Test: Sets - JEE MCQ

Concept:

The null set is a subset of every set. (ϕ ⊆ A)

Every set is a subset of itself. (A ⊆ A)

The number of subsets of a set with n elements is 2ⁿ.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2ⁿ

 = 2⁴ = 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2ⁿ - 1

 = 2⁴ - 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

Using the Venn Diagram,
(i) A - B = A - (A ∩ B) is true

(ii) A = (A ∩ B) ∪ (A - B) is true

(iii) A - (B ∪ C) = (A - B) ∪ (A - C) is false

A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1}

⇒ a ∈ (4, 6) and b ∈ (4, 6)

Therefore, A = {(a, b): a ∈ (4, 6) and b ∈ (4, 6)}

Now, B = {a, b) ∈ R × R : 4 (a − 6) ² + 9(b−5) ² ≤ 36}

From the conditions above for set A, the maximum value of B is:

4 (a − 6) ² + 9(b−5) ² = 4 (4 − 6)² + 9(6−5)² = 25 ≤ 36

We can check for other values as well.

Elements of set A satisfy the conditions in Set B.

Hence, A is a subset of B.

n(A) = 4 and n(B) = 2

Therefore, n(A x B) = 8

Number of subsets of A x B having atleast 3 elements = 2⁸ – ⁸C₀ – ⁸C₁ – ⁸C₂

= 256 – 1 – 8 – 28

= 219

► n(Persons travelling by car) = 20 = n(C)
► n(Persons travelling by bus) = 50 = n(B)
► n(Persons travelling by both car and bus) = 10 = n(C ∩ B)
∴ n(C ∪ B) = n(C) + n(B) - n(C ∩ B)
= 20 + 50 - 10
= 60 

The correct answer is a) A

• B' gives us all the elements in U other than 6 and 7 i.e., B' = {1, 2, 3, 4, 5, 8, 9, 10}
• The intersection of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is set A itself.

• This method involves specifying a rule or condition which can be used to decide whether an object can belong to the set.
• This rule is written inside a pair of curly braces and can be written either as a statement or expressed symbolically or written using a combination of statements and symbols.
• {x :  x ≠ x}
  This implies a null set.

If A ∪ B = A ∩ B then A = B.
Example: Let A = {1, 2, 3, 4} & B = {1, 2, 3, 4}
(AUB) = (A⋂B)
⇒{1,2,3,4} = {1,2,3,4}

A = {2, 3, 4, 8, 10}
B = {3, 4, 5, 10, 12}
C = {4, 5, 6, 12, 14}
► (A ∩ B) = {3,4,10}
► (A ∩ C) = {4}
► (A ∩ B) U (A ∩ C) = {3,4,10}

• (A - B) is A but not B which means A excluding the intersection of A and B.
• Similarly, in the figure, the shaded region is A but not C and B which means A excluding B and C. But if B and C both are excluded from A then the intersection of A, B and C get excluded two times.
• So, we take B ∪ C to be excluded from A. Therefore the answer is A - (B ∪ C).

The correct answer is a) 219

⇒ Number of elements in Set A = 4
⇒ Number of elements in Set B = 2
∴ Number of elements in set (A × B) = 8
∴ Total number of subsets of (A×B) = 2⁸ = 256
⇒ Number of subsets having 0 elements = ⁸C₀ = 1
⇒ Number of subsets having 1 element = ⁸C₁ = 8
⇒ Number of subsets having 2 elements = ⁸C₂ = 28
∴ Number of subsets having at least 3 elements = 256 - 1 - 8 - 28 = 219

From Venn-Euler's Diagram

∴ (A ∪ B)′ ∪ (A′ ∩ B) = A′

The correct answer is b) 8

2a² + 3b² = 35

3b² = 35 – 2a²

The maximum value on RHS is 35.

3b² ≤ 35

b² ≤ 11.66

b = 0, ± 1, ± 2, ± 3….

3b² = 35 – 2a²

Any number multiplied by 2 gives an even number. 

3b² is odd.

If b = 0, then 3b² is not odd.

If b = ± 1, then 3b² is odd.

If b = ± 2, then 3b² is even.

If b = ± 3, then 3b² is odd.

3b² = 35 – 2a²

If b = ± 1, then b² = 1,

3 = 35 – 2a²

2a² = 32

a² = 16

a = ± 4

If b = ± 3, then b² = 9 

27 = 35 – 2a²

2a² = 8

a² = 4

a = ± 2

[(4, 1), (4, -1), (-4, 1), (-4, -1), (2, -3), (2, 3), (-2, 3), (-2, -3)]

Therefore there are 8 elements in total.

• n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 12 + 9 − 4 = 17
• Now, n((A ∪ B)^c) = n(U) − n(A ∪ B) = 20 − 17 = 3

Step-by-step Explanation:
Since we have given that,
Percentage of families own a cell phone = 25%
Percentage of families own a scooter = 15%
Percentage of families neither own = 65%
So, it becomes,
P(C ∪ S)^' = 1 - P(C ∪ S)
0.65 = 1 - P(C ∪ S)
1 - 0.65 = P(C ∪ S)
P(C ∪ S) = 0.35
So, it becomes, 
P(C ∪ S) = P(C) + P(S) - P(C ∩ S)
0.35 = 0.25 + 0.15 - x
0.35 = 0.40 - x
x = 0.05
Thus, 0.05 x Total number of families in town = 1500
Total number of families in the town =
1500 / 0.05 = 30,000
Hence, the total number of families in the town is 30,000.

The correct answer is d)

A={2,3,4,8,10},B={3,4,5,10,12},C={4,5,6,12,14}

To find : (A∩B) ∪(A∩C)
A∩B = elements common in sets A and B ⟹A∩B={3,4,10} 
A∩C = elements common in sets A and C ⟹A∩C={4}
Now, 
(A∩B)∪(A∩C)=elements in (A∩B) and (A∩C)
(A∩B)∪(A∩C)={3,4,10}∪{4}
(A∩B)∪(A∩C)={3,4,10}.

The correct answer is c) 9

Given A=[2,4,5],B=[7,8,9]

To find: n(A×B)=?

A×B={(2,7),(2,8),(2,9),(4,7),(4,8),(4,9),(5,7),(5,8),(5,9)}

∴n(A×B)=9

As n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
But in case of disjoint sets, n(A ∩ B) = 0
∴ n(A ∪ B) = n(A) + n(B)

The number of non-empty subsets = 2ⁿ − 1 = 2⁴ − 1 = 16 − 1 = 15
The subsets are:
[{1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,4}, {1,2,3}, {1,3,4} ,{2,3,4}, {1,2,3,4}]

As the opinions of different person is different about the intelligent student. So, we can't exactly have the same student's name in a set, hence it is not a well-defined collection.

A ∩ (B − A) = ψ,[∵ x ∈ B − A ⇒ x∉ A]

Since A ⊆ B
∴ A ∪ B = B
So, n(A ∪ B) = n(B) = 6

Let A be the set of students who have passed in Mathematics, and B be the set of students who have passed in Physics.
We are given that |A| = 55 and |B| = 67, where |A| and |B| represent the number of students in sets A and B, respectively.
We are also given that there are 100 students in total. We need to find the number of students who have passed in Physics only, which means we need to find |B - A|.
First, let's find the number of students who have passed in both Mathematics and Physics, which can be represented by |A ∩ B|.
Using the principle of inclusion-exclusion, we have:
|A ∪ B| = |A| + |B| - |A ∩ B|
Since there are 100 students in total, we can say that |A ∪ B| = 100.
Now, we can find |A ∩ B|:
100 = 55 + 67 - |A ∩ B|
100 = 122 - |A ∩ B|
|A ∩ B| = 22
Now, we can find the number of students who have passed in Physics only, which is |B - A|:
|B - A| = |B| - |A ∩ B|
|B - A| = 67 - 22
|B - A| = 45
so, the correct answer is 45.

• (A - B) is nothing but the elements which are present in set A but not present in set B.
• In other words, it is elements of set A excluding the elements which are in set B.
• Here A = {1, 2, 3, 4} and B = {4, 5, 6, 7}
  So, (A - B) = {1, 2, 3}