Operations on Sets MCQ Quiz - Objective Question with Answer for Operations on Sets - Download Free PDF

Given:

n(A – B) = 20 + x

n(B – A) = 3x

n(A ∩ B) = x + 1

n(A) = n(B)

Formula used:

Total elements in set A: n(A) = n(A – B) + n(A ∩ B)

Total elements in set B: n(B) = n(B – A) + n(A ∩ B)

Since n(A) = n(B), we equate the expressions.

Calculation:

n(A) = n(A – B) + n(A ∩ B)

⇒ n(A) = (20 + x) + (x + 1)

⇒ n(A) = 21 + 2x

n(B) = n(B – A) + n(A ∩ B)

⇒ n(B) = (3x) + (x + 1)

⇒ n(B) = 4x + 1

Since n(A) = n(B):

⇒ 21 + 2x = 4x + 1

⇒ 21 – 1 = 4x – 2x

⇒ 20 = 2x

⇒ x = 10

We need to find the value of (2x – 5):

⇒ (2 × 10 – 5) = 20 – 5

⇒ 15

∴ The correct answer is option (4).

Explanation:

⇒ Total student = 240

⇒10 students failed in every subject Hence, remaining total = 230

According to question

⇒ a + b + c + d + e + f + g = 230 ..... (1)

⇒ b + c + d = 110.... (2)

⇒ e+ f + g = 60 ..... (3)

Eq. (i) – Eqs. [(ii) + (iii)], we get

⇒ a = 230 –(110 + 60) = 60  

∴ Option (a) is correct.

Concept Used:

The number of one-one functions from a set with n elements to itself is n!

Calculation:

Given:

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

⇒ The set A has 4 elements.

⇒ The number of one-one functions from A to itself is 4!.

⇒ 4! = 4 × 3 × 2 × 1 = 24

Hence option 1 is correct

Calculation

A : log_2/π|sinx| + log_2/π|cosx| = 2

⇒ log_2/π (|sinx . cosx|) = 2 

⇒ \(|\sin 2 x|=\frac{8}{\pi^{2}}\)

Number of solution 4

B : let √x = t < 2

Then \(\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}-4)+3(\sqrt{\mathrm{x}}-2)+6=0\)

⇒ t² – 4t + 3t – 6 + 6 = 0

⇒ t² – t = 0, t = 0, t = 1

x = 0, x = 1 

again let √x = t > 2

then t² – 4t – 3t + 6 + 6 = 0 

⇒ t² – 7t + 12 = 0

⇒ t = 3, 4

Total number of solutions

n(A ∪ B) = 4 + 4 = 8

Hence option 3 is correct

Calculation

Let \(\frac{y}{x}=λ\)

y = λx

⇒ 10 × (⁹C₀ + ⁹C₁ + ⁹C₂ + ⁹C₃ + …. + ⁹C₉)

⇒ 10 × (2⁹)

⇒ 10 × 512

⇒ 5120

Explanation:

(A - B) ∪ (B - A) =  A Δ B

Symmetric Difference of two Sets: Let A and B be two sets. The symmetric difference of sets A and B is the set (A - B) ∪ (B - A) and is denoted as A Δ B.

i.e A Δ B = (A - B) ∪ (B - A)

The Venn diagram representation of symmetric difference of two sets is shown below

Concept:

Union of the sets: 

Union of two given sets is the set that contains those elements that are either in A or in B, or in both.

The  union of the sets A and B, denoted by A U B

Intersection of Sets:

The intersection of two given sets is the largest set which contains all the elements that are common to both the sets.

The intersection of Sets A and B, denoted by A ∩ B

Formula:  A ∪ B = A + B - A ∩ B

Calculation:

Given: A ∩ B = A

To Find: A ∪ B

As we know, 

A ∪ B = A + B - A ∩ B

⇒ A ∪ B = A + B - A

∴ A ∪ B = B

Concept:

De morgan's law:

(A ∪ B)^c = A^c ∩ B^c

(A ∩ B)^c = A^c ∪ B^c

Distributive law in sets:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Calculation:

Given: A and B are two sets

To Find: A ∩ (B ∪ A)^c

As we know, (A ∪ B)^c = A^c ∩ B^c

Therefore (B ∪ A)^c = B^c ∩ A^c

Now A ∩ (B ∪ A)^c = A ∩ (B^c ∩ A^c)

= (A ∩ B^c) ∩ (A ∩ A^c)             (Using distributive law)

= (A ∩ B^c) ∩ ϕ                        (∵ x ∩ ϕ = ϕ)

= ϕ 

Calculation:

Given: The four sets have 150, 180, 210 and 240 elements respectively

n(A) = 150

n(B) = 180

n(C) = 210

n(D) = 240

Each pair of sets has 15 elements

n(A ∩ B) = 15

n(A ∩ C) = 15

n(A ∩ D) = 15

n(B ∩ C) = 15

n(B ∩ D) = 15

n(C ∩ D) = 15

Each triple of sets has 3 elements

n(A ∩ B ∩ C) = 3

n(A ∩ B ∩ D) = 3

n(A ∩ C ∩ D) = 3

n(B ∩ C ∩ D) = 3

A ∩ B ∩ C ∩ D = ϕ 

n(A ∩ B ∩ C ∩ D) = 0

Now, number of elements in the union of 4 sets A, B, C and D

n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)

⇒ 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0

∴ The required number of elements is 702.

Concept:

A proper subset is one that contains few elements of the original set.

A superset is one that contains all the elements including elements of the original set.

Calculation:

The superset of {1, 2, 3, 4} and supersets of set {4} are

{4}, {1, 4}, {2, 4}, {3, 4}, {1, 2, 4}, {2, 3, 4), (1, 3, 4}, {1, 2, 3, 4}

Hence, there are 8 supersets of the set {4}.

∴ The required number of set is 8.

Concept:

Union of the sets: 

Union of two given sets is the set that contains those elements that are either in A or in B, or in both.

The  union of the sets A and B, denoted by A U B

Intersection of Sets:

The intersection of two given sets is the largest set which contains all the elements that are common to both the sets.

The intersection of Sets A and B, denoted by A ∩ B

Formula:  A ∪ B = A + B - A ∩ B

Calculation:

Given: A ∪ B = A

To Find: A ∩ B

As we know, 

A ∪ B = A + B - A ∩ B

⇒ A = A + B - A ∩ B

∴ A ∩ B = B

Concept:

(A ⋃ B) is the elements present in either of the sets A or B.

(A ⋃ B)' =  U - (A ⋃ B)

Calculation:

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

(A ⋃ B) = {1, 2, 3, 4, 6, 8}

⇒(A ⋃ B)' = U - (A ⋃ B)

= {5, 7, 9}

Hence, option (2) is correct. 

Concept:

The intersection of two sets X and Y is the set of elements that are common to both set X and set Y. It is denoted by X ∩ Y and is read ‘X intersection Y ’

Calculation:

A = {(2n, 3n): n ϵ N}

= {(2, 3), (4, 6), (6, 9), ………} And

B = {(3n, 5n): n ϵ N}

= {(3, 5), (6, 10), (9, 15), ………}

There is no member common to both the sets

∴ A ∩ B = ϕ

Concept:

Let A and B denote two sets of elements.

n(A) and n(B) are the numbers of elements present in set A and B respectively.

n(A - B) is the number of elements present in A but NOT in B.

n(B - A) is the number of elements present in B but NOT in A.

n(A ⋃ B) is the total number of elements present in either set A or B.

n(A ⋂ B) is the number of elements present in both the sets A and B.

• n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)
• n(A - B) = n(A) - n(A ⋂ B)
• n(B - A) = n(B) - n(A ⋂ B)

Calculation:

Let A be the set of people who like coffee and B be the set of people who like tea.

Given that n(A) = 37, n(B) = 52 and n(A ⋃ B) = 70.

Since, every person likes at least one drink (0 elements outside A and B),

we have:

n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

⇒ 70 = 37 + 52 - n(A ⋂ B)

⇒ n(A ⋂ B) = 89 - 70 = 19

People who like coffee and NOT tea is given by n(A - B) = n(A) - n(A ⋂ B) = 37 - 19 = 18.

Concept:

Let A and B denote two sets of elements.

• n(A) and n(B) are the number of elements present in set A and B respectively.
• n(A ⋃ B) is the total number of elements present in either set A or B.
• n(A ⋂ B) is the number of elements present in both the sets A and B.
• n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

Calculations:

Let A be the set of students who read "Hitavad" and B the set of students who read "Hindustan".

From the given information:

n(A ⋃ B) = 50 - 10 = 40.

n(A) = 30.

n(B) = 35.

By using n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B), we get:

40 = 30 + 35 - n(A ⋂ B)

⇒ n(A ⋂ B) = 25.

∴ The number of students who read both "Hitavad" and "Hindustan" newpapers is n(A ⋂ B) = 25.