# Practical Problems based on sets

See the word problems solved here to get the basic ideas on how you should be using properties of set’s union and intersection.

## Word Problems:

Example 1. Let finite sets, Set A and Set B like n (A) = 25, n (B) = 29; n (A ∪ B) = 38, so, find –

n (A ∩ B).

Solution:

Using formula of union

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

then, n (A ∩ B) = n (A) + n (B) – n (A ∪ B)

= 25 + 29 – 38

= 54 – 38

= 16

Example 2. If n (A – B) = 20, n(A ∪ B) = 60 and n(A ∩ B) = 30, then find n(B).

Solution:

Using formula,

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

60 = 20 + 30 + n(B – A)

60 = 50 + n(B – A)

n(B – A) = 60 – 50

n(B – A) = 10

Now, n(B) = n(A ∩ B) + n(B – A)

= 30 + 10

= 40

## Different categories of word problems regarding sets:

Question 3. A group of 70 people is there, 50 people like hot soups, and 28 people like cold soups, and each individual like at least one out of 2 soups. How many people like tea and coffee?

Solution:

Let take,

A = Set of individuals liking cold soups.

B = Set of individuals liking hot soups.

Given,

1. (A ∪ B) = 70n
2. (A) = 50
3. n(B) = 28 then;

As per formula , n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 50 + 28 – 70

= 78 – 70 = 8

= 8

Thus, 8 people like coffee and tea

Another Example:

In a school where 600 students are there, 200 like history subject, 160 like English and 50 like both History and English. Find the number of students who like

• Students who like History; not English
• Students who like English; not History
• Students who either like History or English

Solution

Let the total number of students be U that is, the universal set.

Assuming A = set of a number of students who favor History.

Assuming B = set of a number of students who favor English.

Here, n (U) = 600 n (A) = 200 n (B) = 160 n (A∩B) = 50 Here, you have to determine the number of students who favor History but not English, so to find A-B:

As shown in the Venn diagram,

A = (A-B) ∪ (A∩B)

n (A) = n (A-B) + n (A∩B)

n (A-B) = n (A) – n (A∩B)

= 200 – 50 = 150

Therefore, the numbers of students who favor only History but not English = 150

Here, you have to determine the number of students who favor English but not like History, so to find B-A:

B = (B-A) ∪ (A∩B)

n (B) = n (B-A) +n (A∩B)

n (B-A) = n (B)- n (A∩B)

= 160 – 50 = 110

Therefore, the numbers of students who favor only English but not History are= 110.

Here we have to find the number of students who like either English or History.

n (A∪B) = n (A) + n (B) – n (A∩B)

= 200 + 160 – 50

= 310

## Next Word Problem:

Question: A competition was held in the school where students were given awards in different groups. 46 medals were given in dance, 22 medals in sports, and a total of 15 medals in English debate. If these medals were given to 44 total students and only 5 students got medals in each of the 3 groups, then how many medals are acknowledged in exactly 2 of these groups?

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:

Assuming A = set of students who got medals in dance.

B = set of students who got medals in sports.

C = set of persons who got medals in English Debate.

Given,

n(A) = 46 n(B) = 22 n(C) = 15

n(A ∪ B ∪ C) = 44 n(A ∩ B ∩ C) = 5

As we know the number of elements which belongs to exactly 2 out of 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 ∩ C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C)

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

From (i) the required number

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

= 46 + 22 + 15 + 4 – 44 – 12

= 85 – 12

= 73