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Sets - Venn Diagrams

Grade 11CBSE

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

🔑Concepts

Venn Diagrams are graphical representations of sets where a Universal Set UU is depicted as a large rectangle, and its subsets are shown as circles or closed curves within it. The position and overlap of these circles represent the relationships between the sets.

The Union of two sets ABA \cup B is visually represented by shading the entire area covered by circle AA and circle BB. This includes the overlapping region. It represents elements that belong to either AA, BB, or both.

The Intersection of two sets ABA \cap B is shown as the common area where circle AA and circle BB overlap. If the circles do not overlap, the sets are called Disjoint Sets, and AB=ϕA \cap B = \phi.

The Complement of a set AA, denoted as AA' or AcA^c, is the region inside the universal rectangle UU but outside the circle AA. This represents all elements in the universal set that are not in set AA.

The Difference of sets ABA - B (also called the relative complement) is the region of circle AA that does not overlap with circle BB. Visually, it looks like a crescent moon shape within circle AA. It represents elements belonging to AA but not to BB.

A Subset relationship ABA \subset B is represented visually by placing circle AA entirely inside circle BB. This indicates that every element in AA is also an element of BB.

The Symmetric Difference AΔBA \Delta B, defined as (AB)(BA)(A - B) \cup (B - A), is represented by shading the portions of circles AA and BB that do not overlap. The central intersection area remains unshaded.

De Morgan's Laws can be visualized using Venn Diagrams: (AB)(A \cup B)' is the region outside both circles, which is identical to the intersection of the regions outside AA and outside BB (ABA' \cap B').

📐Formulae

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

If AA and BB are disjoint, n(AB)=n(A)+n(B)n(A \cup B) = n(A) + n(B) since n(AB)=0n(A \cap B) = 0

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)

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

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

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

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

💡Examples

Problem 1:

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:

Let CC be the set of people who like coffee and TT be the set of people who like tea.\nGiven:\nn(CT)=70n(C \cup T) = 70 (since everyone likes at least one drink)\nn(C)=37n(C) = 37\nn(T)=52n(T) = 52\nUsing the formula n(CT)=n(C)+n(T)n(CT)n(C \cup T) = n(C) + n(T) - n(C \cap T):\n70=37+52n(CT)70 = 37 + 52 - n(C \cap T)\n70=89n(CT)70 = 89 - n(C \cap T)\nn(CT)=8970=19n(C \cap T) = 89 - 70 = 19

Explanation:

We use the fundamental addition principle for two sets. The union represents the total group because no one falls outside the two categories. Subtracting the union from the sum of individual sets gives the overlap (intersection).

Problem 2:

In a survey of 60 students, 25 like Physics (PP), 20 like Chemistry (CC), and 10 like both. Find the number of students who like neither Physics nor Chemistry.

Solution:

Step 1: Find the number of students who like at least one subject.\nn(PC)=n(P)+n(C)n(PC)n(P \cup C) = n(P) + n(C) - n(P \cap C)\nn(PC)=25+2010=35n(P \cup C) = 25 + 20 - 10 = 35\nStep 2: Find the number of students who like neither.\nn(PC)=n(U)n(PC)n(P \cup C)' = n(U) - n(P \cup C)\nn(PC)=6035=25n(P \cup C)' = 60 - 35 = 25

Explanation:

First, calculate the total number of students covered by the circles (the union). Then, subtract this value from the total students in the universal set (the rectangle) to find those in the exterior region (the complement).