Set Language and Notation · Study Notes · Edexcel IGCSE 4MA1 Higher

Sets, intersections, unions, complements, Venn diagrams. The notation that powers probability, statistics and many of the trickier exam questions.

At a glance

Set = collection of distinct elements.
$\in$ = is an element. $\notin$ = is not.
$\xi$ = universal set (the rectangle in Venn).
$A \cap B$ = intersection (BOTH).
$A \cup B$ = union (EITHER, possibly both).
$A'$ = complement (everything in $\xi$ NOT in $A$ ).
$|A \cup B| = |A| + |B| - |A \cap B|$ (avoid double count).

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.7 — Use set language and notation including $\xi$ , $\cap$ , $\cup$ , $\in$ , $\notin$ , $\subseteq$ , $\subset$ , $A'$ , $\emptyset$ .
1.7 — Use Venn diagrams for two and three sets.
1.7 — Use the inclusion-exclusion formula.

The set notation symbols

▼

Memorise these. Edexcel question stems use them throughout statistics and probability.

Symbols you must know:

Symbol	Meaning
$\{ \}$ or $\emptyset$	Empty set
$\xi$ or $U$	Universal set (everything in the problem)
$\in$	"Is an element of"
$\notin$	"Is not an element of"
$\subseteq$	"Is a subset of (or equal to)"
$\subset$	"Is a proper subset of (and not equal)"
$A \cap B$	Intersection: elements in BOTH
$A \cup B$	Union: elements in EITHER (or both)
$A'$	Complement: NOT in $A$
$\\|A\\|$ or $n(A)$	Number of elements in $A$

Number sets:

Symbol	Set
$\mathbb{N}$	Natural numbers $\{1, 2, 3, \ldots\}$
$\mathbb{Z}$	Integers $\{\ldots, -1, 0, 1, \ldots\}$
$\mathbb{Q}$	Rational numbers (fractions)
$\mathbb{R}$	Real numbers (everything on the number line)

Set-builder notation.

$\{ x : \text{condition}(x) \}$

Read aloud as 'the set of $x$ such that condition holds'.

Example: $\{ x : x \in \mathbb{Z}, -3 \le x < 5 \} = \{-3, -2, -1, 0, 1, 2, 3, 4\}$ .

Intersection is the overlap, union is everything inside either circle, complement is everything outside A.

Worked qualitative. What's the difference between $\subset$ and $\subseteq$ ?

$\subseteq$ : subset, possibly equal. $\{1, 2\} \subseteq \{1, 2\}$ ✓.
$\subset$ : PROPER subset (smaller). $\{1, 2\} \not\subset \{1, 2\}$ ✗ (equal sets aren't proper).
Use $\subseteq$ unless you specifically need to exclude equality.

Edexcel tip. Always write the universal set $\xi$ first when given a problem. Many questions hinge on what's IN the universe but NOT in $A$ .

$\in$ for membership.
$\xi$ for everything in the problem.
$\cap$ both, $\cup$ either.
$A'$ complement.
$\subset$ proper, $\subseteq$ allows equality.

Venn diagrams — 2 sets and 3 sets

▼

Universal set as rectangle. Each set as a circle. Fill the OVERLAP first.

Two-set Venn diagram.

Two overlapping circles inside a rectangle.

Regions:

$A$ only — left of overlap.
$A \cap B$ — middle overlap.
$B$ only — right of overlap.
$(A \cup B)'$ — outside both circles.

How to fill in. ALWAYS:

Start with the OVERLAP ( $A \cap B$ ).
Then 'A only' = $|A| - |A \cap B|$ .
Then 'B only' = $|B| - |A \cap B|$ .
Finally outside = $|\xi| - |A \cup B|$ .

Worked example. $30$ students. $18$ play football ( $A$ ), $14$ play tennis ( $B$ ), $5$ play both.

+------------------------------+
| ξ = 30                       |
|     +--------+--------+      |
|     |        |        |      |
|     |   13   |   5  9 |      |
|     |  (A)   |   (B)  |      |
|     +--------+--------+      |
|              3 (neither)     |
+------------------------------+

Fill the overlap (5) first, then each 'only' region, then the count outside both circles.

Overlap: $5$ .
A only: $18 - 5 = 13$ .
B only: $14 - 5 = 9$ .
Total in $A \cup B$ : $13 + 5 + 9 = 27$ .
Outside: $30 - 27 = 3$ .

Three-set Venn.

Three overlapping circles. 8 distinct regions:

$A \cap B \cap C$ (centre — all three).
$A \cap B$ only (not $C$ ).
$A \cap C$ only (not $B$ ).
$B \cap C$ only (not $A$ ).
$A$ only (not $B$ or $C$ ).
$B$ only.
$C$ only.
Outside all three.

Order to fill in:

Centre ( $A \cap B \cap C$ ).
Two-set 'only' (subtract centre from each pairwise intersection).
Single-set 'only' (subtract all overlaps from each set total).
Outside (subtract all to get $|\xi|$ ).

Worked qualitative. Why does the order matter?

Each given count usually INCLUDES the overlaps below it.
Filling in the centre first prevents double-counting.
Working outward is mechanical and reliable.

Edexcel tip. Draw the diagram FIRST, label all regions with letters/numbers, fill in centre first. Mark schemes credit each correctly-filled region.

Always fill OVERLAP first.
Then 'only' regions.
Then outside.
3-set Venn: centre → pairs → singles → outside.

Inclusion-exclusion formula

▼

Avoid double-counting elements in both sets.

Two-set formula.

$|A \cup B| = |A| + |B| - |A \cap B|$

Why subtract? Adding $|A|$ and $|B|$ counts the overlap TWICE. Subtract once to count it correctly.

Three-set formula (less commonly examined but worth knowing):

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

Worked example. $|A| = 18$ , $|B| = 14$ , $|A \cap B| = 5$ .

$|A \cup B| = 18 + 14 - 5 = 27$ .

Use cases:

'How many do at least one of these?' → $|A \cup B|$ via formula.
'How many do neither?' → $|\xi| - |A \cup B|$ .

Worked qualitative. A class of $30$ . $20$ like maths, $15$ like science, $10$ like both. Find:

How many like at least one? $|A \cup B| = 20 + 15 - 10 = 25$ .
How many like neither? $30 - 25 = 5$ .

Edexcel tip. This formula is NOT on the Edexcel formula sheet. Memorise it. Quote it explicitly to score the M1 mark.

$|A \cup B| = |A| + |B| - |A \cap B|$ .
Subtract the overlap to avoid double count.
'At least one' → use the formula.
'Neither' → $|\xi|$ minus the union.

Quick recap

Symbols: $\in, \notin, \xi, \subseteq, \cap, \cup, A'$ .
Universal set = the rectangle.
Always fill Venn from centre outward.
$|A \cup B| = |A| + |B| - |A \cap B|$ .
'Neither' = $|\xi| - |A \cup B|$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$A \cap B$ — intersection (BOTH).
$A \cup B$ — union (EITHER).
$A'$ — complement (NOT in $A$ ).
Empty set $\emptyset$ or $\{\}$ .
Inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$ .

How it’s examined

Set notation appears Paper 2H. Either standalone (3-5 marks: list elements, identify regions) or embedded in probability questions (5-7 marks: Venn → probability calculation). Examiner reports flag (1) wrong order in Venn diagram filling, (2) confusing $\subset$ with $\in$ , and (3) forgetting to subtract the overlap.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on set language and notation, written exactly the way a tutor would explain them at the board.

1Read set notation
Foundation• set notation
▼
Question
$\xi = \{1, 2, 3, \ldots, 12\}$ , $A = \{$ prime numbers $\}$ , $B = \{$ even numbers $\}$ . List the elements of (a) $A$ , (b) $B$ , (c) $A \cap B$ , (d) $A \cup B$ , (e) $A'$ .
Step-by-step solution
1. Step 1
  (a) $A$ = primes from $1$ - $12$ : $A = \{2, 3, 5, 7, 11\}$ .
2. Step 2
  (b) $B$ = even numbers from $1$ - $12$ : $B = \{2, 4, 6, 8, 10, 12\}$ .
3. Step 3
  (c) $A \cap B$ = elements in BOTH $A$ AND $B$ : $\{2\}$ (the only even prime).
4. Step 4
  (d) $A \cup B$ = elements in $A$ OR $B$ (or both): $\{2, 3, 4, 5, 6, 7, 8, 10, 11, 12\}$ .
5. Step 5
  (e) $A'$ = complement of $A$ = elements in $\xi$ but NOT in $A$ : $\{1, 4, 6, 8, 9, 10, 12\}$ .
Answer
(a) $\{2,3,5,7,11\}$ (b) $\{2,4,6,8,10,12\}$ (c) $\{2\}$ (d) $\{2,3,4,5,6,7,8,10,11,12\}$ (e) $\{1,4,6,8,9,10,12\}$
Examiner tip
Always show the universal set $\xi$ first. Edexcel mark schemes accept lists or set-builder notation but require COMMAS between elements.
2Use a 2-set Venn diagram with overlap
Higher• Adapted from 4MA1/2H May/Jun 2024 Q14• Venn diagram
▼
Question
$30$ students were asked which sport they play. $18$ play football, $14$ play tennis, $5$ play both. Draw a Venn diagram and find (a) $|A \cap B|$ , (b) $|A \cup B|$ , (c) $|(A \cup B)'|$ where $A$ = football players, $B$ = tennis.
Step-by-step solution
1. Step 1
  Start with the OVERLAP: $|A \cap B| = 5$ (in middle of Venn).
2. Step 2
  Football only: $18 - 5 = 13$ . Tennis only: $14 - 5 = 9$ .
3. Step 3
  Total in $A \cup B$ : $13 + 5 + 9 = 27$ .
4. Step 4
  Outside both circles: $30 - 27 = 3$ students play neither.
5. Step 5
  Answers: (a) $|A \cap B| = 5$ . (b) $|A \cup B| = 27$ . (c) $|(A \cup B)'| = 3$ .
Answer
(a) $5$ (b) $27$ (c) $3$
Examiner tip
ALWAYS fill in the overlap first, then 'only' regions, then outside. Edexcel mark schemes give M1 for the correct overlap and A1 for each region. Working backwards is the most common error.
3Three-set Venn diagram
Higher• Venn diagram, three sets
▼
Question
In a class of $50$ students: $25$ study Maths (M), $20$ study Physics (P), $15$ study Chemistry (C). $8$ study M and P, $6$ study M and C, $5$ study P and C. $3$ study all three. Find how many study NONE.
Step-by-step solution
1. Step 1
  Centre (all three): $3$ .
2. Step 2
  Two-set intersections (subtract centre): $|M \cap P|$ only $= 8 - 3 = 5$ . $|M \cap C|$ only $= 6 - 3 = 3$ . $|P \cap C|$ only $= 5 - 3 = 2$ .
3. Step 3
  Single-set 'only' regions: $|M$ only $| = 25 - 5 - 3 - 3 = 14$ . $|P$ only $| = 20 - 5 - 2 - 3 = 10$ . $|C$ only $| = 15 - 3 - 2 - 3 = 7$ .
4. Step 4
  Total in any set: $14 + 10 + 7 + 5 + 3 + 2 + 3 = 44$ .
5. Step 5
  Outside (none): $50 - 44 = 6$ .
Answer
$6$ study none.
Examiner tip
Three-set Venns are best filled in the order: centre → two-set overlaps → single-set 'only' → outside. The 'subtract overlaps' step trips up many candidates.
4Read set-builder notation
Higher• set-builder
▼
Question
Describe the set $S = \{ x : x \in \mathbb{Z}, -3 \le x < 5 \}$ in words and list its elements.
Step-by-step solution
1. Step 1
  Read aloud: 'The set of $x$ such that $x$ is an integer and $x$ is greater than or equal to $-3$ and $x$ is less than $5$ .'
2. Step 2
  List: $S = \{-3, -2, -1, 0, 1, 2, 3, 4\}$ (note: $-3$ included, $5$ excluded).
Answer
$S = \{-3, -2, -1, 0, 1, 2, 3, 4\}$
Examiner tip
Pay attention to inclusive ( $\le$ ) vs strict ( $<$ ) inequalities. Mark schemes deduct for missing or extra elements.
5Recognising the empty set [1 mark]
Getting started• Paper 1H/2H style — 1 mark• empty set, set notation
▼
Question
$G = \{x : x \text{ is an even prime number greater than } 2\}$

Write down $n(G)$ . [1]
Step-by-step solution
1. Step 1
  Every even number greater than $2$ has $2$ as a factor as well as $1$ and itself, so it cannot be prime. $G$ has no elements.
  $G = \varnothing$
2. Step 2
  B1 The empty set has no elements.
  $n(G) = 0$
Answer
$0$
Examiner tip
Examiner tip: A set can be described in words and still be empty. $n(\varnothing) = 0$ .

Common mistake: Writing $n(G) = 1$ because $2$ is the even prime. The description says greater than $2$ , so $2$ is excluded.
6A universal set from the letters of a word [2 marks]
Getting started• Paper 1H/2H style — 2 marks• listing elements, complement
▼
Question
$\mathscr{E} = \{\text{letters of the word MATHEMATICS}\}$ and $V = \{\text{vowels}\}$

(a) Write down $n(\mathscr{E})$ . [1]

(b) List the members of $V'$ . [1]
Step-by-step solution
1. Step 1
  B1 (a) Each different letter is listed once, even though M, A and T appear twice in the word.
  $\mathscr{E} = \{\text{M, A, T, H, E, I, C, S}\}, \qquad n(\mathscr{E}) = 8$
2. Step 2
  B1 (b) $V = \{\text{A, E, I}\}$ , so $V'$ is the consonants.
  $V' = \{\text{M, T, H, C, S}\}$
Answer
(a) $8$ (b) $\{\text{M, T, H, C, S}\}$
Examiner tip
Examiner tip: A set never contains the same element twice, so repeated letters are written once.

Common mistake: Giving $n(\mathscr{E}) = 11$ by counting every letter in the word.
7Reading elements from a described Venn diagram [2 marks]
Building confidence• Paper 1H/2H style — 2 marks• venn diagram, complement
▼
Question
In a Venn diagram for sets $A$ and $B$ inside $\mathscr{E}$ :

the region $A$ only contains $2, 5, 9$

the region $A \cap B$ contains $4, 7$

the region $B$ only contains $1, 8$

the region outside both circles contains $3, 6$

(a) List the members of $B'$ . [1]

(b) List the members of $A \cup B'$ . [1]
Step-by-step solution
1. Step 1
  B1 (a) $B'$ is everything not in circle $B$ : the $A$ only region and the outside.
  $B' = \{2, 3, 5, 6, 9\}$
2. Step 2
  B1 (b) Join $A = \{2, 4, 5, 7, 9\}$ to $B'$ ; every region except $B$ only.
  $A \cup B' = \{2, 3, 4, 5, 6, 7, 9\}$
Answer
(a) $\{2, 3, 5, 6, 9\}$ (b) $\{2, 3, 4, 5, 6, 7, 9\}$
Examiner tip
Examiner tip: For a union with a complement, it is often quicker to find what is left out. Here only the $B$ only region, $\{1, 8\}$ , is missing.

Common mistake: Leaving $3$ and $6$ out of $B'$ . The outside region is part of every complement.
8Complement, intersection and the empty set [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• complement, empty set
▼
Question
$\mathscr{E} = \{x : x \text{ is an integer}, 1 \leqslant x \leqslant 10\}$

$P = \{x : x > 6\}$ and $Q = \{\text{factors of } 30\}$

(a) List the members of $P' \cap Q$ . [2]

(b) Is it true that $P \cap Q = \varnothing$ ? Give a reason for your answer. [1]
Step-by-step solution
1. Step 1
  M1 List the sets. $P' = \{1, 2, 3, 4, 5, 6\}$ (the numbers not greater than $6$ ).
  $Q = \{1, 2, 3, 5, 6, 10\}$
2. Step 2
  A1 Keep the elements of $Q$ that are also in $P'$ .
  $P' \cap Q = \{1, 2, 3, 5, 6\}$
3. Step 3
  B1 (b) No, because $10$ is in both $P$ and $Q$ .
  $P \cap Q = \{10\} \neq \varnothing$
Answer
(a) $\{1, 2, 3, 5, 6\}$ (b) No, $10 \in P$ and $10 \in Q$
Examiner tip
Examiner tip: Name the element that lies in both sets. That is the complete reason for "No".

Common mistake: Including $4$ in (a) because it is in $P'$ . It is not a factor of $30$ , so it fails the intersection.
9Finding an unknown from an intersection [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• intersection, algebra
▼
Question
$A = \{3, 7, 3x - 4\}$ and $B = \{x + 6, 5, 7\}$

$A \cap B = \{7, 11\}$

(a) Find the value of $x$ . Show that your value gives $A \cap B = \{7, 11\}$ . [3]

(b) Find $n(A \cup B)$ . [1]
Step-by-step solution
1. Step 1
  M1 $11$ is in both sets, and the only unknown element of $A$ is $3x - 4$ .
  $3x - 4 = 11$
2. Step 2
  A1 Solve.
  $3x = 15 \;\Rightarrow\; x = 5$
3. Step 3
  B1 Check in $B$ : $x + 6 = 11$ , so $B = \{11, 5, 7\}$ and $A = \{3, 7, 11\}$ . The intersection is $\{7, 11\}$ , as given.
4. Step 4
  B1 (b) List each element once.
  $A \cup B = \{3, 5, 7, 11\}, \qquad n(A \cup B) = 4$
Answer
(a) $x = 5$ (b) $4$
Examiner tip
Examiner tip: Always substitute back into both sets. A value that makes one set right but the other wrong is not the answer.

Common mistake: Giving $n(A \cup B) = 6$ by adding $n(A) + n(B)$ without removing the two shared elements.
10Numbers of elements in three-set regions [5 marks]
Stretch• Paper 1H/2H style — 5 marks• venn diagram, three sets
▼
Question
A Venn diagram for sets $A$ , $B$ and $C$ shows these numbers of elements:

$A$ only: $8$ , $B$ only: $6$ , $C$ only: $5$

$A$ and $B$ only: $3$ , $A$ and $C$ only: $4$ , $B$ and $C$ only: $2$

in all three: $1$ , outside all three: $7$

Find

(a) $n((A \cup B)' \cap C)$ [1]

(b) $n(A \cap (B \cup C))$ [2]

(c) $n((A \cap B) \cup C)$ [2]
Step-by-step solution
1. Step 1
  B1 (a) In $C$ but in neither $A$ nor $B$ : the $C$ only region.
  $n((A \cup B)' \cap C) = 5$
2. Step 2
  M1 (b) The parts of $A$ that are also in $B$ or $C$ : $A$ and $B$ only, $A$ and $C$ only, and the centre.
3. Step 3
  A1 Add them.
  $3 + 4 + 1 = 8$
4. Step 4
  M1 (c) All of $C$ ( $5 + 4 + 2 + 1$ ) together with the part of $A \cap B$ outside $C$ ( $3$ ).
5. Step 5
  A1 Add them.
  $5 + 4 + 2 + 1 + 3 = 15$
Answer
(a) $5$ (b) $8$ (c) $15$
Examiner tip
Examiner tip: Translate each expression into regions before adding. In (c) the centre belongs to both $A \cap B$ and $C$ but is counted only once.

Common mistake: In (b), working out $(A \cap B) \cup C$ instead of $A \cap (B \cup C)$ . The brackets change which regions count.

Question 1
Paper 1H/2H style — 1 mark1 mark
$\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $S = \{1, 4, 9\}$

Describe the set $S$ in words. [1]
Model answer
$S$ is the set of square numbers in $\mathscr{E}$ .
Why this scores
Mark scheme:
- B1 for square numbers oe
Why this scores full marks: One precise property that fits every element and no other element of $\mathscr{E}$ . "Odd numbers" fails because of $4$ , and "numbers with one digit" describes all of $\mathscr{E}$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
$\mathscr{E} = \{x : x \text{ is an integer}, 1 \leqslant x \leqslant 12\}$

$D = \{\text{factors of } 7\}$ and $M = \{\text{multiples of } 7\}$

(a) List the members of $D \cap M$ . [1]

(b) List the members of $D \cup M$ . [1]
Model answer
$D = \{1, 7\}$ and $M = \{7\}$ (the next multiple, $14$ , is not in $\mathscr{E}$ ).

(a) $D \cap M = \{7\}$

(b) $D \cup M = \{1, 7\}$
Why this scores
Mark scheme:
- B1 for $\{7\}$
- B1 for $\{1, 7\}$
Why this scores full marks: Listing $D$ and $M$ first shows that $7$ is both a factor and a multiple of itself, and the check against $\mathscr{E}$ stops $14$ appearing in the union.
Question 3
Paper 1H/2H style — 2 marks2 marks
$\mathscr{E} = \{x : x \text{ is an integer}, 1 \leqslant x \leqslant 15\}$

$T = \{\text{multiples of } 3\}$ and $F = \{\text{multiples of } 5\}$

Find $n((T \cup F)')$ . [2]
Model answer
$T = \{3, 6, 9, 12, 15\}$ , $F = \{5, 10, 15\}$

$T \cup F = \{3, 5, 6, 9, 10, 12, 15\}$ , so $n(T \cup F) = 7$

$n((T \cup F)') = 15 - 7 = 8$
Why this scores
Mark scheme:
- M1 for $T \cup F$ listed or $n(T \cup F) = 7$ or $15 - 7$
- A1 for $8$ cao
Why this scores full marks: $15$ is counted once in the union. Adding $5 + 3 = 8$ and then $15 - 8 = 7$ is the usual wrong route.
Question 4
Paper 1H/2H style — 3 marks3 marks
$X = \{x : x \text{ is an integer}, -4 < 2x \leqslant 7\}$

(a) List the members of $X$ . [2]

(b) Write down $n(X)$ . [1]
Model answer
(a) $-4 < 2x \leqslant 7$

$-2 < x \leqslant 3.5$

$X = \{-1, 0, 1, 2, 3\}$

(b) $n(X) = 5$
Why this scores
Mark scheme:
- M1 for $-2 < x$ or $x \leqslant 3.5$
- A1 for $\{-1, 0, 1, 2, 3\}$ cao
- B1 ft for $5$ (number of elements in their list)
Why this scores full marks: $-2$ is excluded by the strict sign and $3.5$ is not an integer, so the list runs from $-1$ to $3$ . $0$ is included, which weaker answers forget.
Question 5
Paper 1H/2H style — 3 marks3 marks
In a survey of $120$ people, $45\%$ own a bicycle, $30\%$ own a car and $20\%$ own both.

Work out how many of the people own neither a bicycle nor a car. [3]
Model answer
Bicycle: $0.45 \times 120 = 54$

Car: $0.30 \times 120 = 36$

Both: $0.20 \times 120 = 24$

Bicycle or car: $54 + 36 - 24 = 66$

Neither: $120 - 66 = 54$
Why this scores
Mark scheme:
- M1 for converting at least two percentages to numbers, e.g. $54$ and $36$
- M1 for $54 + 36 - 24$ or $30 + 24 + 12$ oe
- A1 for $54$ cao
Why this scores full marks: The candidate subtracts the $24$ who own both, because they are already inside the $54$ and the $36$ . Working in percentages first ( $45 + 30 - 20 = 55\%$ , so $45\%$ neither) is equally valid.
Question 6
Paper 1H/2H style — 3 marks3 marks
$\mathscr{E} = \{\text{positive integers}\}$

$A = \{\text{multiples of } 10\}$ , $B = \{\text{multiples of } 5\}$ , $C = \{\text{odd numbers}\}$

(a) Is it true that $A \subset B$ ? Give a reason for your answer. [1]

(b) Is it true that $A \cap C = \varnothing$ ? Give a reason for your answer. [1]

(c) Describe the members of $B \cap C$ in words. [1]
Model answer
(a) Yes. $10 = 2 \times 5$ , so every multiple of $10$ is also a multiple of $5$ .

(b) Yes. Every multiple of $10$ is even, so $A$ and $C$ have no elements in common.

(c) Odd multiples of $5$ (the numbers ending in $5$ ).
Why this scores
Mark scheme:
- B1 for Yes with a reason, e.g. every multiple of $10$ is a multiple of $5$
- B1 for Yes with a reason, e.g. multiples of $10$ are even, so no elements in common
- B1 for odd multiples of $5$ oe
Why this scores full marks: Each reason is a general property, not one example. One example can disprove a statement, but it cannot prove one about infinitely many numbers.
Question 7
Paper 1H/2H style — 4 marks4 marks
For the sets $A$ and $B$ ,

$n(A \cap B') = 2x + 1$ , $n(A \cap B) = x$ , $n(A' \cap B) = 13$ and $n((A \cup B)') = x - 2$

Given that $n(A) = n(B)$ , find $n(\mathscr{E})$ . [4]
Model answer
$n(A) = (2x + 1) + x = 3x + 1$

$n(B) = x + 13$

$3x + 1 = x + 13$

$2x = 12 \qquad x = 6$

Regions: $13$ , $6$ , $13$ and $6 - 2 = 4$

$n(\mathscr{E}) = 13 + 6 + 13 + 4 = 36$
Why this scores
Mark scheme:
- M1 for $n(A) = 3x + 1$ or $n(B) = x + 13$
- M1 for $3x + 1 = x + 13$ oe
- A1 for $x = 6$
- A1 for $36$ cao
Why this scores full marks: $n(A)$ includes the overlap, so it is $3x + 1$ , not $2x + 1$ . The candidate then remembers the region outside both sets when totalling $n(\mathscr{E})$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
$\mathscr{E} = \{x : x \text{ is an integer}, 1 \leqslant x \leqslant 25\}$

$S = \{\text{square numbers}\}$ , $O = \{\text{odd numbers}\}$ , $T = \{\text{multiples of } 3\}$

(a) List the members of $S \cap O \cap T$ . [1]

(b) Find $n(S \cap O')$ . [1]

(c) List the members of $(S \cup T)' \cap O$ . [2]
Model answer
$S = \{1, 4, 9, 16, 25\}$ , $T = \{3, 6, 9, 12, 15, 18, 21, 24\}$

(a) $S \cap O \cap T = \{9\}$

(b) Even squares: $4$ and $16$ , so $n(S \cap O') = 2$

(c) Odd numbers: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25$

Remove squares ( $1, 9, 25$ ) and multiples of $3$ ( $3, 15, 21$ ):

$(S \cup T)' \cap O = \{5, 7, 11, 13, 17, 19, 23\}$
Why this scores
Mark scheme:
- B1 for $\{9\}$
- B1 for $2$
- M1 for listing odd numbers with at least four of the squares or multiples of $3$ removed
- A1 for $\{5, 7, 11, 13, 17, 19, 23\}$ cao
Why this scores full marks: The candidate starts from $O$ and strikes out anything in $S$ or $T$ , which is exactly what $(S \cup T)' \cap O$ means. Crossing out $9$ once is enough, although it is in both.
Question 9
Paper 1H/2H style — 5 marks5 marks
Each of $100$ students was asked whether they belong to the art club, the book club and the chess club.

$50$ belong to the art club, $45$ to the book club and $40$ to the chess club.

$20$ belong to exactly two of the clubs and $15$ belong to none of them.

(a) Find the number of students who belong to all three clubs. [4]

(b) Find the number of students who belong to exactly one club. [1]
Model answer
(a) Students in at least one club: $100 - 15 = 85$

Let $x$ belong to all three. Then exactly one club $= 85 - 20 - x = 65 - x$ .

Adding the three club totals counts a student in exactly one club once, in exactly two clubs twice and in all three clubs three times:

$(65 - x) + 2 \times 20 + 3x = 50 + 45 + 40$

$105 + 2x = 135$

$x = 15$

(b) Exactly one club $= 65 - 15 = 50$
Why this scores
Mark scheme:
- B1 for $85$
- M1 for exactly one $= 65 - x$ oe
- M1 for $(65 - x) + 40 + 3x = 135$ oe
- A1 for $15$ cao
- B1 ft for $50$
Why this scores full marks: The candidate explains why the club totals add to $135$ , not $85$ : students in several clubs are counted several times. That reasoning replaces the need for the individual two-club regions, which the question does not give.
Question 10
Paper 1H/2H style — 5 marks5 marks
$\mathscr{E} = \{x : x \text{ is an integer}, 0 \leqslant x \leqslant 20\}$

$A = \{x : 4x - 7 > 21\}$ , $B = \{\text{factors of } 60\}$ , $C = \{x : x^{2} \leqslant 50\}$

(a) Show that $A \cap C = \varnothing$ . [2]

(b) List the members of $B \cap C'$ . [2]

(c) Is it true that $B \subset A \cup C$ ? Give a reason for your answer. [1]
Model answer
(a) $4x - 7 > 21 \Rightarrow 4x > 28 \Rightarrow x > 7$ , so $A = \{8, 9, \ldots, 20\}$

$7^{2} = 49 \leqslant 50$ but $8^{2} = 64$ , so $C = \{0, 1, \ldots, 7\}$

Every element of $A$ is at least $8$ and every element of $C$ is at most $7$ , so they have no elements in common: $A \cap C = \varnothing$ .

(b) $B = \{1, 2, 3, 4, 5, 6, 10, 12, 15, 20\}$ , $C' = \{8, 9, \ldots, 20\}$

$B \cap C' = \{10, 12, 15, 20\}$

(c) Yes. $A \cup C = \{0, 1, \ldots, 20\} = \mathscr{E}$ , and every element of $B$ is in $\mathscr{E}$ .
Why this scores
Mark scheme:
- M1 for $x > 7$ or $C = \{0, 1, \ldots, 7\}$
- A1 for a complete argument that no element is in both $A$ and $C$
- M1 for $B$ listed within $\mathscr{E}$ or $C'$ identified
- A1 for $\{10, 12, 15, 20\}$ cao
- B1 for Yes with a reason, e.g. $A \cup C = \mathscr{E}$
Why this scores full marks: In (a) the candidate finds both sets and states the reason in words; " $A \cap C = \varnothing$ " written without them scores nothing. In (c), seeing that $A$ and $C$ together cover $\mathscr{E}$ settles the question without checking elements one by one.

4MA1 Set Language and Notation

At a glance

What you’ll learn

The set notation symbols

Venn diagrams — 2 sets and 3 sets

Inclusion-exclusion formula

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Read set notation

2Use a 2-set Venn diagram with overlap

3Three-set Venn diagram

4Read set-builder notation

5Recognising the empty set [1 mark]

6A universal set from the letters of a word [2 marks]

7Reading elements from a described Venn diagram [2 marks]

8Complement, intersection and the empty set [3 marks]

9Finding an unknown from an intersection [4 marks]

10Numbers of elements in three-set regions [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10