Word Problems – Set

Example 1: In a group of 60 people, 27 like tea, 42 likes coffee, and each person likes at least one of the two drinks. 
How many people like both coffee and tea.

Solution: Let T denote the set of people who likes tea

and C denote the set of people who likes coffee.

Here,  n(C) = 42,   n(T) = 27  and  n(CUT) = 60

We know the formula 

n(CUT) = n(C) + n(T) – n(C ∩ T)

put the value in formula

60 = 42 + 27 – n(C ∩ T)

60 = 69 –  n(C ∩ T)

n(C ∩ T) = 69 – 60

 n(C ∩ T) = 9

Thus, 9 people like both tea and coffee.

Example 2: In a group of 90 people, 60 like tea, 46 likes coffee, and each person likes at least one of the two drinks. 
How many people like both coffee and tea.

Solution: Let T denote the set of people who likes tea and
C denote the set of people who likes coffee.

Here,  n(C) = 46,  n(T) = 60  and n(CUT) = 90

We know the formula 

n(CUT) = n(C) + n(T) – n(C ∩ T)

put the value in formula

90 = 46 + 60 – n(C ∩ T)

90 = 106 –  n(C ∩ T)

 n(C ∩ T) = 106 – 90

 n(C ∩ T) = 16

Thus, 16 people like both tea and coffee.

Example 3: In a group of 70 people, 52 like tea, 37 likes coffee, and each person likes at least one of the two drinks. 
How many people like both coffee and tea.

Solution: Let T denotes the set of people who like tea and
C denotes the set of people who likes coffee.

Here,  n(C) = 37,  n(T) = 52  and n(CUT) = 70

We know the formula 

n(CUT) = n(C) + n(T) – n(C ∩ T)

put the value in formula

70 = 37 + 52 – n(C ∩ T)

70 = 89 –  n(C ∩ T)

 n(C ∩ T) = 89 – 70

 n(C ∩ T) = 19

Thus, 19 people like both tea and coffee.

Example 4: In a school, 30 teachers teach mathematics or physics, 18 teach mathematics and 6 teach both physics and mathematics. How many teach physics only?

Solution : Let M denote the teachers who teach mathematics.

Let P denote the teachers who teach physics.

Total number of teachers who teach mathematics or physics n(M∪P)= 30

Number of teachers who teach mathematics n(M) = 18

Number of teachers who teach both mathematics and physics n(M ⋂ P) = 6

n(M) = 18, n(M⋂P) = 6, and n(M∪P) = 30

put the value in formula

n(M∪P) = n(M) + n(P) – n(M⋂P)

30 = 18 + n(P) – 6

30 = 12 + n(P)

n(P) = 30 – 12

n(P) = 18

Number of teachers who teach both physics only = n(P – M)

n(P – M) = n(P) – n(M⋂P)

= 18 – 6

n(P – M) = 12

Number of teachers who teach only physics = 12

Example 5: In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each students knows either Hindi or English. How many students are there in the group.

Solution: Let U be the set of all students in the group.
Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

(HUE) = U

Here, n(H) = 100,  n(E) = 50  and n(H ∩ E) = 25

We know the formula 

n(HUE) = n(H) + n(E) – n(H ∩ E)

put the value in formula

 n(HUE) = 100 + 50 – 25

n(HUE) = 150 – 25

 Thus, 125 students in the group.

Example 6: In a group of 100 students, 72 students know English, and 43 know Hindi. How many students know only English. How many students know only Hindi and how many know both.

Solution: Let A be the set of students who know English.
Let B be the set of students who know Hindi.

A – B be the set of students who know English and not  Hindi.
B – A be the set of students who know Hindi and not English.

(A ∩ B) be the set of students who know both Hindi and English 

Given n(A) = 72,  n(B) = 43,  n(A U B) = 100,  We know the formula 

n(A U B) = n(A) + n(B) – n(A ∩ B)

put the value in formula

 100 = 72 + 43 – n(A ∩ B)

100 = 115 – n(A ∩ B)

n(A ∩ B) = 115 – 100,

n(A ∩ B) = 15 

Thus, 15 students in the group know both English and Hindi.

n(A) = n(A – B) + n(A ∩ B) 

n(A – B) = n(A) – n(A ∩ B) 

= 72 – 15 = 57

and n(B – A) = n(B) – n(A ∩ B)

 n(B – A) = 43 – 15 = 28

Therefore, number of students know only English = 57 

number of students know only Hindi = 28

Example 7: In an exam 50 students passed in English. 60 students passed in science. 40 students passed in both subjects. None of them fail in both subjects. Find the number of students who passed at least in one of the subjects.

Solution: Let A be the set of students who passed in English.

Let B be the set of students who passed in science.

Given,  n(A) = 50, n(B) = 60,  n(A∩B) = 40,  n(A U B) = ?

We know the formula 

n(A U B) = n(A) + n(B) – n(A ∩ B)

put the value in formula

  = 50 + 60 – 40

n(A U B) = 110 – 40

n(A U B) = 70

Thus, number of students who passed at least in one of the subjects is 70.

Example 8: If R is a set of real numbers and P is a set of rational numbers. Find set of (R – P)

Solution : R is a set of real numbers and P is a set of rational numbers.

Therefore, set (R – P ) is a set of irrational numbers.