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# SETS NCERT Solutions Class 11 Maths Chapter 1 Exercise 1.1 1.2 1.3 1.4 1.5 1.6 Miscellaneous Free pdf Notes Study Material download-Anand Classes

SETS Chapter 1 Class 11 NCERT Solutions PDF material available for all the students of Class 11. The solution material can be essential for last-minute revision. Download it and get accustomed to the critical techniques to solve questions related to Sets.

NCERT Solutions for Class 11 Maths Chapter 1 Sets are prepared by our expert faculty at ANAND CLASSES’S according to the latest update on the CBSE Syllabus. These NCERT Class 11 Solutions of Maths help the students in solving the problems adroitly and efficiently. Also, ANAND CLASSES’S focuses on building step-by-step solutions for all NCERT problems in such a way that it is easy for the students to understand.

# (ix) A collection of most dangerous animals of the world.

Solution:

(i) The collection consists of months January, June and July. It is well-defined and therefore, it is a set.

(ii) A writer of India may be most talented for one person but not for another person. Opinion varies from person to person. So, the given collection is not well-defined and therefore, not a set.

(iii) The term ‘best cricket batsman’ is vague. The same batsman may be one of the best for one person but not for another. Opinion varies from person to person. So, the given collection is not well-defined and therefore, not a set.

(iv) Any boy is either in your class or not in your class. There is no ambiguity. The given collection is well-defined and therefore, it is a set.

(v) The collection consists of first 99 natural numbers. It is well-defined and therefore, it is a set.

(vi) It is a well-defined collection and therefore, it is a set.

(vii) It is a well-defined collection and therefore, it is a set.

(viii) It is a well-defined collection and therefore, it is a set.

(ix) The criterion for determining an animal as most dangerous varies from person to person. For some people, even a lizard is very dangerous. So, the given collection is not well defined and therefore, it is not a set.

Solution:

(i) 5 ∈ A

(ii) 8 ∉ A

(iii) 0 ∉ A

(iv) 4 ∈ A

(v) 2 ∈ A

(vi) 10 ∉ A

# (vi) F = The set of all letters in the word BETTER.

Solution:

(i) The required integers are – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6 (not 7).
∴ In roster form, A = { – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6 }

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

(iii) The required numbers are 17, 26, 35, 44, 53, 62, 71, 80.
∴ In roster form, C = {17, 26, 35, 44, 53, 62, 71, 80 }

(iv) Divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Among them, prime numbers are 2, 3, 5.
∴ In roster form, D = { 2, 3, 5 }

(v) In the word TRIGONOMETRY, the letters T, R and O are repeated. Dropping the repetitions, in roster form
E = { T, R, I, G, O, N, M, E, Y }

(vi) In the word BETTER, the letters E and T are repeated.
Dropping the repetitions, in roster form F = { B, E, T, R }

# (v) {1, 4, 9 … 100}

Solution:

(i) {3, 6, 9, 12}

The given set can be written in the set-builder form as {x: x = 3n∈ N and 1 ≤ n ≤ 4}

(ii) {2, 4, 8, 16, 32}

We know that 2 = 21, 4 = 22, 8 = 23, 16 = 24, and 32 = 25.

Therefore, the given set {2, 4, 8, 16, 32} can be written in the set-builder form as {x: x = 2n∈ N and 1 ≤ n ≤ 5}.

(iii) {5, 25, 125, 625}

We know that 5 = 51, 25 = 52, 125 = 53, and 625 = 54.

Therefore, the given set {5, 25, 125, 625} can be written in the set-builder form as {x: x = 5n∈N and 1 ≤ n ≤ 4}.

(iv) {2, 4, 6 …}

{2, 4, 6 …} is a set of all even natural numbers

Therefore, the given set {2, 4, 6 …} can be written in the set-builder form as {x: x is an even natural number}.

(v) {1, 4, 9 … 100}

We know that 1 = 12, 4 = 22, 9 = 32 …100 = 102.

Therefore, the given set {1, 4, 9… 100} can be written in the set-builder form as {x: x = n2∈ N and 1 ≤ n ≤ 10}.

# (vi) F = {x: x is a consonant in the English alphabet which proceeds k}.

Solution:

(i) A = {xx is an odd natural number}

So the elements are A = {1, 3, 5, 7, 9 …..}

(ii) B = {xx is an integer, -1/2 < x < 9/2}

We know that – 1/2 = – 0.5 and 9/2 = 4.5

So the elements are B = {0, 1, 2, 3, 4}.

(iii) C = {xx is an integer, x2 ≤ 4}

We know that

(–1)2 = 1 ≤ 4; (–2)2 = 4 ≤ 4; (–3)2 = 9 > 4

Here

02 = 0 ≤ 4, 12 = 1 ≤ 4, 22 = 4 ≤ 4, 32 = 9 > 4

So we get

C = {–2, –1, 0, 1, 2}

(iv) D = {xx is a letter in the word “LOYAL”}

So the elements are D = {L, O, Y, A}

(v) E = {xx is a month of a year not having 31 days}

So the elements are E = {February, April, June, September, November}

(vi) F = {xx is a consonant in the English alphabet which proceeds k}

So the elements are F = {b, c, d, f, g, h, j}

# (iv) {1, 3, 5, 7, 9} (d) {x: x is a letter of the word MATHEMATICS}

Solution:

(i) Here the elements of this set are natural number as well as divisors of 6. Hence, (i) matches with (c).

(ii) 2 and 3 are prime numbers which are divisors of 6. Hence, (ii) matches with (a).

(iii) The elements are the letters of the word MATHEMATICS. Hence, (iii) matches with (d).

(iv) The elements are odd natural numbers which are less than 10. Hence, (v) matches with (b).

Exercise 1.2 page: 8

# (iv) {y: y is a point common to any two parallel lines}

Solution:

(i) There is no odd natural number which is divisible by 2. So, the given set is a null set.

(ii) Set of even prime numbers = { 2 } ≠ φ. So, the given set is not a null set. It is a singleton set.

Note: A natural number > 1 is said to be prime if it has only two divisors 1 and itself.

The set of prime numbers is {2, 3, 5, 7, 11,…}.

(iii) There is no natural number which is both less than 5 and greater than 7. So, the given set is a null set.

(iv) Two parallel have no common point. So, the given set is a null set.

Hence (i), (iii) and (iv) are examples of the null set.

# (v) The set of prime numbers less than 99

Solution:

(i) Since there are 12 months (i.e., a definite number of months) in a year, the given set is finite.

(ii) Since the number of elements in the set is infinite, the given set is infinite.

(iii) Since the number of elements in the set is 100 (i.e., a definite number), the given set is finite.

(iv) Since there are infinitely many numbers greater than 100, the given set is infinite.

(v) Since the number of primes less than 99 is a definite number, the given set is finite.

# (v) The set of circles passing through the origin (0, 0)

Solution:

(i) Since there are infinite number of lines parallel to the x-axis, the given set is infinite.

(ii) Since there are 26 letters, i.e., a definite number of letters, in the English alphabet, the given set is finite.

(iii) Since there are infinitely many multiples of 5, the given set is infinite.

(iv) The process of counting the animals living on the earth is terminating. Thus, a definite number of animals live on the earth and hence the given set is finite.

(v) There is no end to the number of circles passing through the origin (0, 0). Hence, the given set is infinite.

# (iv) A = {x: x is a multiple of 10}; B = {10, 15, 20, 25, 30 …}

Solution:

(i) A and B have exactly same elements, though not in the same order.

∴ A = B

(ii) 12 ∈ A but 12 ∉ B

∴ A ≠ B

(iii) In roster form B = { 2, 4, 6, 8, 10 }

Since A and B have exactly same elements, therefore, A = B.

(iv) In roster form A = { 10, 20, 30, … }

Since 15 ∈ B but 15 ∉ A, therefore A ≠ B.

# (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};

B = {xis solution of x2 + 5+ 6 = 0}

x2 + 5x + 6 = 0 can be written as

x(x + 3) + 2(x + 3) = 0

(x + 2) (x + 3) = 0

So we get

x = –2 or x = –3

Hence, B = {–2, –3}

Here

A = {2, 3}; B = {–2, –3}

Therefore, A ≠ B

(ii) A = {xis a letter in the word FOLLOW} = {F, O, L, W}

B = {yis a letter in the word WOLF} = {W, O, L, F}

Order in which the elements of a set which are listed is not significant.

Therefore, A = B.

# E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1}

Solution:

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}

We know that

8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H

A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H

It can be written as

2 ∈ A, 2 ∉ C

Therefore, A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H

B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

It can be written as

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H

Therefore, C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H

Therefore, D ≠ E, D ≠ F, D ≠ G, D ≠ H

Here, E ≠ F, E ≠ G, E ≠ H

F ≠ G, F ≠ H, G ≠ H

Order in which the elements of a set are listed is not significant.

B = D and E = G

Therefore, among the given sets, B = D and E = G.

Exercise 1.3 page: 12

# (vii) {x: x is an even natural number} … {x: x is an integer}

Solution:

(i) Every element of the set {2, 3, 4} is also an element of the set {1, 2, 3, 4, 5 }

∴ { 2, 3, 4 } ⊂ { 1, 2, 3, 4, 5 }.

(ii) a ∈ { a, b, c } but a ∉ { b, c, d }

∴ { a, b, c } ⊄ { b, c, d }.

(iii) Every student of class XI of your school is a student of your school.

∴ { x : x is a student of class XI of your school } ⊂ { x : x is a student of your school }.

(iv) Every circle in a plane is not a circle with radius 1 unit, as it can have any radius r, (r > 0).

∴ { x : x is a circle in the plane } ⊄ { x : x is a circle in the same plane with radius 1 unit }.

(v) Any triangle is never a rectangle.

∴ { x : x is a triangle in the plane } ⊄ { x : x is a rectangle in the plane }.

(vi) Every equilateral triangle in a plane is a triangle in the plane.

∴ { x : x is an equilateral triangle in a plane } ⊂ { x : x is a triangle in the same plane }.

(vii) Every even natural number is an integer.

∴ { x : x is an even natural number } ⊂ { x : x is an integer }.

# (vi) {x: x is an even natural number less than 6} ⊂ {x: x is a natural number which divides 36}

Solution:

(i) False, since every element of the set { a, b } is also an element of the set { b, c, a }, therefore, { a, b } ⊂ {b, c, a}.

(ii) True, since every element of the set { a, e } is also an element of the set of vowels { a, e, i, o, u }, therefore, {a, e} ⊂ { a, e, i, o, u }.

(iii) False, since 2 ∈ { 1, 2, 3 } but 2 ∉ { 1, 3, 5 }.

(iv) True, since a ∈ { a, b, c }.

(v) False, since { a } is subset of the set { a, b, c } but not an element of the set { a, b, c }.

(vi) True, since { x : x is an even natural number less than 6 } = { 2, 4 } and { x : x is a natural number which divides 36} = { 1, 2, 3, 4, 6, 9, 12, 18, 36 }.

Clearly, { 2, 4 } ⊂ { 1, 2, 3, 4, 6, 9, 12, 18, 36 }.

# (xi) {Φ} ⊂ A

Solution:

(i) False, since 3 ∉ A, 4 ∉ A, therefore, { 3, 4 } ⊄ A.

(ii) True, since { 3, 4 } is an element of A.

(iii) True, since { 3, 4 } ∈ A, therefore, {{ 3, 4 }} ⊂ A.

(iv) True, since 1 is an element of A.

(v) False, since 1 is not a set. Only a set can be a subset of another set.

(vi) True, since 1, 2, 5 are elements of A.

(vii) False, since { 1, 2, 5 } is not an element of A.

(viii) False, since 3 ∉ A.

(ix) False, since φ is not an element of A.

(x) True, since φ is a subset of every set.

(xi) False, since φ ⊂ Α and hence {φ} ∈ P(A), power set of set A.

# (iv) Φ

Solution:

(i) Let A = { a }, then A has one element.

∴ Number of subsets of the set A = 2n = 21 = 2.

Subset of A will have either no element or one element.

Subset of A having no element is φ

Subset of A having one element is { a }

∴ The subsets of A are φ, { a }.
(ii) Let A = { a, b }, then A has 2 elements.

∴ Number of subsets of A = 2n = 22 = 4.

Subset of A having no element is φ

Subsets of A having one element are { a }, { b }

Subset of A having two elements is { a, b }

∴ The subsets of A are φ, { a }, { b }, { a, b }.

(iii) Let A = { 1, 2, 3 }, then A has 3 elements.

∴ Number of subsets of A = 2n = 23 = 8.

Subsets of A having no element is φ

Subsets of A having one element are { 1 }, { 2 }, { 3 }

Subsets of A having two elements are { 1, 2 }, { 1, 3 }, { 2, 3 }

Subset of A having three elements is { 1, 2, 3 }

∴The subsets of A are φ, { 1 }, { 2 }, { 3 }, { 1, 2 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 }.

(iv) The only subset of the empty set φ is φ itself.

# 5. How many elements has P (A), if A = Φ?

Solution:

If A is a set with m elements

(A) = m then [P (A)] = 2m

If A = Φ we get n (A) = 0

[P(A)] = 20 = 1

Therefore, P (A) has one element.

# (iv) {x: x ∈ R, 3 ≤ x ≤ 4}

Solution:

(i) {x∈ R, –4 < x ≤ 6} = (–4, 6], since – 4 is not included while 6 is included.

(ii) {x∈ R, –12 < x < –10} = (–12, –10), since both end points are excluded.

(iii) {x∈ R, 0 ≤ x < 7} = [0, 7), since 0 is included while 7 is excluded

(iv) {x∈ R, 3 ≤ x ≤ 4} = [3, 4], since both end points are included

# (iv) [–23, 5)

Solution:

(i) (–3, 0) = {x∈ R, –3 < x < 0}

(ii) [6, 12] = {x∈ R, 6 ≤ x ≤ 12}

(iii) (6, 12] ={x∈ R, 6 < x ≤ 12}

(iv) [–23, 5) = {x∈ R, –23 ≤ x < 5}

# (ii) The set of isosceles triangles

Solution:

(i) Among the set of right triangles, the universal set is the set of triangles or the set of polygons.

(ii) Among the set of isosceles triangles, the universal set is the set of triangles or the set of polygons or the set of two-dimensional figures.

# (iv) {1, 2, 3, 4, 5, 6, 7, 8}

Solution:

(i) We know that A ⊂ {0, 1, 2, 3, 4, 5, 6}

B ⊂ {0, 1, 2, 3, 4, 5, 6}

So C ⊄ {0, 1, 2, 3, 4, 5, 6}

Hence, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ

Hence, Φ cannot be the universal set for the sets A, B, and C.

(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Hence, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

So C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

Hence, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.

Exercise 1.4 page: 17

# (v) A = {1, 2, 3}, B = Φ

Solution:

(i) X = {1, 3, 5} Y = {1, 2, 3}

So the union of the pairs of set can be written as

X ∪ Y= {1, 2, 3, 5}

(ii) A = {aeiou} B = {abc}

So the union of the pairs of set can be written as

A∪ B = {abceiou}

(iii) A = {xx is a natural number and multiple of 3} = {3, 6, 9 …}

B = {xx is a natural number less than 6} = {1, 2, 3, 4, 5, 6}

So the union of the pairs of set can be written as

A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}

Hence, A ∪ B = {xx = 1, 2, 4, 5 or a multiple of 3}

(iv) A = {xx is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {xx is a natural number and 6 < x < 10} = {7, 8, 9}

So the union of the pairs of set can be written as

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

Hence, A∪ B = {x: x ∈ N and 1 < x < 10}

(v) A = {1, 2, 3}, B = Φ

So the union of the pairs of set can be written as

A ∪ B = {1, 2, 3}

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

Solution:

It is given that

A = {ab} and B = {abc}

Yes, A ⊂ B

So the union of the pairs of set can be written as

A∪ B = {abc} = B

# 3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Solution:

If A and B are two sets such that A ⊂ B, then A ∪ B = B.

A ∪ B = { x : x ∈ A or x ∈ B } = { x : x ∈ B }
( A ⊂ B ∴ x ∈ A ⇒ x ∈ B)
A ∪ B = B

# (vii) B ∪ C ∪ D

Solution:

It is given that

A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i) A ∪ B = { 1, 2, 3, 4 } ∪ { 3, 4, 5, 6 } = { 1, 2, 3, 4, 5, 6 }

(ii) A ∪ C = { 1, 2, 3, 4 } ∪ { 5, 6, 7, 8 } = { 1, 2, 3, 4, 5, 6, 7, 8 }

(iii) B ∪ C = { 3, 4, 5, 6 } ∪ { 5, 6, 7, 8 } = { 3, 4, 5, 6, 7, 8 }

(iv) B ∪ D = { 3, 4, 5, 6 } ∪ { 7, 8, 9, 10 } = { 3, 4, 5, 6, 7, 8, 9, 10 }

(v) A ∪ B ∪ C = (A ∪ B) ∪ C = { 1, 2, 3, 4, 5, 6 } ∪ { 5, 6, 7, 8 } = { 1, 2, 3, 4, 5, 6, 7, 8 }

(vi) A ∪ B ∪ D = (A ∪ B) ∪ D = { 1, 2, 3, 4, 5, 6 } ∪ { 7, 8, 9, 10 } = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = (B ∪ C) ∪ D = { 3, 4, 5, 6, 7, 8 } ∪ { 7, 8, 9, 10 } = { 3, 4, 5, 6, 7, 8, 9, 10 }

# (v) A = {1, 2, 3}, B = Φ

Solution:

(i) X = {1, 3, 5}, Y = {1, 2, 3}

So the intersection of the given set can be written as

X ∩ Y = {1, 3}

(ii) A = {aeiou}, B = {abc}

So the intersection of the given set can be written as

A ∩ B = {a}

(iii) A = {xx is a natural number and multiple of 3} = (3, 6, 9 …}

B = {xx is a natural number less than 6} = {1, 2, 3, 4, 5}

So the intersection of the given set can be written as

A ∩ B = {3}

(iv) A = {xx is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {xx is a natural number and 6 < x < 10} = {7, 8, 9}

So the intersection of the given set can be written as

A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ

So the intersection of the given set can be written as

A ∩ B = Φ

# (x) (A ∪ D) ∩ (B ∪ C)

Solution:

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = {A ∩ C} ∩ D

= {11} ∩ {15, 17}

= Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11}

= {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪ Φ

= {7, 9, 11}

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15}

= {7, 9, 11}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

# (vi) C ∩ D

Solution:

It can be written as

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

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

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

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

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

# (iii) {x: x is an even integer} and {x: x is an odd integer}

Solution:

(i) {1, 2, 3, 4}

{xx is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

So we get

{1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Hence, this pair of sets is not disjoint.

(ii) {aeiou} ∩ (cdef} = {e}

Hence, {aeiou} and (cdef} are not disjoint.

(iii) {xx is an even integer} ∩ {xx is an odd integer} = Φ

Hence, this pair of sets is disjoint.

# (xii) D – C

Solution:

(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 }.

# (iii) X ∩ Y

Solution:

(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 }.

# 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

R – Set of real numbers

Q – Set of rational numbers

Hence, R – Q is a set of irrational numbers.

R – Q = { x : x ∈ R and x ∉ Q }
= { x : x is a real number and x is not a rational
number }
= { x : x is an irrational number }
(Every real number is either rational or irrational but not both)

# (iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

Solution:

(i) False

If 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

So we get {2, 3, 4, 5} ∩ {3, 6} = {3} ≠ φ

(ii) False

If a ∈ {aeiou}, a ∈ {abcd}

So we get {aeiou} ∩ {abcd} = {a} ≠ φ

(iii) True

Here {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) True

Here {2, 6, 10} ∩ {3, 7, 11} = Φ

Exercise 1.5 Page: 20

# (vi) (B – C)’

Solution:

It is given that

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

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) A′ = U – A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } – { 1, 2, 3, 4 }

= { 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 } = A

(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 }.

# (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 }.

# (xi) {x: x ∈ N and 2x + 1 > 10}

Solution:

We know that

U = N: Set of natural numbers

(i) {xx is an even natural number}´ = {xx is an odd natural number}

(ii) {xx is an odd natural number}´ = {xx is an even natural number}

(iii) {xx is a positive multiple of 3}´ = {xx ∈ N and x is not a multiple of 3}

(iv) {xx is a prime number}´ ={xx is a positive composite number and x = 1}

Def: Composite number: A natural number > 1 is said to be composite if it is not prime, i.e., if it has at least one divisor other than 1 and itself. For example, 4, 6, 8, 9, … are composite.
Note. In fact N is the union of (set of primes, set of composites and {1})

(v) {xx is a natural number divisible by 3 and 5}´ = {xx is a natural number that is not divisible by 3 or 5}

(vi) {xx is a perfect square}´ = {xx ∈ N and is not a perfect square}

(vii) {xx is a perfect cube}´ = {xx ∈ N and is not a perfect cube}

(viii) {xx + 5 = 8}´ = {xx ∈ N and x ≠ 3}

(ix) {x: 2x + 5 = 9}´ = {xx ∈ N and x ≠ 2}

(x) {xx ≥ 7}´ = {xx ∈ N and x < 7}

(xi) {xx ∈ N and 2x + 1 > 10}´ = {xx ∈ N and ≤ 9/2}

# (ii) (A ∩ B)’ = A’ U B’

Solution:

It is given that

U = {1, 2, 3, 4, 5,6,7,8, 9}

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

Here A′= U – A = { 1, 3, 5, 7, 9 }

and B′ = U – B = { 1, 4, 6, 8, 9 }

(i) A ∪ B = { 2, 3, 4, 5, 6, 7, 8 }

(A ∪ B)′= U – (A ∪ B) = { 1, 9 }

A′ ∩ B′ = { 1, 3, 5, 7, 9 } ∩ { 1, 4, 6, 8, 9 } = { 1, 9 }

∴ (A ∪ B)′= A′ ∩ B′.

(ii) A ∩ B = {2}

(A ∩ B)′= U – (A ∩ B) = { 1, 3, 4, 5, 6, 7, 8, 9 }

A′ ∪ B′ = { 1, 3, 5, 7, 9 } ∪ { 1, 4, 6, 8, 9 }

= { 1, 3, 4, 5, 6, 7, 8, 9 }

∴ (A ∩ B)′= A′ ∪ B′.

Solution:

(i) (A U B)’

(ii) A’ ∩ B’

(iii) (A ∩ B)’

(iv) A’ U B’

# 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:

A′ = U – A
= Set of all triangles with no angle different from 60°
= Set of all triangles with each angle 60°
= Set of all equilateral triangles

# (iv) U’ ∩ A = …….

Solution:

(i) A U A’ = U

(ii) Φ′ ∩ A = U ∩ A = A

So we get

Φ′ ∩ A = A

(iii) A ∩ A’ = Φ

(iv) U’ ∩ A = Φ ∩ A = Φ

So we get

U’ ∩ A = Φ

Exercise 1.6 page: 24

# 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:

Given

n (X) = 17

n (Y) = 23

n (X U Y) = 38

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

38 = 17 + 23 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 40 – 38 = 2

So we get

n (X ∩ Y) = 2

# 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:

Given

n (X U Y) = 18

n (X) = 8

n (Y) = 15

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

18 = 8 + 15 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 23 – 18 = 5

So we get

n (X ∩ Y) = 5

# 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:

Consider H as the set of people who speak Hindi

E as the set of people who speak English

We know that

n(H ∪ E) = 400

n(H) = 250

n(E) = 200

It can be written as

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

By substituting the values

400 = 250 + 200 – n(H ∩ E)

By further calculation

400 = 450 – n(H ∩ E)

So we get

n(H ∩ E) = 450 – 400

n(H ∩ E) = 50

Therefore, 50 people can speak both Hindi and English.

# 4. 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?

Solution:

We know that

n(S) = 21

n(T) = 32

n(S ∩ T) = 11

It can be written as

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

Substituting the values

n (S ∪ T) = 21 + 32 – 11

So we get

n (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

# 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 Y have?

Solution:

We know that

n(X) = 40

n(X ∪ Y) = 60

n(X ∩ Y) = 10

It can be written as

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

By substituting the values

60 = 40 + n(Y) – 10

On further calculation

n(Y) = 60 – (40 – 10) = 30

Therefore, the set Y has 30 elements.

# 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:

Consider C as the set of people who like coffee

T as the set of people who like tea

n(C ∪ T) = 70

n(C) = 37

n(T) = 52

It is given that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

70 = 37 + 52 – n(C ∩ T)

By further calculation

70 = 89 – n(C ∩ T)

So we get

n(C ∩ T) = 89 – 70 = 19

Therefore, 19 people like both coffee and tea.

# 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:

Consider C as the set of people who like cricket

T as the set of people who like tennis

n(C ∪ T) = 65

n(C) = 40

n(C ∩ T) = 10

It can be written as

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

65 = 40 + n(T) – 10

By further calculation

65 = 30 + n(T)

So we get

n(T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that,

(T – C) ∪ (T ∩ C) = T

So we get,

(T – C) ∩ (T ∩ C) = Φ

Here

n (T) = n (T – C) + n (T ∩ C)

Substituting the values

35 = n (T – C) + 10

By further calculation

n (T – C) = 35 – 10 = 25

Therefore, 25 people like only tennis.

# 8. 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?

Solution:

Consider F as the set of people in the committee who speak French

S as the set of people in the committee who speak Spanish

n(F) = 50

n(S) = 20

n(S ∩ F) = 10

It can be written as

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

By substituting the values

n(S ∪ F) = 20 + 50 – 10

By further calculation

n(S ∪ F) = 70 – 10

n(S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages.

Miscellaneous Exercise Page 26

# D = {6}.

Solution:

According to the question,

We have,

A = {x: x ∈ R and x satisfies x2 – 8x + 12 =0}

A = {x: x ∈ R and x satisfies (x – 2)(x – 6) = 0}

2 and 6 are the only solutions of x2 – 8x + 12 = 0.

Hence, A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}

Hence, D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

# (vi) If A ⊂ B and x ∉ B, then x ∉ A

Solution:

(i) False

According to the question,

A = {1, 2} and B = {1, {1, 2}, {3}}

Now, we have,

2 ∈ {1, 2} and {1, 2} ∈ {1, {1, 2}, {3}}

Hence, we get,

A ∈ B

We also know,

{2} ∉ {1, {1, 2}, {3}}

(ii) False

According to the question

Let us assume that,

A {2}

B = {0, 2}

And, C = {1, {0, 2}, 3}

From the question,

A ⊂ B

Hence,

B ∈ C

But, we know,

A ∉ C

(iii) True

According to the question

A ⊂ B and B ⊂ C

Let us assume that,

x ∈ A

Then, we have,

x ∈ B

And,

x ∈ C

Therefore,

A ⊂ C

(iv) False

According to the question

A ⊄ B

Also,

B ⊄ C

Let us assume that,

A = {1, 2}

B = {0, 6, 8}

And,

C = {0, 1, 2, 6, 9}

∴ A ⊂ C

(v) False

According to the question,

x ∈ A

Also,

A ⊄ B

Let us assume that,

A = {3, 5, 7}

Also,

B = {3, 4, 6}

We know that,

A ⊄ B

∴ 5 ∉ B

(vi) True

According to the question,

A ⊂ B

Also,

x ∉ B

Let us assume that,

x ∈ A,

We have,

x ∈ B,

From the question,

We have, x ∉ B

∴ x ∉ A

# 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:

According to the question,

A ∪ B = A ∪ C

And,

A ∩ B = A ∩ C

To show,

B = C

Let us assume,

x ∈ B

So,

x ∈ A ∪ B

x ∈ A ∪ C

Hence,

x ∈ A or x ∈ C

When x ∈ A, then,

x ∈ B

∴ x ∈ A ∩ B

As, A ∩ B = A ∩ C

So, x ∈ A ∩ C

∴ x ∈ A or x ∈ C

x ∈ C

∴ B ⊂ C

Similarly, it can be shown that C ⊂ B

Hence, B = C

# (iv) A ∩ B = A

Solution:

According to the question,

To prove, (i) ⬌ (ii)

Here, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

To prove, A – B ≠ ϕ

Let A – B ≠ ϕ

Hence, there exists X ∈ A, X ≠ B, but since A⊂ B, it is not possible

∴ A – B = ϕ

And A⊂ B ⇒ A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

To prove: A ⊂ B

Let X∈ A

So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)

Hence, A – B = ϕ => A ⊂ B

∴(i) ⬌ (ii)

Let us assume that A ⊂ B

To prove, A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ∈ A∪ B

⇒ X ∈ A or X ∈ B

Taking Case I: X ∈ B

A ∪ B = B

Taking Case II: X ∈ A

⇒ X ∈ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ∈ A

⇒ X ∈ A ∪ B (A ⊂ A ∪ B)

⇒ X ∈ B (A ∪ B = B)

∴A⊂ B

Hence, (i) ⬌ (iii)

To prove (i) ⬌ (iv)

Let us assume that A ⊂ B

A ∩ B ⊂ A

Let X ∈ A

To prove, X ∈ A∩ B

Since, A ⊂ B and X ∈ B

Hence, X ∈ A ∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Let us assume that A ∩ B = A

Let X ∈ A

⇒ X ∈ A ∩ B

⇒ X ∈ B and X ∈ A

⇒ A ⊂ B

∴ (i) ⬌ (iv)

∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)

Hence, proved

# 5. Show that if A ⊂ B, then C – B ⊂ C – A.

Solution:

To show,

C – B ⊂ C – A

According to the question,

Let us assume that x is any element such that X ∈ C – B

∴ x ∈ C and x ∉ B

Since, A ⊂ B, we have,

∴ x ∈ C and x ∉ A

So, x ∈ C – A

∴ C – B ⊂ C – A

Hence, Proved.

# 6. Assume that P (A) = P (B). Show that A = B

Solution:

To show,

A = B

According to the question,

P (A) = P (B)

Let x be any element of set A,

x ∈ A

Since, P (A) is the power set of set A, it has all the subsets of set A.

A ∈ P (A) = P (B)

Let C be an element of set B

For any C ∈ P (B),

We have, x ∈ C

C ⊂ B

∴ x ∈ B

∴ A ⊂ B

Similarly, we have:

B ⊂ A

so, we get,

If A ⊂ B and B ⊂ A

∴ A = B

# 7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.

Solution:

It is not true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)

Justification:

Let us assume,

A = {0, 1}

And, B = {1, 2}

∴ A ∪ B = {0, 1, 2}

According to the question,

We have,

P (A) = {ϕ, {0}, {1}, {0, 1}}

P (B) = {ϕ, {1}, {2}, {1, 2}}

∴ P (A ∪ B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

Also,

P (A) ∪ P (B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}}

∴ P (A) ∪ P (B) ≠ P (A ∪ B)

Hence, the given statement is false

# 8. Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)

Solution:

To Prove,

A = (A ∩ B) ∪ (A – B)

Proof: Let x ∈ A

To show,

X ∈ (A ∩ B) ∪ (A – B)

In Case I,

X ∈ (A ∩ B)

⇒ X ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

In Case II,

X ∉A ∩ B

⇒ X ∉ B or X ∉ A

⇒ X ∉ B (X ∉ A)

⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴A ⊂ (A ∩ B) ∪ (A – B) (i)

It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A

Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)

Equating (i) and (ii),

A = (A ∩ B) ∪ (A – B)

We also have to show,

A ∪ (B – A) ⊂ A ∪ B

Let us assume,

X ∈ A ∪ (B – A)

X ∈ A or X ∈ (B – A)

⇒ X ∈ A or (X ∈ B and X ∉A)

⇒ (X ∈ A or X ∈ B) and (X ∈ A and X ∉A)

⇒ X ∈ (B ∪A)

∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)

According to the question,

To prove:

(A ∪ B) ⊂ A ∪ (B – A)

Let y ∈ A∪B

Y ∈ A or y ∈ B

(y ∈ A or y ∈ B) and (X ∈ A and X ∉A)

⇒ y ∈ A or (y ∈ B and y ∉A)

⇒ y ∈ A ∪ (B – A)

Thus, A ∪ B ⊂ A ∪ (B – A) (iv)

∴From equations (iii) and (iv), we get:

A ∪ (B – A) = A ∪ B

# 9. Using properties of sets, show that: (i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

Solution:

(i) To show: A ∪ (A ∩ B) = A

We know that,

A ⊂ A

A ∩ B ⊂ A

∴ A ∪ (A ∩ B) ⊂ A (i)

Also, according to the question,

We have:

A⊂ A ∪ (A ∩ B) (ii)

Hence, from equation (i) and (ii)

We have:

A ∪ (A ∩ B) = A

(ii) To show,

A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A

# 10. Show that A ∩ B = A ∩ C need not imply B = C.

Solution:

Let us assume,

A = {0, 1}

B = {0, 2, 3}

And, C = {0, 4, 5}

According to the question,

A ∩ B = {0}

And,

A ∩ C = {0}

∴ A ∩ B = A ∩ C = {0}

But,

2 ∈ B and 2 ∉ C

Therefore, B ≠ C

# 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:

According to the question,

Let A and B be two sets such that A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X.

To show, A = B

Proof:

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

= (A ∩ B) ∪ (A ∩ X) [Distributive law]

= (A ∩ B) ∪ Φ [A ∩ X = Φ]

= A ∩ B (i)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

= (B ∩ A) ∪ (B ∩ X) … [Distributive law]

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= A ∩ B (i)

Hence, from equations (i) and (ii), we obtain A = B.

# 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:

Let us assume, A {0, 1}

B = {1, 2}

And, C = {2, 0}

According to the question,

A ∩ B = {1}

B ∩ C = {2}

And,

A ∩ C = {0}

∴ A ∩ B, B ∩ C and A ∩ C are not empty sets

Hence, we get,

A ∩ B ∩ C = Φ

# 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 us assume that,

U = the set of all students who took part in the survey

T = the set of students taking tea

C = the set of the students taking coffee

Total number of students in a school, n (U) = 600

Number of students taking tea, n (T) = 150

Number of students taking coffee, n (C) = 225

Also, n (T ∩ C) = 100

Now, we have to find that number of students taking neither coffee nor tea i.e. n (T ∩ C’)

∴ According to the question,

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

∴ Number of students taking neither coffee nor tea = 325 students

# 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 us assume that,

U = the set of all students in the group

E = the set of students who know English

H = the set of the students who know Hindi

∴ H ∪ E = U

Given that,

Number of students who know Hindi n (H) = 100

Number of students who knew English, n (E) = 50

Number of students who know both, n (H ∩ E) = 25

We have to find the total number of students in the group i.e. n (U)

∴ According to the question,

n (U) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

∴ Total number of students in the group = 125 students

# (i) The number of people who read at least one of the newspapers. (ii) The number of people who read exactly one newspaper.

Solution:

(i) Let us assume that,

A = the set of people who read newspaper H

B = the set of people who read newspaper T

C = the set of people who read newspaper I

According to the question,

Number of people who read newspaper H, n (A) = 25

Number of people who read newspaper T, n (B) = 26

Number of people who read the newspaper I, n (C) = 26

Number of people who read both newspaper H and I, n (A ∩ C) = 9

Number of people who read both newspaper H and T, n (A ∩ B) = 11

Number of people who read both newspaper T and I, n (B ∩ C) = 8

And, Number of people who read all three newspaper H, T and I, n (A ∩ B ∩ C) = 3

Now, we have to find the number of people who read at least one of the newspaper

∴, we get.

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 80 – 28

= 52

∴ There are a total of 52 students who read at least one newspaper.

(ii) Let us assume that,

a = the number of people who read newspapers H and T only

b = the number of people who read newspapers I and H only

c = the number of people who read newspapers T and I only

d = the number of people who read all three newspapers

According to the question,

D = n(A ∩ B ∩ C) = 3

Now, we have:

n(A ∩ B) = a + d

n(B ∩ C) = c + d

And,

n(C ∩ A) = b + d

∴ a + d + c +d + b + d = 11 + 8 + 9

a + b + c + d = 28 – 2d

= 28 – 6

= 22

∴ Number of people read exactly one newspaper = 52 – 22

= 30 people

# 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:

Let
X = set of people who liked product A
Y = set of people who liked product B
Z = set of people who liked product C
Let a denote the number of people who liked products A and B only, b denote the number of people who liked products B and C only, c denote the number of people who liked products C and A only and d denote the number of people who liked all the three products.
Let x, y and z denote the number of people who respectively liked products A, B and C only. Then
b + d = 14, c + d = 12 and d = 8
⇒ b = 6 and c = 4
Since n(Z) = 29
∴ z + b + c + d = 29
⇒ z + 6 + 4 + 8 = 29
⇒ z = 29 – 18 = 11
Hence, 11 people liked product C only.

## Important topics and subtopics of NCERT Solutions for Class 11 Maths Chapter 1 – SETS

ANAND CLASSES’S provides all the concepts and solutions for Chapter 1 of Class 11 Maths in a creative and logical way. The PDF of Maths  NCERT Solutions for Class 11 Chapter 1 includes the following topics and sub-topics.

### 1.1 Introduction

This section contains the origin, basic definition, and applications of sets.

### 1.2 Sets and their Representations

Here, students are able to know what a set is exactly and how it can be represented in roster/ set-builder, and denoted using the English alphabets.

### 1.3 The Empty Set

This topic explains when a set is called an empty set.

### 1.4 Finite and Infinite Sets

Students can learn the definitions and examples of finite and infinite sets.

### 1.5 Equal Sets

This section gives an idea about equal and unequal sets along with solved examples.

### 1.6 Subsets

The main concepts covered under this section are:

1.6.1 Subsets of a set of real numbers

1.6.2 Intervals as subsets of R

### 1.7 Power Set

In this section, students are able to know the definition of power set, which has been derived from the concept of subsets.

### 1.8 Universal Set

This section is well explained with real-life examples such as population studies to define the universal set and representation using a letter.

### 1.9 Venn Diagrams

This section has the origin and definition of a Venn diagram. Also, it is explained using all the above topics and sub-topics.

### 1.10 Operations on Sets

We can perform some basic operations on sets similar to the addition or subtraction of numbers. These can be defined as below:

1.10.1 Union of sets

1.10.2 Intersection of sets

1.10.3 Difference of sets

### 1.11 Complement of a Set

After reading this section, students are able to understand the definition and properties of the complement of a set, along with the examples as well as practice problems.

### 1.12 Practical Problems on Union and Intersection of Two Sets

Practice problems are provided under this section to help the students understand in a better way about the union and intersection of two sets. These will help in their further studies to get thorough with the related concepts.

• Exercise 1.1 Solutions 6 Questions
• Exercise 1.2 Solutions 6 Questions
• Exercise 1.3 Solutions 9 Questions
• Exercise 1.4 Solutions 12 Questions
• Exercise 1.5 Solutions 7 Questions
• Exercise 1.6 Solutions 8 Questions
• Miscellaneous Exercise On Chapter 1 Solutions 16 Questions

## Features of NCERT Solutions for Class 11 Maths Chapter 1 – Sets

There are 6 exercises and a miscellaneous exercise in this chapter to help students understand the concepts related to Sets of the Class 11 Maths CBSE Syllabus (2023-24) in detail. The summary of the topics explained in Chapter 1 of NCERT Solutions for Class 11 Maths is listed below:

1. A well-defined collection of objects is called a set.
2. If a set does not contain any element, it is called an empty set.
3. Finite set: A set that consists of a definite number of elements
Infinite set:  A set that consists of an infinite number of elements
4. Two sets A and B are said to be equal if they have the same elements.
5. A set A is said to be a subset of a set B if every element of A is also an element of B. Intervals are subsets of R.
6. A power set of a set A is a collection of all subsets of A. It is denoted by P(A).
7. The union of two sets A and B is the set of all those elements which are either in A or B.
8. The intersection of two sets A and B is the set of all elements which are common. The difference of two sets A and B in this order is the set of elements that belong to A but not to B.
9. The complement of a subset A of universal set U is the set of all elements of U which are not the elements of A.
10. For any two sets A and B, (A ∪ B)’ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′.

## Frequently Asked Questions (FAQs) on NCERT Solutions for Class 11 Maths Chapter 1

Q1

### What are the topics covered in Chapter 1 of NCERT Solutions for Class 11 Maths?

The topics covered in Chapter 1 of NCERT Solutions for Class 11 Maths are:
1. Introduction
2. Sets and their representations
3. The Empty Set
4. Finite and Infinite Sets
5. Equal Sets
6. Subsets
7. Power Set
8. Universal Set
9. Venn Diagrams
10. Operations on Sets
11. Complement of a Set
12. Practical problems on Union and Intersection of Two Sets
Q2

### How many problems are there in each exercise of NCERT Solutions for Class 11 Maths Chapter 1?

Regular practice of the exercise-wise problems helps students to understand the concepts covered in a better way. The number of questions in each exercise of Chapter 1 is as follows:
Exercise 1.1 – 6 questions
Exercise 1.2 – 6 questions
Exercise 1.3 – 9 questions
Exercise 1.4 – 12 questions
Exercise 1.5 – 7 questions
Exercise 1.6 – 8 questions
Miscellaneous Exercise – 16 questions
Q3

### What applications of Sets are discussed in Chapter 1 of NCERT Solutions for Class 11 Maths?

Sets have numerous applications in both real life and Mathematics. From formulating calculus, topology, and geometry to the formation of algebra around fields, rings and groups, sets play a very important role. It is also used in other fields like Physics, Chemistry, Electrical Engineering, Computer Science and Biology.

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