Keralanotes brings to you the CBSE solved questions of **Ex 1.6 Chapter1 Sets**. This **NCERT solution** is specially prepared for the students of class 11 Maths. In this set, we will give you 1.6 and all its questions with their solutions. In these solutions, the working of every step is clearly illustrated to help you better understand it and to improve your understanding of the topics.

This solution is for exercise 1.6 Chapter1 of NCERT Solution for Class 11 Maths textbook solution for the CBSE students. You can free download it on our websites.

Board |
SCERT, Kerala |

Text Book |
NCERT Based |

Class |
Plus One |

Subject |
Math's Textbook Solution |

Chapter |
Chapter 1 |

Exercise |
Ex 1.6 |

Chapter Name |
Sets |

Category |
Plus One Kerala |

## Kerala Syllabus Plus One Math's Textbook Solution Chapter 1 Sets Exercises 1.6

### Chapter 1 Sets Textbook Solution

- Exercises 1.1
- Exercises 1.2
- Exercises 1.3
- Exercises 1.4
- Exercises 1.5
- Exercises 1.6
- Miscellaneous Exercise Chapter 1

### Chapter 1 Sets Exercise 1.6

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?

n(S)=21

n(T)=32

n(S ∩ T)=11

n(S ∪ T)=n(T)+n(S)-n(S ∩ T)

=21+32-11=42

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?

n(X)= 40

n(XUY)= 60

n(X∩Y)= 10

n(XUY)= 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.

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?

C-coffee T-tea

n(C)=37

n(T)=52

n(CUT)=70

n(CUT)=n(C)+n(T)-n(C∩T)

70=37+52-n(C∩T)

n(C∩T)=37+52-70=19

since each person likes at least one of the two drinks,n(U)=n(CUT)

number of peoples like both tea and coffee=19

If X and Y are two sets such that *n*(X) = 17, *n*(Y) = 23 and *n*(X ∪ Y) = 38, find *n*(X ∩Y).

It is given that:

*n*(X) = 17, *n*(Y) = 23, *n*(X ∪ Y) = 38

*n*(X ∩ Y) = ?

We know that:

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?

It is given that:

*n*(X ∩ Y) = ?

We know that:

In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

number of people speak hindi, n(H)=250

number of people speak english, n(E)=200

n (HUE) = 400

n(HUE)= n(H) +n(E) -n(H∩E)

400 = 250+200 - n(H∩E)

=450-n(H∩E)

n(H∩E)=450-400= 50

Thus, 50 people can speak both Hindi and English.

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?

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) = Î¦

∴ *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.

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?

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.

### PDF Download

#### Chapter 1: Sets EX 1.6 Solution

#### Chapter 1: Sets EX 1.6 Solution- Preview

#### Plus One Math's Chapter Wise Textbook Solution PDF Download

- Chapter 1: Sets
- Chapter 2: Relations and Functions
- Chapter 3: Trigonometric Functions
- Chapter 4: Principle of Mathematical Induction
- Chapter 5: Complex Numbers and Quadratic Equations
- Chapter 6: Linear Inequalities
- Chapter 7: Permutation and Combinations
- Chapter 8: Binomial Theorem
- Chapter 9: Sequences and Series
- Chapter 10: Straight Lines
- Chapter 11: Conic Sections
- Chapter 12: Introduction to Three Dimensional Geometry
- Chapter 13: Limits and Derivatives
- Chapter 14: Mathematical Reasoning
- Chapter 15: Statistics
- Chapter 16: Probability

##### Plus One Math's Part I

##### Plus One Math's Part II

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