# NCERT Solutions for Class 11 Maths Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 1 Sets समुच्चय in PDF form for CBSE, UP Board, MP Board, Gujrat Board and other board’s students following Offline Apps as well as NCERT Solutions on Latest CBSE Syllabus 2019-20 for their final exams.

 Class: 11 Subject: Maths Chapter 1: Sets

## NCERT Solutions for Class 11 Maths Chapter 1

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### 11 Maths Chapter 1 Sets Solutions

#### Important Extra 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]

###### Try These
• 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.
• 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.

#### List all the elements of the following set: A = {x: x is an odd natural number}.

A = {x: x is an odd natural number}

= {1, 3, 5, 7, 9 …}

#### Write down all the subsets of the following set: {1, 2, 3}.

The subsets of {1, 2, 3} are

Φ,
{1},
{2},
{3},
{1, 2},
{2, 3},
{1, 3}
{1, 2, 3}

#### Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?

Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
A ∪ B = {a, b, c} = B

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

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.

#### 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?

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.

#### 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?

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.

#### 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?

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.

#### 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) = 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.

#### 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.

#### In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Let U be the set of all students who took part in the survey.
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.

#### In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Let U be the set of all students in the group.
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.

## 2 thoughts on “NCERT Solutions for Class 11 Maths Chapter 1 Sets”

1. pratik says:

the whole solution should be in one PDF rather than different and thanks for publishing such a sources as i cant afford coaching so i study on u r web for prep of jee mains

2. sobh nath sahu says: