Chapter 1: Sets

According to the question,

There are two sets A and B

To prove: (A ∪ B) – B = A – B

L.H.S = (A ∪ B) – B

Since, A – B = A ∩ B’, we get,

= (A ∪ B) ∩ B’

Since, Distributive property of set: (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,

= (A ∩ B’) ∪ (B ∩ B’)

Since, A ∩ A’ = Φ, we get,

= (A ∩ B’) ∪ Φ

= A ∩ B’

Since, A – B = A ∩ B’, we get,

= A – B

= R.H.S

Hence Proved

According to the question,

To check whether T is an empty set or not,

We solve,

⇒ -(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

Long Answer Type

According to the question,

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 that:

P ⊂ Q and Q ⊂ P ⇒ P = Q

From equations (i) and (ii), we have,

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Hence Proved

According to the question,

Total number of students = 100

Number of students who passed in English = 15

Number of students who passed in Mathematics = 12

Number of students who passed in Science = 8

Number of students who passed in English and Mathematics = 6

Number of students who passed in Mathematics and Science = 7

Number of students who passed in English and Science = 4

Number of students who passed in all three = 4

Let the total number of students = U

Let the number of students passed in English = E

Let the number of students passed in Mathematics = M

Let the number of students passed in Science = S

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

Therefore, from the above equations, we get that,

(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

According to the question,

Total number of students = 60

Students who play cricket = 25

Students who play tennis = 20

Students who play both the games = 10

To find: number of students who play neither

Let the total number of students = S

Let the number of students who play cricket = C

Let the number of students who play tennis = T

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

So, Number of students who play either of them,

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10

= 35

Hence, Number of student who play neither = Total – n(C ∪ T)

= 60 – 35

= 25

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

According to the question,

Total number of students = n(U) = 200

Number of students who study Mathematics = n(M) = 120

Number of students who study Physics = n(P) = 90

Number of students who study Chemistry = n(C) = 70

Number of students who study Mathematics and Physics = n(M ∩ P) = 40

Number of students who study Mathematics and Chemistry = n(M ∩ C) = 50

Number of students who study Physics and Chemistry = n(P ∩ C) = 30

Number of students who study none of them = 20

Let the total number of students = U

Let the number of students who study Mathematics = M

Let the number of students who study Physics = P

Let the number of students who study Chemistry = C

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

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

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(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 equation (i) and (ii), we get,

160 + n(P ∩ M ∩ C) = 180

⇒ n(P ∩ M ∩ C) = 180 – 160

⇒ n(P ∩ M ∩ C) = 20

Therefore, there are 20 students who study all the three subjects.

According to the question,

Total number of families = 10,000

Number of families buying newspaper A = n(A) = 40%

Number of families buying newspaper B = n(B) = 20%

Number of families buying newspaper C = n(C) = 10%

Number of families buying newspaper A and B = n(A ∩ B) = 5%

Number of families buying newspaper B and C = n(B ∩ C) = 3%

Number of families buying newspaper A and C = n(A ∩ C) = 4%

Number of families buying all three newspapers = n(A ∩ B ∩ C) = 2%

Let the total number of families = U

Let the number of families buying newspaper A = A

Let the number of families buying newspaper B = B

Let the number of families buying newspaper C = C

(a) Number of families which buy newspaper A only

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

= ((33/100)×10000)

= 3300

Hence, there are 3300 families which buy newspaper A only

(b) Number of families which buy none of A, B and C

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

= ((40/100)×10000)

= 4000

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

According to the question,

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 the total number of students = U

Let the number of students studying English = E

Let the number of students studying French = F

Let the number of students studying Sanskrit = S

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

Therefore, from the above equations, we get that,

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

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

(iii) Number of students studying 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

According to the question,

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

Now, since each element is used 10 times

We get,

10 × S = 30 × 5

⇒ 10 × S = 150

⇒ S = 15

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

Now, since each element is used 9 times

We get,

9 × S = 3 × n

⇒ 9 × S = 3n

⇒ S = n/3

⇒ n/3 = 15

⇒ n = 45

Therefore, the value of n is 45

Hence, Option (C) 45, is the correct answer.