**NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6**

NCERT Solutions for Class
11 Maths Chapter 1 Sets Ex 1.6 are the part of NCERT Solutions for Class 11
Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 1 Sets
Ex 1.6.

**Ex 1.6 Class 11 Maths 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).

**Solution.**

Given n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38

We know that

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

⇒ 38 = 17 + 23 – n(X ∩ Y)

∴ n(X ∩ Y) = 40 – 38 = 2

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Given n(X ∪ Y) = 18, n(X) = 8 and n(Y) = 15

We know that

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

⇒ 18 = 8 + 15 – n(X ∩ Y)

∴ n(X ∩ Y) = 23 – 18 = 5

Hence, X ∩ Y has 5 elements.

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Let H be the set of people speaking Hindi and E be the set of people speaking
English.

∴ n(H) = 250, n(E) = 200 and n(H ∪ E) = 400

We know that

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

400 = 250 + 200 – n(H ∩ E)

∴ n(H ∩ E) = 450 – 400 = 50

Hence, 50 people can speak
both Hindi and English.

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Given n(S) = 21, n(T) = 32 and n(S ∩ T) = 11

We know that

n(S ∪ T) = n(S) + n(T) – n(S ∩ T)

= 21 + 32 – 11 = 42

Hence, S ∪ T has 42 elements.

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Given n{X) = 40, n(X ∪ Y) = 60 and 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 – 30 = 30

Hence, Y has 30 elements.

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Let C be the set of persons who like coffee and T be the set of persons who
like tea.

∴ n(C) = 37, n(T) = 52 and n(C ∪ T) = 70

We know that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

⇒ 70 = 37 + 52 – n(C ∩ T)

∴ n(C ∩ T) = 89 – 70 = 19

Hence, 19 people like both
coffee and tea.

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Let C be the set of people who like cricket and T be the set of people who like
tennis. Given, n(C) = 40, n(C ∩ T) = 10 and n(C ∪ T) = 65

We know that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

⇒ 65 = 40 + n(T) – 10

⇒ n(T) = 65 – 30 = 35

∴ Number of people who like tennis = 35

Now, the number of people who
like tennis only and not cricket

= n(T – C) = n(T) – n(C ∩ T) = 35 – 10 = 25

**Ex 1.6 Class 11
Maths 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?

**Solution.**

Let F be the set of the people who speak French and S be the set of the people
who speak Spanish.

Given, n(F) = 50, n(S) = 20 and n(F ∩ S) = 10

We know that

n(F ∪ S) = n(F) + n(S) – n(F ∩ S)

n(F ∪ S) = 50 + 20 – 10 = 60

∴ Number of people who speak at least one
of these two languages is 60.

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