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

# Exercise 1.1 page: 4

**1. Which of the following are sets? Justify your answer.**

**(i) The collection of all 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 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.

**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) 5 ∈ A

(ii) 8 ∉ A

(iii) 0 ∉ A

(iv) 4 ∈ A

(v) 2 ∈ A

(vi) 10 ∉ A

**3. Write the following sets in roster form:**

**(i) A = {***x*: *x* is an integer and –3 < *x *< 7}.

*x*:

*x*is an integer and –3 <

*x*< 7}.

**(ii) B = {***x*: *x* is a natural number less than 6}.

*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}

*x*:

*x*is a two-digit natural number such that the sum of its digits is 8}

**(iv) D = {***x*: *x* is a prime number which is divisor of 60}.

*x*:

*x*is a prime number which is divisor of 60}.

**(v) E = The set of all letters in the word TRIGONOMETRY.**

**(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 }

# 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) {3, 6, 9, 12}

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

(ii) {2, 4, 8, 16, 32}

We know that 2 = 2^{1}, 4 = 2^{2}, 8 = 2^{3}, 16 = 2^{4}, and 32 = 2^{5}.

Therefore, the given set {2, 4, 8, 16, 32} can be written in the set-builder form as {*x*:* x* = 2* ^{n}*,

*n*∈ N and 1 ≤

*n*≤ 5}.

(iii) {5, 25, 125, 625}

We know that 5 = 5^{1}, 25 = 5^{2}, 125 = 5^{3}, and 625 = 5^{4}.

Therefore, the given set {5, 25, 125, 625} can be written in the set-builder form as {*x*:* x* = 5* ^{n}*,

*n*∈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 = 1^{2}, 4 = 2^{2}, 9 = 3^{2} …100 = 10^{2}.

Therefore, the given set {1, 4, 9… 100} can be written in the set-builder form as {*x*:* x* = *n*^{2}, *n *∈ N and 1 ≤ *n* ≤ 10}.

**5. List all the elements of the following sets:**

**(i) A = {***x*: *x* is an odd natural number}

*x*:

*x*is an odd natural number}

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

*x*:

*x*is an integer, -1/2 < x < 9/2}

**(iii) C = {***x*: *x* is an integer, x^{2} ≤ 4}

*x*:

*x*is an integer, x

^{2}≤ 4}

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

*x*:

*x*is a letter in the word “LOYAL”}

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

*x*:

*x*is a month of a year not having 31 days}

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

*x*:

*x*is a consonant in the English alphabet which proceeds

*k*}.

**Solution:**

(i) A = {*x*: *x* is an odd natural number}

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

(ii) B = {*x*: *x* 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 = {*x*: *x* is an integer, x^{2} ≤ 4}

We know that

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

Here

0^{2} = 0 ≤ 4, 1^{2} = 1 ≤ 4, 2^{2} = 4 ≤ 4, 3^{2} = 9 > 4

So we get

C = {–2, –1, 0, 1, 2}

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

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

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

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

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

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

**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 a natural number and divisor of 6}**

**(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

**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 numbers, *x *< 5 and *x *> 7}

*x*:

*x*is a natural numbers,

*x*< 5 and

*x*> 7}

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

*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.

**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) 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.

**3. State whether each of the following set is finite or infinite:**

**(i) The set of lines which are parallel to the ***x*-axis

*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) 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.

**4. In the following, state whether A = B or not:**

**(i) A = {***a*, *b*, *c*, *d*}; B = {*d*, *c*, *b*, *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}

*x*:

*x*is positive even integer and

*x*≤ 10}

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

*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.

**5. Are the following pair of sets equal? Give reasons.**

**(i) A = {2, 3}; B = {***x*: *x *is solution of *x*^{2} + 5*x *+ 6 = 0}

*x*:

*x*is solution of

*x*

^{2}+ 5

*x*+ 6 = 0}

**(ii) A = {***x*: *x *is a letter in the word FOLLOW}; B = {*y*: *y *is a letter in the word WOLF}

*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 = {*x*: *x *is solution of *x*^{2} + 5*x *+ 6 = 0}

*x*^{2} + 5*x* + 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 = {*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}

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

Therefore, A = B.

**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}

*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

**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*}

*a*,

*b*,

*c*} … {

*b*,

*c*,

*d*}

**(iii) {***x*: *x* is a student of Class XI of your school} … {*x*: *x* student of your school}

*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}

*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}

*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}

*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}

*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 }**.**

**2. Examine whether the following statements are true or false:**

**(i) {***a*, *b*} ⊄ {*b*, *c*, *a*}

*a*,

*b*} ⊄ {

*b*,

*c*,

*a*}

**(ii) {***a*, *e*} ⊂ {*x*: *x* is a vowel in the English alphabet}

*a*,

*e*} ⊂ {

*x*:

*x*is a vowel in the English alphabet}

**(iii) {1, 2, 3} ⊂ {1, 3, 5}**

**(iv) {***a*} ⊂ {*a*. *b*, *c*}

*a*} ⊂ {

*a*.

*b*,

*c*}

**(v) {***a*} ∈ (*a*, *b*, *c*)

*a*} ∈ (

*a*,

*b*,

*c*)

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

*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 }.

**3. Let A= {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?**

**(i) {3, 4} ⊂ A**

**(ii) {3, 4}}∈ A**

**(iii) {{3, 4}} ⊂ A**

**(iv) 1 ∈ A**

**(v) 1⊂ A**

**(vi) {1, 2, 5} ⊂ A**

**(vii) {1, 2, 5} ∈ A**

**(viii) {1, 2, 3} ⊂ A**

**(ix) Φ ∈ A**

**(x) Φ ⊂ A**

**(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.

**4. Write down all the subsets of the following sets:**

**(i) {***a*}

*a*}

**(ii) {***a*, *b*}

*a*,

*b*}

**(iii) {1, 2, 3}**

**(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

*n *(A) = *m* then *n *[P (A)] = 2^{m}

If A = Φ we get* n *(A) = 0

*n *[P(A)] = 2^{0} = 1

Therefore, P (A) has one element.

**6. Write the following as intervals:**

**(i) {***x*: *x* ∈ R, –4 < *x* ≤ 6}

*x*:

*x*∈ R, –4 <

*x*≤ 6}

**(ii) {***x*: *x* ∈ R, –12 < *x* < –10}

*x*:

*x*∈ R, –12 <

*x*< –10}

**(iii) {***x*: *x* ∈ R, 0 ≤ *x* < 7}

*x*:

*x*∈ R, 0 ≤

*x*< 7}

**(iv) {***x*: *x* ∈ R, 3 ≤ *x* ≤ 4}

*x*:

*x*∈ R, 3 ≤

*x*≤ 4}

**Solution:**

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

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

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

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

**7. Write the following intervals in set-builder form:**

**(i) (–3, 0)**

**(ii) [6, 12]**

**(iii) (6, 12]**

**(iv) [–23, 5)**

**Solution:**

(i) (–3, 0) = {*x*: *x *∈ R, –3 < *x* < 0}

(ii) [6, 12] = {*x*: *x *∈ R, 6 ≤ *x* ≤ 12}

(iii) (6, 12] ={*x*: *x *∈ R, 6 < *x* ≤ 12}

(iv) [–23, 5) = {*x*: *x *∈ R, –23 ≤ *x* < 5}

**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) 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.

**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 universals 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:**

(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

**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*}

*a*,

*e*,

*i*,

*o*,

*u*} B = {

*a*,

*b*,

*c*}

**(iii) A = {***x*: *x* is a natural number and multiple of 3}

*x*:

*x*is a natural number and multiple of 3}

**B = {***x*: *x* is a natural number less than 6}

*x*:

*x*is a natural number less than 6}

**(iv) A = {***x*: *x* is a natural number and 1 <* x* ≤ 6}

*x*:

*x*is a natural number and 1 <

*x*≤ 6}

**B = {***x*: *x* is a natural number and 6 <* x* < 10}

*x*:

*x*is a natural number and 6 <

*x*< 10}

**(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 = {*a*, *e*, *i*, *o*, *u*} B = {*a*, *b*, *c*}

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

A∪ B = {*a*, *b*, *c*, *e*, *i*, *o*, *u*}

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

B = {*x*: *x* 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 = {*x*: *x* = 1, 2, 4, 5 or a multiple of 3}

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

B = {*x*: *x* 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?

*a*,

*b*}, B = {

*a*,

*b*,

*c*}. Is A ⊂ B? What is A ∪ B?

**Solution:**

It is given that

A = {*a*, *b*} and B = {*a*, *b*, *c*}

Yes, A ⊂ B

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

A∪ B = {*a*, *b*, *c*} = 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

**4. If A = {1, 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 ∪ D**

**(v) A ∪ B ∪ C**

**(vi) A ∪ B ∪ D**

**(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 }

**5. Find the intersection of each pair of sets:**

**(i) X = {1, 3, 5} Y = {1, 2, 3}**

**(ii) A = {***a*, *e*, *i*, *o*, *u*} B = {*a*, *b*, *c*}

*a*,

*e*,

*i*,

*o*,

*u*} B = {

*a*,

*b*,

*c*}

**(iii) A = {***x*: *x* is a natural number and multiple of 3}

*x*:

*x*is a natural number and multiple of 3}

**B = {***x*: *x* is a natural number less than 6}

*x*:

*x*is a natural number less than 6}

**(iv) A = {***x*: *x* is a natural number and 1 <* x* ≤ 6}

*x*:

*x*is a natural number and 1 <

*x*≤ 6}

**B = {***x*: *x* is a natural number and 6 <* x* < 10}

*x*:

*x*is a natural number and 6 <

*x*< 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 = {*a*, *e*, *i*, *o*, *u*}, B = {*a*, *b*, *c*}

So the intersection of the given set can be written as

A ∩ B = {*a*}

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

B = {*x*: *x* 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 = {*x*: *x* is a natural number and 1 < *x* ≤ 6} = {2, 3, 4, 5, 6}

B = {*x*: *x* 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

**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:**

(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}

**7. If A = {***x*:* x* is a natural number}, B ={*x*:* x* is an even natural number}

*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

*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:**

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}

**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}

*x*:

*x*is a natural number and 4 ≤

*x*≤ 6}

**(ii) {***a*, *e*, *i*, *o*, *u*}and {*c*, *d*, *e*, *f*}

*a*,

*e*,

*i*,

*o*,

*u*}and {

*c*,

*d*,

*e*,

*f*}

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

*x*:

*x*is an even integer} and {

*x: x*is an odd integer}

**Solution:**

(i) {1, 2, 3, 4}

{*x*: *x* 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) {*a*, *e*, *i*, *o*, *u*} ∩ (*c*, *d*, *e*, *f*} = {*e*}

Hence, {*a*, *e*, *i*, *o*, *u*} and (*c*, *d*, *e*, *f*} are not disjoint.

(iii) {*x*: *x* is an even integer} ∩ {*x*: *x* is an odd integer} = Φ

Hence, this pair of sets is disjoint.

**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}, 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:**

(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 }.

**10. If X = {***a*, *b*, *c*, *d*} and Y = {*f*, *b*, *d, g*}, find

*a*,

*b*,

*c*,

*d*} and Y = {

*f*,

*b*,

*d, g*}, find

**(i) X – Y**

**(ii) Y – X**

**(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)

**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.

*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 sets.**

**(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* ∈ {*a*, *e*, *i*, *o*, *u*}, *a* ∈ {*a*, *b*, *c*, *d*}

So we get {*a*, *e*, *i*, *o*, *u*} ∩ {*a*, *b*, *c*, *d*} = {*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

**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 U C)’**

**(iv) (A U B)’**

**(v) (A’)’**

**(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 }.

**2. If U = {***a, b, c, d, e, f, g, h*}, find the complements of the following sets:

*a, b, c, d, e, f, g, h*}, find the complements of the following sets:

**(i) A = {***a, b, c*}

*a, b, c*}

**(ii) B = {***d, e, f, g*}

*d, e, f, g*}

**(iii) C = {***a, c, e, g*}

*a, c, e, g*}

**(iv) D = {***f*, *g*, *h*, *a*}

Solution:

*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 }.

**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}

*x*:

*x*is an even natural number}

**(ii) {***x*: *x* is an odd natural number}

*x*:

*x*is an odd natural number}

**(iii) {***x*: *x* is a positive multiple of 3}

*x*:

*x*is a positive multiple of 3}

**(iv) {***x*: *x* is a prime number}

*x*:

*x*is a prime number}

**(v) {***x*: *x* is a natural number divisible by 3 and 5}

*x*:

*x*is a natural number divisible by 3 and 5}

**(vi) {***x*: *x* is a perfect square}

*x*:

*x*is a perfect square}

**(vii) {***x*: *x* is perfect cube}

*x*:

*x*is perfect cube}

**(viii) {***x*: *x* + 5 = 8}

*x*:

*x*+ 5 = 8}

**(ix) {***x*: 2*x* + 5 = 9}

*x*: 2

*x*+ 5 = 9}

**(x) {***x*: *x* ≥ 7}

*x*:

*x*≥ 7}

**(xi) {***x*: *x* ∈ N and 2*x* + 1 > 10}

*x*:

*x*∈ N and 2

*x*+ 1 > 10}

**Solution:**

We know that

U = N: Set of natural numbers

(i) {*x*: *x* is an even natural number}´ = {*x*: *x* is an odd natural number}

(ii) {*x*: *x* is an odd natural number}´ = {*x*: *x* is an even natural number}

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

(iv) {*x*: *x* is a prime number}´ ={*x*: *x* 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) {*x*: *x* is a natural number divisible by 3 and 5}´ = {*x*: *x* is a natural number that is not divisible by 3 or 5}

(vi) {*x*: *x* is a perfect square}´ = {*x*: *x* ∈ N and *x *is not a perfect square}

(vii) {*x*: *x* is a perfect cube}´ = {*x*: *x* ∈ N and *x *is not a perfect cube}

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

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

(x) {*x*: *x* ≥ 7}´ = {*x*: *x* ∈ N and *x* < 7}

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

**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 U B)’ = A’ ∩ B’**

**(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′.

**5. Draw appropriate Venn diagram for each of the following:**

**(i) (A U B)’**

**(ii) A’ ∩ B’**

**(iii) (A ∩ B)’**

**(iv) A’ U 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’ is the set of all equilateral triangles.

**7. Fill in the blanks to make each of the following a true statement:**

**(i) A U A’ = ……..**

**(ii) Φ′ ∩ A = …….**

**(iii) A ∩ A’ = …….**

**(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

**1. Decide, among the following sets, which sets are subsets of one and another:**

**A= {x: x ∈ R and x satisfy x ^{2} – 8x + 12 = 0},**

**B = {2, 4, 6},**

**C = {2, 4, 6, 8…},**

**D = {6}.**

**Solution:**

According to the question,

We have,

A = {x: x ∈ R and x satisfies x^{2} – 8x + 12 =0}

2 and 6 are the only solutions of x^{2} – 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

**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:**

(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

**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:**

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

**15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 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:**

(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 A, B and C = the set of people who like product A, product B and product C respectively.

Now, according to the question,

Number of students who like product A, n (A) = 21

Number of students who like product B, n (B) = 26

Number of students who like product C, n (C) = 29

Number of students who like both products A and B, n (A ∩ B) = 14

Number of students who like both products A and C, n(C ∩ A) = 12

Number of students who like both product C and B, n (B ∩ C) = 14

Number of students who like all three product, n (A ∩ B ∩ C) = 8

From the Venn diagram, we get,

Number of students who only like product C = {29 – (4 + 8 + 6)}

= {29 – 18}

= 11 students