ 30,\ 33\}

A' \cap C = \{27,\ 33\}

(Reason: (Reason:

A'

is everything in

\mathcal{E}

that is not even, which leaves the odd numbers. The intersection then keeps only those members that appear in both lists.))

Step 4: part (b), test

B

against

C

member by member

23 = 3 \times 7 + 2

29 = 3 \times 9 + 2

31 = 3 \times 10 + 1

B \cap C = \varnothing

(Reason: (Reason: every member of

B

leaves a remainder when divided by

3

, so not one of them is a multiple of

3

. The two sets therefore share no member, and an intersection with no members is the empty set, so the statement is true. Tick Yes.))

Step 5: part (c), list

A \cup C

A \cup C = \{24,\ 26,\ 28,\ 30,\ 32,\ 34\} \cup \{24,\ 27,\ 30,\ 33\}

A \cup C = \{24,\ 26,\ 27,\ 28,\ 30,\ 32,\ 33,\ 34\}

(Reason: (Reason:

24

and

30

are in both lists but are written once each, so this union has

8

members.))

Step 6: take what is left of

\mathcal{E}

to get

D

(A \cup C)' = \{23,\ 25,\ 29,\ 31\}

D = \{23,\ 25,\ 29,\ 31\}

(Reason: (Reason:

D

shares nothing with

A \cup C

, so every member of

D

must come from the

4

numbers of

\mathcal{E}

that the union left behind. That is exactly the

4

members the question asks for, so there is only one possible set.))

(a)(i)

B \cup C = \{23,\ 24,\ 27,\ 29,\ 30,\ 31,\ 33\}

(a)(ii)

A' \cap C = \{27,\ 33\}

(b) Yes - no member of

B

is a multiple of

3

, so

B \cap C

has no members(c)

D = \{23,\ 25,\ 29,\ 31\}

Verification

Check 1: Count the union in (a)(i) instead of reading it.

B

has

3

members and

C

has

4

, and no number is in both, so the union must hold

3 + 4 = 7

members. The listed answer has exactly

7

members

Check 2: Test the two conditions in (a)(ii) separately on every multiple of

3

. Of

24

27

30

and

33

, the even ones are barred by

A'

24

and

30

are even, so only

27

and

33

pass both tests

Check 3: Count (c) a second way, and test each member.

\mathcal{E}

has

12

members and

A \cup C

has

8

, so

12 - 8 = 4

are left, and each should be odd and not a multiple of

3

23

25

29

and

31

are all odd, none is a multiple of

3

, and there are

4

of them, as the question states

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $B \cup C = \{23,\ 24,\ 27,\ 29,\ 30,\ 31,\ 33\}$	B1	All seven members listed, in any order, with no repeats.	✓
(a)(ii) $A' \cap C = \{27,\ 33\}$	B1	Both members listed, in any order, with no repeats.	✓
(b) Yes, together with a reason such as: no member of $B$ is a multiple of $3$ , so the sets have nothing in common	B1	B1 for Yes and a statement showing the correct meanings of intersection and of the empty set. If no box is ticked, the word Yes must appear in the written answer. Members, numbers, values or elements are all accepted wording.	✓
(c) $D = \{23,\ 25,\ 29,\ 31\}$	B2	B2 for the four correct numbers and no additions. B1 for three correct values with no more than one incorrect, or for four correct values with no more than one incorrect.	✓

Full marks: 5/5

A cylindrical metal drum is stood upright on a workbench.

The volume of the drum is $1575$ cm³.
The force the drum exerts on the workbench is $84$ newtons.

$\text{pressure} = \dfrac{\text{force}}{\text{area}}$

Work out the pressure on the workbench due to the drum. [3 marks]

newtons/cm²

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

A cylindrical drum standing upright, of volume

1575

cm³.

Its height is

21

cm, marked on the figure.

The force it exerts on the workbench is

84

newtons.

\text{pressure} = \dfrac{\text{force}}{\text{area}}

Find

The pressure on the workbench, in newtons/cm². The area in that formula is the area of the face touching the workbench, so the circular end of the drum is what has to be found first.

Plan the Solution

A cylinder is a prism, so its volume is the area of its circular end multiplied by its height.
So the area of that end comes from dividing $1575$ by $21$ , and the radius is never needed.
Then put the force and that area into the pressure formula and divide.

Worked Solution [3 marks]

Rule - Cylinder and pressure:

\text{volume} = \text{cross-sectional area} \times \text{height}

, so

\text{area} = \dfrac{\text{volume}}{\text{height}}

, and then

\text{pressure} = \dfrac{\text{force}}{\text{area}}

Step 1: Find the area of the circular end

1575 = \text{area} \times 21

\text{area} = \dfrac{1575}{21} = 75

(Reason: The drum is a prism, so its volume is the area of the end multiplied by the height. Reversing that gives an end of area

75

cm², and no radius is needed to get there.)

Step 2: Put the force and the area into the formula

\text{pressure} = \dfrac{\text{force}}{\text{area}} = \dfrac{84}{75}

(Reason: The whole

84

newtons is carried by the

75

cm² of circle that touches the workbench.)

Step 3: Work out the division

\dfrac{84}{75} = 1.12

(Reason: This is the share of the force carried by each square centimetre of the workbench, which is what pressure measures.)

1.12

newtons/cm²

Verification

Check 1: Multiply back. If the pressure is right, multiplying it by the area must return the force the question gives.

1.12 \times 75 = 84

newtons, which is the force in the question.

Check 2: Take the long way round, through the radius:

r = \sqrt{\dfrac{1575}{21\pi}} \approx 4.886

cm.

\pi \times 4.886^2 \approx 75

cm², the same area, so the pressure is again

\dfrac{84}{75} = 1.12

Check 3: A size check.

84

newtons spread over

75

cm² is a little more than one newton for each square centimetre, because

84

is a little more than

75

. An answer of

1.12

sits just above

1

, which is the size expected.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1575 = \text{area} \times 21$ , or $\text{area} = 75$ , or $1575 = \pi \times r^2 \times 21$ , or $r^2 = \dfrac{1575}{21\pi} \approx 23.87$ , or $r = \sqrt{\dfrac{1575}{21\pi}} \approx 4.886$	M1	for finding the area using $\text{volume} = \text{cross-sectional area} \times \text{height}$ , or for finding $r$ or $r^2$ using $\text{volume} = \pi r^2 h$ . Note that $r^2$ and $r$ may be rounded or truncated.	✓
$\dfrac{84}{75}$ , or $\dfrac{84}{\pi \times 4.88^2}$ , or $\dfrac{84}{\pi \times 23.8}$	M1	for $\dfrac{84}{\text{area of circle}}$ , worked on their own area from the first mark.	✓
$1.12$	A1	accept $1.06$ to $1.121$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

The table gives the amount of maize produced by each of two countries in $2020$

$\begin{array}{|c|c|}\hline \textbf{Country} & \textbf{Amount of maize (tonnes)} \\ \hline \text{Ukraine} & 3.5 \times 10^{7} \\ \hline \text{Portugal} & 8.2 \times 10^{5} \\ \hline \end{array}$

(a) Write $3.5 \times 10^{7}$ as an ordinary number. [1 mark]

In $2020$ , Romania produced $6\,780\,000$ more tonnes of maize than Portugal.

(b) Work out the amount of maize Romania produced in $2020$
Give your answer in standard form. [2 marks]

(a)(b) tonnes

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A table of the maize produced by two countries in

2020

: Ukraine

3.5 \times 10^{7}

tonnes, Portugal

8.2 \times 10^{5}

tonnes.

Romania produced

6\,780\,000

more tonnes than Portugal, so Romania's amount is not in the table.

Find

(a) The value

3.5 \times 10^{7}

written as an ordinary number. (b) Romania's amount, written in standard form.

Plan the Solution

Part (a) is a conversion, not a calculation: $\times 10^{7}$ moves every digit seven places to the left, so write the digits down and fill the gap with zeros.
Part (b) mixes a standard form number with an ordinary one, so put both into the same form before adding anything.
Add, then convert the total back to standard form, checking the front number really does sit between $1$ and $10$ .

Worked Solution [3 marks]

Rule - Standard form:

a \times 10^{n}

means the digits of

a

shifted

n

places to the left, and a number is only in standard form when

1 \leq a < 10

Step 1: read what the power of ten is worth

10^{7} = 10\,000\,000

(Reason: (Reason: the index

7

counts the zeros, so

10^{7}

is ten million.))

Step 2: multiply the front number by that power

3.5 \times 10\,000\,000 = 35\,000\,000

(Reason: (Reason: multiplying by

10^{7}

shifts

3.5

seven places to the left, which turns the

3.5

into

35

followed by six zeros.))

Step 3: write Portugal's amount as an ordinary number

8.2 \times 10^{5} = 820\,000

(Reason: (Reason: the extra tonnes are given as an ordinary number, so Portugal's amount has to be an ordinary number too before the two can be added.))

Step 4: add the extra tonnes Romania produced

820\,000 + 6\,780\,000 = 7\,600\,000

(Reason: (Reason: Romania produced

6\,780\,000

tonnes more than Portugal, and more means add.))

Step 5: put the total back into standard form

7\,600\,000 = 7.6 \times 10^{6}

(Reason: (Reason: the point moves six places to sit after the

7

, and

7.6

is between

1

and

10

, so this is standard form. Writing

76 \times 10^{5}

would be the same number but not standard form.))

(a)

35\,000\,000

(b)

7.6 \times 10^{6}

tonnes

Verification

Check 1: Work part (b) backwards. Take the extra

6\,780\,000

tonnes off the answer and the table's figure for Portugal should come back.

7\,600\,000 - 6\,780\,000 = 820\,000 = 8.2 \times 10^{5}

Check 2: Undo part (a) by dividing the ordinary number by

10^{7}

. A correct conversion gives back the

3.5

the table prints.

\dfrac{35\,000\,000}{10^{7}} = 3.5

Check 3: Add in hundred-thousands instead, so neither number is ever written out in full:

6\,780\,000

67.8

lots of

10^{5}

(8.2 + 67.8) \times 10^{5} = 76 \times 10^{5} = 7.6 \times 10^{6}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write $3.5 \times 10^{7}$ as an ordinary number	B1	$35\,000\,000$ cao	✓
(b) Add the extra tonnes to Portugal's amount	M1	$8.2 \times 10^{5} + 6\,780\,000$ oe, or $820\,000 + 6\,780\,000$ oe, or $7\,600\,000$ or $76 \times 10^{5}$ oe, or $7.6 \times 10^{n}$ where $n \neq 6$ . Allow a correct mixture of ordinary numbers and standard form numbers.	✓
(b) Give the total in standard form	A1	$7.6 \times 10^{6}$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Simplify $(2p)^{0}$ , given that $p > 0$ . [1 mark]

$y^{9} \times y^{-3} = y^{n}$

(b) Work out the value of $n$ . [1 mark]

(a)(b) n =(c)

[Total 4 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

(a) the expression

(2p)^{0}

, with

p > 0

, so the bracket is never zero

(b)

y^{9} \times y^{-3} = y^{n}

, a power of

y

multiplied by another power of

y

(5a^{4}c^{2})^{3}

, a product of three factors inside one bracket

Find

(a)

(2p)^{0}

in its simplest form (b) the value of the index

n

(c)

(5a^{4}c^{2})^{3}

written as one simplified product

Plan the Solution

Each part is one index law, so name the law first and then apply it.
(a) uses the zero index. The condition $p > 0$ is there to guarantee the bracket is not zero, which is the one case the law excludes.
(b) is a multiplication of powers of the same letter, so the indices are ADDED. The second index is negative, so adding it makes the result smaller.
(c) is a bracket raised to a power, so every factor inside is raised to that power: the number $5$ and both letters. The indices on the letters are MULTIPLIED, not added.
Finish (c) by writing the number first, then $a$ , then $c$ , as a single product.

Worked Solution [4 marks]

Index laws:

x^{0} = 1

for any

x \neq 0

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

(x^{m})^{n} = x^{mn}

and

(kx)^{n} = k^{n}x^{n}

Step 1: part (a), use the zero index

(2p)^{0} = 1

(Reason: (Reason: anything that is not zero raised to the power

0

equals

1

. Here

p > 0

, so

2p

is a positive number and the law applies. The

2

does not survive: the whole bracket carries the index, not just the

p

.))

Step 2: part (b), add the indices

y^{9} \times y^{-3} = y^{9 + (-3)} = y^{6}

n = 6

(Reason: (Reason: multiplying powers of the same letter adds the indices. Adding

-3

is the same as subtracting

3

, so

9 + (-3) = 6

. Comparing

y^{6}

with

y^{n}

gives the index

n

directly.))

Step 3: part (c), give every factor inside the bracket the power

3

(5a^{4}c^{2})^{3} = 5^{3} \times (a^{4})^{3} \times (c^{2})^{3}

(Reason: (Reason: the bracket is a product of three things,

5

a^{4}

and

c^{2}

, and cubing a product cubes each of them. Forgetting to cube the

5

is the commonest way to lose a mark here.))

Step 4: work out the three factors

5^{3} = 5 \times 5 \times 5 = 125

(a^{4})^{3} = a^{4 \times 3} = a^{12}

(c^{2})^{3} = c^{2 \times 3} = c^{6}

(Reason: (Reason: a power raised to another power multiplies the indices, because

a^{4}

is being written down

3

times. The number is different:

5

has no index of its own to combine, so it is simply cubed to

125

.))

Step 5: write it as one product

(5a^{4}c^{2})^{3} = 125a^{12}c^{6}

(Reason: (Reason: the number goes first, then the letters in alphabetical order. Nothing can be added or cancelled here because

125

a^{12}

and

c^{6}

are multiplied, not added, so this is fully simplified.))

(a)

1

(b)

n = 6

(c)

125a^{12}c^{6}

Verification

Check 1: Part (a) with a number. Put

p = 5

, so the bracket is

2 \times 5 = 10

, which is not zero.

10^{0} = 1

, and the same happens for every positive

p

Check 2: Part (b) with a number. Put

y = 2

. Then

y^{9} = 512

and

y^{-3} = \dfrac{1}{8}

512 \times \dfrac{1}{8} = 64 = 2^{6}

, so

n = 6

Check 3: Part (c) with numbers. Put

a = 2

and

c = 3

. The bracket is

5 \times 16 \times 9 = 720

, so the left-hand side is

720^{3}

720^{3} = 373\,248\,000

and

125 \times 2^{12} \times 3^{6} = 373\,248\,000

, so the two forms agree.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $1$	B1	cao. The whole bracket carries the index, so $2$ on its own scores nothing.	✓
(b) $n = 6$	B1	Accept $y^{6}$ .	✓
(c) $125a^{12}c^{6}$	B2	Fully simplified. Multiplication signs between the terms are accepted, so $125 \times a^{12} \times c^{6}$ also scores B2.	✓
(c) a product in the form $ka^{p}c^{q}$ with $2$ of $k$ , $p$ , $q$ correct	B1	Partial credit, eg $5a^{12}c^{6}$ where the coefficient was not cubed. Allow $125a^{12}$ or $125c^{6}$ or $a^{12}c^{6}$ , as long as they are not added to any other terms.	✓

Full marks: 4/4

The diagram shows a wooden frame that supports a large advertising sign.

The frame is made from four lengths of timber, $AB$ , $AC$ , $BC$ and $MC$
$AC = BC = 9$ m $AB = 12$ m
angle $AMC = 90^\circ$

Martin is going to buy lengths of timber to make the frame.

The timber costs $21.50$ euros per metre.
Each length of timber he buys has to be a whole number of metres.

Work out the total cost of the timber Martin needs to buy.
Show your working clearly. [4 marks]

euros

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A triangular frame

ABC

with

AC = BC = 9

m and

AB = 12

A strut

MC

from the apex to the point

M

AB

, with angle

AMC = 90^\circ

Timber costs

21.50

euros per metre, and every length bought is a whole number of metres.

Find

The total cost, in euros, of the four lengths

AB

AC

BC

and

MC

Plan the Solution

The two sloping sides are equal and the strut is perpendicular to the base, so $M$ is the midpoint of $AB$ and triangle $AMC$ is right-angled.
Use Pythagoras' theorem in triangle $AMC$ to find the length of the strut $MC$ .
Round that length UP to a whole number of metres, because a shorter length would not reach.
Add the four lengths, then multiply the total by the price of one metre.

Worked Solution [4 marks]

Rule - Pythagoras' theorem: in a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides, so a shorter side is

\sqrt{\text{hypotenuse}^2 - \text{other short side}^2}

Step 1: Find

AM

AM = MB = \dfrac{12}{2} = 6 \text{ m}

(Reason:

AC = BC

and angle

AMC = 90^\circ

, so triangles

AMC

and

BMC

are congruent and

M

is the midpoint of

AB

Step 2: Use Pythagoras' theorem to find

MC

MC^2 = 9^2 - 6^2 = 81 - 36 = 45

MC = \sqrt{45} = 3\sqrt{5} = 6.7082\ldots \text{ m}

(Reason: Triangle

AMC

is right-angled at

M

, and the

9

m side

AC

is its hypotenuse, so the strut is the square root of

81 - 36

Step 3: Round the strut up to a whole number of metres

6 < 6.7082\ldots < 7

(Reason: Every length bought is a whole number of metres. A

6

m length would be too short to reach, so Martin has to buy a

7

m length for the strut.)

Step 4: Add the four lengths

7 + 9 + 9 + 12 = 37 \text{ m}

(Reason: The four lengths he buys are the strut

MC

, the two sloping sides

AC

and

BC

, and the base

AB

Step 5: Work out the cost

37 \times 21.50 = 795.50

(Reason: Each metre of timber costs

21.50

euros, so multiply the total length by

21.50

795.50

euros

Verification

Check 1: Put the strut back into Pythagoras' theorem. Its square is

45

, and half the base squared is

36

, so

6^2 + 45 = 36 + 45 = 81

and

81 = 9^2

. The hypotenuse comes back to

AC = 9

m, so

MC = \sqrt{45}

is right.

Check 2: Reach the strut by trigonometry instead. The base angle is

\cos^{-1}\left(\dfrac{6}{9}\right) = 48.1896\ldots^\circ

, and then

MC = 9\sin(48.1896\ldots^\circ)

MC = 6.7082\ldots

m, the same length the Pythagoras route gave, so it still rounds up to

7

Check 3: Split the final multiplication.

37 \times 21 = 777

and

37 \times 0.50 = 18.50

777 + 18.50 = 795.50

, which matches the cost worked out in one go.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Set up Pythagoras' theorem in triangle $AMC$	M1	for $9^2 - \left(\dfrac{12}{2}\right)^2 (= 81 - 36 = 45)$ or $(MC)^2 + \left(\dfrac{12}{2}\right)^2 = 9^2$ oe	✓
Find the length of the strut $MC$	M1	for $\sqrt{9^2 - \left(\dfrac{12}{2}\right)^2}$ oe, that is $\sqrt{81 - 36} = \sqrt{45} = 3\sqrt{5} = 6.7(08\ldots)$	✓
Cost the four whole-number lengths	M1	for their whole-number strut used as $(7 + 9 + 9 + 12) \times 21.5(0)$ or $37 \times 21.5(0)$	✓
State the total cost	A1	for $795.5(0)$ . Working required.	✓
Special case - the strut never rounded up	SC	SC B3 for awrt $789$ if no other marks are earned. That is the total costed from $6.7\ldots$ m of strut instead of the $7$ m length that has to be bought.	✓
Alternative method for the first two marks	Note	M2 for $\cos^{-1}\left(\dfrac{6}{9}\right) = 48.1(896\ldots)^\circ$ and $MC = 6\tan(48.1\ldots^\circ)$ or $MC = 9\sin(48.1\ldots^\circ)$ , each $= 6.7\ldots$	✓

Full marks: 4/4

(a) Factorise fully $6y^2 - 5y - 4$ [2 marks]

(b) Express $\dfrac{2x + 1}{4x} + \dfrac{7 - 5x}{3x}$ as a single fraction in its simplest form. [3 marks]

(a)(b)

[Total 5 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

The quadratic expression

6y^2 - 5y - 4

The sum of two algebraic fractions

\dfrac{2x + 1}{4x} + \dfrac{7 - 5x}{3x}

Find

(a)

6y^2 - 5y - 4

written as a product of two brackets, with nothing left to take out (b) The two fractions written as one fraction, in its simplest form

Plan the Solution

(a) Multiply the coefficient of $y^2$ by the constant term. Look for two numbers with that product whose sum is the coefficient of $y$ , use them to split the middle term, then factorise in pairs.
(b) The denominators are $4x$ and $3x$ , so both fractions go over $12x$ . Add the numerators, expand the brackets, collect like terms, then look for anything that cancels.

Worked Solution [5 marks]

To factorise

ay^2 + by + c

, find two numbers whose product is

ac

and whose sum is

b

. To add algebraic fractions, write each one over the lowest common denominator, then add the numerators.

Step 1: the two numbers for part (a)

6 \times (-4) = -24

-8 \times 3 = -24

-8 + 3 = -5

(Reason: The coefficient of

y^2

times the constant term is

-24

. Of all the pairs multiplying to

-24

, only

-8

and

3

add to

-5

, which is the coefficient of

y

Step 2: split the middle term

6y^2 - 5y - 4 = 6y^2 - 8y + 3y - 4

(Reason:

-5y

is written as

-8y + 3y

. That is the same quantity in two pieces, so the expression has not been changed, but it can now be grouped in pairs.)

Step 3: factorise in pairs

6y^2 - 8y + 3y - 4 = 2y(3y - 4) + 1(3y - 4)

2y(3y - 4) + 1(3y - 4) = (2y + 1)(3y - 4)

(Reason: The first two terms have

2y

in common and the last two have

1

. Both pairs leave

(3y - 4)

, so that bracket is a common factor of the whole expression and comes out. Neither bracket has a factor left inside it, so this is the full factorisation.)

Step 4: a common denominator for part (b)

\dfrac{2x + 1}{4x} + \dfrac{7 - 5x}{3x} = \dfrac{3(2x + 1)}{12x} + \dfrac{4(7 - 5x)}{12x}

(Reason: The lowest common multiple of

4x

and

3x

12x

. The first fraction is multiplied top and bottom by

3

, the second top and bottom by

4

, so neither fraction changes in value.)

Step 5: add the numerators and expand

\dfrac{3(2x + 1) + 4(7 - 5x)}{12x} = \dfrac{6x + 3 + 28 - 20x}{12x}

(Reason: Two fractions over the same denominator add by adding their numerators. Every term inside a bracket is multiplied by the number in front of it:

3

multiplies both

2x

and

1

, and

4

multiplies both

7

and

-5x

Step 6: collect like terms

3 + 28 = 31

\dfrac{6x + 3 + 28 - 20x}{12x} = \dfrac{31 - 14x}{12x}

(Reason: The number terms give

31

and the

x

terms give

6x - 20x = -14x

. Since

31

and

14

share no factor, and the numerator has a number term so no

x

can be cancelled, the fraction is already in its simplest form.)

(a)

(2y + 1)(3y - 4)

(b)

\dfrac{31 - 14x}{12x}

Verification

Check 1: Multiply the brackets in part (a) back out and collect the

y

terms.

6y^2 - 8y + 3y - 4 = 6y^2 - 5y - 4

Check 2: Put

y = 2

into the original expression, and into the factorised form. Both must give the same number.

6 \times 2^2 - 5 \times 2 - 4 = 10

and

(2 \times 2 + 1) \times (3 \times 2 - 4) = 5 \times 2 = 10

Check 3: Put

x = 1

into the two fractions in part (b), and into the single fraction.

\dfrac{3}{4} + \dfrac{2}{3} = \dfrac{17}{12}

and

\dfrac{31 - 14}{12} = \dfrac{17}{12}

Check 4: Repeat with

x = 2

. One value agreeing could be luck, two is evidence.

\dfrac{5}{8} - \dfrac{3}{6} = \dfrac{1}{8}

and

\dfrac{31 - 28}{24} = \dfrac{3}{24} = \dfrac{1}{8}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A pair of brackets of the right form, or the middle term split and factorised in pairs	M1	For $(2y \pm 1)(3y \pm 4)$ or $(2y \pm 4)(3y \pm 1)$ , or for $2y(3y - 4) + 1(3y - 4)$ or $3y(2y + 1) - 4(2y + 1)$ . The factors must be of the form $(ay + b)$ where $a$ and $b$ are integers. Condone the use of a different letter in place of $y$ .	✓
(a) The full factorisation	A1	For $(2y + 1)(3y - 4)$ , or $(3y - 4)(2y + 1)$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working. Ignore further working if the quadratic is then solved to find roots.	✓
(b) Both fractions over a common denominator, with the intention to add	M1	For $\dfrac{3(2x + 1)}{12x} + \dfrac{4(7 - 5x)}{12x}$ or $\dfrac{3x(2x + 1)}{12x^2} + \dfrac{4x(7 - 5x)}{12x^2}$ , or for a single correct fraction such as $\dfrac{3(2x + 1) + 4(7 - 5x)}{12x}$ or $\dfrac{3x(2x + 1) + 4x(7 - 5x)}{12x^2}$ oe. For this mark $12x$ may be written as $(3)(4x)$ or $(4)(3x)$ , and $12x^2$ as $(3x)(4x)$ .	✓
(b) A correct single fraction with every bracket expanded	M1	For $\dfrac{6x + 3 + 28 - 20x}{12x}$ oe, or $\dfrac{6x^2 + 3x + 28x - 20x^2}{12x^2}$ oe, or $\dfrac{31x - 14x^2}{12x^2}$ oe.	✓
(b) The single fraction in its simplest form	A1	For $\dfrac{31 - 14x}{12x}$ , or $\dfrac{14x - 31}{-12x}$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 5/5

Ravinder has two boxes of buttons.

In box $A$ , there are $3$ white buttons and $7$ black buttons.
In box $B$ , there are $5$ white buttons and $4$ black buttons.

Ravinder takes at random a button from box $A$ and a button from box $B$

(a) Complete the probability tree diagram. [2 marks]

(b) Work out the probability that Ravinder takes two buttons of the same colour. [3 marks]

(b)

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 13 - Exam Solution

Understanding the Question

Given

Box

A

holds

3

white buttons and

7

black buttons, so

10

buttons altogether.

Box

B

holds

5

white buttons and

4

black buttons, so

9

buttons altogether.

One button is taken from each box, so what happens in one box cannot change the other.

Find

(a) The six probabilities that belong on the branches of the tree. (b) The probability that the two buttons match in colour.

Plan the Solution

Write each probability as the number of buttons of that colour over the total in that box.
Check that each fork adds to $1$ before going any further.
Two buttons match in only two ways: white then white, or black then black.
Multiply along each of those two routes, then add the two results.

Worked Solution [5 marks]

Rule - Tree diagrams: multiply ALONG a pair of branches, add BETWEEN separate routes. For two picks that do not affect each other,

P(\text{both}) = P(\text{first}) \times P(\text{second})

Step 1: the box

A

probabilities

P(\text{white}) = \dfrac{3}{10}

P(\text{black}) = \dfrac{7}{10}

(Reason: Box

A

holds

10

buttons and

3

of them are white, so

3

of the

10

equally likely picks give white. The pair adds to

1

, as every fork must.)

Step 2: the box

B

probabilities

P(\text{white}) = \dfrac{5}{9}

P(\text{black}) = \dfrac{4}{9}

(Reason: Box

B

holds

9

buttons. This same pair goes on BOTH box

B

forks, because taking a button out of box

A

leaves box

B

exactly as it was.)

Step 3: multiply along the two same-colour routes

P(\text{white, white}) = \dfrac{3}{10} \times \dfrac{5}{9} = \dfrac{15}{90}

P(\text{black, black}) = \dfrac{7}{10} \times \dfrac{4}{9} = \dfrac{28}{90}

(Reason: Multiplying along a pair of branches gives the probability of that whole route through the tree. Leaving both products over

90

means they can be added straight away.)

Step 4: add the two same-colour routes

P(\text{same colour}) = \dfrac{15}{90} + \dfrac{28}{90}

P(\text{same colour}) = \dfrac{43}{90}

(Reason: The two routes cannot both happen, so their probabilities add.

\dfrac{43}{90}

will not cancel, because

43

is prime, and as a decimal it is

0.477\ldots

, a little under a half.)

(a)

\dfrac{3}{10}

and

\dfrac{7}{10}

on the box

A

branches;

\dfrac{5}{9}

and

\dfrac{4}{9}

on both box

B

forks(b)

\dfrac{43}{90}

Verification

Check 1: Every fork of a tree must add to

1

. Add each pair that was written on the diagram.

\dfrac{3}{10} + \dfrac{7}{10} = 1

and

\dfrac{5}{9} + \dfrac{4}{9} = 1

Check 2: Work out the opposite outcome, two different colours, and see whether the two probabilities add to

1

\dfrac{3}{10} \times \dfrac{4}{9} + \dfrac{7}{10} \times \dfrac{5}{9} = \dfrac{47}{90}

and

\dfrac{43}{90} + \dfrac{47}{90} = 1

Check 3: Count instead of multiplying. There are

10 \times 9 = 90

equally likely pairs of buttons, of which

3 \times 5 = 15

are both white and

7 \times 4 = 28

are both black.

\dfrac{15 + 28}{90} = \dfrac{43}{90}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) All three pairs on the correct branches: $\dfrac{3}{10}$ , $\dfrac{7}{10}$ on box $A$ ; $\dfrac{5}{9}$ , $\dfrac{4}{9}$ on each box $B$ fork	B2	B2 for all $3$ correct pairs of probabilities on the correct branches. If not B2, then award B1 for $1$ correct pair of probabilities on a correct branch. Allow equivalent fractions or decimals, to $2$ dp truncated or rounded, ie $0.55$ (...) and/or $0.44$ (...)	✓
(b) One correct product, eg $\dfrac{3}{10} \times \dfrac{5}{9}$ or $\dfrac{7}{10} \times \dfrac{4}{9}$	M1ft	Follow through on the candidate's own tree, provided every probability used is less than $1$ . Any one of the four products earns this mark, including a different-colour one such as $\dfrac{3}{10} \times \dfrac{4}{9}$ .	✓
(b) A complete method: $\dfrac{3}{10} \times \dfrac{5}{9} + \dfrac{7}{10} \times \dfrac{4}{9}$ or $1 - \left( \dfrac{3}{10} \times \dfrac{4}{9} + \dfrac{7}{10} \times \dfrac{5}{9} \right)$	M1ft	Both same-colour routes added, or both different-colour routes subtracted from $1$ . Follow through on the candidate's own tree, again with every probability less than $1$ .	✓
(b) $\dfrac{43}{90}$	A1ft	Or equivalent: $0.47$ (77...) to $2$ dp truncated or rounded, or $47$ .(77) per cent to $2$ sf truncated or rounded.	✓
Note	Note	Working is not required in part (b), so a correct answer scores full marks unless it follows obviously incorrect working. This row carries no mark of its own.	✓

Full marks: 5/5

Owen, Bilal and Devika each raised money for a village library appeal.

The combined amount raised by Owen and Bilal is $15\,435$ dirhams.

The amount raised by Bilal is $45\%$ more than the amount raised by Owen.

The amount raised by Bilal is $\dfrac{3}{2}$ times the amount raised by Devika.

Work out the amount raised by Devika. [5 marks]

dirhams

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 14 - Exam Solution

Understanding the Question

Given

Owen and Bilal together raised

15\,435

dirhams.

Bilal raised

45\%

more than Owen.

Bilal raised

\dfrac{3}{2}

times as much as Devika.

Devika's amount is not linked to the total directly, only through Bilal.

Find

The amount Devika raised, in dirhams.

Plan the Solution

Call the amount Owen raised $x$ dirhams, so the amount Bilal raised is $1.45x$ dirhams.
Add the two amounts, put the sum equal to $15\,435$ , and solve for $x$ .
Scale Owen's amount by $1.45$ to get Bilal's amount.
Bilal's amount is $\dfrac{3}{2}$ of Devika's, so undo that by multiplying Bilal's amount by $\dfrac{2}{3}$ .

Worked Solution [5 marks]

Rule - a rise of

45\%

is a multiplier of

1 + 0.45 = 1.45

, and a statement of the form

P = \dfrac{3}{2} \times Q

is undone by

Q = \dfrac{2}{3} \times P

Step 1: write both amounts using one letter

\text{Owen} = x

\text{Bilal} = x + 0.45x = 1.45x

(Reason: Bilal's amount is Owen's amount plus another

45\%

of it, and

1 + 0.45 = 1.45

, so one letter now carries both people.)

Step 2: use the combined total to find Owen's amount

x + 1.45x = 15\,435

2.45x = 15\,435

x = \dfrac{15\,435}{2.45} = 6300

(Reason: Adding the two expressions gives

2.45

lots of

x

, so dividing the total by

2.45

leaves one lot. Owen raised

6300

dirhams.)

Step 3: work out Bilal's amount

\text{Bilal} = 15\,435 - 6300 = 9135

(Reason: The pair raised

15\,435

between them, so taking Owen's share off the total leaves Bilal's. Multiplying instead gives the same value, since

1.45 \times 6300 = 9135

Step 4: work back from Bilal to Devika

\text{Bilal} = \dfrac{3}{2} \times \text{Devika}

\text{Devika} = 9135 \times \dfrac{2}{3} = 6090

(Reason: Bilal's amount is the larger one here, so Devika's amount is found by reversing the

\dfrac{3}{2}

and multiplying by

\dfrac{2}{3}

. Devika raised

6090

dirhams.)

6090

dirhams

Verification

Check 1: Add Owen's

6300

to Bilal's

9135

and compare the sum with the combined total in the question.

6300 + 9135 = 15\,435

, which is the combined amount given.

Check 2: Work out how much more Bilal raised than Owen, as a fraction of Owen's amount.

\dfrac{9135 - 6300}{6300} = 0.45

, so Bilal did raise

45\%

more.

Check 3: Scale the answer back up by

\dfrac{3}{2}

and see whether Bilal's amount reappears.

6090 \times \dfrac{3}{2} = 9135

, which is Bilal's amount.

Check 4: Take the whole journey in one fraction instead: Bilal is

\dfrac{29}{49}

of the pair's total, and Devika is

\dfrac{2}{3}

of Bilal.

15\,435 \times \dfrac{58}{147} = 6090

, the same answer by a route that never names Owen.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Set up the link between two of the amounts	B1	Any one of $1 + 1.45 = 2.45$ , $B = 1.45A$ , $A : B = 100 : 145$ , $100 + 145 = 245$ or $B : C = 3 : 2$ . Any letters may be used, provided the ratio is identified with the right pair of people.	✓
A method to find Owen's amount	M1	$x + 1.45x = 15\,435$ , $\dfrac{15\,435}{2.45}$ , $\dfrac{15\,435}{245} \times 100$ or $63 \times 100$ seen, or the value $6300$ seen anywhere.	✓
A method to find Bilal's amount	M1	$15\,435 - 6300$ , $1.45 \times 6300$ or $145 \times 63$ , each using the candidate's own value for Owen, or the value $9135$ seen anywhere.	✓
A method to find Devika's amount	M1	$9135 \times \dfrac{2}{3}$ or $\dfrac{9135}{\dfrac{3}{2}}$ or equivalent, using the candidate's own value for Bilal.	✓
Correct answer	A1	$6090$ , correct answer only.	✓
One-step alternative to the first two method marks	Note	A candidate who goes straight to Bilal's amount with $\dfrac{15\,435}{\dfrac{1}{1.45} + 1} = 9135$ or $15\,435 \times \dfrac{29}{49} = 9135$ earns both method marks at once, so award M2 for either.	✓
Answer with no working	Note	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 5/5

Continue to questions 15 to 25

The remaining 11 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 25 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

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

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

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

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

Keep revising

Once you have worked through this paper, read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Edexcel IGCSE 4MA1/1H, Thursday 16 May 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 14 of 25

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Frequently asked questions

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