# CLASS 11 MATHS CHAPTER-1 SET

### Exercise 1.1

Question1. Which of the following are sets ? Justify your asnwer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter.
(ix) A collection of most dangerous animals of the world.

Solution :
(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identify a month that belongs to this collection.
Hence, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection because the criteria for determining a writer’s talent may vary from person to person.
Hence, this collection is not a set.

(iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person.
Hence, this collection is not a set.

(iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection.
Hence, this collection is a set.

(v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection.
Hence, this collection is a set.

(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection.
Hence, this collection is a set.

(vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection.
Hence, this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter.
Hence, this collection is a set.

(ix) The collection of most dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person.
Hence, this collection is not a set.

Question2. 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

Question3. Write the following sets in roster form:
(i) A = {x : x is an integer and –3 < x < 7}
(ii) B = {x : x is a natural number less than 6}
(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}
(iv) D = {x : x is a prime number which is divisor of 60}
( v) E = The set of all letters in the word
(vi) F = The set of all letters in the word

Solution :
(i) A = {x: x is an integer and –3 < x < 7}
The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.
Therefore, the given set can be written in roster form as
A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x: x is a natural number less than 6}
The elements of this set are 1, 2, 3, 4, and 5 only.
Therefore, the given set can be written in roster form as
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}
The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.
Therefore, this set can be written in roster form as
C = {17, 26, 35, 44, 53, 62, 71, 80}

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

∴60 = 2 × 2 × 3 × 5
The elements of this set are 2, 3, and 5 only.
Therefore, this set can be written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY
There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.
Therefore, this set can be written in roster form as
E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER
There are 6 letters in the word BETTER, out of which letters E and T are repeated.
Therefore, this set can be written in roster form as
F = {B, E, T, R}

Question4. 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} = {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}

(ii) {2, 4, 8, 16, 32}
It can be seen that 2 = 21, 4 = 22, 8 = 23, 16 = 24, and 32 = 25.
∴ {2, 4, 8, 16, 32} = {x: x = 2n, n∈ N and 1 ≤ n ≤ 5}

(iii) {5, 25, 125, 625}
It can be seen that 5 = 51, 25 = 52, 125 = 53, and 625 = 54.
∴ {5, 25, 125, 625} = {x: x = 5n, n∈N and 1 ≤ n ≤ 4}

(iv) {2, 4, 6 …}
It is a set of all even natural numbers.
∴ {2, 4, 6 …} = {x: x is an even natural number}

(v) {1, 4, 9 … 100}
It can be seen that 1 = 12, 4 = 22, 9 = 32 …100 = 102.
∴ {1, 4, 9… 100} = {x: x = n2, n∈N and 1 ≤ n ≤ 10}

Question5. List all the elements of the following sets:
(i) A = { x : x is an odd natural number}

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

(iii) C = { x : x is an integer, x² ≤ 4}

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

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

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

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

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

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

we know that -1/2 = -0.5 and 9/2 = 4.5

B = {0, 1, 2, 3, 4}

(iii) C = { x : x is an integer, x² ≤ 4}}

x² ≤ 4

(-2)² = 4 ≤ 4

(-1)² = 1 ≤ 4

(0)² = 0 ≤ 4

(1)² = 1 ≤ 4

(2)² = 4 ≤ 4

∴ the set  C = {x 😡 is an integer ;x² ≤ 4} contains elements such as

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

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

so the element are D = {L, O, Y, A}

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

so the element are E = {February, April, June, September, November}

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

so the element are F = {B, C, D, F, G, H, J}

Question6. Match each of the set on the left in the roster form with the same set on the right
described in set-builder form:
(i) {1, 2, 3, 6}…….. (a) {x : x is a prime number and a divisor of 6}
(ii) {2, 3}……… (b) {x : x is an odd natural number less than 10}
(iii) {M ,A ,T ,H ,E ,I, C, S}…….. (c) {x : x is natural number and divisor of 6}
(iv) {1, 3, 5, 7, 9}……. (d) {x : x is a letter of the word MATHEMATICS}

Solution :
The sets which are in set-builder form can be written as
(a) { x : x is a prime number and a divisor of 6} = {2, 3}

(b) { x : x is an odd natural number less than 10} = {1, 3, 5, 7, 9}

(c) { x : x is a natural number and divisor of 6} = {1, 2, 3, 6}

(d) { x : x is a letter of word “ MATHEMATICS ”} = {M, A, T, H, E, I, C, S}

Hence the correct matching is:
(i) ⇒(c)
(ii) ⇒(a)
(iii) ⇒(d)
(iv) ⇒(b)

### Exercise 1.2

Question1. Which of the following are examples of the null set:

(i) Set of odd natural numbers divisible by 2.

(ii) Set of even prime numbers.

(iii) {x : x  is a natural number, x < 5 and x > 7 }

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

Solution :
(i) Set of odd natural numbers divisible by 2 is an empty set because odd natural numbers are not divisible by 2.

(ii) Set of even prime numbers is {2} which is not empty set.

(iii) { x : x is a natural number, x< 5 and x > 7 } is an empty set because there is no natural number which satisfies simultaneously x < 5 and  x > 7.

(iv) { y : y  is a point common to any two parallel lines} is an empty set because two parallel lines do not have a common point.

Question2. Which of the following sets are finite or infinite:

(i) The set of months of a year.

(ii) {1, 2, 3, ………..}

(iii) {1, 2, 3, ………….., 99, 100}

(iv) The set of positive integers greater than 100.

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

Solution :
(i) The set of months of a year is finite set because there are 12 months in a year.

(ii) {1, 2, 3, ………..} is an infinite set because there are infinite elements in the set.

(iii) {1, 2, 3, ………….., 99, 100} is a finite set because the set contains finite number of elements.

(iv) The set of positive integers greater than 100 is an infinite set because there are infinite number of positive integers greater than 100.

(v) The set of prime numbers less than 99 is a finite set because the set contains finite number of elements.

Question3. State whether each of the following sets is finite or infinite:

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

(ii) The set of letters in the English alphabet.

(iii) The set of numbers which are multiple of 5.

(iv) The set of animals living on the earth.

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

Solution :
(i) The set of lines which are parallel to the x-axis is an infinite set because we can draw infinite number of lines parallel to x-axis.

(ii) The set of letters in the English alphabet is a finite set because there are 26 letters in the English alphabet.

(iii) The set of numbers which are multiple of 5 is an infinite set because there are infinite multiples of 5.

(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is every large but finite.

(v) The set of circles passing through the origin (0, 0) is an infinite set because we can draw infinite number of circles through origin in different radii.

Question4. In the following, state whether A = B or not:
(i) A = { a, b, c, d },……… B = { d, c, b, a }
(ii) A = { 4, 8, 12, 16 },…….. B = { 8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10},……… B = { x : x is positive even integer and x ≤ 10}
(iv) A = { x : x is a multiple of 10},……… B = { 10, 15, 20, 25, 30, . . . }

Solution :
(i) A = {a, b, c, d}; B = {d, c, b, a}
The order in which the elements of a set are listed is not significant.
∴A = B

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
It can be seen that 12 ∈ A but 12 ∉ B.
∴A ≠ B

(iii) A = {2, 4, 6, 8, 10}
B = {x: x is a positive even integer and x ≤ 10}
= {2, 4, 6, 8, 10}
∴A = B

(iv) A = {x: x is a multiple of 10}
B = {10, 15, 20, 25, 30 …}
It can be seen that 15 ∈ B but 15 ∉ A.
∴A ≠ B

Question5. Are the following pairs of sets equal? Give reason.

(i) A = {2, 3} and B = {x : x is a solution of x2 + 5x + 6 = 0 }

(ii) A = { x : x is a letter in the word FOLLOW}

B = { y : y is a letter in the word WOLF}

Solution :
(i) A = {2, 3} and B

= {x:x  is a solution of x² +5x +6 = 0}

Here  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

A = { 2 ,3 }; B={-2, -3}

A = {2,3};  B = {-2,-3}

There for  A

Therefore, A and B are not equal sets.

(ii) A = :{x : x is a letter in the word FOLLOW} = {F, O, L, W};

B = {y : y  is a letter in the word WOLF} = {W, O, L, F}

Therefore, A = B = {F, O, L, W}

Question6. From the sets given below, select equal sets :
A = { 2, 4, 8, 12},
B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14},
D = { 3, 1, 4, 2}
E = {–1, 1}, F = { 0, a},
G = {1, –1}, H = { 0, 1 }

Solution :
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}

It can be seen 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
Also, 2 ∈ A, 2 ∉ C
∴ A ≠ C
3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H
∴ B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H
12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H
∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H
4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H
∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H
Similarly, E ≠ F, E ≠ G, E ≠ H
F ≠ G, F ≠ H, G ≠ H
The order in which the elements of a set are listed is not significant.
∴ B = D and E = G
Hence, among the given sets, B = D and E = G.

### Exercise 1.3

Question1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:

(i) {2, 3, 4} … {1, 2, 3, 4, 5}

(ii) {a, b, c} … {b, c, d}

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

(iv) {x: x is a circle in the plane} … {x: x is a circle in the same plane with radius 1 unit}

(v) {x: x is a triangle in a plane}…{x: x is a rectangle in the plane}

(vi) {x: x is an equilateral triangle in a plane}… {x: x is a triangle in the same plane}

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

Solution :
(i) {2,3,4} ⊂ {1,2,3,4,5}

(ii) {a,b,c} ⊄ {b,c,d}

(iii) {x: x is a student of class XI of your school}⊂ {x: x is student of your school}

(iv) {x: x is a circle in the plane} ⊄ {x: x is a circle in the same plane with radius 1 unit}

(v) {x: x is a triangle in a plane} ⊄ {x: x is a rectangle in the plane}

(vi) {x: x is an equilateral triangle in a plane}⊂ {x: x in a triangle in the same plane}

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

Question2. Examine whether the following statements are true or false:
(i) { a, b } ⊄ { b, c, a }
(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet}
(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }
(iv) { a } ⊂ { a, b, c }
(v) { a } ∈ { a, b, c }
(vi) { x : x is an even natural number less than 6}  ⊂  { x : x is a natural number
which divides 36}

Solution :
(i) Let A = { a, b } and B = { b, c, a }

Here, every element of set A is an element of set B.

∴ A ⊂ B

Therefore, statement is false.

(ii) Let A = { a, e } and B

= {x: x  is a vowel in the English alphabet}

Here, every element of set A is an element of set B.

= ∴{a,e}⊂{x:x is an vowel in English alpahbet}

∴A ⊂B

Therefore, statement is true.

(iii) Let A = {1, 2, 3} and B = {1, 3, 5}

Here, 2 ∈ { 1, 2, 3 } but 2 ∉ B

∴ A ∉ B

Therefore, statement is false.

(iv) Let A = { a }and B = { a, b, c }

Here, every element of set A is an element of set B.

∵ A ∈ B

Therefore, statement is true.

(v) Let A = { a } and B = { a, b, c }

Here, A⊂B

Therefore, statement is false.

(vi) Let A = {x: x is an even natural number less than 6}

= {2, 4}

And B = }⊂ {x: x  is a natural number which divide 36}

= {1, 2, 3, 4, 6, 12, 18, 36]

Here, every element of set A is an element of set B.

∴ A ⊂ B

Therefore, statement is true.

Question 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) {3, 4} is a member of set A.

∵ {3, 4} ⊂ A

Therefore, {3, 4} ⊂ A is incorrect.

(ii) {3, 4} is a member of set A. Therefore, {3, 4} ∈ A is incorrect.

(iii) {3, 4} is a member of set A.

∵ {{3, 4}} is a set.

Therefore, {{3, 4}} ∈ A is incorrect

(iv) 1 is a member of set A. Therefore 1 ∈ A is correct.

(v) 1 is not a set, it is a member of set A. Therefore, 1 ⊂ A is incorrect.

(vi) 1, 2, 5 are the members of set A.

∵ {1, 2, 5} is a subset of set A.

Therefore, {1, 2, 5} ⊂ A is correct.

(vii) 1, 2, 5 are the members of set A.

∵ {1, 2, 5} is a subset of set A.

Therefore, {1, 2, 5} ⊂ A is incorrect.

(viii) 3 is not a member of set A.

∴ {1, 2, 3} is not a subset of set A.

Therefore, {1, 2, 3} ⊂ A is incorrect.

(ix) ∅∈A is not a member of set A. Therefore, ∅ ∈ A is correct.

(x) {φ} ⊂ A is not a member of set A. Therefore, {φ} ⊂ A is incorrect.

Question4. Write down all the subsets of the following sets:

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

(iii) {1, 2, 3},(iv) Φ

Solution :
(i) The subsets of {a} are Φ and {a}.

(ii) The subsets of {a, b} areΦ, {a}, {b}, and {a, b}.

(iii) The subsets of {1, 2, 3} areΦ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and

{1, 2, 3}

(iv) The only subset of Φ isΦ.

Question5. How many elements has P(A), if A = φ?

Solution :
We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2m

If A = Φ, then n(A) = 0.

∴ n[P(A)] =  20

= 1

Hence, P(A) has one element.

Question6. Write the following as intervals :

(i) {x : x ∈ R, – 4

Solution :
(i) {x: x ∈ R, –4

Question7. 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) {[−23,5)={x:x∈R,−23≤x<5}

Question8. What universal set(s) would you propose for each of the following:

(i) The set of right triangles

(ii) The set of isosceles triangles

Solution :
(i) Right triangle is a type of triangle. Therefore, the set of triangles contain all types of triangles.

∴ U = { x: x  is a triangle in plane}

(ii) Isosceles triangle is a type of triangle. Therefore, the set of triangles contain all types of triangles.

∴ U = { x: x is a triangle in plane}

Question9. Given the set A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set(s) for all the three sets A, B and C:

(i) {0, 1, 2, 3, 4, 5, 6}

(ii) Φ

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

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

Solution :
(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6}

B ⊂ {0, 1, 2, 3, 4, 5, 6}

However, C ⊄ {0, 1, 2, 3, 4, 5, 6}

Therefore, 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

Therefore, 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}

Therefore, 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}

However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

Therefore, 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

Question1. Find the union of each of the following pairs of sets:

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

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

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

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

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

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

(v) A = {1, 2, 3}, B = Φ

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

X∪Y={1,3,5}∪{1,2,3}

∴X∪Y={1,2,3,5}

(ii) A {a, e, i, o, u}  B = {a, b, c}

A∪B={a,e,i,o,u}∪{a,b,c}

∴ A∪B={a,b,c,e,i,o,u}

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

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

To find the union of two sets

Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and B.

A={x:x is a natural number an multiple of 3},

={3,6,9,…}

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

={1,2,3,4,5,6}

A∪B={3,6,9,…}∪{1,2,3,4,5,6}

={1,2,3,4,5,6,9,12,15…}

∴A∪B
={1,2,3,4,5,6,9,12,15…}

(iv) A={x:x is a natural number 1x≤6},  B={x:x is a natural number 6×10}

To find the union of two sets

Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and B.

A={x:x is a natural number 1x≤6}={2,3,4,5,6}

B={x:x is a natural number 6×10}={7,8,9}

A∪B={2,3,4,5,6}∪{7,8,9}

∴A∪B={2,3,4,5,6,7,8,9}

(v) Given that,

A={1,2,3},B=∅

To find the union of two sets

Let A and B be any two sets. The union of A and B is the set that consists of all the elements of A and B.

A∪B={1,2,3}∪∅

∴A∪B={1,2,3}

Question2. Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?

Solution :
Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

A∪ B = {a, b, c} = B

Question3. 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.

Question4. 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 :
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, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

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

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

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

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

Question 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}

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

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

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

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

(v) A = {1, 2, 3}, B = Φ

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

X ∩ Y = {1, 3}

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

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}

∴ 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}

A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ

A ∩ B = Φ

Question 6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find

(i) A ∩ B

(ii) B ∩ C

(iii) A ∩ C ∩ D

(iv) A ∩ C

(v) B ∩ D

(vi) A ∩ (B ∪ C)

(vii) A ∩ D

(viii) A ∩ (B ∪ D)

(ix) (A ∩ B) ∩ (B ∪ C)

(x) (A ∪ D) ∩ (B ∪ C)

Solution :
(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}

Question7. If A = {x: x is a natural number}, B ={x: x is an even natural number}

C = {x: x is an odd natural number} and D = {x: x is a prime number}, find

(i) A ∩ B

(ii) A ∩ C

(iii) A ∩ D

(iv) B ∩ C

(v) B ∩ D

(vi) C ∩ D

Solution :
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}

Question 8. Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}

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

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

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

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

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {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

Therefore, this pair of sets is disjoint.

Question9. 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 :
Given: A = {3, 6, 9, 12, 15, 18, 21},

B = {4, 8, 12, 16, 20},

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

D = {5, 10, 15, 20};

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

= {3, 6, 9, 15, 18, 21}

(ii) A – B = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12 , 14, 16}

= {3, 9, 15, 18, 21}

(iii) A – B = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}

= {3, 6, 9, 15, 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, 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}

Question10. If X = {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, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

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

Solution :
R: set of real numbers

Q: set of rational numbers

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

Question 12. State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

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

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

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

Solution :
(i) False

As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) False

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

⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) True

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

(iv) True

As {2, 6, 10} ∩ {3, 7, 11} = Φ

### Exercise 1.5

Question1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find

(i) A′, (ii) B′,

(iii) (A ∪ C)′,

(iv) (A ∪ B)′,

(v) (A′)′, (vi) (B – C)′

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

A = {1, 2, 3, 4},

B = {2, 4, 6, 8} and C = {3, 4, 5, 6}.

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

= {5, 6, 7, 8, 9}

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

= {1, 3, 5, 7, 9}

(iii) (A ∪ C)’ = U – (A ∪ C)

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

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

(iv) (A ∪ B)’ = U – (A ∪ B)

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

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

(v) (A’)’ = U – A’ = U – (U – A)

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

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

= {1, 2, 3, 4} = A

(vi) (B – C)’ = U – (B – C)

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

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

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

Question2. If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :

(i) A = {a, b, c}, (ii) B = {d, e, f, g},

(iii) C = {a, c, e, g}, (iv) D = { f, g, h, a}

Solution :
U = {a, b, c, d, e, f, g, h}

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

A′={d,e,f,g,h}

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

∴B′={a,b,c,h}

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

∴C′={b,d,f,h}

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

∴D′={b,c,d,e}

Question3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) {x : x is an even natural number}, (ii) { x : x is an odd natural number }

(iii) {x : x is a positive multiple of 3}, (iv) { x : x is a prime number }

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

(vi) { x : x is a perfect square }, (vii) { x : x is a perfect cube}

(viii) { x : x + 5 = 8 }, (ix) { x : 2x + 5 = 9}

(x) { x : x ≥ 7 }, (xi) { x : x ∈ N and 2x + 1 > 10 }

Solution :
Given: U = N ; set of Natural numbers

(i) Let A = {x : x   is an even natural number}

∴  A’ = U – A = U– {x : x   is an even natural number}

= {x : x  is an odd natural number}

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

∴ A’ = U – A = u– {x : x is an odd natural number}

= {x : x is an even natural number}

(iii) Let A = {x : x  is a positive multiple of 3}

∴ A’ = U – A = U– {x : x  is a positive multiple of 3}

= {x : x is not a positive multiple of 3}

(iv) Let A = {x : x  is a prime number}

∴ A’ = U – A = U– {x : x is a prime number}

= {x : x is not a prime number}

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

∴ A’ = U – A = U– {x : x is a natural number divisible by 15}

= {x : x is not divisible by 15}

(vi) Let A = {x : x is a perfect square}

∴ A’ = U – A = U– {x : x is a perfect square}

= {x : x  is not a perfect square}

(vii) Let A = {x : x is a perfect cube}

∴A’ = U – A = u– {x : x is a perfect cube}

= {x : x is not a perfect cube}

(viii) Let A = {x : x  + 5 = 8} = {3}

To find the complement of {x:x+5=8}

x+5=8

x=3

The complement of set A is the set of all elements of U which are not the elements of A.

∴{x:x+5=8}′={x:x∈N and x≠3}

(ix) Let A = {x : 2x + 5 = 9 } = {2}

Given that,

The set of natural number is the universal set

To find the complement of the

{x:2x+5=9}

The complement of set A is the set of all elements of U which are not the elements of A.

2x+5=9

2x=4

x=2

∴{x:2x+5=9}′={x:x∈N and x≠2 }

(x) Given that,

The set of natural number is the universal set

To find the complement of

{x:x≥7}

The complement of set A is the set of all elements of U which are not the elements of A.

∴{x:x≥7}′={x:x∈N and x7}

(xi) Given that,

The set of natural number is the universal set

To find the complement of the

{x:x∈N and 2x+110}

The complement of set A is the set of all elements of U which are not the elements of A.

2x+1>10

2x>9

x>92

∴{x:x∈N and 2x+110}′
={x:x∈N and x≤92}

Question4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that

(i) (A ∪ B)′ = A′ ∩ B′,

(ii) (A ∩ B)′ = A′ ∪ B′

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

A = {2, 4, 6, 8} and B = {2, 3, 5, 7}

(i) L.H.S. = (A ∪ B)′  = U – (A ∪ B)

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

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

R.H.S. = A′ ∩ B′ = (U – A) ∩ (U – B)

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

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

L.H.S. = R. H. S.

(A∪B)′=A′∩B′

(ii) L.H.S. = (A ∩ B)′

= U – (A ∪ B)

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

= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2}

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

R.H.S. = A′ ∪ B′ = (U – A) ∪  (U – B)

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

= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}

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

L.H.S. = R. H. S.

Hence it has been proved that (A∩B)′=A′∪B′

Question5. Draw appropriate Venn diagram for each of the following :

(i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′

Solution :
(i) In the diagrams, shaded portion represents (A ∪ B)′

(ii) In the diagrams, shaded portion represents A′ ∩ B′

(iii) In the diagrams, shaded portion represents (A ∩ B)′

(iv) In the diagrams, shaded portion represents A′ ∪ B′

Question6. 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.

Question7. Fill in the blanks to make each of the following a true statement :

(i) A ∪ A′ = . . ., (ii) φ′ ∩ A = . . .,

(iii) A ∩ A′ = . . . ,(iv) U′ ∩ A = . . .

Solution :
(i) A∪A′=U

(ii) Φ′ ∩ A = U ∩ A = A

∴ Φ′ ∩ A = A

(iii) A ∩ A′ = Φ

(iv) U′ ∩ A = Φ ∩ A = Φ

∴ U′ ∩ A = Φ

### Exercise 1.6

Question 1. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ).

Solution :
It is given that:

n(X) = 17, n(Y) = 23,

n(X ∪ Y) = 38

n(X ∩ Y) = ?

We know that:

n(X∪Y)= n(X) + n(Y) − n(X∩Y)

∴38=17 + 23 − n(X∩Y)

⇒n(X∩Y) = 40−38=2

∴n(X∩Y) =2

Question 2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X ∩ Y have?

Solution :
It is given that:

n(X∪Y) =18,

n(X)=8, n(Y)=15

n(X ∩ Y) = ?

We know that:

n(X∪Y) =n(X)+ n(Y) −n(X∩Y)

∴18=8+15 −n(X∩Y)

⇒n(X∩Y) =23−18=5

∴n(X∩Y) =5

Question 3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Solution :
Let H be the set of people who speak Hindi, and

E be the set of people who speak English

∴ n(H ∪ E) = 400,

n(H) = 250,

n(E) = 200

n(H ∩ E) = ?

We know that:

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

∴ 400 = 250  +  200 – n(H ∩ E)

⇒ 400 = 450 –  n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

Question 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 :
It is given that:

n(S) = 21, n(T) = 32, n(S ∩ T) = 11

We know that:

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

∴ n (S ∪ T) = 21 + 32 – 11 = 42

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

Question 5. If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?

Solution :
It is given that:

n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10) = 30

Thus, the set Y has 30 elements.

Question 6. In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?

Solution :
Let C denote the set of people who like coffee, and T denote the set of people who like tea

n(C ∪ T) = 70,

n(C) = 37, n(T) = 52

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 70 = 37 + 52 – n(C ∩ T)

⇒ 70 = 89 – n(C ∩ T)

⇒ n(C ∩ T) = 89 – 70 = 19

Thus, 19 people like both coffee and tea.

Question7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Solution :
Let C denote the set of people who like cricket, and T denote the set of people who like tennis

∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 65 = 40 + n(T) – 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 – 30 = 35

Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T Also,

(T – C) ∩ (T ∩ C) = Φ

∴ n (T) = n (T – C) + n (T ∩ C)

⇒ 35 = n (T – C) + 10

⇒ n (T – C) = 35 – 10 = 25

Thus, 25 people like only tennis.

Question8. 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 :
Let F be the set of people in the committee who speak French, and S be the set of people in the committee who speak Spanish

∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10

We know that:

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

= 20 + 50 – 10

= 70 – 10 = 60

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

### Miscellaneous Exercise

Question1. Decide, among the following sets, which sets are subsets of one and another:

A = { x : x ∈ R and x satisfy x2−8x+12=0}

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

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

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

∴ A = {2, 6}

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

∴ D ⊂ A ⊂ B ⊂ C

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

Question2. 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  Let A= {1,2} and  B={1, {1,2},{3}} Now, 2∈ {1,2}  and  {1,2}∈ {{3},1,{1,2}}

∴A∈B However, 2∉{{3}, 1,{1,2}}

(ii) False  Let A= {2}, B={0,2},  and C ={1,{0,2},3}

As A⊂B

B∈C However, A∉C

(iii) True

Let A ⊂ B and B ⊂ C.

Let x ∈ A

⇒x∈B  [∵A⊂B]

⇒x∈C  [∵B⊂C]

∴ A ⊂ C

(iv) False

Let A={1,2},B = {0,6,8}, and C= {0,1,2,6,9}

Accordingly,  A⊄B and B⊄C

However, A ⊂ C

(v) False

Let A = {3, 5, 7} and B = {3, 4, 6}

Now, 5 ∈ A and A ⊄ B

However, 5 ∉ B

(vi) True

Let A ⊂ B and x ∉ B.

To show: x ∉ A

If possible, suppose x ∈ A.

Then, x ∈ B, which is a contradiction as x ∉ B

∴x ∉ A

Question3. Let A, B, and C be the sets such that A ∪ B =  A ∪ C and A ∩ B =  A∩ C. Show

that B = C.

Solution :
Let, A, B and C be the sets such that  A∪B=A∪C  and A∩B = A∩C

To show: B = C

Let x ∈ B

⇒x∈A∪B [B⊂A∪B]

⇒x∈A∪C [A∪B=A∪C]

⇒x∈A or x∈C

Case I

x ∈ A

Also, x ∈ B

∴x∈A∩B

⇒x∈A∩C [∵A∩B =A∩C]

∴ x ∈ A and x ∈ C

∴ x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B.

∴ B = C

Question4. Show that the following four conditions are equivalent :

(i) A ⊂ B (ii) A – B = φ  (iii) A ∪ B = B  (iv) A ∩ B = A

Solution :
First, we have to show that (i) ⇔ (ii).

Let A ⊂ B

To show: A – B ≠ Φ

If possible, suppose A – B ≠ Φ

This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.

∴ A – B = Φ

∴ A ⊂ B ⇒ A – B = Φ

Let A – B = Φ

To show: A ⊂ B

Let x ∈ A

Clearly, x ∈ B because if x ∉ B, then A – B  ≠  Φ

∴ A – B = Φ ⇒ A ⊂ B

∴ (i) ⇔ (ii)

Let A ⊂ B

To show:  A∪B =B Clearly, B⊂A∪B

Let x∈A∪B

⇒x∈A or x∈B

case l:  x∈A ⇒ x∈B      [∵A⊂B]

∴A∪B⊂B

Case  II: x∈B Then, A∪B=B Conversely, let A∪B=B Let x∈A

⇒x∈A∪B  [∵A⊂A∪B]

⇒x∈B [∵A∪B=B]

∴A⊂B Hence, (i) ⇔(iii)

Now, we have to show that (i)⇔(iv)

Let A⊂B Clearly A∩B ⊂ A

Let x∈A We have to show that  x ∈ A∩B

Let A⊂B Clearly A∩B⊂A Let x∈A We have to show that x∈A∩B

As  A ⊂ B,x ∈ B

∴A⊂A ∩ B

∴A ⊂ A ∩ B

Hence, A = A ∩ B Conversely, suppose A ∩ B  = A

Let x∈A

⇒x∈A∩B

⇒x∈A and x∈B

⇒A∈B

∴A⊂B Hence, (i) ⇔ (iv).

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

Solution :
Let A ⊂ B To show: C – B ⊂ C – A

Let x ∈ C – B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A [A ⊂ B]

⇒ x ∈ C – A

∴ C – B ⊂ C – A

Question6. Assume that P ( A ) = P ( B ). Show that A = B

Solution :
Let P(A) = P(B) To show: A = B

Let x ∈ A

A ∈ P(A) = P(B)

∴ x ∈ C, for some C ∈ P(B)

Now, C ⊂ B

∴ x ∈ B

∴ A ⊂ B

Similarly, B ⊂ A

∴ A = B

Question7. Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your answer.

Solution :
False

Let A = {0, 1} and B = {1, 2}

∴ A ∪ B = {0, 1, 2}

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}}

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

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Question8. Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )

Solution :
To show: A = (A ∩ B) ∪ (A – B)

Let x ∈ A

We have to show that x ∈ (A ∩ B) ∪ (A – B)

Case I

x ∈ A ∩ B

Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II

x ∉ A ∩ B

⇒ x ∉ A or x ∉ B

∴ x ∉ B [x ∉ A]

∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)

It is clear that

A ∩ B ⊂ A and (A – B) ⊂ A

∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)

From (1) and (2), we obtain

A = (A ∩ B) ∪ (A – B)

To prove: A ∪ (B – A) ⊂ A ∪ B

Let 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 or x ∉ A)

⇒ x ∈ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)

Next, we show that (A ∪ B) ⊂ A ∪ (B – A).

Let y ∈ A ∪ B

⇒ y ∈ A or y ∈ B

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

⇒ y ∈ A or (y ∈ B and y ∉ A)

⇒ y ∈ A ∪ (B – A)

∴ A ∪ B ⊂ A ∪ (B – A) … (4)

Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.

Question9. 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 … (1)

Also,  A ⊂ A ∪ (A ∩ B) … (2)

∴ From (1) and (2), A ∪ (A ∩ B) = A

(ii) To show: A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A {from (1)}

Question10. Show that A ∩ B = A ∩ C need not imply B = C.

Solution :
Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}

Accordingly, A ∩ B = {0} and A ∩ C = {0}

Here, A ∩ B = A ∩ C = {0}

However, B ≠ C [2 ∈ B and 2 ∉ C]

Question 11. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

(Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )

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

To show: A = B, It can be seen that

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 … (1)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

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

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= B ∩ A

= A ∩ B … (2)

Hence, from (1) and (2), we obtain A = B.

Question12. 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 A = {0, 1}, B = {1, 2}, and C = {2, 0}.

Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.

∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.

However, A ∩ B ∩ C = Φ

Question13. 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 U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T’ ∩ C’).

n(T’ ∩ C’) = n(T ∪ C)’

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275 = 325

Hence, 325 students were taking neither tea nor coffee.

Question14. 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 U be the set of all students in the group.

Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

n(H∩E)=25n(U)=n(H)+n(E)−n(H∩E)=100+50−25 =125

Hence, there are 125 students in the group.

Question15. 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 :
Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26

n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8

n(A ∩ B ∩ C) = 3

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

(i) Accordingly, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 52

Hence, 52 people read at least one of the newspapers.

(ii) Let a be the number of people who read newspapers H and T only.

Let b denote the number of people who read newspapers I and H only.

Let c denote the number of people who read newspapers T and I only.

Let d denote the number of people who read all three newspapers.

Accordingly, d = n(A ∩ B ∩ C) = 3

Now, n(A ∩ B) = a + d

n(B ∩ C) = c + d

n(C ∩ A) = b + d

∴ a + d + c + d + b + d = 11 + 8 + 9 = 28

⇒ a + b + c + d = 28 – 2d = 28 – 6 = 22

Hence, (52 – 22) = 30 people read exactly one newspaper

Question16. 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 be the set of people who like product A, product B, and product C respectively.

Accordingly, n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12,

n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

The Venn diagram for the given problem can be drawn as

It can be seen that number of people who like product C only is

{29 – (4 + 8 + 6)} = 11

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