# 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.

https://gmath.in/ex-1-5-sets-ncert-maths-solution-class-11/

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