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Sets and Functions - Venn Diagrams and Operations on Sets

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Venn diagram represents sets as regions inside a rectangle (the Universal Set UU) and circles (individual sets). Operations like union and intersection are visualized by shading specific regions.

Venn diagram showing two intersecting circles A and B within a universal set rectangle U.
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The Union of sets AA and BB, denoted as A∪BA \cup B, represents the set of all elements belonging to AA or BB or both. In a Venn diagram, this is the entire region covered by both circles.

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The Intersection of sets AA and BB, denoted as A∩BA \cap B, represents elements common to both sets. In a Venn diagram, this is the overlapping region between the circles.

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The Difference of sets A−BA - B represents elements that belong to AA but not to BB. This is often called the 'only A' region.

Intersection diagram showing regions for A minus B, B minus A, and the intersection.
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The Complement of a set A′A', denoted as AcA^c, consists of all elements in the universal set UU that are not in AA.

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Symmetric Difference AΔBA \Delta B is the set of elements in either AA or BB but not in their intersection: (A−B)∪(B−A)(A - B) \cup (B - A).

📐Formulae

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

De Morgan's First Law: (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'

De Morgan's Second Law: (A∩B)′=A′∪B′(A \cap B)' = A' \cup B'

n(A−B)=n(A)−n(A∩B)n(A - B) = n(A) - n(A \cap B)

n(AΔB)=n(A−B)+n(B−A)=n(A∪B)−n(A∩B)n(A \Delta B) = n(A - B) + n(B - A) = n(A \cup B) - n(A \cap B)

n(A′)=n(U)−n(A)n(A') = n(U) - n(A)

💡Examples

Problem 1:

In a class of 50 students, 30 study Mathematics, 25 study Physics, and 10 study both subjects. Find the number of students who study: (i) either Mathematics or Physics, and (ii) neither of the two subjects.

Solution:

Let MM be the set of students studying Mathematics and PP be the set of students studying Physics. Given: n(U)=50n(U) = 50, n(M)=30n(M) = 30, n(P)=25n(P) = 25, and n(M∩P)=10n(M \cap P) = 10. (i) To find those studying either subject, we find the union: n(M∪P)=n(M)+n(P)−n(M∩P)n(M \cup P) = n(M) + n(P) - n(M \cap P) n(M∪P)=30+25−10=45n(M \cup P) = 30 + 25 - 10 = 45. (ii) To find those studying neither, we find the complement of the union: n((M∪P)′)=n(U)−n(M∪P)n((M \cup P)') = n(U) - n(M \cup P) n((M∪P)′)=50−45=5n((M \cup P)') = 50 - 45 = 5.

Explanation:

We use the Principle of Inclusion-Exclusion to find the number of students in at least one set, then subtract from the total universal set to find those outside both sets.

Problem 2:

If U={x:x∈N,x≤10}U = \{x : x \in \mathbb{N}, x \leq 10\}, A={1,3,5,7,9}A = \{1, 3, 5, 7, 9\}, and B={2,3,5,7}B = \{2, 3, 5, 7\}, verify De Morgan's First Law: (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'.

Solution:

Step 1: Find A∪BA \cup B. A∪B={1,2,3,5,7,9}A \cup B = \{1, 2, 3, 5, 7, 9\}. \nStep 2: Find the LHS (A∪B)′(A \cup B)' relative to U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. (A∪B)′={4,6,8,10}(A \cup B)' = \{4, 6, 8, 10\}. \nStep 3: Find A′A' and B′B'. A′={2,4,6,8,10}A' = \{2, 4, 6, 8, 10\} B′={1,4,6,8,9,10}B' = \{1, 4, 6, 8, 9, 10\}. \nStep 4: Find the RHS A′∩B′A' \cap B'. A′∩B′={4,6,8,10}A' \cap B' = \{4, 6, 8, 10\}. \nSince LHS = RHS, the law is verified.

Explanation:

This demonstrates De Morgan's Law by calculating the complement of a union and showing it equals the intersection of the individual complements.

Problem 3:

In a survey of 100 families, 60 use Brand A detergent, 45 use Brand B, and 20 use both. Represent this on a Venn diagram and find the number of families using neither brand.

Venn diagram with 40 in Only A, 20 in intersection, 25 in Only B, and 15 in the outer rectangle.

Solution:

  1. Let n(U)=100n(U) = 100, n(A)=60n(A) = 60, n(B)=45n(B) = 45, and n(A∩B)=20n(A \cap B) = 20.
  2. Families using only Brand A: n(A)−n(A∩B)=60−20=40n(A) - n(A \cap B) = 60 - 20 = 40.
  3. Families using only Brand B: n(B)−n(A∩B)=45−20=25n(B) - n(A \cap B) = 45 - 20 = 25.
  4. Total families using at least one brand: n(A∪B)=40+20+25=85n(A \cup B) = 40 + 20 + 25 = 85.
  5. Families using neither brand: n(U)−n(A∪B)=100−85=15n(U) - n(A \cup B) = 100 - 85 = 15.

Explanation:

We subtract the intersection from individual sets to find the 'only' regions. The sum of all regions within the circles subtracted from the universal set gives the 'neither' category.

Problem 4:

Given U={1,2,3,4,5,6,7,8}U = \{1, 2, 3, 4, 5, 6, 7, 8\}, A={2,4,6,8}A = \{2, 4, 6, 8\}, and B={1,2,3,4}B = \{1, 2, 3, 4\}. Find and illustrate A−BA - B and B−AB - A.

Venn diagram showing elements 6 and 8 in A-B, 2 and 4 in intersection, 1 and 3 in B-A, and 5 and 7 in U.

Solution:

  1. Intersection A∩B={2,4}A \cap B = \{2, 4\}.
  2. A−B=elements in A not in B={6,8}A - B = \text{elements in A not in B} = \{6, 8\}.
  3. B−A=elements in B not in A={1,3}B - A = \text{elements in B not in A} = \{1, 3\}.
  4. Elements in UU not in AA or BB: {5,7}\{5, 7\}.

Explanation:

Elements common to both sets are placed in the overlap. A−BA - B contains elements found exclusively in AA, and B−AB - A contains elements found exclusively in BB.