Cardinality of a Set | Word Problems based on Sets and Venn Diagrams

Cardinality of a Set | Word Problems based on Sets and Venn Diagrams

Cardinality of a Set

Let, A be a finite set. The number of elements in set A is called the cardinal number of the set A and it is denoted by n(A). If A = {a, b, c, d, e} then n(A) = 5 since the number of elements contained in set A is 5. The properties based on the cardinal number is called the cardinality of a set.

Let, U be the universal set, A and B be any two finite and overlapping subsets of U. Then,

          n(U) -----------> Number of elements in set U

          n(A) -----------> Number of elements in set A

          n(B) -----------> Number of elements in set B

          n(A∪B) -------> Number of elements in set A∪B

          n(A∩B) -------> Number of elements in A∩B

          n_o(A) ----------> Number of elements in set A only

          n_o(B) ----------> Number of elements in set B only

          n(A∪B)^c ------> Number of elements in set (A∪B)^c

List of formulae based on cardinal numbers of sets (Two sets)

Let, A and B are two subsets of the universal set U then,

(a)   n(A∪B) = n(A) + n(B) – n(A∩B)

(b)   n(A∩B) = n(A) + n(B) – n(A∪B)

(c)    n(A∪B)^c = n(U) – n(A∪B)

(d)   n_o(A) = n(A) – n(A∩B)

(e)   n_o(B) = n(B) – n(A∩B)

(f)    n(A∪B) = n_o(A) + n_o(B) + n(A∩B)

(g)   n(U) = n_o(A) + n_o(B) + n(A∩B) + n(A∪B)^c

(h)   n(U) = n(A) + n(B) + n(A∪B)^c – n(A∪B)

Word Problems based on Sets and Venn Diagrams

Workout Examples

Example 1: Out of 50 students of class IX, 25 students liked to play football, 35 liked to play basketball and 15 liked to play both the games. How many students do not like to play any games? Show the above information in a Venn-diagram.

Solution: Let U, F and B be the universal set, set of students who like to play football and set of students who like to play basketball respectively.

∴     n(U) = 50

       n(F) = 25

       n(B) = 35

       n(F ∩ B) = 15

       n(F ∪ B)^c = ?

Venn diagram,

 We Know,

       n(F ∪ B) = n(F) + n(B) – n(F ∩ B)

                      = 25 + 35 – 15

                      = 45

Now,

       n(F ∪ B)^c = n(U) – n(F∪ B)

                        = 50 – 45

                        = 5

Hence 5 students do not like to play any games.

Example 2: Out of 100 students, 80 passed the science, 71 the mathematics, 10 failed the both subjects and 7 did not appear the examination. Find the number of students who passed the both subjects by representing the above information in a Venn-diagram.

Solution: Let U, S and M be the universal set, set of students passed in science and set of students passed in mathematics respectively.

∴     n(U) = 100

       n(S) = 80

       n(M) = 71

       n(S∪M)^c = 10 + 7 = 17

       n(S∩M) = ?

Venn diagram,

 We Know,

       n(S ∪ M) = n(U) – n(S ∪ M)^c

                       = 100 – 17

                       = 83

Now,

       n(S ∩ M) = n(S) + n(M) – n(S ∪ M)

                       = 80 + 71 – 83

                       = 68

Hence 68 students passed both the subjects.

Example 3: In a survey, it was found that 80% people liked oranges, 85% liked mangoes and 75% liked both. But 45 people liked none of them. By drawing a Venn diagram, find the number of people who were in the survey.

Solution: Let U, O and M be the universal set, set of people who liked oranges and set of people who liked mangoes respectively.

∴     n(U) = 100%

       n(O) = 80%

       n(M) = 85%

       n(O ∩ M) = 75%

       n(O ∪ M)^c = 45

Venn diagram,

We Know,

       n(O ∪ M) = n(O) + n(M) – n(O ∩ M)

                       = 80% + 85% – 75%

                       = 90%

Now,

       n(O ∪ M)^c = n(U) – n(O ∪ M)

                         = 100% – 90%

                         = 10%

Let, the total number of people in the survey be ‘x’.

∴      10% of x = 45

or,    10/100 × x = 45

or,    1/10 × x = 45

or,    x = 450

Hence, the number of people who were in the survey is 450.

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