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Sets - Operations on Sets: Union and Intersection (using Venn diagrams)

Grade 7ICSE

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

πŸ”‘Concepts

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Union of Sets (AβˆͺBA \cup B): The union of two sets AA and BB is the set of all elements that belong to set AA, or to set BB, or to both. In a Venn diagram, this is represented by shading the entire region covered by both circle AA and circle BB within the universal set rectangle.

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Intersection of Sets (A∩BA \cap B): The intersection of two sets AA and BB is the set of elements that are common to both AA and BB. Visually, this corresponds to the overlapping middle region where circle AA and circle BB cross each other.

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Disjoint Sets: Two sets are disjoint if they have no elements in common, meaning A∩B=βˆ…A \cap B = \emptyset. In a Venn diagram, disjoint sets are represented by two separate circles that do not touch or overlap at any point.

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Overlapping Sets: Sets that have at least one common element are called overlapping sets. On a Venn diagram, these sets are drawn with an overlapping area in the center. The union AβˆͺBA \cup B includes the elements in the left circle, the right circle, and the overlap, while the intersection A∩BA \cap B is only the overlap.

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Universal Set (ΞΎ\xi or UU): The universal set contains all the elements under consideration for a particular discussion and is represented by a large rectangle. All other sets (represented by circles) are subsets of this rectangle.

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Cardinal Number of Union: The number of elements in the union of two sets, denoted as n(AβˆͺB)n(A \cup B), is found by adding the number of elements in AA and BB, then subtracting the number of elements in their intersection to avoid double-counting. Visually, if you count all dots in circle AA and all dots in circle BB, the dots in the overlapping area were counted twice and must be subtracted once.

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Subsets in Venn Diagrams: If BβŠ‚AB \subset A, then every element of BB is also an element of AA. In a Venn diagram, this is shown as circle BB being entirely contained inside circle AA. In this case, AβˆͺB=AA \cup B = A and A∩B=BA \cap B = B.

πŸ“Formulae

n(AβˆͺB)=n(A)+n(B)βˆ’n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)

If A∩B=βˆ…A \cap B = \emptyset (disjoint), then n(AβˆͺB)=n(A)+n(B)n(A \cup B) = n(A) + n(B)

AβˆͺB=BβˆͺAA \cup B = B \cup A (Commutative Law)

A∩B=B∩AA \cap B = B \cap A (Commutative Law)

Aβˆͺβˆ…=AA \cup \emptyset = A

Aβˆ©βˆ…=βˆ…A \cap \emptyset = \emptyset

πŸ’‘Examples

Problem 1:

Given two sets P={2,4,6,8,10}P = \{2, 4, 6, 8, 10\} and Q={1,2,3,4,5}Q = \{1, 2, 3, 4, 5\}, find: (i) PβˆͺQP \cup Q, (ii) P∩QP \cap Q, and (iii) Verify the cardinality formula.

Solution:

Step 1: To find PβˆͺQP \cup Q, list all elements from both sets without repeating: PβˆͺQ={1,2,3,4,5,6,8,10}P \cup Q = \{1, 2, 3, 4, 5, 6, 8, 10\} Step 2: To find P∩QP \cap Q, identify the elements common to both: P∩Q={2,4}P \cap Q = \{2, 4\} Step 3: Verify the formula n(PβˆͺQ)=n(P)+n(Q)βˆ’n(P∩Q)n(P \cup Q) = n(P) + n(Q) - n(P \cap Q). \ n(P)=5n(P) = 5 \ n(Q)=5n(Q) = 5 \ n(P∩Q)=2n(P \cap Q) = 2 \ n(PβˆͺQ)=5+5βˆ’2=8n(P \cup Q) = 5 + 5 - 2 = 8. \ Since the set PβˆͺQP \cup Q has 8 elements, the formula is verified.

Explanation:

We first combine the sets for the union and find the common elements for the intersection. Then we use the counts (cardinal numbers) of these sets to check if the sum of individual sets minus their overlap equals the total union count.

Problem 2:

In a class of 30 students, 18 like Cricket (CC) and 15 like Football (FF). If 7 students like both sports, find the number of students who like either Cricket or Football using a Venn diagram approach.

Solution:

Step 1: Identify the given values: \ n(C)=18n(C) = 18 \ n(F)=15n(F) = 15 \ n(C∩F)=7n(C \cap F) = 7 \ Step 2: Calculate students who like only Cricket: n(CΒ only)=18βˆ’7=11n(C \text{ only}) = 18 - 7 = 11. \ Step 3: Calculate students who like only Football: n(FΒ only)=15βˆ’7=8n(F \text{ only}) = 15 - 7 = 8. \ Step 4: Find the union n(CβˆͺF)n(C \cup F) by adding only Cricket, only Football, and both: n(CβˆͺF)=11+8+7=26n(C \cup F) = 11 + 8 + 7 = 26 Alternatively, using the formula: n(CβˆͺF)=18+15βˆ’7=26n(C \cup F) = 18 + 15 - 7 = 26

Explanation:

This problem uses the inclusion-exclusion principle. By subtracting the intersection from the individual totals, we find the number of people in the 'only' sections of the Venn diagram. Adding these to the intersection gives the total for the union.