NCERT Solutions for Class 11 Maths Chapter 1 – Sets

(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, x2 ≤ 4}

We know that

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

Here

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

So we get

C = {–2, –1, 0, 1, 2}

(iv) D = {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}

 

(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).

 

---

 

 

(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects as one can identify a month which belongs to this collection.

Therefore, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection as the criteria to determine a writer’s talent may differ from one person to another.

Therefore, this collection is not a set.

(iii) A team of eleven best-cricket batsmen of the world is not a well-defined collection as the criteria to determine a batsman’s talent may vary from one person to another.

Therefore, this collection is not a set.

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

Therefore, this collection is a set.

(v) The collection of all natural numbers less than 100 is a well-defined collection as one can find a number which belongs to this collection.

Therefore, this collection is a set.

(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection as one can find any book which belongs to this collection.

Therefore, this collection is a set.

(vii) The collection of all even integers is a well-defined collection as one can find an integer which belongs to this collection.

Therefore, this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection as one can find a question which belongs to this chapter.

Therefore, this collection is a set.

(ix) A collection of most dangerous animals of the world is not a well-defined collection as the criteria to find the dangerousness of an animal can differ from one animal to another.

Therefore, this collection is not a set.

 

(i) 5 ∈ A

(ii) 8 ∉ A

(iii) 0 ∉ A

(iv) 4 ∈ A

(v) 2 ∈ A

(vi) 10 ∉ A

 

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

–2, –1, 0, 1, 2, 3, 4, 5, and 6 only are the elements of this set.

Hence, 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}

1, 2, 3, 4, and 5 only are the elements of this set

Hence, 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}

17, 26, 35, 44, 53, 62, 71, and 80 only are the elements of this set

Hence, the given 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 divisor of 60}

Here 60 = 2 × 2 × 3 × 5

2, 3 and 5 only are the elements of this set

Hence, the given set can be written in roster form as

D = {2, 3, 5}

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

TRIGONOMETRY is a 12 letters word out of which T, R and O are repeated.

Hence, the given 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

BETTER is a 6 letters word out of which E and T are repeated.

Hence, the given set can be written in roster form as

F = {B, E, T, R}

 

(i) {3, 6, 9, 12}

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

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

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

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

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

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

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

(iv) {2, 4, 6 …}

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

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

(v) {1, 4, 9 … 100}

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

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

 

(i) Set of odd natural numbers divisible by 2 is a null set as odd numbers are not divisible by 2.

(ii) Set of even prime numbers is not a null set as 2 is an even prime number.

(iii) {x: x is a natural number, x < 5 and x > 7} is a null set as a number cannot be both less than 5 and greater than 7.

(iv) {y: y is a point common to any two parallel lines} is a null set as the parallel lines do not intersect. Therefore, they have no common point.

 

(i) The set of months of a year is a finite set as it contains 12 elements.

(ii) {1, 2, 3 …} is an infinite set because it has infinite number of natural numbers.

(iii) {1, 2, 3 …99, 100} is a finite set as the numbers from 1 to 100 are finite.

(iv) The set of positive integers greater than 100 is an infinite set as the positive integers which are greater than 100 are infinite.

(v) The set of prime numbers less than 99 is a finite set as the prime numbers which are less than 99 are finite.

 

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

(ii) The set of letters in the English alphabet is a finite set as it contains 26 elements.

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

(iv) The set of animals living on the earth is a finite set as the number of animals living on the earth is finite.

(v) The set of circles passing through the origin (0, 0) is an infinite set as infinite number of circles can pass through the origin.

 

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

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

Therefore, A = B.

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

We know that 12 ∈ A but 12 ∉ B.

Therefore, 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}

Therefore, A = B

(iv) A = {x: x is a multiple of 10}

B = {10, 15, 20, 25, 30 …}

We know that 15 ∈ B but 15 ∉ A.

Therefore, A ≠ B

 

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

x2 + 5x + 6 = 0 can be written as

x(x + 3) + 2(x + 3) = 0

By further calculation

(x + 2) (x + 3) = 0

So we get

x = –2 or x = –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.

 

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.

 

 

 

(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}

 

(i) False.

Here each element of {a, b} is an element of {b, c, a}.

(ii) True.

We know that a, e are two vowels of the English alphabet.

(iii) False.

2 ∈ {1, 2, 3} where, 2∉ {1, 3, 5}

(iv) True.

Each element of {a} is also an element of {a, b, c}.

(v) False.

Elements of {a, b, c} are a, b, c. Hence, {a} ⊂ {a, b, c}

(vi) True.

{x: x is an even natural number less than 6} = {2, 4}

{x: x is a natural number which divides 36}= {1, 2, 3, 4, 6, 9, 12, 18, 36}

 

It is given that A= {1, 2, {3, 4}, 5}

(i) {3, 4} ⊂ A is incorrect

Here 3 ∈ {3, 4}; where, 3∉A.

(ii) {3, 4} ∈A is correct

{3, 4} is an element of A.

(iii) {{3, 4}} ⊂ A is correct

{3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) 1∈A is correct

1 is an element of A.

(v) 1⊂ A is incorrect

An element of a set can never be a subset of itself.

(vi) {1, 2, 5} ⊂ A is correct

Each element of {1, 2, 5} is also an element of A.

(vii) {1, 2, 5} ∈ A is incorrect

{1, 2, 5} is not an element of A.

(viii) {1, 2, 3} ⊂ A is incorrect

3 ∈ {1, 2, 3}; where, 3 ∉ A.

(ix) Φ ∈ A is incorrect

Φ is not an element of A.

(x) Φ ⊂ A is correct

Φ is a subset of every set.

(xi) {Φ} ⊂ A is incorrect

Φ∈ {Φ}; where, Φ ∈ A.

 

(i) Subsets of {a} are

Φ and {a}.

(ii) Subsets of {a, b} are

Φ, {a}, {b}, and {a, b}.

(iii) Subsets of {1, 2, 3} are

Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}.

(iv) Only subset of Φ is Φ.

 

If A is a set with m elements

n (A) = m then n [P (A)] = 2m

If A = Φ we get n (A) = 0

n [P(A)] = 20 = 1

Therefore, P (A) has one element.

 

(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]

 

(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}

 

(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.

 

(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.

 

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

 

A’ is the set of all equilateral triangles.

 

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, 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}

 

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

(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

 

(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

 

(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}

 

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}

 

(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.

 

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

 

(i) X – Y = {a, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

 

We know that

R – Set of real numbers

Q – Set of rational numbers

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

 

(i) False

If 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

So we get {2, 3, 4, 5} ∩ {3, 6} = {3}

(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}

(iii) True

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

(iv) True

Here {2, 6, 10} ∩ {3, 7, 11} = Φ

 

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’ = {5, 6, 7, 8, 9}

(ii) B’ = {1, 3, 5, 7, 9}

(iii) A U C = {1, 2, 3, 4, 5, 6}

So we get

(A U C)’ = {7, 8, 9}

(iv) A U B = {1, 2, 3, 4, 6, 8}

So we get

(A U B)’ = {5, 7, 9}

(v) (A’)’ = A = {1, 2, 3, 4}

(vi) B – C = {2, 8}

So we get

(B – C)’ = {1, 3, 4, 5, 6, 7, 9}

 

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

So we get

A’ = {d, e, f, g, h}

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

So we get

B’ = {a, b, c, h}

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

So we get

C’ = {b, d, f, h}

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

So we get

D’ = {b, c, d, e}

 

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}

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

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

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

 

It is given that

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

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

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

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

Therefore, (A U B)’ = A’ ∩ B’.

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

A’ U B’ = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}

Therefore, (A ∩ B)’ = A’ U B’.

 

(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}

 

(i) (A U B)’

(ii) A’ ∩ B’

(iii) (A ∩ B)’

(iv) A’ U B’

 

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

 

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

 

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.

 

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.

 

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.

 

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.

 

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.

 

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.

 

According to the question,

We have,

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

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

Hence, A = {2, 6}

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

Hence, D ⊂ A ⊂ B ⊂ C

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

 

(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

 

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

 

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

 

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.

 

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

 

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

 

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

 

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

 

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

 

(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

 

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.

 

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

 

Let us assume that,

U = the set of all students in the group

E = the set of students who know English

H = the set of the students who know Hindi

∴ H ∪ E = U

Given that,

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

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

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

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

∴ According to the question,

n (U) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

∴ Total number of students in the group = 125 students

 

(i) 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

 

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