Ex 1.6 sets ncert maths solution class 11

           EXERCISE 1.6

Ex 1.6 sets ncert maths solution class 11

1. If X and Y are two sets, such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).

Solution:  Given,

n (X) = 17,  n (Y) = 23,

n (X U Y) = 38

We know that

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values,

38 = 17 + 23 – n (X ∩ Y)

⇒ n (X ∩ Y) = 40 – 38 = 2

Now,  n (X ∩ Y) = 2

2. If X and Y are two sets, such that X ∪Y has 18 elements, X has 8 elements, and Y has 15 elements, how many elements does X ∩ Y have?

Solution:  Given,

n (X U Y) = 18, n (X) = 8,

n (Y) = 15

We know that

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values,

18 = 8 + 15 – n (X ∩ Y)

⇒ n (X ∩ Y) = 23 – 18 = 5

Now, n (X ∩ Y) = 5

3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Solution:  Let  H =  the set of people who speak Hindi and

E = the set of people who speak English.

Given,

n(H ∪ E) = 400, n(H) = 250

n(E) = 200

We know that

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

By substituting the values,

400 = 250 + 200 – n(H ∩ E)

⇒ 400 = 450 – n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

Now, n(H ∩ E) = 50

Therefore, 50 people can speak both Hindi and English.

4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

Solution: We know that

n(S) = 21, n(T) = 32

n(S ∩ T) = 11

We know that

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

Substituting the values,

n (S ∪ T) = 21 + 32 – 11

⇒ n (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

5. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements, and X ∩Y has 10 elements, how many elements does Y have?

Solution: Given

n(X) = 40, n(X ∪ Y) = 60

n(X ∩ Y) = 10

We know that

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

By substituting the values,

60 = 40 + n(Y) – 10

⇒ n(Y) = 60 – (40 – 10) = 30

Therefore, set Y has 30 elements.

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

Solution: Let C  =  the set of people who like coffee and

T  =  the set of people who like tea.

n(C ∪ T) = 70, n(C) = 37

n(T) = 52

We know that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values,

70 = 37 + 52 – n(C ∩ T)

⇒ 70 = 89 – n(C ∩ T)

⇒ n(C ∩ T) = 89 – 70 = 19

Therefore, 19 people like both coffee and tea.

7. In a group of 65 people, 40 like cricket, and 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Solution: Let C =  the set of people who like cricket and

T  =  the set of people who like tennis.

n(C ∪ T) = 65, n(C) = 40

n(C ∩ T) = 10

We know that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values,

65 = 40 + n(T) – 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that

n(only tennis not cricket) = n(T) – n(T∩C)

= 35 – 10 = 25

Therefore, 25 people like only tennis.

8. In a committee, 50 people speak French, 20 speak Spanish, and 10 speak both Spanish and French. How many speak at least one of these two languages?

Solution:Let  F =  the set of people in the committee who speak French and

S =  the set of people in the committee who speak Spanish.

n(F) = 50, n(S) = 20

n(S ∩ F) = 10

We know that

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

By substituting the values,

n(S ∪ F) = 20 + 50 – 10

⇒ n(S ∪ F) = 70 – 10

⇒ n(S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages.


Ex 1.5 sets ncert maths solution class 11

Ex 1.4 sets ncert maths solution class 11

Team Gmath

Leave a Reply

Your email address will not be published. Required fields are marked *