Sets

Q1: If A = {1}, which of the following statements is correct:

Step 1: A is a set containing the number 1. So, A = {1}.
Step 2: Let’s examine each option:
a. A = 1 → This is false because A is a set, not a number.
b. \(1 \subset A\) → This is incorrect because 1 is not a set; only sets can be subsets of sets.
c. \(\left\{1\right\} \in A\) → This is false because the set {1} is not an element of A. A contains the number 1, not a set.
d. \(1 \in A\) → This is true because the element 1 is inside the set A = {1}.
Answer: d. \(1 \in A\)

Q2: If A = {1, 2, {3, 4}}, which of the following is a correct statement?

Step 1: Understand the elements of A.
A = {1, 2, {3, 4}}
So, A has 3 elements:
– 1
– 2
– The set {3, 4}
Step 2: Now evaluate each option.
a. \(3 \in A\): This is false because 3 is not directly in A. Only {3, 4} is in A, and 3 is an element inside that inner set.
b. \(\left\{1\right\} \subset A\): This is true because {1} is a subset of A — 1 is an element in A.
c. \(\left\{2\right\} \in A\): This is false. The set {2} is not an element of A — only the number 2 is.
d. \(1 \subseteq A\): This is false because 1 is not a set. Only sets can be subsets.
Answer: b. \(\left\{1\right\} \subset A\)

Q3: Consider the following statements:

1. Any set A is comparable with itself.
2. {0} is a singleton set.
3. {\(\phi\)} is an empty set.
Of these statements, the correct ones are:

Step 1: Analyze statement 1:
“Any set A is comparable with itself.”
This is true because every set is always a subset of itself, i.e., \(A \subseteq A\).
Step 2: Analyze statement 2:
“{0} is a singleton set.”
This is true because it contains only one element — 0 — hence it’s a singleton set.
Step 3: Analyze statement 3:
\(\phi\) is an empty set. But {\(\phi\)} is not the correct representation.
This is False.
Answer: a. 1 & 2

Q4: Total number of elements in the power set of a set A containing n elements is:

Step 1: Understand what a power set is.
A power set is the set of all subsets of a given set A.
This includes the empty set \(\phi\), all proper subsets, and the set itself.
Step 2: If A has \(n\) elements, the total number of subsets (including empty set and A itself) is given by: \[ \text{Number of subsets} = 2^n \]Step 3: So, the power set of a set with \(n\) elements contains \(2^n\) elements.
Step 4: Let’s compare the options:
a. \(n^2\) — Incorrect.
b. \(2^n\) — ✅ Correct.
c. \(2^n – 1\) — Incorrect, this gives number of non-empty subsets only.
d. none of these — Incorrect, since (b) is correct.
Answer: b. \(2^n\)

Q5: If a finite set S contains n elements, then the number of non-empty proper subsets of S is:

Step 1: Total number of subsets of a finite set with \(n\) elements is: \[ 2^n \]Step 2: These include:
• One empty set
• One set itself
• All other subsets are proper and non-empty
Step 3: So, number of proper subsets (excluding the set itself) = \[ 2^n – 1 \]Step 4: To find number of non-empty proper subsets, we must remove the empty set also: \[ (2^n – 1) – 1 = 2^n – 2 \]Step 5: Let’s simplify and verify the options:
a. \({2 \cdot 2}^{n-1} = 2^n\) → ❌ gives total subsets
b. \(2^n\) → ❌ includes empty and full set
c. \(2^n – 1\) → ❌ excludes only the full set, not empty
d. \(2\left(2^{n-1} – 1\right)\) → ✅ let’s check: \[ 2 \cdot (2^{n-1} – 1) = 2^n – 2 \] ✔️ This is the correct number of non-empty proper subsets
Answer: d. \(2\left(2^{n-1} – 1\right)\)

Q6: The number of all possible proper subsets of {2, 3, 5} are:

Step 1: Total number of elements in the set = 3
Set: {2, 3, 5}
Step 2: Total number of subsets of a set with \(n\) elements is given by: \[ \text{Total subsets} = 2^n = 2^3 = 8 \]Step 3: Proper subsets exclude the set itself.
So, number of proper subsets = \(2^n – 1 = 8 – 1 = 7\)
Step 4: Let’s list them for verification:
Subsets of {2, 3, 5} are:
\(\phi\), {2}, {3}, {5}, {2, 3}, {2, 5}, {3, 5}, {2, 3, 5}
Removing the full set → 7 proper subsets ✅
Answer: c. 7

Q7: If A = {a, b}, then the power set of A is:

Step 1: Power set of a set A is the set of all possible subsets of A.
Given: \(A = \left\{a, b\right\}\)
Number of subsets = \(2^n = 2^2 = 4\)
Step 2: Let us list all subsets of A:
• \(\phi\) (empty set)
• \(\left\{a\right\}\)
• \(\left\{b\right\}\)
• \(\left\{a, b\right\}\)
Step 3: Therefore, power set of A is: \[ \mathcal{P}(A) = \left\{\phi,\ \left\{a\right\},\ \left\{b\right\},\ \left\{a,b\right\}\right\} \]Answer: d. \(\phi,\ \left\{a\right\},\ \left\{b\right\},\ \left\{a,b\right\}\)

Q8: Which one of the following is a correct statement?

Step 1: Understand the symbol \(\phi\):
\(\phi\) represents the empty set, i.e., a set that contains no elements.
Step 2: Now examine each option:
a. \(\phi = 0\) → ❌ Incorrect.
0 is a number, whereas \(\phi\) is a set with no elements.
b. \(\phi = \left\{0\right\}\) → ❌ Incorrect.
\(\left\{0\right\}\) is a set with one element, which is 0.
But \(\phi\) has no elements at all.
c. \(\phi = \left\{\phi\right\}\) → ❌ Incorrect.
\(\left\{\phi\right\}\) is a set containing the empty set as an element.
So, it has one element: \(\phi\).
d. \(\phi = \ \) → ✅ Correct.
This option indicates that the empty set is simply an empty set (i.e., contains nothing).
Step 3: Hence, the only correct representation of the empty set is: \[ \phi = \{\ \} \]Answer: d. \(\phi = \ \)

Q9: Which one of the following is a correct statement?

Step 1: Let’s examine option (a):
\(\left\{a\right\} \in \left\{a,b,c\right\}\)
Here, \(\left\{a\right\}\) is a set, but all elements inside \(\left\{a,b,c\right\}\) are just elements, not sets.
So, this is False.
Step 2: Examine option (b):
\(a \subseteq \left\{a,b,c\right\}\)
This is also False because \(a\) is an element, not a set. Only sets can be subsets.
Step 3: Examine option (c):
\(a \in \left\{\left\{a\right\},b\right\}\)
The set \(\left\{\left\{a\right\},b\right\}\) contains two elements: a set \(\left\{a\right\}\) and an element b.
But it does not contain a directly. So this is also False.
Step 4: Since none of the options (a), (b), or (c) are correct,
Answer: d. None of these

Q10: Which one of the following is a correct statement?

Step 1: Consider option (a):
“Every subset of an infinite set is infinite.”
This is False because infinite sets can have finite subsets.
For example, \(\mathbb{N} = \{1,2,3,\ldots\}\) is infinite but \(\{1,2\}\) is a finite subset.
Step 2: Consider option (b):
“Every set has a proper subset.”
This is False for the empty set \(\phi\), which has no proper subsets.
Step 3: Consider option (c):
“\{a, b, c, 1, 2, 3, a, b, c, 1, 2, 3, ….\} is an infinite set.”
Repeating elements do not add new elements in a set.
But the pattern continues infinitely, so this is indeed an infinite set.
Hence, option (c) is True.
Step 4: Consider option (d):
“Every subset of a finite set is finite.”
This is True, as subsets of finite sets cannot be infinite.
Step 5: Therefore, both (c) and (d) are correct.
Since the question asks “Which one,” the best answer is the most precise true statement.
Option (d) is a well-known fundamental fact in set theory.
Answer: d. Every subset of a finite set is finite.

Q11: Which of the following is a singleton set?

Step 1: Understand what a singleton set is:
A singleton set contains exactly one element.
Step 2: Analyze each option:
a. Solve \(x^2 = x\) in \(\mathbb{R}\): \[ x^2 = x \\ x^2 – x = 0 \\ x(x – 1) = 0 \] Thus, \[ x = 0 \text{ or } x = 1 \] So, the set contains two elements: \(\{0,1\}\).
Hence, option (a) is not a singleton.
b. Solve \(3x = 4\) in \(\mathbb{N}\): \[ x = \frac{4}{3} \] But \(\frac{4}{3}\) is not a natural number.
Therefore, this set is empty (\(\phi\)), not singleton.
c. Solve \(x^2 = -1\) in \(\mathbb{R}\):
No real solution exists because square of any real number is \(\geq 0\).
Hence, the set is empty (\(\phi\)), not singleton.
d. “x is an integer which is neither positive nor negative.”
The only integer satisfying this is \(0\).
So the set is \(\{0\}\), which is a singleton set.
Step 3: Conclusion:
Only option (d) is a singleton set.
Answer: d. {x : x is an integer which is neither positive nor negative}

Q12: Which one of the following is an infinite set?

Step 1: Analyze option (a):
Set of natural numbers less than 50 means numbers from 1 to 49.
Number of elements is finite (49).
So, (a) is a finite set.
Step 2: Analyze option (b):
Set of integers less than 50 means all integers less than 50, including negative numbers, zero, and positive numbers less than 50.
Since integers extend infinitely in the negative direction, this set contains infinitely many elements.
So, (b) is an infinite set.
Step 3: Analyze option (c):
Set of integer factors of 500.
Factors of 500 are finite in number.
So, (c) is a finite set.
Step 4: Analyze option (d):
Set of whole numbers less than 1000 means numbers from 0 to 999.
Number of elements is finite (1000).
So, (d) is a finite set.
Step 5: Conclusion:
Only option (b) is an infinite set.
Answer: b. \(\left\{x : x \text{ is an integer}, x < 50 \right\}\)

Q13: Which one of the following is a finite set?

Step 1: Analyze option (a):
Set of integers less than 1 means \(\{…, -3, -2, -1, 0\}\).
This set extends infinitely in the negative direction.
So, (a) is an infinite set.
Step 2: Analyze option (b):
Set of rational numbers between 0 and 1.
Between any two numbers, there are infinitely many rational numbers.
So, (b) is an infinite set.
Step 3: Analyze option (c):
Set of natural numbers greater than 5 means \(\{6, 7, 8, 9, …\}\).
This set continues infinitely.
So, (c) is an infinite set.
Step 4: Analyze option (d):
Set of even prime numbers.
The only even prime number is 2.
So, the set is \(\{2\}\), which contains exactly one element.
Therefore, (d) is a finite set.
Step 5: Conclusion:
Only option (d) is a finite set.
Answer: d. \(\left\{x : x \text{ is an even prime} \right\} = \{2\} \)

Q14: If A = {1, 2, 3, 4} and B = {5, 6, 7}, then \(A \cap B\) = ?

Step 1: Recall the definition of intersection:
\(A \cap B = \{x | x \in A \text{ and } x \in B\}\)
Step 2: Check common elements between A and B:
A = {1, 2, 3, 4}
B = {5, 6, 7}
No element is present in both sets.
Step 3: Since there are no common elements,
the intersection \(A \cap B\) is the empty set \(\phi\).
Answer: d. \(\phi\)

Q15: Let A = {0, 1, 3, 4}, B = {5, 6, 1, 3, 9} and C = {0, 1, 2, 3, 9, 13}. Then, \((A \cap B) \cup C\) is:

Step 1: Find the intersection \(A \cap B\):
\(A = \{0, 1, 3, 4\}\)
\(B = \{5, 6, 1, 3, 9\}\)
Common elements in both sets are 1 and 3.
So, \(A \cap B = \{1, 3\}\).
Step 2: Find the union \((A \cap B) \cup C\):
\(C = \{0, 1, 2, 3, 9, 13\}\)
Union of \(\{1, 3\}\) and \(C\) is: \[ \{1, 3\} \cup \{0, 1, 2, 3, 9, 13\} = \{0, 1, 2, 3, 9, 13\} \]Step 3: So, \[ (A \cap B) \cup C = \{0, 1, 2, 3, 9, 13\} \]Answer: a. {0, 1, 2, 3, 9, 13}

Q16: Let A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Then, \((A \cup B) \cap C\) is:

Step 1: Find the union \(A \cup B\):
\(A = \{1, 2, 3, 4, 5\}\)
\(B = \{2, 4, 6, 8\}\)
Union includes all unique elements: \[ A \cup B = \{1, 2, 3, 4, 5, 6, 8\} \]Step 2: Find the intersection \((A \cup B) \cap C\):
\(C = \{3, 4, 5, 6\}\)
Common elements between \(\{1, 2, 3, 4, 5, 6, 8\}\) and \(\{3, 4, 5, 6\}\) are: \[ \{3, 4, 5, 6\} \]Step 3: Therefore, \[ (A \cup B) \cap C = \{3, 4, 5, 6\} \]Answer: b. {3, 4, 5, 6}

Q17: If A has 3 elements and B has 6 elements, then the minimum and maximum number of elements in \(A \cup B\) are respectively:

Step 1: Understand the union of sets \(A\) and \(B\): \[ |A \cup B| = |A| + |B| – |A \cap B| \]Step 2: Given: \[ |A| = 3, \quad |B| = 6 \] The size of intersection \(|A \cap B|\) can vary from:
Minimum intersection: 0 (when no common elements)
Maximum intersection: 3 (since \(A\) has only 3 elements)
Step 3: Calculate minimum number of elements in \(A \cup B\):
When \(|A \cap B| = 3\), all elements of \(A\) are in \(B\). So, \[ |A \cup B| = 3 + 6 – 3 = 6 \]Step 4: Calculate maximum number of elements in \(A \cup B\):
When \(|A \cap B| = 0\), no elements are common. So, \[ |A \cup B| = 3 + 6 – 0 = 9 \]Step 5: Therefore,
Minimum number of elements = 6
Maximum number of elements = 9
Answer: c. 6 and 9

Q18: If A has 3 elements and B has 6 elements, then the minimum and maximum number of elements in \(A \cap B\) are respectively:

Step 1: Understand the intersection of sets \(A\) and \(B\):
The intersection \(A \cap B\) contains elements common to both sets.
Step 2: Given: \[ |A| = 3, \quad |B| = 6 \] The number of common elements \(|A \cap B|\) can be:
– Minimum: 0 (if no element is common)
– Maximum: 3 (since \(A\) has only 3 elements, intersection can’t have more than 3 elements)
Step 3: So,
Minimum number of elements in \(A \cap B = 0\)
Maximum number of elements in \(A \cap B = 3\)
Answer: a. 0 and 3

Q19: Consider the following statements:
1. \(\left(A \cup B\right)^\prime = A^\prime \cup B^\prime\)
2. \(\left(\phi^\prime\right)^\prime = \xi\)
3. \(A \cap \left(B \cup C\right) = \left(A \cap B\right) \cup \left(A \cap C\right)\)
4. \(\xi^\prime = \phi\)
Of these statements:

Step 1: Verify statement 1: \[ \left(A \cup B\right)^\prime = A^\prime \cup B^\prime \] By De Morgan’s law: \[ \left(A \cup B\right)^\prime = A^\prime \cap B^\prime \] So, statement 1 is FALSE.
Step 2: Verify statement 2: \[ \phi^\prime = \xi \Rightarrow (\phi^\prime)^\prime = \xi^\prime = \phi \] But statement 2 says: \[ (\phi^\prime)^\prime = \xi \] This is FALSE.
Step 3: Verify statement 3: \[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \] This is the Distributive Law of sets, which is TRUE.
Step 4: Verify statement 4: \[ \xi^\prime = \phi \] Complement of the universal set is the empty set, so statement 4 is TRUE.
Conclusion:
Statements 3 and 4 are TRUE.
Answer: d. 3 and 4 are correct

Q20: If \(A \subseteq B\), then \(A \cap B\) is equal to

Step 1: Understand the given condition:
\(A \subseteq B\) means every element of set \(A\) is also in set \(B\).
Step 2: Find the intersection \(A \cap B\): \[ A \cap B = \{x \mid x \in A \text{ and } x \in B\} \] Since all elements of \(A\) are in \(B\), the intersection contains all elements of \(A\).
Step 3: Therefore, \[ A \cap B = A \]Answer: a. A

Q21: If \(A \subseteq B\), then \(A \cup B\) is equal to

Step 1: Understand the given condition:
\(A \subseteq B\) means every element of set \(A\) is also an element of set \(B\).
Step 2: Find the union \(A \cup B\): \[ A \cup B = \{ x \mid x \in A \text{ or } x \in B \} \] Since all elements of \(A\) are already in \(B\), the union will contain all elements of \(B\).
Step 3: Therefore, \[ A \cup B = B \]Answer: b. B

Q22: Consider the following sets:
\(A = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \right\}\),
\(B = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 5 \right\}\),
\(C = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 9 \right\}\).
Find \(A \cap B\):

Step 1: List the elements of sets \(A\) and \(B\): \[ A = \{1, 3\} \quad \text{(factors of 3)}
B = \{1, 5\} \quad \text{(factors of 5)} \]Step 2: Find the intersection \(A \cap B\): \[ A \cap B = \{x : x \in A \text{ and } x \in B\} = \{1\} \]Step 3: Analyze the options:
– Option (a) describes union-like property, not intersection.
– Option (b) states factors common to both 3 and 5.
– Option (c) states factors of 15 (LCM of 3 and 5). Factors of 15 are \(\{1, 3, 5, 15\}\).
– Option (d) says both (b) and (c).
Since common factors of 3 and 5 are \(\{1\}\) only,
– Option (b) implies factors of both 3 and 5 which is \(\{1\}\).
– Option (c) contains all factors of 15.
Answer: d. Both (b) and (c)

Q23: Given the sets:
\(A = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \right\}\) and
\(B = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 5 \right\}\),
Find \(A \cup B\):

a. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \text{ or } 5 \right\}\)
b. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \text{ or } 5 \text{ or both} \right\}\)
c. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of } 15 \right\}\)
d. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of both } 3 \text{ and } 5 \right\}\)
Step 1: Recall the definition of union: \[ A \cup B = \{x : x \in A \text{ or } x \in B\} \]Step 2: Elements of the sets: \[ A = \{1, 3\} \quad \text{(factors of 3)}
B = \{1, 5\} \quad \text{(factors of 5)} \]Step 3: Find union: \[ A \cup B = \{1, 3, 5\} \]Step 4: Analyze options:
– Option (a) means factors of 3 or factors of 5, which is union.
– Option (b) is similar and correct (including both or either).
– Option (c) factors of 15 are \(\{1,3,5,15\}\), which is not the union.
– Option (d) factors common to both 3 and 5 (intersection), not union.
Since (a) and (b) express the union correctly,
Answer: b. \(\left\{x:x \in \mathbb{N}, x \text{ is a factor of } 3 \text{ or } 5 \text{ or both}\right\}\)

Q24: Given the sets:
\(A = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \right\}\) and
\(C = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 9 \right\}\),
Find \(A \cup C\):

Step 1: List the elements of each set: \[ A = \{1, 3\} \quad \text{(factors of 3)}
C = \{1, 3, 9\} \quad \text{(factors of 9)} \]Step 2: Find the union: \[ A \cup C = \{1, 3, 9\} \]Step 3: Compare with options:
– Option (a) contains only factors of 3, which is a subset of union.
– Option (b) contains factors of 9, which is exactly the union.
– Option (c) contains factors of 27: \(\{1, 3, 9, 27\}\), which is not the union.
– Option (d) is incorrect as option (b) is correct.
Answer: b. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of } 9 \right\}\)

Q25: Given the sets:
\(A = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \right\}\) and
\(C = \left\{x : x \in \mathbb{N}, x \text{ is a factor of } 9 \right\}\),
Find \(A \cap C\):

Step 1: List the elements of each set: \[ A = \{1, 3\} \quad \text{(factors of 3)}
C = \{1, 3, 9\} \quad \text{(factors of 9)} \]Step 2: Find the intersection: \[ A \cap C = \{1, 3\} \]Step 3: Compare with options:
– Option (a) is factors of 3, which is \(\{1,3\}\), exactly the intersection.
– Option (b) is factors of 9, which includes 9, so not the intersection.
– Option (c) is factors of 27, which is larger.
– Option (d) is incorrect.
Answer: a. \(\left\{x : x \in \mathbb{N}, x \text{ is a factor of } 3 \right\}\)

Q26: How many of the following statements are true:
I. \(A⊂B\)
II. \(A⊆C\)
III. \(B⊆A\)
IV. \(C⊆A\)

Step 1: Analyze each statement:
– \(A = \{1,3\}\), \(B = \{1,5,9\}\), \(C = \{1,3,9\}\)
Check 1: \(A \subset B\)?
Is every element of A in B?
Elements of A: 1, 3
B contains 1 but not 3.
So, False.
Check 2: \(A \subseteq C\)?
Is every element of A in C?
A: 1,3
C: 1,3,9
Yes, all elements of A are in C.
So, True.
Check 3: \(B \subseteq A\)?
B: 1,5,9
A: 1,3
B has 5 and 9 which are not in A.
So, False.
Check 4: \(C \subseteq A\)?
C: 1,3,9
A: 1,3
C has 9 which is not in A.
So, False.
Number of True statements = 1
Answer: b. One