# Word Problems on Sets

Word problems on sets are solved here to get the basic ideas how to use the  properties of union and intersection of sets.

Solved basic word problems on sets:

1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Solution:

Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

then n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

= 20 + 28 - 36

= 48 - 36

= 12

2. If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).

Solution:

Using the formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)

70 = 18 + 25 + n(B - A)

70 = 43 + n(B - A)

n(B - A) = 70 - 43

n(B - A) = 27

Now n(B) = n(A ∩ B) + n(B - A)

= 25 + 27

= 52

Different types on word problems on sets:

3. In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea?

Solution:

Let A = Set of people who like cold drinks.

B = Set of people who like hot drinks.

Given

(A ∪ B) = 60            n(A) = 27       n(B) = 42 then;

n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

= 27 + 42 - 60

= 69 - 60 = 9

= 9

Therefore, 9 people like both tea and coffee.

4. There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class.

When two classes meet at different hours and 12 students are enrolled in both activities.

When two classes meet at the same hour.

Solution:

n(A) = 35,       n(B) = 57,       n(A ∩ B) = 12

(Let A be the set of students in art class.
B be the set of students in dance class.)

(i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

= 35 + 57 - 12

= 92 - 12

= 80

(ii) When two classes meet at the same hour, A∩B = ∅ n (A ∪ B) = n(A) + n(B) - n(A ∩ B)

= n(A) + n(B)

= 35 + 57

= 92

Further concept to solve word problems on sets:

5. In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Solution:

Let A be the set of people who speak English.

B be the set of people who speak French.

A - B be the set of people who speak English and not French.

B - A be the set of people who speak French and not English.

A ∩ B be the set of people who speak both French and English.

Given,

n(A) = 72       n(B) = 43       n(A ∪ B) = 100

Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

= 72 + 43 - 100

= 115 - 100

= 15

Therefore, Number of persons who speak both French and English = 15

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)

= 43 - 15

= 28

Therefore, Number of people speaking English only = 57

Number of people speaking French only = 28

Word problems on sets using the different properties (Union & Intersection):

6. In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?

Solution:

Let A = set of persons who got medals in dance.

B = set of persons who got medals in dramatics.

C = set of persons who got medals in music.

Given,

n(A) = 36                              n(B) = 12       n(C) = 18

n(A ∪ B ∪ C) = 45       n(A ∩ B ∩ C) = 4

We know that number of elements belonging to exactly two of the three sets A, B, C

= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C)

= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4       ……..(i)

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)

Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C)

From (i) required number

= n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12

= 36 + 12 + 18 + 4 - 45 - 12

= 70 - 57

= 13

Apply set operations to solve the word problems on sets:

7. Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

Solution:

Let A be the set of students who play chess

B be the set of students who play scrabble

C be the set of students who play carrom

Therefore, We are given n(A ∪ B ∪ C) = 40,

n(A) = 18,         n(B) = 20         n(C) = 27,

n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4

We have

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4

40 = 69 – 19 - n(C ∩ A)

40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40

n(C ∩ A) = 10

Therefore, Number of students who play chess and carrom are 10.

Also, number of students who play chess, carrom and not scrabble.

= n(C ∩ A) - n(A ∩ B ∩ C)

= 10 – 4

= 6

Therefore, we learned how to solve different types of word problems on sets without using Venn diagram.

Set Theory

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles

1. ### Fraction as a Part of Collection | Pictures of Fraction | Fractional

Feb 24, 24 04:33 PM

How to find fraction as a part of collection? Let there be 14 rectangles forming a box or rectangle. Thus, it can be said that there is a collection of 14 rectangles, 2 rectangles in each row. If it i…

2. ### Fraction of a Whole Numbers | Fractional Number |Examples with Picture

Feb 24, 24 04:11 PM

Fraction of a whole numbers are explained here with 4 following examples. There are three shapes: (a) circle-shape (b) rectangle-shape and (c) square-shape. Each one is divided into 4 equal parts. One…

3. ### Identification of the Parts of a Fraction | Fractional Numbers | Parts

Feb 24, 24 04:10 PM

We will discuss here about the identification of the parts of a fraction. We know fraction means part of something. Fraction tells us, into how many parts a whole has been

4. ### Numerator and Denominator of a Fraction | Numerator of the Fraction

Feb 24, 24 04:09 PM

What are the numerator and denominator of a fraction? We have already learnt that a fraction is written with two numbers arranged one over the other and separated by a line.