# NCERT Solutions for Class 11 Maths Chapter 1

NCERT Solutions for Class 11 Maths Chapter 1 Sets समुच्चय in PDF form free download for CBSE, UP Board, MP Board, Gujrat Board and other board’s students following latest CBSE Syllabus 2020-2021.

NCERT Solutions 2020-21 and Offline Apps are based on latest NCERT Books and Latest CBSE Syllabus 2020-21 for their final exams.## NCERT Solutions for Class 11 Maths Chapter 1

Class: | 11 |

Subject: | Mathematics |

Chapter 1: | Sets |

### 11th Maths Chapter 1 Solutions

NCERT Solutions for Class 11 Maths Chapter 1 Sets in English Medium is given below to free download. These solutions and Offline apps 2020-21 are based on latest NCERT Books 2020-21 for all students who are following CBSE Syllabus 2020-21.

### Class 11 Maths Chapter 1 Sets Solutions

- Class 11 Maths Exercise 1.1 Solutions
- Class 11 Maths Exercise 1.2 Solutions
- Class 11 Maths Exercise 1.3 Solutions
- Class 11 Maths Exercise 1.4 Solutions
- Class 11 Maths Exercise 1.5 Solutions
- Class 11 Maths Exercise 1.6 Solutions
- Class 11 Maths Miscellaneous Exercise 1 Solutions
- Class 11 Maths Exercise 1.1 Download
- Class 11 Maths Exercise 1.2 Download
- Class 11 Maths Exercise 1.3 Download
- Class 11 Maths Exercise 1.4 Download
- Class 11 Maths Exercise 1.5 Download
- Class 11 Maths Exercise 1.6 Download
- Class 11 Maths Miscellaneous Exercise 1 Download
- Class 11 Maths Solutions Main Page

#### Important Questions on Sets

1. In a survey of 450 people, it was found that 110 play cricket, 160 play tennis and 70 play both cricket as well as tennis. How many play neither cricket nor tennis? [Answer: 23]

2. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families by newspaper C. 5% families buy A and B, 3%, buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, find the no of families which buy(1) A only (2) B only (3) none of A, B and C (4) exactly two newspapers (5) exactly one newspaper (6) A and C but not B (7) at least one of A,B, C. [Answer: {2, 3, 5}]

3. Two finite sets have m and n elements. The total number of subsets of first set is 56 more than the total number of subsets of the second set. Find the value of m and n. [Answer: m = 6 and n = 3]

##### Class 11 Maths Chapter 1 Important Questions for Practice

1. In a group of 84 persons, each plays at least one game out of three viz. tennis, badminton and cricket. 28 of them play cricket, 40 play tennis and 48 play badminton. If 6 play both cricket and badminton and 4 play tennis and badminton and no one plays all the three games, find the number of persons who play cricket but not tennis.

2. In a class, 18 students took Physics, 23 students took Chemistry and 24 students took Mathematics of these 13 took both Chemistry and Mathematics, 12 took both Physics and Chemistry and 11 took both Physics and Mathematics. If 6 students offered all the three subjects, find: (1) The total number of students. (2) How many took Maths but not Chemistry. (3) How many took exactly one of the three subjects.

### Important Questions on 11th Maths Chapter 1

= {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}

= {1, 3, 5, 7, 9 …}

Φ,

{1},

{2},

{3},

{1, 2},

{2, 3},

{1, 3}

{1, 2, 3}

Yes, A ⊂ B.

A ∪ B = {a, b, c} = B

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.

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.

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.

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.

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.

∴ 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.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e.,

we have to find n(T’ ∩ C’).

n(T’ ∩ C’) = n(T ∪ C)’

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

n( H U E ) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

Hence, there are 125 students in the group.