Intersection of Sets MCQ Quiz - Objective Question with Answer for Intersection of Sets - Download Free PDF

X = {8, 14, 20, .......368,374}

8 = 2 + 6 (1) and 374 = 2 + 6 (62)

Therefore, there are 62 elements in X. If we form pairs like (8,374), (14, 368), there will be 31 pairs. From each pair, if we can choose only one number, the sum of no two numbers selected will be 382. Thus, we can choose at the most 31 numbers so that the given condition is satisfied.

Calculation:

R ∩ 3  = [3, 6] ∩ 3 = 3 

R ∩ 3 × P = 3× [4, 5]  = {(3, 4), (3, 5)} 

Hence, The Correct Answer is option 3.

Answer : 1

Solution :

To find n(XY), the number of elements in the intersection of X and Y, we first need to determine the sets X and Y themselves. These correspond to the sets of all positive divisors of 400 and 1000, respectively.

First, let's factorize 400 and 1000:

• 400 = 2⁴.5²

• 1000 = 2³.5³

The set of divisors for a number n = p^a. q^b is given by varying the powers of p and q from 0 to their maximum in the factorization. Here, to find X ∩ Y, we need the maximum powers of primes that occur in both 400 and 1000.

The greatest common divisor (GCD) of 400 and 1000 incorporates the lowest powers of the common prime factors:

• For prime factor 2: the lower of 4 (from 400) and 3 (from 1000) is 3.

• For prime factor 5: the lower of 2 (from 400) and 3 (from 1000) is 2.

Thus, GCD(400, 1000) = 2³.5² = 200,

The set of divisors of 200 (which represents X ∩ Y) is formed by taking all combinations of 2⁰, 2¹, 2², 2³ and 5⁰, 5¹, 5²:

• Divisors of 200:

。2⁰.5⁰ = 1

。2⁰.5¹ = 5

。2⁰.5² = 25

。2¹.5⁰ = 2

。2¹.5¹ = 10

。2¹.5² = 50

。2².5⁰ = 4

。2².5¹ = 20

。2².5² = 100

。2³.5⁰ = 8

。2³.5¹ = 40

。2³.5² = 200

Counting these, there are 12 divisors. Therefore, n(X ∩ Y) = 12.

This corresponds to Option 1 : 12.

\(\Longrightarrow { 4 }^{ n }-3n-1\)

\(\Longrightarrow { \left( 1+3 \right) }^{ n }-3n-1\)

\(\Longrightarrow 1+3n+{ n }_{ { C }_{ 2 } }{ 3 }^{ 2 }+\ldots+{ 3 }^{ n }-3n-1\)

\(\Longrightarrow 9k\)

So, the elements of \(X\) are multiples of \(9\),

The elements of the set \(Y\) are also multiples of \(9\),

But if we extend the domain of \(X\) to the set of real numbers there would be other elements which are not present in \(Y\) as the elements of \(Y\) are only natural numbers whereas the elements of \(X\) can also be rational and irrational,

So the elements of \(Y\) are contained in \(X\) for all natural numbers,

\(\therefore X\cap Y=Y\)

we have,

(-1, 0) ∈ A x A and (0, 1) ∈ A x A ⇒ -1, 0, 1 ∈ A

But, A x A has 9 elements. Therefore, A has 3 elements.

Hence, A = {-1, 0, 1}.

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:

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

Calculation:

The set X is given by

{2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98}.

We want to find the maximum size of a subset S of X such that no two elements sum to 100.

The pairs in X that sum to 100 are

(2, 98), (6, 94), (10, 90), (14, 86), (18, 82), (22, 78), (26, 74), (30, 70), (34, 66), (38, 62), (42, 58), (46, 54), (50, 50){note: 50 appears only once in X }

Therefore, 

To maximize the number of elements in S while ensuring no two elements sum to 100:

• Choose one element from each of the 12 pairs (but not both)
• Additionally, include the element 50

The maximum possible number of elements in S = 13

∴ The maximum possible number of elements in S be 13.

Concept:

(A ⋂ B) are the elements present in both the sets (A and B)

Calculation:

Here, A = {x: x is a natural number and 1 < x ≤ 4},

A = {2, 3, 4}

B = {x: x is a natural number and 4 < x ≤ 7}

B = {5, 6, 7}

Here, not a single element is common in both sets.

So, (A ∩ B) = ϕ 

Hence, option (4) is correct.

Concept:

• P - Q = P ∩ Q^C      ----(1)
• De Morgan's Laws: (P ∪ Q)^C = P^C ⋂ Q^C and (P ⋂ Q)^C = P^C ∪ Q^C      ----(2)
• Law of Double complementation: (P^C)^C = P      ----(3)

• Distributivity: P ⋂ (Q ∪ R) = (P ⋂ Q) ∪ (P ⋂ R)      ----(4)

Calculation:

A - (A - B)

⇒ A - (A ∩ B^C)      [using (1)]

⇒ A ∩ (A ∩ BC)^C      [using (1)]

⇒ A ∩ (A^C ∪ B)      [using (2) & (3)]

⇒ (A ∩ A^C) ∪ (A ∩ B)      [using (4)]

⇒ A ∩ B

Hence, A - (A - B) = A ∩ B.

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 '

Cardinality is the number of elements present in the set.

Calculation:

Here, A = {x : x is a square of a natural number and x is less than 100}

So, A = {1, 4, 9, 16, 25, 36, 49, 64, 81} and

B is a set of even natural numbers.

So, B = {2, 4, 6, 8, ....., }

Now, A ∩ B = {4, 16, 36, 64}

∴ Cardinality = n(A ∩ B) = 4

Hence, option (1) 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 = {(n, 2n): n ϵ N} when N is 1, 2, 3, ..........

= {(1, 2), (2, 4), (3, 6), ………} 

B = {(2n, 3n): n ϵ N} when N is 1, 2, 3, ..........

= {(2, 3), (4, 6), (6, 9), ………}

There is no member common to both the sets

 A ∩ B = ϕ

∴ A ∩ B has no set common.

Calculation:

Given:

A = {x ∈ R : x² + 6x – 7 < 0}

⇒ x² + 6x – 7 < 0

⇒ x² + 7x - x – 7 < 0

⇒ x (x + 7) – 1 (x + 7) < 0

⇒ (x – 1) (x + 7) < 0

∴ A ∈ (-7, 1)

Now, B = {x ∈ R : x² + 9x + 14 > 0}

⇒ x² + 9x + 14 > 0

⇒ x² + 7x + 2x + 14 > 0

⇒ x (x + 7) + 2(x + 7) > 0

⇒ (x + 7) (x + 2) > 0

∴ B ∈ (-∞, -7) ∪ (-2, ∞)

Now, A ∩ B

∴ A ∩ B = (-2, 1) ⇔ A ∩ B = {x ∈ R : - 2 < x < 1}

So, statement 1 is true,

Now, A ∪ B = R – {-7}

So, statement 2 is wrong,

Given

n(Q) = 36, n(R) = 40 and n(Q U R) = 50

Calculation

n(Q U R) = n(Q) + n(R) - n(Q ∩ R)

⇒ 50 = 36 + 40 - n(Q ∩ R)

⇒ 50 = 76 - n(Q ∩ R)

n(Q ∩ R) = 26

CONCEPT:

Intersection:

Let A and B be two sets. The intersection of A and B is the set of all those elements which are present in both the sets A and B.

The intersection of A and B is denoted by A ∩ B i.e A ∩ B = {x : x ∈ A and x ∈ B}

CALCULATION:

Given: A = {1, 2, 3, 4} and B = {x ∈ N : x ≤ 5}

The set B can be re-written as B = {1, 2, 3, 4, 5}

As we know that, A ∩ B = {x : x ∈ A and x ∈ B}

⇒ A ∩ B = {1, 2, 3, 4} = A

Hence, correct option is 2.