## NCERT Solutions for Class 11 Maths Chapter 1 – Sets Ex 1.6

#### Page No 24:

#### Question 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).#### Answer:

It is given that:

*n*(X) = 17,

*n*(Y) = 23,

*n*(X ∪ Y) = 38

*n*(X ∩ Y) = ?

We know that:

#### Question 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?

#### Answer:

It is given that:

*n*(X ∩ Y) = ?

We know that:

#### Question 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?

#### Answer:

Let H be the set of people who speak Hindi, and

E be the set of people who speak English

∴

*n*(H ∪ E) = 400,*n*(H) = 250,*n*(E) = 200*n*(H ∩ E) = ?

We know that:

*n*(H ∪ E) =

*n*(H) +

*n*(E) –

*n*(H ∩ E)

∴ 400 = 250 + 200 –

*n*(H ∩ E)
⇒ 400 = 450 –

*n*(H ∩ E)
⇒

*n*(H ∩ E) = 450 – 400
∴

*n*(H ∩ E) = 50
Thus, 50 people can speak both Hindi and English.

#### Question 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?

#### Answer:

It is given that:

*n*(S) = 21,

*n*(T) = 32,

*n*(S ∩ T) = 11

We know that:

*n*(S ∪ T) =

*n*(S) +

*n*(T) –

*n*(S ∩ T)

∴

*n*(S ∪ T) = 21 + 32 – 11 = 42
Thus, the set (S ∪ T) has 42 elements.

#### Question 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?

#### Answer:

It is given that:

*n*(X) = 40,

*n*(X ∪ Y) = 60,

*n*(X ∩ Y) = 10

We know that:

*n*(X ∪ Y) =

*n*(X) +

*n*(Y) –

*n*(X ∩ Y)

∴ 60 = 40 +

*n*(Y) – 10
∴

*n*(Y) = 60 – (40 – 10) = 30
Thus, the set Y has 30 elements.

#### Question 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?

#### Answer:

Let C denote the set of people who like coffee, and

T denote 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)

∴ 70 = 37 + 52 –

*n*(C ∩ T)
⇒ 70 = 89 –

*n*(C ∩ T)
⇒

*n*(C ∩ T) = 89 – 70 = 19
Thus, 19 people like both coffee and tea.

#### Question 7:

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

#### Answer:

Let C denote the set of people who like cricket, and

T denote 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)

∴ 65 = 40 +

*n*(T) – 10
⇒ 65 = 30 +

*n*(T)
⇒

*n*(T) = 65 – 30 = 35
Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T

Also,

(T – C) ∩ (T ∩ C) = Φ

∴

*n*(T) =*n*(T – C) +*n*(T ∩ C)
⇒ 35 =

*n*(T – C) + 10
⇒

*n*(T – C) = 35 – 10 = 25
Thus, 25 people like only tennis.

#### Question 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?

#### Answer:

Let F be the set of people in the committee who speak French, and

S be 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)

= 20 + 50 – 10

= 70 – 10 = 60

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

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