Sets

(i) Given: A = {x : x ∈ R, 2x + 11 = 15}

To find: Roster form of given set

2x + 11 = 15

⇒ 2x = 15 – 11

⇒ 2x = 4

⇒ x = 2

So, A = {2}

(ii) Given: B = {x | x² = x, x ∈ R}

To find: Roster form of given set

x² = x

⇒ x² – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 or 1

So, B = {0, 1}

(iii) Given: C = {x | x is a positive factor of a prime number p}

To find: Roster form of given set

Only possible positive factors of a prime number p are 1 and p itself.

Hence,

x = 0 or p

So, C = {0, p}

(i) Given: D = {t | t³ = t, t ∈ R}

To find: Roster form of given set

t³ = t

⇒ t³ – t = 0

⇒ t(t² – 1) = 0

⇒ t(t – 1)(t + 1) = 0

⇒ t = 0, -1 or 1

So, D = {-1, 0, 1}

(ii) Given:

To find: Roster form of given set

⇒ w – 2 = 3(w + 3)

⇒ w – 2 = 3w + 9

⇒ 3w – w = - 9 – 2

⇒ 2w = –11

(iii) Given: F = {x | x⁴ – 5x² + 6 = 0, x ∈ R}

To find: Roster form of given set

x⁴ – 5x² + 6 = 0

⇒ x⁴ – 3x² – 2x² + 6 = 0

⇒ x²(x² – 3) – 2(x² – 3) = 0

⇒ (x² – 3) (x² – 2) = 0

⇒ x² = 3 or 2

Given: Y = {x | x is a positive factor of the number 2^{p – 1}(2^p – 1), where 2^p – 1 is a prime number}.

To find: Roster form of given set

Only possible positive factors of a prime number p are 1 and p itself.

Possible factors of 2^{p – 1} (2^p – 1) are all possible factors of 2^{p – 1} and 2^p – 1 individually.

Possible factors of 2^{p – 1} are 2⁰, 2¹,………, 2^{p – 1} and that of 2^p – 1 are 1 and 2^p – 1 {∵ 2^p – 1 is prime number}

Hence,

x = 1, 2¹,………, 2^{p – 1}, 2^p – 1

So, Y = {1, 2¹,………, 2^{p – 1}, 2^p – 1}

(i) Given: 35 ∈ {x | x has exactly four positive factors}

To check: 35 belongs to given set or not

The possible positive factors of 35 are 1, 5, 7, 35

Since, 35 has exactly four positive factors

⇒ The given statement is true.

(ii) Given: 128 ∈ {y | the sum of all the positive factors of y is 2y}

To check: 128 belongs to given set or not

The possible positive factors of 128 are 1, 2, 4, 8, 16, 32, 64, 128

The sum of them

= 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128

= 255

2y = 2 * 128 = 256

Since, the sum of all the positive factors of y is not equal to 2y

⇒ The given statement is false.

(iii) Given: 3 ∉ {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0}

To check: 128 belongs to given set or not

x⁴ – 5x³ + 2x² – 112x + 6 = 0

On putting x = 3 in LHS:

(3)⁴ – 5(3)³ + 2(3)² – 112(3) + 6

= 81 – 135 + 18 – 336 + 6

= –366

So, 3 does not belong to given set

⇒ The given statement is true.

(iv) Given: 496 ∉ {y | the sum of all the positive factors of y is 2y}

To check: 128 belongs to given set or not

The possible positive factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248, 496

The sum of them

= 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496

= 996

2y = 2 * 496 = 992

Since, the sum of all the positive factors of y is equal to 2y

496 belongs to given set

⇒ The given statement is false.

Given: L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}

To verify: L – (M ∪ N) = (L – M) ∩ (L – N)

Formula used:

The union of two sets is a set containing all elements that are in both sets.

For example: {1, 2, 3} ∪ {2, 4} = {1, 2, 3, 4}

The difference (subtraction) is defined as: The set A – B consists of elements that are in A but not in B.

For example: if A = {1, 2, 3} and B = {3, 5}, then A−B = {1, 2}

The intersection of two sets A and B, consists of all elements that are both in A and B.

For example: {1, 2} ∩ {2, 3} = {2}

Therefore,

M = {3, 4, 5, 6}, N = {1, 3, 5} ⇒ M ∪ N = {1, 3, 4, 5, 6}

L = {1, 2, 3, 4} and M ∪ N = {1, 3, 4, 5, 6}

⇒ L – (M ∪ N) = {2}………………(1)

L = {1, 2, 3, 4} and M = {3, 4, 5, 6} ⇒ L – M = {1, 2}

L = {1, 2, 3, 4} and N = {1, 3, 5} ⇒ L – N = {2, 4}

L – M = {1, 2} and L – N = {2, 4}

⇒ (L – M) ∩ (L – N) = {2}………………(2)

Clearly, from (1) and (2):

L – (M ∪ N) = (L – M) ∩ (L – N)

Hence verified

(i) Given: A and B are two subsets

To prove: A ⊂ A ∪ B

Let x ∈ A

⇒ x ∈ A or x ∈ B

⇒ x ∈ A ∪ B

⇒ A ⊂ A ∪ B

Hence Proved

(ii) Given: A and B are two sets

To prove: A ⊂ B ⇔ A ∪ B = B

Let x ∈ A ∪ B

⇒ x ∈ A or x ∈ B

⇒ x ∈ B {∵ A ⊂ B}

⇒ A ∪ B ⊂ B…………(1)

We know,

B ⊂ A ∪ B {this is always true}…………(2)

From (1) and (2):

A ∪ B = B

Now,

Let y ∈ A

⇒ y ∈ A ∪ B

⇒ y ∈ B {∵ A ∪ B = B}

⇒ A ⊂ B

So,

A ⊂ B ⇔ A ∪ B = B

Hence Proved

(iii) Given: A and B are two subsets

To prove: (A ∩ B) ⊂ A

Let x ∈ A ∩ B

⇒ x ∈ A and x ∈ B

⇒ x ∈ A

⇒ A ∩ B ⊂ A

Hence Proved

(i) Given: N = {1, 2, 3 ,..., 100}

To find: subset of N whose elements are even numbers

A set A is a subset of a set B, if A is "contained" inside B, that is, all elements of A are also elements of B.

N = {1, 2, 3 ,..., 100}

Hence, subset of N whose elements are even numbers

= {2, 4, 6, 8,………,100}

(ii) Given: N = {1, 2, 3 ,..., 100}

To find: subset of N whose elements are perfect square numbers

A set A is a subset of a set B, if A is "contained" inside B, that is, all elements of A are also elements of B.

N = {1, 2, 3 ,..., 100}

Hence, subset of N whose elements are perfect square numbers = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

(i) Given: X = {1, 2, 3} where n represents any member of X

To find: sets containing all numbers represented by 4n

X = {1, 2, 3}

{4n | n ∈ x}

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

= {4, 8, 12}

(ii) Given: X = {1, 2, 3} where n represents any member of X

To find: sets containing all numbers represented by n + 6

X = {1, 2, 3}

{n + 6 | n ∈ x}

= {1 + 6, 2 + 6, 3 + 6}

= {7, 8, 9}

(iii) Given: X = {1, 2, 3} where n represents any member of X

To find: sets containing all numbers represented by

X = {1, 2, 3}

(iv) Given: X = {1, 2, 3} where n represents any member of X

To find: sets containing all numbers represented by n – 1

X = {1, 2, 3}

{n – 1 | n ∈ x}

= {1 – 1, 2 – 1, 3 – 1}

= {0, 1, 2}

(i) Given: Y = {1, 2, 3,..., 10} where a represents any element of Y

To find: sets containing all numbers represented by a ∈ Y but a²∉ Y

Y = {1, 2, 3,..., 10}

1² = 1, 2² = 4, 3² = 9

1, 4, 9 ∈ Y ⇒ 1, 2, 3 does not satisfy given condition

Therefore,

{a: a ∈ Y and a²∉ Y}

= {4, 5, 6, 7, 8, 9, 10}

(ii) Given: Y = {1, 2, 3,..., 10} where a represents any element of Y

To find: sets containing all numbers represented by a + 1 = 6, a ∈ Y

Y = {1, 2, 3,..., 10}

a + 1 = 6 ⇒ a = 5

⇒ 5 satisfies given condition

Therefore,

{a: a + 1 = 6, a ∈ Y }

= {5}

(iii) Given: Y = {1, 2, 3,..., 10} where a represents any element of Y

To find: sets containing all numbers represented by a is less than 6 and a ∈ Y

Y = {1, 2, 3,..., 10}

a is less than 6

⇒ 1, 2, 3, 4, 5 satisfy given condition

Therefore,

{a: a is less than 6, a ∈ Y }

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

Given: A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20}

To draw: Venn diagram showing relation of U, A, B and C sets

Here, U is a universal set

The intersection of two sets A and B, consists of all elements that are both in A and B.

For example: {1, 2} ∩ {2, 3} = {2}

Therefore,

A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15} and C = {5, 10, 15, 20}

⇒ A ∩ B = {6, 12}, B ∩ C = {15}, A ∩ C = {20}, A ∩ B ∩ C = {Φ}

The venn diagram showing relation of given sets is as follows:

Given: There are four sets U, G, B, S

To draw: Venn diagram showing relation among sets U, G, B and S sets

Here, U is a universal set showing set of all boys and girls in a school

G is the set of all girls in the school

B is the set of all boys in the school

S is the set of all students in the school who take swimming

Girls cannot be boys and boys cannot be girls

Any student girl or boy can swim.

The venn diagram showing relation of given sets is as follows:

Given: There are three sets A, B and C

To prove: (A – B) ∩ (A – C) = A – (B ∪ C)

Let x ∈ (A – B) ∩ (A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

⇒ (x ∈ A and x ∉ B) and (x ∈ A and x ∉ C)

⇒ x ∈ A and (x ∉ B and x ∉ C)

⇒ x ∈ A and x ∉ (B ∪ C)

⇒ x ∈ A – (B ∪ C)

⇒ (A – B) ∩ (A – C) ⊂ A – (B ∪ C)………(i)

Let y ∈ A – (B ∪ C)

⇒ y ∈ A and y ∉ (B ∪ C)

⇒ y ∈ A and (y ∉ B and y ∉ C)

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

⇒ y ∈ (A – B) and y ∈ (A – C)

⇒ y ∈ (A – B) ∩ (A – C)

⇒ A – (B ∪ C) ⊂ (A – B) ∩ (A – C) ………(ii)

We know:

P ⊂ Q and Q ⊂ P ⇒ P = Q

From (i) and (ii):

A – (B ∪ C) = (A – B) ∩ (A – C)

Given: There are two sets A and B

To check: (A – B) ∪ (A ∩ B) = A is true or false

Take L.H.S

(A – B) ∪ (A ∩ B)

= (A ∩ B’) ∪ (A ∩ B) {∵ A – B = A ∩ B’}

= A ∩ (B’ ∪ B)

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= A ∩ U

= A

= R.H.S

Hence, the given statement is true

Given: There are three sets A, B and C

To check: A – (B – C) = (A – B) – C is true or not

Step 1:

B – C

Step 2:

A – (B – C)

Step 3:

A – B

Step 4:

(A – B) – C

Clearly, the Venn diagram in Step 2 and Step 4 are not equal

Hence, A – (B – C)(A – B) – C

Hence, the given statement is not true

Given: There are three sets A, B and C

To check: if A ⊂ B, then A ∩ C ⊂ B ∩ C is true or false

Let x ∈ A ∩ C

⇒ x ∈ A and x ∈ C

⇒ x ∈ B and x ∈ C {∵ A ⊂ B}

⇒ x ∈ B ∩ C

⇒ A ∩ C ⊂ B ∩ C

Hence, the given statement is true

Given: There are three sets A, B and C

To check: if A ⊂ B, then A ∪ C ⊂ B ∪ C is true or false

Let x ∈ A ∪ C

⇒ x ∈ A or x ∈ C

⇒ x ∈ B or x ∈ C {∵ A ⊂ B}

⇒ x ∈ B ∪ C

⇒ A ∪ C ⊂ B ∪ C

Hence, the given statement is true

Given: There are three sets A, B and C

To check: if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C is true or false

Let x ∈ A ∪ B

⇒ x ∈ A or x ∈ C

⇒ x ∈ C or x ∈ C {∵ A ⊂ C and B ⊂ C}

⇒ x ∈ C

⇒ A ∪ B ⊂ C

Hence, the given statement is true

Given: There are two sets A and B

To prove: A ∪ (B – A) = A ∪ B

Take L.H.S

A ∪ (B – A)

= A ∪ (B ∩ A’)

{∵ A – B = A ∩ B’}

= (A ∪ B) ∩ (A ∪ A’)

{∵ Distributive property of set:

(A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C)}

= (A ∪ B) ∩ U

{∵ A ∪ A’ = U}

= A ∪ B

= R.H.S

Hence Proved

Given: There are two sets A and B

To prove: A – (A – B) = A ∩ B

Take L.H.S

A – (A – B)

= A – (A ∩ B’)

{∵ A – B = A ∩ B’}

= A ∩ (A ∩ B’)’

= A ∩ [A’ ∪ (B’)’]

{∵ (A ∩ B)’ = A’ ∪ B’}

= A ∩ (A’ ∪ B)

{∵ (B’)’ = B}

= (A ∩ A’) ∪ (A ∩ B)

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= Φ ∪ (A ∩ B)

{∵ A ∩ A’ = Φ}

= A ∩ B

= R.H.S

Hence Proved

Given: There are two sets A and B

To prove: A – (A ∩ B) = A – B

Take L.H.S

A – (A ∩ B)

= A ∩ (A ∩ B)’

{∵ A – B = A ∩ B’}

= A ∩ (A ∩ B’)’

= A ∩ (A’ ∪ B’)

{∵ (A ∩ B)’ = A’ ∪ B’}

= (A ∩ A’) ∪ (A ∩ B’)

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= Φ ∪ (A ∩ B’)

{∵ A ∩ A’ = Φ}

= A ∩ B’

= A – B

{∵ A – B = A ∩ B’}

= R.H.S

Hence Proved

Given: There are two sets A and B

To prove: (A ∪ B) – B = A – B

Take L.H.S

(A ∪ B) – B

= (A ∪ B) ∩ B’

{∵ A – B = A ∩ B’}

= (A ∩ B’) ∪ (B ∩ B’)

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= (A ∩ B’) ∪ Φ

{∵ A ∩ A’ = Φ}

= A ∩ B’

= A – B

{∵ A – B = A ∩ B’}

= R.H.S

Hence Proved

Given:

To check: T is an empty set or not

⇒ -(4x – 40)(13 – x) = (4x – 40)(x – 7)

⇒ (4x – 40)(x – 7) + (4x – 40)(13 – x) = 0

⇒ (4x – 40)(x – 7 + 13 – x) = 0

⇒ 6(4x – 40) = 0

⇒ 24(x – 10) = 0

⇒ x – 10 = 0

⇒ x = 10

So, T = {10}

⇒T is not an empty set

Given: A, B and C are three given sets

To prove: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Let x ∈ A ∩ (B ∪ C)

⇒ x ∈ A and x ∈ (B ∪ C)

⇒ x ∈ A and (x ∈ B or x ∈ C)

⇒ (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)

⇒ x ∈ A ∩ B or x ∈ A ∩ C

⇒ x ∈ (A ∩ B) ∪ ( A ∩ C)

⇒ A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ ( A ∩ C)………(i)

Let y ∈ (A ∩ B) ∪ (A ∩ C)

⇒ y ∈ A ∩ B or x ∈ A ∩ C

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

⇒ y ∈ A and (y ∈ B or y ∈ C)

⇒ y ∈ A and y ∈ (B ∪ C)

⇒ y ∈ A ∩ (B ∪ C)

⇒ (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)………(ii)

We know:

P ⊂ Q and Q ⊂ P ⇒ P = Q

From (i) and (ii):

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Hence Proved

Given:

Total number of students = 100

Number of students passed in English = 15

Number of students passed in Mathematics = 12

Number of students passed in Science = 8

Number of students passed in English and Mathematics = 6

Number of students passed in Mathematics and Science = 7

Number of students passed in English and Science = 4

Number of students passed in all three = 4

Let U be the total number of students, E, M and S be the number of students passed in English, Mathematics and Science respectively

n(M ∩ S ∩ E) = a = 4

n(M ∩ S) = a + d = 7

⇒ 4 + d = 7

⇒ d = 3

n(M ∩ E) = a + b = 6

⇒ 4 + b = 6

⇒ b = 2

n(S ∩ E) = a + c = 4

⇒ 4 + c = 4

⇒ c = 0

n(M) = e + d + a + b = 12

⇒ e + 4 + 3 + 2 = 12

⇒ e + 9 = 12

⇒ e = 3

n(E) = g + c + a + b = 15

⇒ g + 0 + 4 + 2 = 15

⇒ g + 6 = 15

⇒ g = 9

n(S) = f + c + a + d = 8

⇒ f + 0 + 4 + 3 = 8

⇒ f + 7 = 8

⇒ f = 1

(i) Number of students passed in English and Mathematics but not in Science = b = 2

(ii) Number of students in Mathematics and Science but not in English = d = 3

(iii) Number of students in Mathematics only = e = 3

(iv) Number of students in more than one subject only

= a + b + c + d

= 4 + 3 + 2 + 0

= 9

Given:

Total number of students are 60

Students who play cricket and tennis are 25 and 20 respectively

Students who play both the games are 10

To find: number of students who play neither

Let S be the total number of students, C and T be the number of students who play cricket and tennis respectively

n(S) = 60, n(C) = 25, n(T) = 20, n(C ∩ T) = 10

Number of students who play either of them = n(C ∪ T)

= n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10

= 35

Number of student who play neither

= Total – n(C ∪ T)

= 60 – 35

= 25

Hence, there are 25 students who play neither cricket nor tennis.

Given:

Total number of students = 200

Number of students study Mathematics = 120

Number of students study Physics = 90

Number of students study Chemistry = 70

Number of students study Mathematics and Physics = 40

Number of students study Mathematics and Chemistry = 50

Number of students study Physics and Chemistry = 30

Number of students study none of them = 20

Let U be the total number of students, P, M and C be the number of students study Physics, Mathematics and Chemistry respectively

To find: number of students who study all the three subjects n(M ∩ P ∩ C)

n(U) = 200, n(M) = 120, n(P) = 90, n(C) = 70, n(M ∩ P) = 40

n(M ∩ C) = 50, n(P ∩ C) = 30

Number of students who play either of them = n(P ∪ M ∪ C)

= Total – none of them

= 200 – 20

= 180……………(i)

Number of students who play either of them = n(P ∪ M ∪ C)

= n(C) + n(P) + n(M) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(P ∩ M ∩ C)

= 120 + 90 + 70 – 40 – 30 – 50 + n(P ∩ M ∩ C)

= 160 + n(P ∩ M ∩ C)……………(ii)

From (i) and (ii):

160 + n(P ∩ M ∩ C) = 180

⇒ n(P ∩ M ∩ C) = 180 – 160

⇒ n(P ∩ M ∩ C) = 20

Hence, there are 20 students who study all three subjects.

Given:

Total number of families = 10,000

Number of families buy newspaper A = 40%

Number of families buy newspaper B = 20%

Number of families buy newspaper C = 10%

Number of families buy newspaper A and B = 5%

Number of families buy newspaper B and C = 3%

Number of families buy newspaper A and C = 4%

Number of families buy all three newspapers = 2%

Let U be the total number of families, A, B and C be the number of families buy newspaper A, B and C respectively

(i) To find: number of families which buy newspaper A only

n(A) = 40%, n(B) = 20%, n(C) = 10%, n(A ∩ B) = 5%

n(B ∩ C) = 3%, n(A ∩ C) = 4%, n(A ∩ B ∩ C) = 2%

Percentage of families which buy newspaper A only

= n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)

= 40 – 5 – 4 + 2

= 33%

Number of families which buy newspaper A only

= 3300

Hence, there are 3300 families which buy newspaper A only

(b) To find: number of families which buy none of A, B and C

n(A) = 40%, n(B) = 20%, n(C) = 10%, n(A ∩ B) = 5%

n(B ∩ C) = 3%, n(A ∩ C) = 4%, n(A ∩ B ∩ C) = 2%

Percentage of families which buy either of A, B and C

= n(A ∪ B ∪ C)

= n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

= 40 + 20 + 10 – 5 – 3 – 4 + 2

= 60%

Percentage of families which buy none of A, B and C

= Total percentage – Number of students who play either

= 100% – 60%

= 40%

Number of families which buy none of A, B and C

= 4000

Hence, there are 4000 families which buy none of A, B and C

Given:

Total number of students = 50

Number of students studying English = 13

Number of students studying French = 17

Number of students studying Sanskrit = 15

Number of students studying English and French = 9

Number of students studying French and Sanskrit = 5

Number of students studying English and Sanskrit = 4

Number of students studying all three subjects = 3

Let U be the total number of students, E, F and S be the number of students passed in English, French and Sanskrit respectively

n(F ∩ S ∩ E) = a = 3

n(F ∩ S) = a + d = 5

⇒ 3 + d = 5

⇒ d = 2

n(F ∩ E) = a + b = 9

⇒ 3 + b = 9

⇒ b = 6

n(S ∩ E) = a + c = 4

⇒ 3 + c = 4

⇒ c = 1

n(F) = e + d + a + b = 17

⇒ e + 2 + 3 + 6 = 17

⇒ e + 11 = 17

⇒ e = 6

n(E) = g + c + a + b = 13

⇒ g + 1 + 3 + 6 = 13

⇒ g + 10 = 13

⇒ g = 3

n(S) = f + c + a + d = 15

⇒ f + 1 + 3 + 2 = 15

⇒ f + 6 = 15

⇒ f = 9

(i) Number of students study French only = e = 6

(ii) Number of students study English only = g = 3

(iii) Number of students study Sanskrit only = f = 9

(iv) Number of students studying English and Sanskrit but not French = c = 1

(v) Number of students studying French and Sanskrit but not English = d = 2

(vi) Number of students studying French and English but not Sanskrit = b = 6

(vii) Number of students studying at least one of the three languages

= a + b + c + d + e + f + g

= 3 + 6 + 1 + 2 + 6 + 9 + 3

= 30

(viii) Number of students studying none of the three languages = Total – (a + b + c + d + e + f + g)

= 50 – (3 + 6 + 1 + 2 + 6 + 9 + 3)

= 50 – 30

= 20

Given:

To find: value of n

Since elements are not repeating, number of elements in A₁∪ A₂∪ A₃∪ ………∪ A₃₀ is 30 × 5

But each element is used 10 times

So, 10 × S = 30 × 5

⇒ 10 × S = 150

⇒ S = 15

Since elements are not repeating, number of elements in B₁∪ B₂∪ B₃∪ ………∪ B_n is 3 × n

But each element is used 9 times

So, 9 × S = 3 × n

⇒ 9 × S = 3n

⇒ n = 45

Hence, the value of n is 45

Given: Two finite sets have m and n elements.

To find: value of m and n

Formula used:

The number of subsets of a set containing x elements is given by 2^x

According to question:

2^m – 2ⁿ = 112

⇒ 2ⁿ (2^m-n – 1) = 16 × 7

⇒ 2ⁿ (2^m-n – 1) = 2⁴ × 7

On comparing:

2ⁿ = 2⁴ and 2^m-n – 1 = 7

⇒ n = 4 and 2^m-n = 8

⇒ 2^m-n = 2³

⇒ m – n = 3

⇒ m – 4 = 3

⇒ m = 7

Hence, value of m and n is 7 and 4 respectively

To find: (A ∩ B′)′ ∪ (B ∩ C)

(A ∩ B′)′ ∪ (B ∩ C)

= [A’ ∪ (B′)′] ∪ (B ∩ C)

{∵ (A ∩ B)’ = A’ ∪ B’}

= (A’ ∪ B) ∪ (B ∩ C)

{∵ (B’)’ = B}

= A’ ∪ (B ∪ (B ∩ C))

= A’ ∪ B

Hence, (A∩B′)′∪(B∩C) = A’∪B

We know,

Every rectangle, square and rhombus is a parallelogram

But, no trapezium is a paralleogrm

Hence, F₁ = F₂∪F₃∪F₄∪F₁

Given: S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle

Since, the triangle and circle intersect each other and are contained in a square.

Clearly, S∪T∪C = S

Given: R be set of points inside a rectangle of sides a and b

Since, a, b > 1

a and b cannot be equal to 0

Therefore, R = {(x, y) : 0 < x < a, 0 < y < b}

Given:

Total number of students are 60

Students who play cricket and tennis are 25 and 20 respectively

Students who play both the games are 10

To find: number of students who play neither

Let S be the total number of students, C and T be the number of students who play cricket and tennis respectively

n(S) = 60, n(C) = 25, n(T) = 20, n(C ∩ T) = 10

Number of students who play either of them = n(C ∪ T)

= n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10

= 35

Number of student who play neither

= Total – n(C ∪ T)

= 60 – 35

= 25

Hence, there are 25 students who play neither cricket nor tennis.

Given:

Total number of persons are 840

Persons who read Hindi and English are 450 and 300 respectively

Persons who read both are 200

To find: number of persons who read neither

Let U be the total number of persons, H and E be the number of persons who read Hindi and English respectively

n(U) = 840, n(H) = 450, n(E) = 300, n(H ∩ E) = 200

Number of persons who read either of them = n(H ∪ E)

= n(H) + n(E) – n(H ∩ E)

= 450 + 300 – 200

= 550

Number of persons who read neither

= Total – n(H ∪ E)

= 840 – 550

= 290

Hence, there are 290 persons who read neither Hindi nor English.

Given: X = {8ⁿ – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}

X = {8ⁿ – 7n – 1 | n ∈ N}

On putting n = 1, 2, 3….

X = {0, 49, 490………}

Y = {49n – 49 | n ∈ N}

On putting n = 1, 2, 3, 4,….,11,…

Y = {0, 49, 98, 147,………, 490………}

Clearly, X⊂Y

Given: 63% of the people watch a News Channel whereas 76% watch another channel

To find: x% of the people watch both channel, then value of x

Let p% and q% of people watch a news channel and another channel respectively

n(p)= 63, n(q) = 76, n(p ∩ q) = x, n(p ∪ q) 100

We know,

n(p ∪ q) n(p) + n(q) – n(p ∩ q)

⇒ 100 63 + 76 – x

⇒ x 139 – 100

⇒ x 39

Now,

n(p ∪ q) n(p) and n(p ∪ q) n(q)

⇒ x 63 and x 76

Hence, 39x63

Given:

To find: A ∩ B or A ∪ B

For any element of A ∩ B, A and B will have same value of y

⇒ -x² = 1

⇒ x² = -1

Square of any value cannot be negative

Hence, there is no value of x for which A and B will have same value of y

⇒ A∩B =ϕ

Given: A and B are two given sets

To find: A ∩ (A ∪ B)

Let x ∈ A ∩ (A ∪ B)

⇒ x ∈ A and x ∈ (A ∪ B)

⇒ x ∈ A and (x ∈ A or x ∈ B)

⇒ (x ∈ A and x ∈ A) or (x ∈ A and x ∈ B)

⇒ x ∈ A or x ∈ A ∩ B

⇒ x ∈ A

Hence, A∩(A∪B) = A

Given: A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18}

To find: A′ ∪ (A ∪ B) ∩ B′

A′ ∪ (A ∪ B) ∩ B′

= A′ ∪ [(B’ ∩ A) ∪ (B’ ∩ B)]

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= A′ ∪ [(A ∩ B’) ∪ Φ]

{∵ (B’ ∩ B) = Φ}

= A′ ∪ (A ∩ B’)

= (A’ ∪ A) ∩ (A’ ∪ B’)

{∵ Distributive property of set:

(A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C)}

= Φ ∩ (A’ ∪ B’)

{∵ (A’ ∩ A) = Φ}

= (A’ ∪ B’)

= (A ∩ B)’

{∵ (A’ ∪ B’) = (A ∩ B)’}

A′ ∪ (A ∪ B) ∩ B′ = (A ∩ B)’

A contains all odd numbers and B contains all even numbers

Hence, A ∩ B = Φ

⇒ A′ ∪ (A ∪ B) ∩ B′ = {Φ}’

⇒ A′∪(A∪B)∩B′ = N

Given: S = {x | x is a positive multiple of 3 less than 100} and P = {x | x is a prime number less than 20}

To find: n(S) + n(P)

S = {x | x is a positive multiple of 3 less than 100}

⇒ S = {3, 6, 9, 12, 15,……, 99}

P = {x | x is a prime number less than 20}

⇒ P = {2, 3, 5, 7, 11, 13, 17, 19}

⇒ n(P) = 8

n(S) + n(P) = 33 + 8 = 41

Hence, answer is 41

Given: X and Y are two sets

To find: X ∩ (X ∪ Y)′

X ∩ (X ∪ Y)′

{∵ A’ ∪ B’ = (A ∩ B)’}

= X ∩ (X’ ∩ Y′)

= (X ∩ X’) ∩ (X ∩ Y’)

= Φ ∩ (X ∩ Y’)

{∵ A’ ∩ A = Φ}

= Φ

{∵ Φ ∩ A = Φ}

Hence, X∩(X∪Y)′=Φ

{1, 2}

Given: {x ∈ R: 1 ≤ x < 2}

To find: roster form of given set

Let A = {x ∈ R: 1 ≤ x < 2}

⇒ A = {1, 2}

1

Given: A = ϕ

To find: n[P(A)]

A = ϕ ⇒ n(A) = 0

P(A) is a power set and number of elements in power set is 2^x where x is number of elements in set A

Therefore,

n[P(A)] = 2⁰ = 1

n(B)

A ⊂ B ⇒ A ∪ B = B

⇒ n(A∪B) = n(B)

A ∩ B’

From Venn diagram you can clearly see,

A – B = A∩B’

{{1}, {2}, {1, 2}, Φ}

Given: set A = {1, 2}

To find: power set of A

A power set is a set of all the subsets of set A

Subsets of a are {1}, {2}, {1, 2}, Φ

Hence, power set of A = {{1}, {2}, {1, 2}, Φ}

{0, 1, 2, 3, 4, 5, 6, 8}

Given: A = {1, 3, 5}. B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

To find: universal set of all the three sets

Universal set is a set which contain elements of all sets

⇒ U = {0, 1, 2, 3, 4, 5, 6, 8}

(i) {1, 5, 9, 10} (ii) {1, 2, 3, 5, 6, 7, 9, 10}

Given: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}

To find: (B ∪ C)′ and (C – A)′

Formula used:

The union of two sets is a set containing all elements that are in both sets.

For example: {1, 2, 3} ∪ {2, 4} = {1, 2, 3, 4}

The difference (subtraction) is defined as: The set A – B consists of elements that are in A but not in B.

For example: if A = {1, 2, 3} and B = {3, 5}, then A−B = {1, 2}

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

B ∪ C = {2, 3, 4, 6, 7, 8}

(B ∪ C)’ = U – (B ∪ C)

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 3, 4, 6, 7, 8}

= {1, 5, 9, 10}

A = {1, 2, 3} and C = {2, 3, 4, 8}

C – A = {4, 8}

(C – A)’ = U – (C – A)

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {4, 8}

= {1, 2, 3, 5, 6, 7, 9, 10}

A ∩ B’

Given: A and B are two given sets

To find: A – (A ∩ B)

A – (A ∩ B)

= A ∩ (A ∩ B)’

{∵ A – B = A ∩ B’}

= A ∩ (A’ ∪ B’)

{∵ (A ∩ B)’ = A’ ∪ B’}

= (A ∩ A’) ∪ (A ∩ B’)

{∵ Distributive property of set:

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}

= Φ ∪ (A ∩ B’)

{∵ A ∩ A’ = Φ}

= A ∩ B’

A – (A∩B) = A∩B’

(i) ((A′ ∪ B′) – A)′

[(A′ ∪ B′) – A]′

= [(A′ ∪ B′) ∩ A’]’

{∵ A – B = A ∩ B’}

= [(A ∩ B)’ ∩ A’]’

{∵ (A ∩ B)’ = A’ ∪ B’}

= [(A ∩ B)’]’ ∪ (A’)’

= (A ∩ B) ∪ A

= A

((A′∪B′) – A)′ = A

(ii) [B′ ∪ (B′ – A)]′

[B′ ∪ (B′ – A)]′

= (B’)’ ∩ (B’ – A)’

{∵ (A ∪ B)’ = A’ ∩ B’}

= B ∩ (B’ ∩ A’)’

{∵ A – B = A ∩ B’}

= B ∩ [(B’)’ ∪ (A’)’]

{∵ (A ∩ B)’ = A’ ∪ B’}

= B ∩ (B ∪ A)

= B

[B′∪(B′ – A)]′ = B

(iii) (A – B) – (B – C)

Step 1:

A – B

Step 2:

B – C

Step 3:

(A – B) – (B – C)

Clearly, the Venn diagram in Step 3 shows same region as in Step 1

Hence, (A – B) – (B – C) = A – B

(iv) (A – B) ∩ (C – B)

Step 1:

A – B

Step 2:

C – B

Step 3:

(A – B)∩(B – C)

Clearly, from the Venn diagram in Step 3:

(A – B)∩(B – C) = A∩C – B

(v) A × (B ∩ C)

Let y ∈ A × (B ∩ C)

⇒ y ∈ A and y ∈ (B ∩ C)

⇒ y ∈ A and (y ∈ B and y ∈ C)

⇒ (y ∈ A and y ∈ B) and (y ∈ A and y ∈ C)

⇒ y ∈ (A × B) and y ∈ (A × C)

⇒ y ∈ (A × B) ∩ (A × C)

A × (B∩C) = (A × B)∩(A × C)

(vi) A × (B ∪ C)

Let y ∈ A × (B ∪ C)

⇒ y ∈ A and y ∈ (B ∪ C)

⇒ y ∈ A and (y ∈ B or y ∈ C)

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

⇒ y ∈ (A × B) or y ∈ (A × C)

⇒ y ∈ (A × B) ∪ (A × C)

A × (B∪C) = (A × B)∪(A × C)

True

Every set is subset of itself.

Hence, A⊂A is true

False

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

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

Every element of M and B is same

Hence, M = B and B ⊂ M

So, the given statement is false

False

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

Since, every element of both the sets are not same

They are not equal

So, the given statement is false

True

We know, every integer is a rational number

Hence, Z ⊂ Q and Q ∪ Z = Q

So, the given statement is true

True

R and T can be represented in roster form as follows:

R = {…,-8, -6, -4, -2, 0, 2, 4, 6, 8,….}

T = {…,-18, -12, -6, 0, 6, 12, 18,….}

Since, every element of T is present in R

T ⊂ R

So, the given statement is true

True

A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}

Roster form of B:

B = {0, 1, 2}

Every element of both the sets A and B are same

⇒ A = B

So, the given statement is true