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Table of Contents

## Exercise 1.1

**1. Which of the following are sets? Justify your answer.****(i) The collection of all the months of a year beginning with the letter J**

The collection of all months of a year beginning with letter J is January, June, July which is a well defined collection of objects(words) that’s why it will be a Set.**(ii) The collection of ten most talented writers of India**

Collection of Ten Most Talented writers of India is not a certain list, as everybody in India can be fan of different writers which means if asked to every Indian the list of Most Ten Talented writers would be different for each one of Indian. This means that collection of ten most talented writers of India is not a well defined collection hence it’s not a Set.**(iii) A team of eleven best-cricket batsmen of the world**

What makes a batsmen best is not well defined, so if asked to people from different countries they will come up with different teams which they think contain eleven best-cricket batsmen of the world. As team of eleven best-cricket batsmen of the world is not well defined and depends upon personal choices of people. That’s why it’s not a Set.

**(iv) The collection of all boys in your class**

Collection of all boys in a class is well defined that’s why it’s a Set.**(v) The collection of all natural numbers less than 100**

All natural numbers which are less than 100 are 1, 2, 3, 4, 5 ………. 99 which is a well defined list of numbers. That’s why collection of all natural numbers less than 100 is a Set.**(vi) A collection of novels written by the writer Munshi Prem Chand**

Collection of novels written by the writer Munshi Prem Chand is a well defined list. That’s why its a Set.

One thing to note here is that statement “A collection of novels written by Munshi Prem Chand” is not a Set because this sentence clearly does not state which Munshi Prem Chand. Is it famous Munshi Prem Chand or some common person whose name is Munshi Prem Chand. So statement “A collection of novels written by Munshi Prem Chand” is not well defined hence it will not be a Set.*A collection of novels written by the writer Munshi Prem Chand* – is a Set

*A collection of novels written by Munshi Prem Chand*– is not a Set

**(vii) The collection of all even integers**

The collection of all even integers is 2, 4, 6, 8, 10, 12, ……………. which is well defined hence it’s a Set**(viii) The collection of questions in this Chapter**

Collection of questions in a chapter is well defined that’s why it’s a Set**(ix) A collection of most dangerous animals of the world**

Collection of most dangerous animals of the world is not well defined, that’s why it’s not a Set

**2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces:****(i) 5. . .A**

As 5 is in Set A = {1, 2, 3, 4, 5, 6} that’s *5 ∈ A***(ii) 8 . . . A**

As 8 is not in Set A = {1, 2, 3, 4, 5, 6} that’s *5 ∉ A***(iii) 0. . .A**

As 0 is not in Set A = {1, 2, 3, 4, 5, 6} that’s *0 ∉ A***(iv) 4. . . A**

As 4 is in Set A = {1, 2, 3, 4, 5, 6} that’s *4 ∈ A***(v) 2. . .A**

As 2 is in Set A = {1, 2, 3, 4, 5, 6} that’s *2 ∈ A***(vi) 10. . .A**

As 10 is not in Set A = {1, 2, 3, 4, 5, 6} that’s *10 ∉ A*

**3. Write the following sets in roster form: ****(i) A = {x : x is an integer and –3 ≤ x < 7}**

Set A contains x such that x is an integer and it lies between -3 and 7 (-3 included but 7 excluded).

Hence *A = {-3 , -2, -1, 0, 1, 2, 3, 4, 5, 6}*

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

Set B contains x such that x is a natural number less than 6

Hence *B = {1, 2, 3, 4, 5}*

(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}

Set C contains x such that x is a two-digit natural number such that the sum of its digits is 8

Hence *C = {17, 26, 35, 44, 53, 62, 71}*

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

Set D contains x such that x is a prime number which is divisor of 60

Hence *D = {2, 3, 5}*

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

There are some repeated letters in word TRIGONOMETRY but upon forming a set only unique letters should be included.

Hence *E = {T, R, I, G, O, N, M, E, Y}*

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

There are some repeated letters in word BETTER but upon forming a set only unique letters should be included.

Hecen *F = {B, E, T, R}*

**4. Write the following sets in the set-builder form****(i) {3, 6, 9, 12} **

So common thing amongst 3, 6, 9, 12 is that all of these are multiples of 3 but less than 15

Hence {3, 6, 9, 12} in Set Builder Form is *{x | x is a natural number multiple of 3 and x < 15}***(ii) {2,4,8,16,32} **

So common thing amongst 2, 4, 6, 8, 16, 32 is that all of these are powers of 2 but less than and equal to 32 which is 2^{5}

Hence {2, 4, 6, 8, 16, 32} in Set Builder Form is *{x | x = 2 ^{n} where n ∈ N and n ≤ 5}*

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

So common thing amongst 5, 25, 125, 625 is that all of these are powers of 5 but less than and equal to 625 which is 5

^{4}

Hence {5, 25, 125, 625} in Set Builder Form is

*{x | 5*

^{n}where n ∈ N and n ≤ 4}**(iv) {2, 4, 6, . . .}**

So common thing amongst 2, 4, 6, ……. is that all of these numbers are multiples of 2 and even it’s a set of even number.

Hence {2, 4, 6, …. } in Set Builder Form is

*{x | x is an even natural numbers}*

**(v) {1,4,9, . . .,100}**

So common thing amongst 1, 4, 9, ….. , 100 is that all of these numbers are squares of natural numbers up till square of 10.

Hence {1, 4, 9, …, 100} in Set Builder Form is

*{x | x = n2 where n ∈ N and n ≤ 10}*

**5. List all the elements of the following sets: (i) A = {x : x is an odd natural number} **

x is an odd natural number means

*{1, 3, 5, 7, 9, …….}*

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

x is an integer – 1/2 < x < 9/2 means

*{0, 1, 2, 3, 4}*

**(iii) C = {x : x is an integer, x**

^{2}≤ 4}x is an integer x

^{2}≤ 4 means

*{-2, -1, 0, 1, 2}*

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

x is a letter in the word “LOYAL” means

*{L, O, Y, A}*

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

x is a month of a year not having 31 days means

*{February, April, June, September, November}*

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

x is a consonant in the English alphabet which preceded k means

*{b, e, d, f, g, h, j}*

**6. Match each of the set on the left in the roster form with the same set on the rightdescribed in set-builder form:**

(i) {1, 2, 3, 6} | (a) {x : x is a prime number and a divisor of 6} |

(ii) {2, 3} | (b) {x : x is an odd natural number less than 10} |

(iii) {M,A,T,H,E,I,C,S} | (c) {x : x is natural number and divisor of 6} |

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

**(i) {1, 2, 3, 6} → (c) {x : x is natural number and divisor of 6}(ii) {2, 3} → (a) {x : x is a prime number and a divisor of 6}(iii) {M,A,T,H,E,I,C,S} → (d) {x : x is a letter of the word MATHEMATICS}(iv) {1, 3, 5, 7, 9} → (b) {x : x is an odd natural number less than 10}**

## Exercise 1.2

**1. Which of the following are examples of the null set** **(i) Set of odd natural numbers divisible by 2**

Set of odd natural numbers which can be divided by 2 is a null set as there doesn’t exist any odd natural number divisible by 2.** ****(ii) Set of even prime numbers**

Set of even prime numbers is *{2}***(iii) {x : x is a natural numbers, x < 5 and x > 7}**

x is a natural number where x < 5 and x > 7 that’s impossible.

Hence its a null set.**(iv) {y : y is a point common to any two parallel lines}**

As two parallel lines never intersect with each other that’s why there doesn’t exist any point y which is common to any two parallel lines.

**2. Which of the following sets are finite or infinite****(i) The set of months of a year**

Set of months of a year is {Jan, Feb, March, April, June, July, Aug, Sep, Oct, Nov, Dec} which contains just 12 different names of months in a year.

Hence it’s a Finite Set.**(ii) {1, 2, 3, . . .}**

Set {1, 2, 3, ….. } doesn’t have any ending which means it continues up til infinity.

Hence it’s an Infinite Set.**(iii) {1, 2, 3, . . .99, 100}**

Set {1, 2, 3, ….. 99, 100} continues up til only 100. That’s why it’s a Finite Set.**(iv) The set of positive integers greater than 100**

Set of positive integers which are greater than 100 continues up to infinity, so doesn’t have a definite end. Hence it’s an Infinite Set.**(v) The set of prime numbers less than 99**

Set of prime numbers which are less than 99 is a well defined set containing numbers 2,3, 5, 7, 11, ……… 89, 97 that’s why its a Finite Set.

**3. State whether each of the following set is finite or infinite:****(i) The set of lines which are parallel to the x-axis**

Set of lines which are parallel to x-axis is Infinite Set as there can be many many lines which are parallel to x-axis.**(ii) The set of letters in the English alphabet**

Set of letters in the English alphabet contains 26 letters only {a, b, c, ………., w, x, y, z}. Hence it’s a Finite Set.**(iii) The set of numbers which are multiple of 5**

Set of numbers which are multiple of 5 is an Infinite Set because there can be infinite multiples of 5 {5, 10, 15, 20, 25, ………………}**(iv) The set of animals living on the earth**

Set of animals living on the earth is a Finite Set as the animals living on the earth is countable.**(v) The set of circles passing through the origin (0,0)**

Set of circles passing through the origin (0, 0) is an Infinite Set as there can be many circles which passed through origin (0, 0)

**4. In the following, state whether A = B or not:****(i) A = { a, b, c, d } B = { d, c, b, a }**

As Sets A and B contain same elements that’s why A = B**(ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18 }**

As Set A contain 12 but Set B does not hence A **≠** B**(iii) A = {2, 4, 6, 8, 10 } B = { x : x is positive even integer and x ≤ 10}**

Set B can be rewritten as {2, 4, 6, 8, 10}

So we have B = {2, 4, 6, 8, 10} and A = {2, 4, 6, 8, 10}

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

Set A can be rewritten as {10, 20, 30, 40, 50} and B = {10, 15, 20, 25, 30, ……}

It’s clear that there are some values in Set A which are not in Set B

Hence A **≠** B

**5. Are the following pair of sets equal ? Give reasons.****(i) A = {2, 3}, B = {x : x is solution of x ^{2} + 5x + 6 = 0}**

In order to check is Set A and Set B are equal or not firstly let’s simplify set B

On simplifying Quadratic Equation x

^{2}+ 5x + 6 = 0 have solutions – 3 and -2

Hence Set B = {-3, – 2}

But Set A = {2, 3}

Hence A

**≠**B

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

For checking is Set A and Set B are equal, let’s first simplify both of these sets.

Set A = {F, O, L, W}

Set B = {W, O, L, F}

As both sets have same letters hence 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}**

Sets B = D = {1, 2, 3, 4}

Sets E = G = {-1, 1}

## Exercise 1.3

**1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:****(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 } **

{ 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}**

As every student in Class XI is also a student of the school.

Hence

{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}**

As every Equilateral Triangle in a plane is a Triangle

Hence

{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}**

As every even natural number is also an integer

Hence

{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 }**

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

{ x : x is a vowel in the English alphabet} = {a, e, i, o, u}

As a, e are also in Set {a, e, i, o, u}

Hence

{ a, e } ⊂ {a, e, i, o, u}

So { a, e } ⊂ { x : x is a vowel in the English alphabet} is *True***(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }**

False**(iv) { a } ⊂ { a, b, c }**

True**(v) { a } ∈ { a, b, c }**

True**(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} = {2, 4}

{ x : x is a natural number which divides 36} = {2, 4, 6}

As {2, 4} is a subset of {2, 4, 6}

Hence

{ x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36} is TRUE

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

As 3, 4 individually is not in set A.

Hence

{3, 4} ⊂ A is incorrect**(ii) {3, 4} ∈ A **

As {3, 4} as a whole in set A

Hence

{3, 4} ∈ A is correct**(iii) {{3, 4}} ⊂ A**

As {3, 4} as a whole is in set A

Hence

{{3, 4}} ⊂ A is correct**(iv) 1 ∈ A**

As 1 individually is in set A

Hence

1 ∈ A is correct**(v) 1 ⊂ A**

1 not a subset of A

Hence

1 ⊂ A is incorrect**(vi) {1, 2, 5} ⊂ A**

As 1, 2, 5 are in set A

Hence

{1, 2, 5} ⊂ A is correct**(vii) {1, 2, 5} ∈ A **

{1, 2, 5} as a whole is not in set A

Hence

{1, 2, 5} ∈ A is incorrect**(viii) {1, 2, 3} ⊂ A**

As 3 individually is not in set A

Hence

{1, 2, 3} ⊂ A is incorrect**(ix) φ ∈ A**

As φ is not in set A

Hence

φ ∈ A is incorrect**(x) φ ⊂ A**

It’s a general rule that φ is subset of every possible set

Hence φ is subset of A

So φ ⊂ A is correct**(xi) {φ} ⊂ A**

As {φ} is not in set A

Hence

{φ} ⊂ A is incorrect

**4. Write down all the subsets of the following sets****(i) {a}**

Subsets of {a} are *φ, {a}***(ii) {a, b} **

Subsets of {a, b} are *φ, {a}, {b}, {a, b}***(iii) {1, 2, 3}**

Subsets of {1, 2, 3} are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1,2,3}**(iv) φ**

Subset of φ is just φ itself

**5. How many elements has P(A), if A = φ?**

We know that if A is a set with m elements, then P(A) represents the power set of A which means a set containing all subsets of A.

If A = Φ, then n (A) = 0 (as Φ means a null set).

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

Hence, P(A) has **one element** which is Φ

**6. Write the following as intervals: ****(i) {x : x ∈ R, – 4 < x ≤ 6}**

(- 4, 6]**(ii) {x : x ∈ R, – 12 < x < –10}**

(- 12, – 10)**(iii) {x : x ∈ R, 0 ≤ x < 7}**

[0, 7)**(iv) {x : x ∈ R, 3 ≤ x ≤ 4}**

[3, 4]

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

Set of Triangles**(ii) The set of isosceles triangles**

Set of Triangles

**9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C****(i) {0, 1, 2, 3, 4, 5, 6}**

No**(ii) φ**

No**(iii) {0,1,2,3,4,5,6,7,8,9,10}**

Yes**(iv) {1,2,3,4,5,6,7,8}**

No

## Exercise 1.4

**1. Find the union of each of the following pairs of sets:** **(i) X = {1, 3, 5} Y = {1, 2, 3}**

X ∪ Y = {1, 2, 3, 5}**(ii) A = [ a, e, i, o, u} B = {a, b, c}**

A ∪ B = {a, b, c, e, i, o, u}**(iii) A = {x : x is a natural number and multiple of 3} B = {x : x is a natural number less than 6}**

Upon simplifying sets A and B can be rewritten as

A = {3, 6, 9, 12, 15, ………}

B = {1, 2, 3, 4, 5}

A ∪ B = {1, 2, 3, 4, 5, 6, 9, 12, 15, ………..}**(iv) A = {x : x is a natural number and 1 < x ≤6 } B = {x : x is a natural number and 6 < x < 10 } **

Upon simplifying sets A and B can be rewritten as

A = {2, 3, 4, 5, 6}

B = {7, 8, 9}

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}**(v) A = {1, 2, 3}, B = φ**

A ∪ B = {1, 2, 3} ∪ φ = {1, 2, 3}

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

Given

A = {a, b}

B = {a, b, c}

Yes A ⊂ B as every element in set A is also in Set B

A ∪ B = {a, b} ∪ {a, b, c} = {a, b, c}

A ∪ B = {a, b, c}

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

Because A is subset of B

Every element of A is an element of 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 **

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

A ∪ B = {1, 2, 3, 4, 5, 6}

**(ii) A ∪ C**

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

A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

**(iii) B ∪ C**

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

B ∪ C = {3, 4, 5, 6, 7, 8}

**(iv) B ∪ D**

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

B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

**(v) A ∪ B ∪ C**

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

A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

**(vi) A ∪ B ∪ D**

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

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

**(vii) B ∪ C ∪ D**

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

B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

**5. 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 **

A ∩ B = {1, 2, 3, 4} ∩ {3, 4, 5, 6} = {3, 4}

A ∩ B = {3, 4}

**(ii) A ∩ C**

A

**∩**C = {1, 2, 3, 4}

**∩**{5, 6, 7, 8} = φ

A

**∩**C = φ

**(iii) B ∩ C**

B

**∩**C = {3, 4, 5, 6}

**∩**{5, 6, 7, 8} = {5, 6}

B

**∩**C = {5, 6}

**(iv) B ∩ D**

B

**∩**D = {3, 4, 5, 6}

**∩**{7, 8, 9, 10} = φ

B

**∩**D = φ

**(v) A ∩ B ∩ C**

A

**∩**B

**∩**C = {1, 2, 3, 4}

**∩**{3, 4, 5, 6}

**∩**{5, 6, 7, 8} = φ

A

**∩**B

**∩**C = φ

**(vi) A ∩ B ∩ D**

A

**∩**B

**∩**D = {1, 2, 3, 4}

**∩**{3, 4, 5, 6}

**∩**{7, 8, 9, 10} = φ

A

**∩**B

**∩**D = φ

**(vii) B ∩ C ∩ D**

B

**∩**C

**∩**D = {3, 4, 5, 6}

**∩**{5, 6, 7, 8}

**∩**{7, 8, 9, 10} = φ

B

**∩**C

**∩**D = 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 **

A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}

A ∩ B = {7, 9, 11}**(ii) B ∩ C **

B ∩ C = {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}

B ∩ C = {11, 13}**(iii) A ∩ C ∩ D**

A ∩ C ∩ D = {3, 5, 7, 9, 11} ∩ {11, 13, 15} ∩ {15, 17} = φ

A ∩ C ∩ D = φ**(iv) A ∩ C**

A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}

A ∩ C = {11}**(v) B ∩ D**

B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = φ

B ∩ D = φ

(v**i) A ∩ (B ∪ C)**

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

A ∩ (B ∪ C) = {7, 9, 11}**(vii) A ∩ D**

A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17} = φ

A ∩ D = φ**(viii) A ∩ (B ∪ D) **

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

A ∩ (B ∪ D) = {7, 9, 11}**(ix) (A ∩ B) ∩ (B ∪ C)**

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

= {7, 9, 11} ∩ {7, 9, 11, 13, 15}

= {7, 9, 11}

(A ∩ B) ∩ (B ∪ C) = {7, 9, 11}

(**x) (A ∪ D) ∩ ( B ∪ C)**

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

= {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

(A ∪ D) ∩ (B ∪ C) = {7, 9, 11, 15}

**7. If A = {x : x is a natural number}B = {x : x is an even natural number}C = {x : x is an odd natural number}D = {x : x is a prime number}Then Find**

**(i) A ∩ B**

A ∩ B = {x : x is a natural number} ∩ {x : x is an even natural number}

A ∩ B = {x : x is an even natural number}

**(ii) A ∩ C**

A ∩ C = {x : x is a natural number} ∩ {x : x is an odd natural number}

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

**(iii) A ∩ D**

A ∩ D = {x : x is a natural number} ∩ {x : x is a prime number}

A ∩ D = {x : x is a prime number}

**(iv) B ∩ C**

B ∩ C = {x : x is an even natural number} ∩ {x : x is an odd natural number}

B ∩ C = φ

**(v) B ∩ D**

B ∩ D = {x : x is an even natural number} ∩ {x : x is a prime number}

B ∩ D = {2}

**(vi) C ∩ D**

C ∩ D = {x : x is an odd natural number} ∩ {x : x is a prime number}

C ∩ D = {2}

**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}**

Let’s first simplify {x : x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

So {1, 2, 3, 4} and {4, 5, 6} have nothing in common hence these are disjoint**(ii) {a, e, i, o, u} and {c, d, e, f}**

Sets {a, e, i, o, u} and {c, d, e, f} have element *e* in common hence these two sets are not disjoint**(iii) {x : x is an even integer} and {x : x is an odd integer}**

Sets {x : x is an even integer} and {x : x is an odd integer} does not have anything in common that’s why these two disjoint sets.

**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**

A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}

A – B = {3, 6, 9, 15, 18, 21}

**(ii) A – C**

A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16}

A – C = {3, 9, 15, 18, 21}

**(iii) A – D**

A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}

A – D = {3, 6, 9, 12, 18, 21}

**(iv) B – A**

B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}

B – A = {4, 8, 16, 20}

**(v) C – A**

C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}

C – A = {2, 4, 8, 10, 14, 16}

**(vi) D – A**

D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21}

D – A = {5, 10, 20}

**(vii) B – C**

B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16}

B – C = {20}

**(viii) B – D**

B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}

B – D = {4, 8, 12, 16}

**(ix) C – B**

C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}

C – B = {2, 6, 10, 14}

**(x) D – B**

D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}

D – B = {5, 10, 15}

**(xi) C – D**

C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}

C – D = {2, 4, 6, 8, 12, 14, 16}

**(xii) D – C**

D – C = {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}

D – C = {5, 15, 20}

**10. If X= {a, b, c, d} and Y = {f, b, d, g} Find(i) X – Y**

X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}

X – Y = {a, c}

**(ii) Y – X**

Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}

Y – X = {f, g}

**(iii) X ∩ Y**

X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}

X ∩ Y = {b, d}

**11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?**

R = Set of Real Numbers = Set of Rational and Irrational Numbers

Q = Set of Rational Numbers

R – Q = Set of Rational and Irrational Numbers – Set of Rational Numbers*R – Q = Set of Irrational Numbers*

**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**

False**(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets**

False**(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets**

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

False

## Exercise 1.5

**1. Let 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}Find**

**(i) A′**

A′ = {1, 2, 3, 4}’ = {5, 6, 7, 8, 9}

A’ = {5, 6, 7, 8, 9}

**(ii) B′**

B′ = {2, 4, 6, 8}’ = {1, 3, 5, 7, 9}

B′ = {1, 3, 5, 7, 9}

**(iii) (A ∪ C)′**

(A ∪ C)′ = ({1, 2, 3, 4} ∪ {3, 4, 5, 6})’ = {1, 2, 3, 4, 5, 6}’ = {7, 8, 9}

(A ∪ C)′ = {7, 8, 9}

**(iv) (A ∪ B)′**

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

(A ∪ B)′ = {5, 7, 9}

**(v) (A′)′**

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

(A′)′ = A = {1, 2, 3, 4}

**(vi) (B – C)′**

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

(B – C)′ = {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:**

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

Complement of A = A′ = {a, b, c}′ = {d, e, f, g, h}

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

Complement of B = B′ = {d, e, f, g}′ = {a, b, c, h}

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

Complement of C = C′ = {a, c, e, g}′ = {b, d, f, h}

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

Complement of D = D′ = {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 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 is not a positive multiple of 3} **(iv) {x : x is a prime number}**

{x : x is a positive composite number or x = 1}**(v) {x : x is a natural number divisible by 3 and 5}**

{x : x is neither a multiple of 3 nor multiple of 5}**(vi) {x : x is a perfect square} **

{x : x is a Natural Number and its not a perfect square}**(vii) {x : x is a perfect cube}**

{x : x is a Natural Number and its not a perfect cube}**(viii) {x : x + 5 = 8} **

{x : x is a Natural Number and x is not equal to 3}**(ix) {x : 2x + 5 = 9}**

{x : x is a Natural Number and x is not equal to 2}**(x) {x : x ≥ 7}**

{x : x is a Natural Number and it’s less than 7} **(xi) {x : x ∈ N and 2x + 1 > 10}**

{x : x is a Natural Number and it’s greater than 4}

**4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}A = {2, 4, 6, 8}B = {2, 3, 5, 7}Verify that**

**(i) (A ∪ B)′ = A′ ∩ B′**

Let’s solve both sides of this equation and then compare

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

A′ ∩ B′ = {2, 4, 6, 8}′ ∩ {2, 3, 5, 7}′ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}

Hence (A ∪ B)′ = A′ ∩ B′ = {1, 9}

**(ii) (A ∩ B)′ = A′ ∪ B′**

(A ∩ B)′ = ({2, 4, 6, 8} ∩ {2, 3, 5, 7})′ = {2}′ = {1, 3, 4, 5, 6, 7, 8, 9}

A′ ∪ B′ = {2, 4, 6, 8}′ ∪ {2, 3, 5, 7}′ = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}

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

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

**(i) (A ∪ B)′**

**(ii) A′ ∩ B′**

**(iii) (A ∩ B)′**

**(iv) A′ ∪ 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′?**

U = {Set of all triangles in a plane}

A = {Set of all triangles in a plane with at least one angle different from 60°}

A′ = U – A = {Set of all Equilateral Triangles}

**7. Fill in the blanks to make each of the following a true statement:****(i) A ∪ A′ = . . .**

A ∪ A′ = U**(ii) φ′ ∩ A = . . .**

φ′ ∩ A = U ∩ A = A**(iii) A ∩ A′ = . . .**

A ∩ A′ = φ**(iv) U′ ∩ A = . . .**

U′ ∩ A = φ ∩ A = φ

## Exercise 1.6

**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)**

We know that

n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)

Replacing given values in this formula

38 + n(X ∩ Y) = 17 + 23

38 + n(X ∩ Y) = 40

n(X ∩ Y) = 40 – 38

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?**

We know that

n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)

Replacing given values in this formula

18 + n(X ∩ Y) = 8 + 15

18 + n(X ∩ Y) = 23

n(X ∩ Y) = 23 – 18

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?**

Suppose that

X = Set of Hindi speaking people

Y = Set of English speaking people

X ∪ Y = Set of total people

As per question

n(X) = 250

n(Y) = 200

n(X ∪ Y) = 400

We know that

n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)

400 + n(X ∩ Y) = 250 + 200

400 + n(X ∩ Y) = 450

n(X ∩ Y) = 450 – 400 = 50

n(X ∩ Y) = 50

Hence there are 50 people in the group who can speak both English and Hindi.

**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?**

As per question

n(S) = 21

n(T) = 32

n(S ∩ T) = 11

We know that

n(S ∪ T) + n(S ∩ T) = n(S) + n(T)

n(S ∪ T) + 11 = 21 + 32

n(S ∪ T) + 11 = 53

n(S ∪ T) = 53 – 11 = 42

Hence number of elements in S ∪ T is 42

**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?**

As per question

n(X) = 40

n(X ∪ Y) = 60

n(X ∩ Y) = 10

We know that

n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)

60 + 10 = 40 + n(Y)

70 = 40 + n(Y)

70 – 40 = n(Y)

30 = n(Y)

Hence number of elements in Y are 30

**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?**

Let’s suppose that

C = Set of people who like coffee

T = Set of people who like Tea

C ∪ T = Set of people who likes at least one of two drinks = Set of all people in group

n(C) = 37

n(T) = 52

n(C ∪ T) = 70

We know that

n(C ∪ T) + n(C ∩ T) = n(C) + n(T)

70 + n(C ∩ T) = 37 + 52

70 + n(C ∩ T) = 89

n(C ∩ T) = 89 – 70 = 19

n(C ∩ T) = 19

So out of 70 people in the group, there are 19 people who both likes 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?**

Let’s suppose that

C = Set of people who like cricket

T = Set of people who like tennis

C ∪ T = Set of people in the group

C ∩ T = Set of people who like cricket and tennis

As per question

n(C) = 40

n(C ∪ T) = 65

n(C ∩ T) = 10

We know that

n(C ∪ T) + n(C ∩ T) = n(C) + n(T)

65 + 10 = 40 + n(T)

75 = 40 + n(T)

75 – 40 = n(T)

35 = n(T)

So Number of people who like Tennis are 35

Let’s now find out Number of people who like Tennis but not Cricket

In terms of Sets we need to find out n(T ∩ C’)

We know that

n(A ∩ B’) = n(A) – n(A ∩ B) where A, B are two given sets

Similarly

n(T ∩ C’) = n(T) – n(T ∩ C)

Putting in the values

n(T ∩ C’) = 45 – 10 = 35

So number of people who only like Tennis but not Cricket are 35

**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?**

Suppose that

F = Set of people who speak French

S = Set of people who speak Spanish

F ∩ S = Set of people who speak both French and Spanish

As per question

n(F) = 50

n(S) = 20

n(F ∩ S) = 10

We know that

n(F ∪ S) + n(F ∩ S) = n(F) + n(S)

n(F ∪ S) + 10 = 50 + 20 = 70

n(F ∪ S) + 10 = 70

n(F ∪ S) = 70 – 10 = 60

n(F ∪ S) = 60

So number of people who speak at least one of French or Spanish are 60

## Miscellaneous Exercise on Chapter 1

**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}**

Let’s first simplify these sets and then find out which set is subset of another given set.

A = {x : x ∈ R and x satisfy x

^{2}– 8x + 12 = 0}

x

^{2}– 8x + 12 = 0 can be rewritten as (x – 2)(x – 6) = 0 implies x = 2, 6

Hence

A = {2, 6}

B = {2, 4, 6}

C = {2, 4, 6, 8, ……}

D = {6}

As elements of set A are also in sets B and C hence

**A ⊂ B**and

**A ⊂ C**

As elements of set B are also in set C hence

**B ⊂ C**

As elements of set D are also in sets A, B, C hence

**D ⊂ A**,

**D ⊂ B**and

**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**

False**(ii) If A ⊂ B and B ∈ C , then A ∈ C**

False**(iii) If A ⊂ B and B ⊂ C , then A ⊂ C**

True**(iv) If A ⊄ B and B ⊄ C , then A ⊄ C**

False**(v) If x ∈ A and A ⊄ B , then x ∈ B**

False**(vi) If A ⊂ B and x ∉ B , then x ∉ A**

True

**3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C**

**4. Show that the following four conditions are equivalent:****(i) A ⊂ B**

This means A is a subset of B hence all elements of A are in B**(ii) A – B = φ**

This means all elements of A are in B**(iii) A ∪ B = B**

This means that all elements of A are in B**(iv) A ∩ B = A**

This means that all elements of A are in B

Hence all of statements A ⊂ B, A – B = φ, A ∪ B = B, A ∩ B = A are same because all elements of A are in B.

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

Let’s suppose that

x ∈ C – B

=> x ∈ C but x ∉ B

As A ⊂ B hence above equation can be rewritten as x ∈ C but x ∉ A

Using equations

=> x ∈ C but x ∉ A means that x ∈ C – A

Hence we started by assuming that x ∈ C – B and proved that x ∈ C – A

That’s why **C – B ⊂ C – A**

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

Let’s suppose that

x ∈ A

=> {x} ∈ P(A)

As P(A) = P(B)

Hence {x} ∈ P(B) => x ∈ B

So x ∈ A => x ∈ B which means A ⊂ B

Now let’s suppose that

x ∈ B

=> {x} ∈ P(B)

As P(A) = P(B)

Hence {x} ∈ P(A) => x ∈ A

So x ∈ B => x ∈ A which means B ⊂ A

Hence A ⊂ B and B ⊂ A which means A = B

Thus if P(A) = P(B) then A will be equal to B

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

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

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

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

Let’s suppose that

A = {1, 2, 3, 4}

B = {2, 3, 4, 5}

C = {2, 3, 4, 7}

=> A ∩ B = {2, 3, 4}

=> A ∩ C = {2, 3, 4}

This means that *A ∩ B = A ∩ C = {2, 3, 4}* but 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**

**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**

**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?**

Let’s suppose that

C = Set of students taking coffee

T = Set of students taking tea

C ∩ T = Set of students who take both coffee and tea

As per question

n(C) = 225

n(T) = 150

n(C ∩ T) = 100

We know that

n(C ∪ T) + n(C ∩ T) = n(C) + n(T)

n(C ∪ T) + 100 = 225 + 150 = 375

n(C ∪ T) + 100 = 375

n(C ∪ T) = 375 – 100 = 275*n(C ∪ T) = 275*

But total number of students in the group = 600

Number of students who take neither coffee nor tea = Total number of students in the group – n(C ∪ T)

= 600 – 275 = 325*Hence number of students who were taking neither tea nor coffee is 325*

**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?**

H = Set of people who know Hindi

E = Set of people who know English

H ∩ E = Set of people who know both Hindi and English

As per question

n(H) = 100

n(E) = 50

n(H ∩ E) = 25

We know that

n(H ∪ E) + n(H ∩ E) = n(H) + n(E)

n(H ∪ E) + 25 = 100 + 50

n(H ∪ E) + 25 = 150

n(H ∪ E) = 150 – 25 = 125*n(H ∪ E) = 125**Hence total number of students in the group = 125*

**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.**

**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.**