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Sets - Practical Problems on Union and Intersection of Two Sets

Grade 11CBSE

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

🔑Concepts

Cardinal Number of a Set: The number of distinct elements present in a finite set AA is called its cardinal number and is denoted by n(A)n(A). For example, if A={1,2,3,4}A = \{1, 2, 3, 4\}, then n(A)=4n(A) = 4.

Union of Two Sets (ABA \cup B): This represents the collection of elements that belong to set AA, or set BB, or both. Visually, in a Venn diagram, this corresponds to the entire area enclosed by both overlapping circles representing AA and BB.

Intersection of Two Sets (ABA \cap B): This represents elements that are common to both set AA and set BB. In a Venn diagram, this is depicted as the overlapping region where the two circles cross each other.

Disjoint Sets: Two sets AA and BB are called disjoint if they have no common elements, meaning AB=A \cap B = \emptyset. Visually, these are represented as two separate circles that do not touch or overlap within the universal set rectangle.

Difference of Sets (ABA - B and BAB - A): The set ABA - B contains elements present in AA but not in BB. Visually, this is the region of circle AA that excludes the overlapping intersection part, resembling a crescent moon shape. Similarly, BAB - A is the part of circle BB excluding the intersection.

The Inclusion-Exclusion Principle: To find the total elements in the union n(AB)n(A \cup B), we add n(A)n(A) and n(B)n(B), then subtract the intersection n(AB)n(A \cap B) because those elements were counted twice (once in AA and once in BB).

Universal Set and Complements: The universal set UU is represented by a rectangle containing the circles. Elements that belong to neither AA nor BB are located inside the rectangle but outside both circles, denoted by (AB)(A \cup B)'.

📐Formulae

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

If AB=A \cap B = \emptyset, then n(AB)=n(A)+n(B)n(A \cup B) = n(A) + n(B)

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

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

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

n(AB)=n(AB)+n(BA)+n(AB)n(A \cup B) = n(A - B) + n(B - A) + n(A \cap 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 (total people, since everyone likes at least one)\nn(C)=37n(C) = 37\nn(T)=52n(T) = 52\nWe need to find n(CT)n(C \cap T).\nUsing the formula:\nn(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 identify the given values for the union and individual sets, then substitute them into the addition principle formula to solve for the unknown intersection value.

Problem 2:

In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics?

Solution:

Let MM be the set of teachers who teach Mathematics and PP be the set of teachers who teach Physics.\nGiven:\nn(MP)=20n(M \cup P) = 20\nn(M)=12n(M) = 12\nn(MP)=4n(M \cap P) = 4\nUsing the formula:\nn(MP)=n(M)+n(P)n(MP)n(M \cup P) = n(M) + n(P) - n(M \cap P)\n20=12+n(P)420 = 12 + n(P) - 4\n20=8+n(P)20 = 8 + n(P)\nn(P)=208=12n(P) = 20 - 8 = 12.

Explanation:

Here, 'Mathematics or Physics' indicates the union. We plug the known union, individual count for Math, and the shared intersection count into the formula to isolate the count for the Physics set.