Pearson Edexcel International GCSE Mathematics A 4MA1: Set language and notation – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 1.5, Set language and notation, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for papers sat in January and June. Questions 1–5 use Foundation statements 1.5 A–E, which both tiers need. Questions 6–11 use Higher statements 1.5 A–D and are labelled (Higher). A calculator is allowed on every 4MA1 paper but is not needed here.

Before you start, you may want the set language study guide or the set language revision notes. These questions are different from those in the topic 1 practice questions. Course links: the Edexcel IGCSE Mathematics hub and the printable 4MA1 checklist.

Questions

1. ℰ = {1, 2, 3, …, 15}. A = {odd numbers} and B = {multiples of 3}.

(a) List the members of A ∩ B. [1] (b) List the members of A ∪ B. [2] (c) Is 12 ∈ A ∩ B? Give a reason for your answer. [1]

2. P = {2, 4, 6, 8}, Q = {1, 3, 5, 7} and R = {3, 5}.

(a) Write down P ∩ Q, using the correct symbol. [1] (b) Copy and complete each statement using ∈ or ∉: 6 … P; 6 … Q; 7 … Q ∪ R. [2] (c) List the members of Q ∩ R. [1]

3. ℰ = {a, b, c, d, e, f, g, h, i, j}. M = {a, e, i} and N = {a, b, c, d, e}.

(a) List the members of M′. [1] (b) List the members of (M ∪ N)′. [2] (c) List the members of M′ ∩ N. [1]

4. ℰ = {10, 11, 12, …, 20}. A = {even numbers} and B = {multiples of 5}.

(a) Draw a Venn diagram to show ℰ, A and B, with every element of ℰ in the correct region. [3] (b) List the members of A′ ∩ B. [1]

5. A Venn diagram shows two overlapping circles, A and B, inside a rectangle ℰ. Write each described shaded region in set notation.

(a) Only the overlap of the two circles is shaded. [1] (b) Only the region outside both circles is shaded. [1] (c) Only the part of circle A outside circle B is shaded. [1] (d) Both circles are shaded except their overlap. [1]

6. (Higher) ℰ = {x : x is an integer, −4 ≤ x ≤ 6}. P = {x : x² < 9} and Q = {x : 3x − 2 > 4}.

(a) List the members of P. [1] (b) List the members of Q. [2] (c) Write down P ∩ Q. [1] (d) Find n((P ∪ Q)′). [2]

7. (Higher) A = {x : x is a factor of 24}, B = {x : x is a factor of 8} and C = {x : x is a factor of 12}.

(a) Explain why B ⊂ A. [1] (b) Is C ⊂ B? Give a reason. [1] (c) List the members of B ∩ C. [1] (d) Taking ℰ = A, find n(A ∩ C′). [2]

8. (Higher) A hotel has 120 guests. 74 guests used the pool (P), 51 used the gym (G) and 19 used neither.

(a) Find how many guests used both the pool and the gym. [2] (b) Find n(P ∩ G′). [1] (c) Find n(P′ ∪ G′). [2]

9. (Higher) n(ℰ) = 90, n(A) = 50 and n(B) = 30. n(A ∩ B) = x. The number of elements in A only is three times the number of elements in B only.

(a) Write expressions in x for the number of elements in A only and in B only. [2] (b) Form an equation and solve it to find x. [2] (c) Find n((A ∪ B)′). [1]

10. (Higher) 60 students were asked which of three clubs they attend: Chess (C), Drama (D) and Music (M).

• 5 attend all three clubs.
• 9 attend Chess and Drama; 11 attend Drama and Music; 8 attend Chess and Music.
• 22 attend Chess, 27 attend Drama and 25 attend Music.

(a) Draw a Venn diagram showing the number of students in each region. [4] (b) How many students attend exactly one club? [1] (c) Find n((C ∪ D)′). [2] (d) Find n(M ∩ C′). [1]

11. (Higher) ℰ = {x : x is an integer, 1 ≤ x ≤ 30}. A = {x : x is a multiple of 4}, B = {x : x is a multiple of 6} and C = {x : x is a factor of 30}.

(a) List the members of A ∩ B. [1] (b) Describe A ∩ B using the notation {x : …}. [1] (c) Find n(A ∪ B). [2] (d) List the members of B ∩ C. [1] (e) Is (B ∩ C) ⊂ A? Give a reason. [1] (f) Find n((A ∪ B ∪ C)′). [2]

Answers

1. (a) A = {1, 3, 5, 7, 9, 11, 13, 15}, B = {3, 6, 9, 12, 15}, so A ∩ B = {3, 9, 15} [1] (b) Combine both lists, each element once [1]: {1, 3, 5, 6, 7, 9, 11, 12, 13, 15} [1] (c) No: 12 is even, so 12 ∉ A and cannot be in A ∩ B [1] Examiner insight: in (b), one repeated element or one missing element usually limits you to the method mark only, so tick off each element of ℰ against both sets.

2. (a) P has only even numbers and Q only odd numbers, so P ∩ Q = ∅ [1] (b) 6 ∈ P, 6 ∉ Q, 7 ∈ Q ∪ R. Two correct [1]; all three correct [1] (c) {3, 5} [1] Examiner insight: {∅} or {0} for the empty set is not accepted; the mark needs the symbol ∅ on its own, or the words “the empty set”.

3. (a) M′ = {b, c, d, f, g, h, j} [1] (b) M ∪ N = {a, b, c, d, e, i} [1], so (M ∪ N)′ = {f, g, h, j} [1] (c) In N but not in M: {b, c, d} [1] Examiner insight: for (b), showing M ∪ N first earns the method mark even if you slip when taking the complement.

4. (a) A = {10, 12, 14, 16, 18, 20}, B = {10, 15, 20}. Overlap: 10, 20 [1] A only: 12, 14, 16, 18; B only: 15 [1] Outside both circles: 11, 13, 17, 19 [1] (b) Not even, but a multiple of 5: {15} [1] Examiner insight: an element left off the diagram, or the outside region left empty, costs a mark; count the elements on your diagram and check the total is 11.

5. (a) A ∩ B [1] (b) (A ∪ B)′ (or A′ ∩ B′) [1] (c) A ∩ B′ [1] (d) (A ∩ B′) ∪ (A′ ∩ B) (or (A ∪ B) ∩ (A ∩ B)′) [1] Examiner insight: any correct equivalent notation scores, but (A ∩ B)′ for part (b) does not, because it also includes “A only” and “B only”.

6. (a) x² < 9 means −3 < x < 3, so P = {−2, −1, 0, 1, 2} [1] (b) 3x − 2 > 4 gives 3x > 6, so x > 2 [1]; Q = {3, 4, 5, 6} [1] (c) No element is in both, so P ∩ Q = ∅ [1] (d) P ∪ Q = {−2, −1, 0, 1, 2, 3, 4, 5, 6}, so (P ∪ Q)′ = {−4, −3} [1]; n((P ∪ Q)′) = 2 [1] Examiner insight: in (a), including −3 and 3 is a common slip; x² < 9 is a strict inequality, and (−3)² = 9 is not less than 9.

7. (a) B = {1, 2, 4, 8} and every one of these is a factor of 24, so every element of B is in A [1] (b) No: 3 ∈ C but 3 ∉ B [1] (c) {1, 2, 4} [1] (d) A = {1, 2, 3, 4, 6, 8, 12, 24}, C = {1, 2, 3, 4, 6, 12}, so A ∩ C′ = {8, 24} [1]; n(A ∩ C′) = 2 [1] Examiner insight: “No” on its own scores nothing in (b); the mark is for naming one element of C that is not in B.

8. (a) n(P ∪ G) = 120 − 19 = 101 [1]; n(P ∩ G) = 74 + 51 − 101 = 24 [1] (b) Pool only: 74 − 24 = 50 [1] (c) P′ ∪ G′ is everything except P ∩ G [1]: 120 − 24 = 96 [1] Examiner insight: a labelled Venn diagram (50, 24, 27, 19) earns method credit in every part, even after a later slip.

9. (a) A only = 50 − x [1]; B only = 30 − x [1] (b) 50 − x = 3(30 − x) [1]; 50 − x = 90 − 3x, so 2x = 40 and x = 20 [1] (c) n(A ∪ B) = 50 + 30 − 20 = 60, so n((A ∪ B)′) = 90 − 60 = 30 [1] Examiner insight: the equation mark needs “three times B only” written as 3(30 − x) with the bracket; 3 × 30 − x is a different equation and loses the method mark.

10. (a) Centre (all three): 5. Chess and Drama only: 9 − 5 = 4; Drama and Music only: 11 − 5 = 6; Chess and Music only: 8 − 5 = 3 [1] Chess only: 22 − 5 − 4 − 3 = 10; Drama only: 27 − 5 − 4 − 6 = 12; Music only: 25 − 5 − 6 − 3 = 11 [1] In at least one club: 5 + 4 + 6 + 3 + 10 + 12 + 11 = 51 [1]; none: 60 − 51 = 9 [1] Diagram: centre 5; C and D only 4; D and M only 6; C and M only 3; C only 10; D only 12; M only 11; outside 9. (b) 10 + 12 + 11 = 33 [1] (c) n(C ∪ D) = 22 + 27 − 9 = 40 [1]; n((C ∪ D)′) = 60 − 40 = 20 (Music only 11 + none 9) [1] (d) In Music but not Chess: 11 + 6 = 17 [1] Examiner insight: writing 9, 11 and 8 straight into the “two clubs only” regions is the error that loses the most marks here, because every later region follows from them; allow follow-through in (b)–(d) from your own diagram.

11. (a) A = {4, 8, 12, 16, 20, 24, 28}, B = {6, 12, 18, 24, 30}, so A ∩ B = {12, 24} [1] (b) {x : x is a multiple of 12} (with ℰ as given) [1] (c) n(A) = 7, n(B) = 5, n(A ∩ B) = 2 [1]; n(A ∪ B) = 7 + 5 − 2 = 10 [1] (d) C = {1, 2, 3, 5, 6, 10, 15, 30}, so B ∩ C = {6, 30} [1] (e) No: 6 ∈ B ∩ C but 6 is not a multiple of 4, so 6 ∉ A [1] (f) Elements of C not already in A ∪ B: {1, 2, 3, 5, 10, 15}, so n(A ∪ B ∪ C) = 10 + 6 = 16 [1]; n((A ∪ B ∪ C)′) = 30 − 16 = 14 [1] Examiner insight: in (f), adding n(C) = 8 to 10 double-counts 6 and 30, which are already in B; the method mark is for removing the overlap.

Where marks are usually lost

• Repeating shared elements when listing a union, or missing one, so the accuracy mark is lost.
• Writing {∅} or {0} instead of ∅.
• Taking a complement without checking the universal set, so elements outside ℰ appear in the answer.
• Confusing (A ∪ B)′ (outside both circles) with (A ∩ B)′ (everything except the overlap).
• Including the boundary values in strict inequalities, such as ±3 in x² < 9.
• Giving “yes” or “no” for a subset question without naming an element as evidence.
• Putting a set’s total straight into its “only” region in survey questions.
• Forgetting the “none” region, so the counts do not add up to n(ℰ).
• In three-set problems, not subtracting the centre from each pairwise total first.
• Writing n(A) as a list, or a list where the question asks “how many”.

Next steps

• Recap with the set language revision notes.
• Re-read any weak section in the set language study guide.
• Try more questions across topic 1 in the Numbers and the number system practice questions.
• Go to the Edexcel IGCSE Mathematics hub or tick off topics on the 4MA1 checklist.
• Find your weak spots with all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification Issue 2, November 2017, Pearson Education Limited. Topic 1, Numbers and the number system: section 1.5, Set language and notation (Foundation Tier statements A–E; Higher Tier statements A–D), with the notation listed in Appendix 6.