# NCERT Solutions | Class 11 Maths Chapter 1 | Sets

## CBSE Solutions | Maths Class 11

Check the below NCERT Solutions for Class 11 Maths Chapter 1 Sets Pdf free download. NCERT Solutions Class 11 Maths  were prepared based on the latest exam pattern. We have Provided Sets Class 11 Maths NCERT Solutions to help students understand the concept very well.

### NCERT | Class 11 Maths

Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 11th Maths 1 Sets English

#### Sets | Class 11 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Maths Chapter 1 Sets to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1

Ex 1.1 Class 11 Maths Question 1.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter,
(ix) A collection of most dangerous animals of world.
Solution.
(i) The collection of all months of a year beginning with J is {J anuary, June, July}, which is well defined and hence it forms a set.
(ii) The collection of most talented writers of India is not well defined because opinions about ‘most talented writers’ vary from person to person and hence it does not form a set.
(iii) A team of eleven best-cricket batsmen of the world is not well defined because opinions about ‘best-cricket batsmen’ vary from person to person and hence it does not form a set.
(iv) The collection of all boys in your class is well defined and hence it forms a set.
(v) The collection of all natural numbers less than 100 is {1, 2, 3, 4,…………, 99}, which is well
defined and hence it forms a set.
(vi) A collection of novels written by the writer Munshi Prem Chand is well defined and hence it forms a set.
(vii) The collection of all even integers is {…………..,-4, -2, 0, 2, 4,……….. } which is well defined and hence it forms a set.
(viii) The collection of questions in this chapter is well defined and hence it forms a set.
(ix) A collection of most dangerous animals of the world is not well defined because opinions about ‘most dangerous animals’ vary from person to person and hence it does not form a set.

Ex 1.1 Class 11 Maths Question 2.
Let A = {1, 2, 3, 4, 5, 6). Insert the appropriate symbol ∈ or ∉ in the blank spaces:
(i) 5…A
(ii) 8…A
(iii) 0…A
(iv) 4…A
(v) 2…A
(vi) 10…A
Solution.
(i) Since 5 is the element of A. ∴ 5 ∉ A.
(ii) As 8 is not the element of A. ∴ 8 ∉ A
(iii) As 0 is not the element of A ∴ 0 ∈ A.
(iv) 4 is the element of A ∴ 4 ∈ A.
(v) 2 is the element of A ∴ 2 ∈ A.
(vi) 10 is not the element of A ∴ 10 ∉ A.

Ex 1.1 Class 11 Maths Question 3.
Write the following sets in roster form:
(i) A = {x: x is an integer and -3 < x < 7}
(ii) B = {x: x is a natural number less than 6}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word
Solution.
(i) Integers lying between -3 and 7 are -2, -1, 0, 1, 2, ……….. , 6
∴ A = {-2,-1, 6}.

(ii) Natural numbers less than 6 are 1, 2, 3, 4, 5.
∴ B = 11, 2, 3, 4, 5}

(iii) Two digit natural numbers such that the sum of its digits is 8 are 17, 26, 35, 44, 53, 62, 71, 80.
∴ C= (17, 26, 35, 44, 53, 62, 71,80}

(iv) Prime number divisors of 60 are 2, 3, 5.
∴ D = (2, 3, 5}

(v) Word TRIGONOMETRY is formed by using the letters T, R, I, G, O, N, M, E, Y.
∴ E = (T, R, I, G, N, O, M, E, Y}

(vi) Word BETTER is formed by using the letters B, E, T, R
∴ F = (B, E, T, R}

Ex 1.1 Class 11 Maths Question 4.
Write the following sets in the set-builder form:
(i) {3, 6, 9, 12}
(ii) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625}
(iv) {2,4,6,…}
(v) {1,4,9,……..,100}
Solution.
(i) Let A = (3, 6, 9, 12}
All elements of the set are natural numbers that are multiples of 3.
∴ A = (x : x = 3n, n∈N and 1 ≤ n ≤4}

(ii) Let B = (2, 4, 8, 16, 32} = (21, 22, 23, 24, 25}
∴ B = {x : x = 2n, n ∈ N and 1 ≤ n ≤ 5}

(iii) Let C = (5, 25, 125, 625} = (51, 52, 53, 54}
∴ C = {x : x = 5n, n ∈ N and 1 ≤ n ≤ 4}

(iv) Let D = (2, 4, 6,……………..}
All elements of the set are even natural numbers.
∴ D = (x: x is an even natural number)

(v) Let E = {1,4,9,……….,100}
All elements of the set are perfect squares.
∴ E = {x: x = n2, n ∈ N and 1 ≤ n ≤ 10}

Ex 1.1 Class 11 Maths Question 5.
List all the elements of the following sets:
(i) A = {x: x is an odd natural number}
(ii) B = {x: x is an integer, $$-\frac { 1 }{ 2 }$$ < x < $$\frac { 9 }{ 2 }$$}
(iii) C = {x: x is an integer, x2 ≤ 4}
(iv) D = {x: x is a letter in the word “LOYAL”}
(v) E = {x: x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k}.
Solution.
(i) A = {x: x is an odd natural number}
∴ A = {1, 3, 5, 7,……………}

(ii) B = {x: x is an integer, $$-\frac { 1 }{ 2 }$$ < x < $$\frac { 9 }{ 2 }$$}
∴ B = { 0, 1, 2, 3, 4}

(iii) C = {x: x is an integer, x2 ≤ 4}
x2 ≤ 4⇒ -2 ≤ x ≤ 2
∴ C = {-2, -1, 0, 1, 2}

(iv) D = {x: x is a letter in the word “LOYAL”}
∴ D = {L, O, Y, A}

(v) E = {x: x is a month of a year not having 31 days}
∴ E = {February, April, June, September, November}

(vi) F = {x: x is a consonant in the English alphabet which precedes k}
∴ F = {b, c, d, f, g, h, j}

Ex 1.1 Class 11 Maths Question 6.
Match each of the set on the left in the roster form with the same set on the right described in set-builder form:
i. {1,2, 3,6} (a) {x: x is a prime number and a divisor of 6}
ii. {2,3} (b) {x : x is an odd natural number less than 10}
iii. {M, A, T, H, E,I,C, S} (c) {x: x is natural number and divisor of 6} iv. {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

Solution.
(i) → (c),
(ii) → (a),
(iii) → (d),
(iv) → (b).
The sets which are in set-builder form can be written in roster form as follows:
(a)
{x : x is a prime number and a divisor of 6} = {2, 3}
(b) {x: x is an odd natural number less than 10} = {1, 3, 5, 7, 9}
(c) {x : x is natural number and divisor of 6} = {1, 2, 3, 6}
(d) {x: x is a letter of the word MATHEMATICS} = {M, A, T, H, E, I, C, S}

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

Ex 1.2 Class 11 Maths Question 1.
Which of the following are examples of the null set?
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x: x is a natural number, x ≤ 5 and x > 7}
(iv) {y: y is a point common to any two parallel lines}
Solution.
(i) Set of odd natural numbers divisible by 2 is a null set because odd natural numbers are not divisible by 2.
(ii) Set of even prime numbers is {2} which is not a null set.
(iii) {x: x is a natural number, x < 5 and x >7} is a null set because there is no natural number which satisfies x < 5 and x > 7 simultaneously,
(iv) [y: y is a point common to any two parallel lines) is a null set because two parallel lines
do not have any common point.

Ex 1.2 Class 11 Maths Question 2.
Which of the following sets are finite or infinite?
(i) The set of months of a year
(ii) {1,2,3,…}
(iii) {1,2,3, …,99,100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Solution.
(i) The set of months of a year is finite set because there are 12 months in a year.
(ii) {1, 2, 3, …} is an infinite set because there are infinite elements in the set.
(iii) {1, 2, 3, …, 99, 100) is a finite set because the set contains finite number of elements.
(iv) The set of positive integers greater than 100 is an infinite set because there are infinite
number of positive integers greater than 100.
(v) The set of prime numbers less than 99 is a finite set because the set contains finite number of elements.

Ex 1.2 Class 11 Maths Question 3.
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
Solution.
(i) The set of lines which are parallel to the x-axis is an infinite set because we can draw infinite number of lines parallel to x-axis.
(ii) The set of letters in the English alphabet is a finite set because there are 26 letters in the English alphabet.
(iii) The set of numbers which are multiple of 5 is an infinite set because there are infinite multiples of 5.
(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is very large but finite.
(v) The set of circles passing through the origin (0, 0) is an infinite set because we can draw infinite number of circles passing through origin of different radii.

Ex 1.2 Class 11 Maths Question 4.
In the following, state whether A = B or not:
(i) A = {a, b, c, d};B = {d, c, b, a}
(ii) A = {4, 8, 12, 16};B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}
B = {x : x is positive even integer and x≤ 10}
(iv) A = {x: x isa multiple of 10}
B = {10, 15, 20, 25, 30,…}
Solution.
(i) A = {a, b, c, d} and B = {d, c, b, a} are equal sets because order of elements does not changes a set.
∴ A = B = [a, b, c, d}.

(ii) A = {4, 8, 12, 16} and B = {8, 4, 16, 18} are not equal sets because 12 ∈ A but 12 ∉ B and 18 ∉ B but 18 ∉ A.

(iii) A = {2, 4, 6, 8,10} and B = {x: x is a positive even integer and x ≤ 10) which can be written in roster form as B = (2, 4, 6, 8, 10) are equal sets.
∴ A = B = {2, 4, 6, 8,10).

(iv) A = {x: x is a multiple of 10) can be written in roster form as A = {10, 20, 30, 40,…….. } and
B – {10, 15, 20, 25, 30, ………..} are not equal sets because 15 ∈ B but 15 ∉ A.

Ex 1.2 Class 11 Maths Question 5.
Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B={x: x is solution of x2 + 5x + 6 = 0}
(ii) A = {x: x is a letter in the word FOLLOW}
B = {y: y is a letter in the word WOLF}
Solution.
(i) A = (2, 3} and B = {x: x is a solution of x2 + 5x + 6 = 0}
Now, x2 + 5x + 6 = 0 ⇒ x2 + 3x + 2x + 6 = 0 ⇒ (x + 3)(x + 2) = 0 ⇒ x = -3, -2
∴ B = {-2, -3}
Hence, A and B are not equal sets.

(ii) A = {x : x is a letter in the word FOLLOW} = {F, O, L, W}
B = {y: y is a letter in the word WOLF}
= {W, O, L, F}
Hence, A = B = {F, O, L, W}.

Ex 1.2 Class 11 Maths Question 6
From the sets given below, select equal sets:
A = {2, 4, 8, 12),
B = {1, 2, 3, 4},
C = {4, 8, 12, 14},
D ={3,1,4,2},
E ={-1, 1},
F ={0, a},
G ={1, -1},
H ={0, 1}
Solution.
From the given sets, we see that sets B and D have same elements and also sets E and G have same elements.
∴ B = D = {1 ,2, 3, 4} and E = G = {-1, 1}.

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3

Ex 1.3 Class 11 Maths Question 1.
Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4} …{1, 2, 3, 4, 5}
(ii) {a, b, c}… {b, c, d}
(iii) {x: x is a student of Class XI of your school} … {x: x student of your school}
(iv) {x : x is a circle in the plane}… {x: x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane}… {x : x is a rectangle in the plane}
(vi) {x: x is an equilateral triangle in a plane} … {x: x is a triangle in the same plane}
(vii) {x: x is an even natural number}… {x: x is an integer}
Solution.
(i) {2, 3, 4} ⊂ {11, 2, 3, 4, 5}
(ii) [a, b, c) ⊄ {{b, c, d}
(iii) {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}
(vii) {x: x is an even natural number} ⊂ {x: x is an integer}

Ex 1.3 Class 11 Maths Question 2.
Examine whether the following statements are true or false:
(i) {a, b} ⊄{b, c, a}
(ii) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ∈ la, b, c}
(vi) {x: x is an even natural number less than 6} ⊂ {x: x is a natural number which divides 36}
Solution.

Ex 1.3 Class 11 Maths Question 3.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

Solution.

Ex 1.3 Class 11 Maths Question 4.
Write down all the subsets of the following sets
(i) {a}
(ii) {a,b}
(iii) {1,2,3}
(iv) φ
Solution.
(i) Number of elements in given set = 1
Number of subsets of given set = 21 = 2
∴ Subsets of given set are φ , {a}.

(ii) Number of elements in given set = 2
Number of subsets of given set = 212 = 4
∴ Subsets of given set are φ, {a}, {b}, {a, b}.

(iii) Number of elements in given set = 3
Number of subsets of given set = 23 = 8
Subsets of given set are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}.

(iv) Number of elements in given set = 0
Number of subsets of given set = 20= 1
∴ Subset of given set is φ.

Ex 1.3 Class 11 Maths Question 5.
How many elements has P(A), if A = φ?
Solution.
Number of elements in set A = 0
Number of subset of set A = 20 = 1
Hence, number of elements of P(A) is 1.

Ex 1.3 Class 11 Maths Question 6.
Write the following as intervals:
(i) {x: x ∈ R, -4 < x ≤ 6}
(ii) {x: x ∈ R, -12 < x < -10}
(iii) {x: x ∈ R, 0 ≤ x < 7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}
Solution.
(i)Let A = {x: x ∈ R, -4 < x ≤ 6}
It can be written in the form of interval as (-4, 6)
(ii) Let A= {x: x ∈ R, -12 < x < -10}
It can be written in the form of interval as (-12, -10)
(iii) Let A = {x: x ∈ R, 0 ≤ x < 7}
It can be written in the form of interval as (0, 7).
(iv) Let A = {x: x ∈ R, 3 ≤ x ≤ 4}
It can be written in the form of interval as (3,4).

Ex 1.3 Class 11 Maths Question 7.
Write the following intervals in set-builder form:
(i) (-3,0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [-23, 5)
Solution.
(i) The interval (-3, 0) can be written in set-builder form as {x : x ∈ R,-3 < x < 0}.
(ii) The interval [6, 12] can be written in set-builder form as {x : x ∈ R, 6 ≤ x ≤ 12}.
(iii) The interval (6, 12] can be written in set-builder form as {x : x ∈ R, 6 < x ≤ 12}
(iv) The interval [-23,5) can be written in set-builder form as {x : x ∈ R, -23 ≤ x < 5}

Ex 1.3 Class 11 Maths Question 8.
What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Solution.
(i) Right triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = {x : x is a triangle in a plane}

(ii) Isosceles triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = }x : x is a triangle in a plane}

Ex 1.3 Class 11 Maths Question 9.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set(s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Solution.
(iii)

(i) {0, 1, 2, 3, 4, 5, 6} is not a universal set for A, B, C because 8 ∈ C but 8 is not a member of {0, 1, 2, 3, 4, 5, 6}.
(ii) φ is a set which contains no element. So it is not a universal set for A, B, C.
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set for A, B, C because all members of A, B, C are present in {0,1 , 2, 3, 4, 5, 6, 7, 8, 9, 10).
(iv) (1, 2, 3, 4, 5, 6, 7, 8) is not a universal set for A, B, C because 0 ∈ C but 0 is not a member of {1, 2, 3, 4, 5, 6, 7, 8)

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4

Ex 1.4 Class 11 Maths Question 1.
Find the union of each of the following pairs of sets:
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.

Ex 1.4 Class 11 Maths Question 2.
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B ? What is A ∪B?
Solution.
Here A = {a, b} and B = {a, b, c}. All elements of set A are present in set B.
∴ A ⊂ B. Now, A ∪ B = {a, b, c) = B.

Ex 1.4 Class 11 Maths Question 3.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution.
Here A and B are two sets such that A ⊂ B.
Take A = {1, 2} and B = {1, 2, 3}.
A ∪ B = {1, 2, 3) = B.

Ex 1.4 Class 11 Maths Question 4.
If A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}; find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ O
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Solution.
Here A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}

Ex 1.4 Class 11 Maths Question 5.
Find the intersection of each pair of sets of .
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.
(i) Here X = {1, 3, 5} and Y = {1, 2, 3}
∴ X ∩ Y= {1,3}

(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
∴ A ∩ B = {a}

(iii) Here A = {x: x is a natural number and multiple of 3} = {3, 6, 9,12,….} and B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5} ∴ A ∩ B = {3}

(iv) Here A = {x: x is a natural number and 1 < x < 6} ={2, 3, 4, 5, 6} and B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9} ∴ A ∩ B = φ

(v) Here A = {1, 2, 3) and B = φ
∴ A ∩ B = φ

Ex 1.4 Class 11 Maths Question 6.
If A = (3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∪ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution.
Here A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}

Ex 1.4 Class 11 Maths Question 7.
If A = {x: x is a natural number), B = {x: x is an even natural number}, C={x : x is an odd natural number} and D = {x: x is a prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution.
Here A = {x: x is a natural number}
= (1, 2, 3, 4, 5, …….}
B = {x: x is an even natural number}
= 12, 4, 6,………}
C = {x: x is an odd natural number}
= {1, 3, 5, 7,………}
and D = {x: x is a prime number}
= {2, 3, 5, 7,….}

(i) A ∩ B = {x: x is a natural number} ∩ {x: x is an even natural number}
= {x: x is an even natural number} = B.

(ii) A ∩ C = {x: x is a natural number} ∩ {x: x is an odd natural number}
= {x: x is an odd natural number} = C.

(iii) A ∩ D = {x: x is a natural number} ∩ {x: x is a prime number}
= {x: x is a prime number} = D.

(iv) B ∩ C = {x: x is an even natural number} ∩{x: x is an odd natural number} = φ .

(v) B ∩ D = [x: x is an even natural number} ∩ {x: x is a prime number} = {2}.

(vi) C ∩ D = {x: x is an odd natural number} ∩ {x: x is a prime number} = {x: x is an odd prime number}.

Ex 1.4 Class 11 Maths Question 8.
Which of the following pairs of sets are disjoint?
(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, o, u] and {c, d, e, f}
(iii) {x: x is an even integer} and {x: x is an odd integer}
Solution.
(i) Let A = {1,2,3,4}
and B = {x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
∴ A ∩ B = {1,2,3,4} n {4,5, 6} = {4}
Hence A and B are not disjoint sets.

(ii) Let A = {a, e, i, o, u} and B = {c, d, e, f}
∴ A ∩ B = {e}
Hence A and B are not disjoint sets.

(iii) Let A = {x : x is an even integer} and B = {x: x is an odd integer}
∴ A ∩ B = φ. Hence A and B are disjoint sets.

Ex 1.4 Class 11 Maths Question 9.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii)D – C
Solution.
Here A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C ={2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20}
(i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8,12,16, 20} = {3, 6, 9,15,18, 21}
(ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16} = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5,10,15, 20} = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21} = {4, 8,16, 20}
(v) C – A = {2,4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
(vii) B – C={4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C – B = {2,4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20} = {2, 6, 10, 14}
(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20} = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C={5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {5, 15, 20}

Ex 1.4 Class 11 Maths Question 10.
If X= {a, b, c, d} and Y={f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Solution.
Here X = {a, b, c, d} and Y = {f, b, d, g}
(i) X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}
(ii) Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}
(iii) X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}

Ex 1.4 Class 11 Maths Question 11.
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Solution.
We know that set of real numbers contain rational and irrational numbers. So R – Q = set of irrational numbers.

Ex 1.4 Class 11 Maths Question 12.
State whether each of the following statement is true or false. Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint
Solution.

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5

Ex 1.5 Class 11 Maths Question 1.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = { 1, 2, 3, 4}, B = (2,4,6,8} and C = {3,4,5,6}. Find
(i) A’
(ii) B’
(iii) (A ∪ C)’
(iv) (A ∪B)’
(v) (A’)’
(vi) (B – C)’
Solution.
Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C ={3, 4, 5, 6}
(i) A’=U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}

(ii) B’=U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= (1, 2, 3, 4, 5, 6}
(A∪C)’=U-(A∪C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
= {7, 8, 9}

(iv) A ∪ B = {1, 2, 3,4} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
(A∪B)’ = U – (A∪B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}

(v) We know that A’ = {5, 6, 7, 8, 9}
(A’)’ =U – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(B-C)’=U – (B-C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {1, 3, 4, 5, 6, 7, 9}.

Ex 1.5 Class 11 Maths Question 2.
If U = {a,b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution.
(i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e,f, g, h}

(ii) B’ = U – B = {a, b, c, d, e,f, g, h} – {d, e, f, g}
= {a, b, c, h}

(iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}.

Ex 1.5 Class 11 Maths Question 3.
Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x: x is an even natural number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x: x is a perfect square}
(vii) {x: x is a perfect cube}
(viii) {x: x + 5 = 8}
(ix) (x: 2x + 5 = 9)
(x) {x: x ≥ 7}
(xi) {x: x ∈ W and 2x + 1 > 10}
Solution.

Ex 1.5 Class 11 Maths Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)’ = A’∩B’
(ii) (A ∩ B)’ = A’∪B’
Solution.

Ex 1.5 Class 11 Maths Question 5.
Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)’
(ii) A’∩B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Solution.

Ex 1.5 Class 11 Maths Question 6.
Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?
Solution.
Here U = {x : x is a triangle}
A = {x: x is a triangle and has at least one angle different from 60°}
∴ A’ = U – A = {x : x is a triangle} – {x : x is a triangle and has atleast one angle different from 60°}
= {x : x is a triangle and has all angles equal to 60°)
= Set of all equilateral triangles.

Ex 1.5 Class 11 Maths Question 7.
Fill in the blanks to make each of the following a true statement:
(i) A ∪ A’ = …….
(ii) φ’ ∩ A = .…….
(iii) A ∩ A’ = …….
(iv) U’ ∩ A = .…….
Solution.
(i) A ∪ A’= U
(ii) φ’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = φ
(iv) U’ ∩ A = φ ∩ A = φ

## 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

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise

Miscellaneous Exercise Class 11 Maths Question 1.
Decide, among the following sets, which sets are subsets of one and another:
A = {x : x ∈R and x satisfy x2 – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8,…}, D = {6}.
Solution.
Here A = {x : x ∈ R and x satisfies x2 – 8x + 12 = 0}
= {x : x ∈ R and (x – 6)(x – 2) = 0} = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8,…….} and D = {6}
Now, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and D ⊂ C

Miscellaneous Exercise Class 11 Maths Question 2.
In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B,then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution.

Miscellaneous Exercise Class 11 Maths Question 3.
Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B – C.
Solution.
Given that A ∩ B = A ∩C and A ∪ B=A ∪ C

Miscellaneous Exercise Class 11 Maths Question 4.
Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution.
(i) ⇒ (ii)
A – B = {x: x∈ A and x∉B]
Since A⊂B
∴ A – B = φ

(ii) ⇒ (iii)
A – B = φ ⇒ A⊂B ⇒ A∪B = B

(iii) ⇒ (iv)
AuB = B ⇒A⊂B ⇒ A∩B = A

(iv) ⇒ (i)
A∩B = A ⇒ A⊂B
Thus (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).

Miscellaneous Exercise Class 11 Maths Question 5.
Show that if A ⊂ B, then C – B ⊂ C – A.
Solution.
Let x ∈ C – B ⇒x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [∵ A ⊂ B]
⇒ x ∈ C – A
Hence C – B ⊂C – A

Miscellaneous Exercise Class 11 Maths Question 6.
Assume that P(A) = P(B). Show that A = B
Solution.

Miscellaneous Exercise Class 11 Maths Question 7.
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B) Justify your answer.
Solution.
No, it is not true.
Take A = {1, 2} and B = {2,3}

Miscellaneous Exercise Class 11 Maths Question 8.
Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution.
(A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B’)
= A ∩ (B ∪ B’) (By distributive law)
= A ∩ U = A
Hence, A = (A ∩ B) ∪ (A – B)
Also A ∪ (B – A) = A ∪ (B ∩ A’)
= (A ∪ B) ∩ (A ∪ A’) (By distributive law)
= (A ∪ B) ∩ U= A ∪ B
Hence, A ∪ (B – A) = A ∪ B.

Miscellaneous Exercise Class 11 Maths Question 9.
Using properties of sets, show that
(i) A ∪ (A ∩ B)=A
(ii) A ∩ (A ∪ B) = A.
Solution.
(i) We know that if A ⊂ B, then
A ∪ B = B. Also, A ∩ B ⊂ A
∴ A ∪ (A ∩ B) = A.
(ii) We know that if A ⊂ B,
then A ∩ B = A Also, A ⊂ A ∪ B
∴ A ∩ (A ∪ B) = A.

Miscellaneous Exercise Class 11 Maths Question 10.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution.
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6}, C = {2, 3, 4, 9,10}.
∴ A ∩ B = [1, 2,3,4} ∩ {2,3,4, 5, 6]
= {2, 3, 4}
A ∩ C = {1, 2, 3, 4} ∩ {2, 3, 4, 9, 10} = {2, 3, 4}
Now we have A ∩ B = A ∩ C. But B ≠ C.

Miscellaneous Exercise Class 11 Maths Question 11.
Let A and B be sets. If A ∩ X=B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ (A∪X), B = B ∩ (B ∪ X) and use Distributive law)
Solution.
Here
A ∪ X = B ∪ X for some set X

Miscellaneous Exercise Class 11 Maths Question 12.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ
Solution.
Take A = {1, 2}, B = {1, 4} and C = {2, 4}
Now, A ∩ B = {1} ≠ φ, B ∩ C = {4} ≠ φ and
A ∩ C = {2} ≠ φ
But A ∩ B ∩ C = φ.

Miscellaneous Exercise Class 11 Maths Question 13.
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?
Solution.
Let T be the set of students who takes tea and C be the set of students who takes coffee. Here, n(T) = 150, n(C) = 225 and n(C ∩ T) = 100
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 150 + 225 -100 = 275
∴ Number of students taking either tea or coffee = 275
∴ Number of students taking neither tea nor coffee = 600 – 275 = 325.

Miscellaneous Exercise Class 11 Maths Question 14.
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?
Solution.
Let H be the set of students who know Hindi and E be the set of students who know English.
Here n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25 = 125

Miscellaneous Exercise Class 11 Maths Question 15.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper H, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T
and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution.

Miscellaneous Exercise Class 11 Maths Question 16.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution.