Word Problems on Sets | Solved Problems on Sets

# Word Problems on Sets | Solved Problems on Sets

1. Let A and B be two finite sets such that n(A) = 30, n(B) = 18 and n(A ∪ B) = 26, find n(A ∩ B)?

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Solution:

Formula for n(A ∪ B) = n(A) + n(B) – n(A ∩ B).

Rearranging it we get the n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

=30+18 – 26

= 22

Therefore, n(A ∩ B) = 22.

2. If If n(A – B) = 12, n(A ∪ B) = 45 and n(A ∩ B) = 15, then find n(B)?

Solution:

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

45 = 12+15 +n(B-A)

n(B-A) = 45-12-15

= 45-27

= 18

n(B) = n(A ∩ B) + n(B – A)

= 15+18

= 33

3. In a group of 80 people, 37 like cold drinks and 52 like hot drinks and each person likes at least one of the two drinks. Find How many people 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) = 80 n(A) = 37 n(B) = 22 then;

n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 37+52-80

= 89 – 80

= 9

Therefore, 9 people like both tea and coffee.

4. Let S={4, 5, 6}. Write all the possible partitions of S?

Solution:

Remember that partition of S is a collection of nonempty sets that are disjoint and their union is S.

Possible Partitions of S are

{4},{5},{6}
{4,5},{6}
{4,6},{5}
{5,6},{4}
{4,5,6}.

5. In a school, there are 30 teachers who teach Mathematics or Physics. Of these, 18 teach Mathematics and 6 teach both Physics and Mathematics. How many teach Physics only?

Solution:

Total Number of Teachers who teach Mathematics or Physics = 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

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 teach Physics only = n( P – M)

n( P – M) = n(P) – n( M ∩ P )

= 18 – 6

n( P – M) = 12

Number of teachers who teach physics only is 12.

6. In a survey of 80 people, it was found that 35 people read newspaper H, 20 read newspaper T, 15 read the newspaper I, 5 read both H and I, 10 read both H and T, 7 read both T and I, 4 read all three newspapers. Find the number of people who read at least one of the newspapers?

Solution:

n(H) is the Number of People who read the newspaper H i.e. 35

n(I) is the Number of People who read the newspaper I i.e. 15

n(T) is the Number of People who read the newspaper T i.e. 20

n(H ∩ I) is the Number of People who read H and I i.e. 5

n(H ∩ T) is the Number of People who read H and T i.e 10

n(T ∩ I) is the Number of People who read T and I i.e. 7

n(H ∩ T ∩ I) is the Number of People who read all three Newspapers H, T, I i.e. 4

n(H ∪T ∪ I ) is the Number of people who read at least one of the newspapers

n(H ∪T ∪ I ) = n(H) + n(T) + n(I) – n(H ∩ T) – n(T ∩ I) – n(H ∩ I) + n(H ∩ T ∩ I)

n(H ∪T ∪ I )= 35+20+15-10-7-5+4

= 74-22

= 52

Therefore, the number of people who read at least one of the newspapers is 52.

7. In a school, all pupils play either Hockey or Football or both. 400 play Football, 150 play Hockey, and 130 play both the games. Find
(i) The number of pupils who play Football only,
(ii) The number of pupils who play Hockey only,
(iii) The total number of pupils in the school.

H = Hockey and F = Football

n (H ) = 150 n (F)= 400

n ( H ∩ F) = 130

(i) The number of pupils who only play Football = n (F – H )

n (F – H ) = n(F) – n( F ∩ H )

= 400 – 130

= 270

(ii) The number of pupils who only play Hockey = n (H – F )

n (H– F ) = n(H) – n( F ∩ H )

= 150 – 130

= 20

(iii) The total number of pupils in school

= n(H) + n(F) – n (F ∩ H)

= 150 + 400 – 130

= 420 Join Infinity Learn Regular Class Program!

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