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Sets - Operations on Sets: Union, Intersection, Difference

Grade 11CBSE

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

🔑Concepts

Union of Sets (ABA \cup B): The union of two sets AA and BB is the set containing all elements that belong to AA, or to BB, or to both. In a Venn diagram, this is visualized by shading the entire area of both overlapping circles.

Intersection of Sets (ABA \cap B): The intersection consists of only those elements that are common to both AA and BB. On a Venn diagram, this is represented by the central football-shaped overlapping region where the two circles meet.

Disjoint Sets: Two sets AA and BB are said to be disjoint if their intersection is an empty set, i.e., AB=A \cap B = \emptyset. Visually, these are represented as two separate circles that do not touch or overlap in the Venn diagram.

Difference of Sets (ABA - B): The difference ABA - B is the set of elements that belong to AA but do not belong to BB. Visually, it is shown as circle AA with the overlapping part shared with BB removed, appearing like a crescent moon shape.

Complement of a Set (AA'): The complement of a set AA is the set of all elements in the Universal set UU that are not in AA. In a Venn diagram, if UU is the bounding rectangle and AA is a circle inside it, AA' is the entire area inside the rectangle but outside the circle.

Symmetric Difference (AΔBA \Delta B): This operation is defined as (AB)(BA)(A - B) \cup (B - A). It includes elements that are in either AA or BB, but not in both. Visually, it shades the outer portions of both circles while leaving the middle intersection blank.

De Morgan's Laws: These are fundamental rules relating unions and intersections via complements: (AB)=AB(A \cup B)' = A' \cap B' and (AB)=AB(A \cap B)' = A' \cup B'. These laws show how the complement of a union is the intersection of the complements, and vice versa.

📐Formulae

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

n(AB)=n(A)+n(B)n(A \cup B) = n(A) + n(B) (if AA and BB are disjoint)

n(AB)=n(A)n(AB)n(A - B) = n(A) - n(A \cap B)

n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA)+n(ABC)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)

AB=ABA - B = A \cap B'

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

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

💡Examples

Problem 1:

Given two sets A={2,4,6,8,10}A = \{2, 4, 6, 8, 10\} and B={6,8,10,12,14}B = \{6, 8, 10, 12, 14\}, find (i) ABA \cup B, (ii) ABA \cap B, and (iii) ABA - B.

Solution:

Step 1: To find ABA \cup B, combine all elements from both sets and list each unique element once: AB={2,4,6,8,10,12,14}A \cup B = \{2, 4, 6, 8, 10, 12, 14\}.\nStep 2: To find ABA \cap B, identify the elements that appear in both AA and BB: AB={6,8,10}A \cap B = \{6, 8, 10\}.\nStep 3: To find ABA - B, list elements that are in AA but are not present in BB. Since 6,8,106, 8, 10 are in BB, we remove them from AA: AB={2,4}A - B = \{2, 4\}.

Explanation:

Union represents the 'all-inclusive' set, Intersection represents 'common' elements, and Difference represents 'exclusive' elements of the first set.

Problem 2:

In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Solution:

Step 1: Let CC be the set of people who like coffee and TT be the set of people who like tea.\nStep 2: We are given n(CT)=70n(C \cup T) = 70, n(C)=37n(C) = 37, and n(T)=52n(T) = 52.\nStep 3: Use the formula n(CT)=n(C)+n(T)n(CT)n(C \cup T) = n(C) + n(T) - n(C \cap T).\nStep 4: Substitute the values: 70=37+52n(CT)70 = 37 + 52 - n(C \cap T).\nStep 5: 70=89n(CT)70 = 89 - n(C \cap T).\nStep 6: n(CT)=8970=19n(C \cap T) = 89 - 70 = 19.

Explanation:

This problem uses the Principle of Inclusion-Exclusion for two sets to find the size of the intersection based on the total union and individual set sizes.