# 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.**