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IGCSE Maths · Edexcel 4XMA1 · Unit 1 Higher · Number

Sets, Notation and Venn Diagrams — Worked Solutions

Full worked solutions to all 12 questions (40 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 4–5)
  1. (2 marks)ξ = {1, 2, 3, …, 12}. A = {multiples of 3}. B = {factors of 12}.
    (a) List the members of A ∩ B.
    (b) Find n(A ∪ B).
    Solution
    1. A = {3, 6, 9, 12}; B = {1, 2, 3, 4, 6, 12}.
    2. (a) In both: {3, 6, 12}.
    3. (b) A ∪ B = {1, 2, 3, 4, 6, 9, 12}, which has 7 members.
    (a) {3, 6, 12} (b) 7
  2. (2 marks)Using the sets in Question 1:
    (a) List the members of A′ ∩ B.
    (b) Is it true that 9 ∈ B? Give a reason.
    Solution
    1. A′ = {1, 2, 4, 5, 7, 8, 10, 11}.
    2. (a) The members of A′ that are also in B: {1, 2, 4}.
    3. (b) 9 is not a factor of 12, so 9 ∉ B. The statement is false.
    (a) {1, 2, 4} (b) No — 9 is not a factor of 12
  3. (2 marks)Use set notation to describe the shaded region in each diagram.
    ξAB
    (a)
    ξAB
    (b)
    Solution
    1. (a) The shaded region is inside B but outside A: A′ ∩ B.
    2. (b) The shaded region is outside both circles: (A ∪ B)′, which can also be written A′ ∩ B′.
    (a) A′ ∩ B (b) (A ∪ B)′
  4. (4 marks)ξ = {x : x is an integer, 1 ≤ x ≤ 15}. P = {prime numbers}. E = {even numbers}.
    (a) List the members of P ∩ E.
    (b) Find n(P′ ∩ E′).
    (c) A number is chosen at random from ξ. Find the probability that it is in P ∪ E.
    Solution
    1. P = {2, 3, 5, 7, 11, 13}; E = {2, 4, 6, 8, 10, 12, 14}.
    2. (a) P ∩ E = {2} (2 is the only even prime).
    3. (b) Neither prime nor even: {1, 9, 15}, so n(P′ ∩ E′) = 3.
    4. (c) n(P ∪ E) = 15 − 3 = 12, so P = 1215 = 45.
    (a) {2} (b) 3 (c) 4/5
Section B — Core (grades 5–7)
  1. (4 marks)50 people were asked whether they like coffee or tea. 32 like coffee, 21 like tea and 9 like neither.
    (a) Draw a Venn diagram to show this information.
    (b) One of the people is chosen at random. Find the probability that they like tea but not coffee.
    Solution
    1. At least one: 50 − 9 = 41. Both: 32 + 21 − 41 = 12.
    2. Coffee only = 32 − 12 = 20; tea only = 21 − 12 = 9; neither = 9.
    ξCT201299
    1. (b) Tea only = 9, so P = 950.
    (a) 20, 12, 9, outside 9 (b) 9/50
  2. (4 marks)n(ξ) = 60. The Venn diagram shows the number of elements in each region.
    ξAB2xx + 33x − 110
    (a) Find the value of x.
    (b) Find n(B).
    Solution
    1. Regions add to 60: 2x + (x + 3) + (3x − 1) + 10 = 60.
    2. 6x + 12 = 60, so 6x = 48 and x = 8.
    3. (b) n(B) = (x + 3) + (3x − 1) = 4x + 2 = 4(8) + 2 = 34.
    (a) x = 8 (b) n(B) = 34
  3. (5 marks)100 students study at least one of French (F), German (G) and Spanish (S), or none of them.
    5 study all three. 12 study French and German. 15 study French and Spanish. 10 study German and Spanish.
    45 study French, 38 study German and 40 study Spanish.
    (a) Complete a three-set Venn diagram.
    (b) How many students study none of the three languages?
    (c) Find the probability that a student chosen at random studies exactly two languages.
    Solution
    1. Centre = 5. F∩G only = 12 − 5 = 7; F∩S only = 15 − 5 = 10; G∩S only = 10 − 5 = 5.
    2. F only = 45 − 7 − 10 − 5 = 23; G only = 38 − 7 − 5 − 5 = 21; S only = 40 − 10 − 5 − 5 = 20.
    ξFGS232120710559
    1. (b) Inside the circles: 23 + 21 + 20 + 7 + 10 + 5 + 5 = 91, so none = 100 − 91 = 9.
    2. (c) Exactly two = 7 + 10 + 5 = 22, so P = 22100 = 1150.
    (b) 9 (c) 11/50
  4. (3 marks)A = {x : x is an integer, x² < 10}. B = {x : x is an integer, 0 < x < 4}.
    (a) List the members of A.
    (b) Explain why B ⊂ A.
    Solution
    1. (a) x² < 10 for the integers −3, −2, −1, 0, 1, 2, 3 (since 3² = 9 < 10 but 4² = 16). A = {−3, −2, −1, 0, 1, 2, 3}.
    2. (b) B = {1, 2, 3}. Every element of B is also an element of A, so B ⊂ A.
    (a) {−3, −2, −1, 0, 1, 2, 3} (b) every element of B is in A
  5. (4 marks)The Venn diagram shows the number of students in a group of 40 who have a cat (C) or a dog (D).
    ξCD1461010
    A student is chosen at random. Find
    (a) P(C)
    (b) P(C ∩ D′)
    (c) the probability that the student has a dog, given that they have a cat.
    Solution
    1. (a) n(C) = 14 + 6 = 20, so P(C) = 2040 = 12.
    2. (b) C ∩ D′ is “cat but no dog” = 14, so P = 1440 = 720.
    3. (c) Restrict to the 20 cat owners: 6 of them also have a dog, so P = 620 = 310.
    (a) 1/2 (b) 7/20 (c) 3/10
Section C — Stretch (grades 8–9)
  1. (3 marks)ξ = {x : x is an integer, 1 ≤ x ≤ 30}. A = {square numbers}. B = {multiples of 4}.
    Find n((A ∪ B)′).
    Solution
    1. A = {1, 4, 9, 16, 25}, so n(A) = 5. B = {4, 8, 12, 16, 20, 24, 28}, so n(B) = 7.
    2. A ∩ B = {4, 16}, so n(A ∩ B) = 2.
    3. n(A ∪ B) = 5 + 7 − 2 = 10, so n((A ∪ B)′) = 30 − 10 = 20.
    20
  2. (3 marks)A and B are any two sets. For each statement, write always true, sometimes true or never true.
    (a) A ∩ B ⊂ A ∪ B
    (b) n(A ∪ B) = n(A) + n(B)
    (c) A ∪ A′ = ∅
    Solution
    1. (a) Anything in both sets is certainly in at least one of them: always true.
    2. (b) Only when A and B have no elements in common (A ∩ B = ∅): sometimes true.
    3. (c) A ∪ A′ contains everything in ξ, so it equals ξ, not ∅ (unless ξ itself is empty, which isn’t the case in practice): never true.
    (a) always (b) sometimes (c) never
  3. (4 marks)n(ξ) = 48, n(A) = 25, n(B) = 20.
    n(A ∩ B) = x and n(A′ ∩ B′) = 2x.
    (a) Find the value of x.
    (b) An element is chosen at random from ξ. Find the probability that it is in B but not in A.
    Solution
    1. n(A ∪ B) = 25 + 20 − x = 45 − x.
    2. Everything adds to 48: (45 − x) + 2x = 48, so 45 + x = 48 and x = 3.
    3. (b) B only = 20 − 3 = 17, so P = 1748.
    (a) x = 3 (b) 17/48
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