NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6 is part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6.

- Sets Class 11 Ex 1.1
- Sets Class 11 Ex 1.2
- Sets Class 11 Ex 1.3
- Sets Class 11 Ex 1.4
- Sets Class 11 Ex 1.5
- Sets Class 11 Miscellaneous Exercise

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Sets |

Exercise |
Ex 1.6 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## 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.**

Here 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.**

Here 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.

**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.

**Ex 1.6 Class 11 Maths Question 4.**

If 5 and Tare two sets such that 5 has 21 elements, T has 32 elements, and S ∩T has 11 elements, how many elements does S ∪ T have?

**Solution.**

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

We know that

n(S ∪ T) = n(S) + n(T) – n(S ∩ T) n(S ∪ T)

= 21 + 32 – 11 = 42.

**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 X have?

**Solution.**

Here 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.

**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.

**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. Here n(Q) = 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 j Now 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 f 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 people who speak French and S be the set of people who speak Spanish.

Here 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 = 60

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