Number - Set Theory and Venn Diagrams

Grade 12IGCSE

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

🔑Concepts

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Universal Set (ξ): The set containing all possible elements under consideration.

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Null/Empty Set (\emptyset or {}): A set containing no elements.

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Subsets: A is a subset of B (A \subseteq B) if every element of A is also in B. A is a proper subset (A \subset B) if A is in B but A \neq B.

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Union (A \cup B): The set of elements that are in A, or in B, or in both.

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Intersection (A \cap B): The set of elements that are in both A and B.

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Complement (A'): The set of elements in the universal set that are not in A.

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Cardinality (n(A)): The number of elements in set A.

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Disjoint Sets: Sets that have no elements in common (A \cap B = \emptyset).

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De Morgan's Laws: Relationship between the complement of unions and intersections.

📐Formulae

n(A \cup B) = n(A) + n(B) - n(A \cap B)

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)

n(A) + n(A') = n(\xi)

(A \cup B)' = A' \cap B'

(A \cap B)' = A' \cup B'

💡Examples

Problem 1:

In a group of 100 students, 60 like Math, 45 like Physics, and 20 like both. How many students like neither subject?

Solution:

15 students

Explanation:

Let M be the set of students who like Math and P be the set who like Physics. We are given n(M) = 60, n(P) = 45, and n(M \cap P) = 20. Using the formula n(M \cup P) = n(M) + n(P) - n(M \cap P), we get 60 + 45 - 20 = 85. The number of students who like neither is n(ξ) - n(M \cup P) = 100 - 85 = 15.

Problem 2:

List the elements of the set A \cap B' where ξ = {x : x is an integer, 1 \leq x \leq 10}, A = {prime numbers}, and B = {even numbers}.

Solution:

{3, 5, 7}

Explanation:

First, define ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Set A (primes) = {2, 3, 5, 7}. Set B (evens) = {2, 4, 6, 8, 10}. Set B' (complements of B in ξ) = {1, 3, 5, 7, 9}. The intersection A \cap B' represents elements that are prime AND odd. Looking at both sets, the common elements are {3, 5, 7}.

Problem 3:

Given n(ξ) = 30, n(A) = 15, n(B) = 12, and n(A \cup B)' = 8. Find n(A \cap B).

Solution:

5

Explanation:

First, find n(A \cup B). Since n(A \cup B)' = 8, then n(A \cup B) = n(ξ) - 8 = 30 - 8 = 22. Now use the inclusion-exclusion principle: n(A \cup B) = n(A) + n(B) - n(A \cap B). Substituting the values: 22 = 15 + 12 - n(A \cap B). This simplifies to 22 = 27 - n(A \cap B), so n(A \cap B) = 5.