Q.If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

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1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Given the sets $A=\{1,2,3\} ; B=\{2,3,5\}$ and $C=\{4,5,6\}$ identify which of the following statement is incorrect. (A) $B-C=\{2,3\}$ (B) $n[(A \cup B) \cap(B \cap C)]=1$ (C) $(A \cap B) \cup C=A \cap(B \cup C)$ (D) $(A \cap B) \cap C=\emptyset$

   ›Reveal solutionSolution

   Checking each statement, three are true and only option (C) asserts an equality that fails, so the incorrect statement is (C).

   Concept

   The problem tests basic set operations — union, intersection, difference and cardinality. Work each statement out explicitly; for an equality, compute both sides separately and compare. A frequent trap is to assume set operations distribute like ordinary arithmetic, which they do not.

   Solution

   Using $A = \{1,2,3\},\; B = \{2,3,5\},\; C = \{4,5,6\}$:

   1. (A) $B - C$: keep elements of $B$ not in $C$, giving $\{2,3\}$ — statement correct.
   2. (B) $A \cup B = \{1,2,3,5\}$ and $B \cap C = \{5\}$; their intersection is $\{5\}$, so $n = 1$ — statement correct. …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.If $A=\{x : x \text{ is the first three odd numbers}\}$, $B=\{2x+3 : 0\le x<5,\ x\in\mathbb{N}\}$, then which of the following is true (A) $A\subset B$ (B) $n(B)=5$ (C) $A\cap B=\emptyset$ (D) $A\cap B$ is a singleton set

   ›Reveal solutionSolution

   [!TLDR]

   Writing out $A=\{1,3,5\}$ and $B=\{3,5,7,9,11\}$ shows $B$ has exactly five elements, so $n(B)=5$ is the true statement.

   Concept

   This CBSE Class 11 Sets question is answered by roster-forming each set and testing subset/intersection/cardinality claims directly.

   Solution

   • $A$: the first three odd numbers are $1,3,5$, so $A=\{1,3,5\}$.
   • $B$: for $x=0,1,2,3,4$ (with $0\le x<5$), $2x+3$ gives $3,5,7,9,11$, so $B=\{3,5,7,9,11\}$ and $n(B)=5$.
   • Check (A): $A\subset B$ is false because $1\in A$ but $1\notin B$.
   • Check (C): $A\cap B=\{3,5\}\ne\emptyset$, so false.
   • Check (D): $A\cap B=\{3,5\}$ has two elements, so it is not a singleton, false. …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.The set expression $A \cup\left(B \cap\left(A^{\prime} \cup B^{\prime}\right)\right)$ is equivalent to (A) $\left(A^{\prime} \cup B^{\prime}\right)^{\prime}$ (B) $\xi$ (Universal set) (C) $A \cup B$ (D) $A \cap B^{\prime}$

   ›Reveal solutionSolution

   The expression simplifies to $A \cup B$ by applying De Morgan’s law and the distributive law. The correct option is (C).

   We start with the expression $A \cup \left( B \cap (A' \cup B') \right)$. The goal is to simplify it step by step using set identities. The key insight is to notice that $A' \cup B'$ is the complement of $A \cap B$ (by De Morgan’s law), so the inner part $B \cap (A' \cup B')$ is “everything in $B$ that is not in $A$”. Then unioning with $A$ gives all of $A$ plus that part of $B$ outside $A$, which is exactly $A \cup B$.

   Let’s verify this carefully.

   1. Apply De Morgan’s law to the innermost complement expression: $A' \cup B' = (A \cap B)'$. So the expression becomes:

   $A \cup \left( B \cap (A \cap B)' \right).$

   2. Interpret $B \cap (A \cap B)'$: This is the set of elements that are in $B$ but not in $A \cap B$. Since $A \cap B$ is the overlap of $A$ and $B$, removing it from $B$ leaves exactly the part of $B$ that is not in $A$. That is:

   $B \cap (A \cap B)' = B \cap A' = B - A.$

   3. Now the whole expression is:

   $A \cup (B \cap A').$

   4. Use the distributive law (or simply think: $A$ plus the part of $B$ outside $A$): $A \cup (B \cap A') = (A \cup B) \cap (A \cup A').$ …

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4. KCET 2025Set A-11 markMCQ
   Q.A and B are two sets having 3 and 6 elements respectively. Consider the following statements. Statement (I): Minimum number of elements in $A \cup B$ is 3 Statement (II): Maximum number of elements in $A \cap B$ is 3 Which of the following is correct? (A) Statement (I) is true, statement (II) is false (B) Statement (I) is false, statement (II) is true (C) Both statements (I) and (II) are true (D) Both statements (I) and (II) are false

   ›Reveal solutionSolution

   Both extremes occur when the smaller set sits inside the bigger one: then $A\cup B=B$ (size 6, so the minimum union is 6, not 3) and $A\cap B=A$ (size 3, so the maximum intersection is 3).

   Step 1 — The governing bounds.

   For finite sets,

   $\max\{n(A),n(B)\}\;\le\; n(A\cup B)\;\le\; n(A)+n(B)$

   $0 \;\le\; n(A\cap B)\;\le\; \min\{n(A),n(B)\}$

   Both follow from the inclusion–exclusion identity

   $n(A\cup B)=n(A)+n(B)-n(A\cap B)$

   A union can never be smaller than either of the sets it contains, and an intersection can never be bigger than the smaller set.

   Step 2 — Test Statement (I): "minimum $n(A\cup B)$ is 3".

   Here $n(A)=3$, $n(B)=6$. The union must contain all of $B$, so

   $n(A\cup B)\ge \max\{3,6\}=6$

   The minimum 6 is attained when $A\subseteq B$ (then $A\cup B=B$). It can never be 3.

   $\Rightarrow \textbf{Statement (I) is FALSE}$

   Step 3 — Test Statement (II): "maximum $n(A\cap B)$ is 3". …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.Let $A=\{x : x=4n+1,\ n\in\mathbb{Z},\ 0\le n<4\}$, $B=\{x : x=15n+4,\ n\in\mathbb{N},\ n\le 3\}$, $C=\{x : x \text{ is a prime number},\ x\in A\cup B\}$. Then the cardinal number of set $C$ is (A) $\emptyset$ (B) $3$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   Common terms of $A$ ($x=4n+1$) and $B$ ($x=5m+2$) satisfy $x\equiv17\pmod{20}$, i.e. $17,37,57,\dots$; within the stated ranges there are exactly 3 of them — option (B).

   A member of both sets needs $4n+1=5m+2$, i.e. $x\equiv1\pmod4$ and $x\equiv2\pmod5$.

   Combine by the Chinese Remainder Theorem. The two congruences (moduli 4 and 5, $\gcd=1$) give a unique residue mod $\operatorname{lcm}(4,5)=20$. Testing, $x=17$ satisfies $17=4(4)+1=5(3)+2$, so

   $x\equiv17\pmod{20}\quad\Rightarrow\quad x=17,\,37,\,57,\,77,\dots$

   The common terms form an AP with first term 17 and common difference 20. …

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6. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $A=\{1,2,4\} \quad B=\{2,4,5\} \quad C=\{2,5\}$ then $(A-B) \cap(B-C)=$ (A) $\{2,4,5\}$ (B) $\{1,2,4,5\}$ (C) $\emptyset$ (D) $\{4,5\}$

   ›Reveal solutionSolution

   The key idea is to compute the set differences first, then intersect them. The result is the empty set, so the correct option is (C).

   We start by recalling what set difference means: $A - B$ is the set of elements that are in $A$ but not in $B$. Similarly, $B - C$ is the set of elements in $B$ but not in $C$. The intersection $(A - B) \cap (B - C)$ then collects elements that belong to both of these difference sets. Intuitively, an element would have to be in $A$ but not in $B$, and also in $B$ but not in $C$ — which is impossible because if it’s not in $B$, it can’t be in $B$. So we expect the answer to be empty.

   Let’s verify step by step.

   1. Compute $A - B$

      $A = \{1, 2, 4\}$, $B = \{2, 4, 5\}$.

      Remove from $A$ any element that also appears in $B$.

      • $1$ is in $A$ but not in $B$ → keep.
      • $2$ is in both → remove.
      • $4$ is in both → remove. So $A - B = \{1\}$.

   2. Compute $B - C$

      $B = \{2, 4, 5\}$, $C = \{2, 5\}$.

      Remove from $B$ any element that also appears in $C$.

      • $2$ is in both → remove.
      • $4$ is in $B$ but not in $C$ → keep.
      • $5$ is in both → remove. So $B - C = \{4\}$.

   3. Intersect the results

      $(A - B) \cap (B - C) = \{1\} \cap \{4\}$. …

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7. COMEDK 2025Set 2025-E1 markMCQ
   Q.If $P=\{5 m: m \in N\}$ and $Q=\left\{5^m: m \in N\right\}$, where $N$ is set of natural numbers, then (A) $P=Q$ (B) $P \subset Q$ (C) $Q \subset P$ (D) $P \cup Q=N$

   ›Reveal solutionSolution

   The set $P$ contains all multiples of 5, while $Q$ contains only powers of 5. Since every power of 5 is a multiple of 5 but not every multiple of 5 is a power of 5, we have $Q \subset P$. The correct option is (C).

   Concept and intuition:

   We are comparing two sets defined over the natural numbers $N = \{1,2,3,\dots\}$.

   • $P = \{5m : m \in N\}$ means $P$ is the set of all positive multiples of 5: $5, 10, 15, 20, 25, 30, \dots$
   • $Q = \{5^m : m \in N\}$ means $Q$ is the set of all positive powers of 5: $5^1=5,\;5^2=25,\;5^3=125,\dots$

   The key insight: every power of 5 is certainly a multiple of 5, but not every multiple of 5 is a power of 5 (e.g., 10 is a multiple of 5 but not a power of 5). This tells us that $Q$ is a proper subset of $P$.

   Step-by-step reasoning:

   1. Understand the definitions

      $N$ is the set of natural numbers, typically $\{1,2,3,\dots\}$.

      • $P = \{5, 10, 15, 20, 25, 30, \dots\}$ — all numbers divisible by 5.
      • $Q = \{5, 25, 125, 625, \dots\}$ — numbers that are 5 raised to a natural exponent.

   2. Check if $P = Q$

      For equality, every element of $P$ must be in $Q$ and vice versa.

      Take $10 \in P$. Is $10$ a power of 5? No, because $5^1=5$ and $5^2=25$, and 10 is between them. So $10 \notin Q$. Hence $P \neq Q$. Option (A) is false.

   3. Check if $P \subset Q$

      This would mean every multiple of 5 is a power of 5. But as we just saw, 10 is a multiple of 5 but not a power of 5. So $P \not\subset Q$. Option (B) is false.

   4. Check if $Q \subset P$ …

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8. COMEDK 2025Set 2025-M1 markMCQ
   Q.If $n(A)=3$ and $n(B)=7$ and $A \subseteq B$ then the number of elements in $A \cap B$ is equal to (A) 7 (B) 10 (C) 0 (D) 3

   ›Reveal solutionSolution

   Since $A \subseteq B$, every element of $A$ is also in $B$, so the intersection $A \cap B$ is exactly $A$ itself. With $n(A)=3$, the number of elements in $A \cap B$ is $3$, making option (D) correct.

   The key idea here is the meaning of subset and intersection. When one set is contained entirely inside another, their overlap is just the smaller set. Many students mistakenly think intersection always means "common elements" in a vague sense, but here "common" means all of $A$ is common because $A$ lives inside $B$.

   Let’s walk through it step by step.

   1. Understand the given information.

      We have $n(A) = 3$ (set $A$ has 3 elements) and $n(B) = 7$ (set $B$ has 7 elements). Crucially, $A \subseteq B$ means every element of $A$ is also an element of $B$. So $A$ is a subset of $B$.

   2. What is $A \cap B$?

      The intersection $A \cap B$ is the set of all elements that belong to both $A$ and $B$. Since every element of $A$ is in $B$ (by the subset condition), every element of $A$ qualifies. Are there any elements of $B$ that are also in $A$? Only those that are already in $A$ — no extra ones. So $A \cap B = A$.

   3. Count the elements.

      If $A \cap B = A$, then the number of elements in the intersection is exactly $n(A) = 3$.

   4. Check the options.

      • (A) 7 — that would be $n(B)$, but the intersection can’t be larger than $A$. …

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9. COMEDK 2025Set 2025-M1 markMCQ
   Q.Identify the correct statement (A) $A \cup A^{\prime}=\emptyset$ (B) $A-B=A^{\prime} \cap B$ (C) $(A \cup B)^{\prime}=A^{\prime} \cup B^{\prime}$ (D) $A \subseteq B \Rightarrow B^{\prime} \subseteq A^{\prime}$

   ›Reveal solutionSolution

   The key idea is to test each set-theory statement using definitions and simple examples. Only statement (D) is always true: if $A \subseteq B$, then $B' \subseteq A'$.

   We need to check each option carefully. The best way is to recall the definitions of complement, difference, and subset, and test with a small universal set.

   1. Option (A): $A \cup A' = \emptyset$

      • $A'$ is the complement of $A$ (everything not in $A$).
      • The union of a set and its complement is always the universal set $U$, not the empty set.
      • For example, let $U = \{1,2,3\}$ and $A = \{1\}$. Then $A' = \{2,3\}$, so $A \cup A' = \{1,2,3\} = U \neq \emptyset$.
      • False.

   2. Option (B): $A - B = A' \cap B$

      • $A - B$ means elements in $A$ but not in $B$.
      • $A' \cap B$ means elements not in $A$ but in $B$.
      • These are completely different: one is inside $A$, the other is outside $A$.
      • Example: $U = \{1,2,3\}$, $A = \{1,2\}$, $B = \{2,3\}$. Then $A - B = \{1\}$, but $A' \cap B = \{3\}$.
      • False.

   3. Option (C): $(A \cup B)' = A' \cup B'$

      • This is a common mistake. De Morgan’s laws say: $(A \cup B)' = A' \cap B'$, not union.
      • Example: $U = \{1,2,3\}$, $A = \{1\}$, $B = \{2\}$. Then $(A \cup B)' = \{3\}$, but $A' \cup B' = \{2,3\} \cup \{1,3\} = \{1,2,3\}$.
      • False.

   4. Option (D): $A \subseteq B \Rightarrow B' \subseteq A'$ …

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10. CA Foundation 2025Set may-20251 markMCQ
    Q.If $A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$ and $C = \{3, 4, 5, 6\}$, the value of $A - \{B \cup C\}$ is (A) {1, 2, 3} (B) {2, 3, 4, 5} (C) {1} (D) {0}

    ›Reveal solutionSolution

    $B\cup C=\{2,3,4,5,6,8\}$; removing these from $A$ leaves $\{1\}$.

    Step 1 — Compute the union $B\cup C$

    $\{2,4,6,8\}\cup\{3,4,5,6\}=\{2,3,4,5,6,8\}$

    Step 2 — Compute the difference $A-(B\cup C)$

    Keep elements of $A=\{1,2,3,4\}$ NOT in the union. Elements $2,3,4$ are all present in the union; only $1$ survives.

    $A-(B\cup C)=\{1\}$

    Why the other options are wrong: (A) {1,2,3} and (B) {2,3,4,5} keep elements that ARE in the union; (D) {0} introduces $0$, which is in no set. …

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11. KCET 2024Set A-11 markMCQ
    Q.The negation of the statement “For every real number $x$ ; $x^2 + 5$ is positive” is (A) For every real number $x$ ; $x^2 + 5$ is not positive (B) For every real number $x$ ; $x^2 + 5$ is negative (C) There exists at least one real number $x$ such that $x^2 + 5$ is not positive (D) There exists at least one real number $x$ such that $x^2 + 5$ is positive

    ›Reveal solutionSolution

    Negating a "for every" statement flips the quantifier to "there exists" and negates the predicate — both changes must happen.

    Step 1 — Identify the logical form of the statement.

    The given statement is:

    $\forall x \in \mathbb{R},\ p(x) \qquad \text{where } p(x): \text{“}x^2 + 5 \text{ is positive”}$

    This is a universally quantified statement.

    Step 2 — Apply the rule for negating a quantifier.

    The standard rule of quantifier negation is

    $\sim\left(\forall x,\ p(x)\right) \;\equiv\; \exists x \text{ such that } \sim p(x)$

    Why: "It is not true that $p$ holds for every $x$" means precisely that $p$ fails for at least one $x$. It does not mean $p$ fails for all $x$ — that is a much stronger claim, and it is the classic trap in this question.

    Step 3 — Negate the predicate.

    $\sim p(x): \text{“}x^2 + 5 \text{ is NOT positive”}$

    Note carefully: the negation of "positive" is "not positive" (i.e. $\le 0$), not "negative" (i.e. $< 0$), because zero is neither.

    Step 4 — Assemble the negation.

    $\text{“There exists at least one real number } x \text{ such that } x^2 + 5 \text{ is not positive.”}$

    Step 5 — Eliminate the distractors.

    • (A) keeps "for every" — the quantifier was not flipped. ✗ …

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12. COMEDK 2024Set 2024-A1 markMCQ
    Q.

    [!FORMULA] $\text { If } a \mathcal{N}=\{a x: x \in \mathcal{N}\} \text {, then } 3 \mathcal{N} \cap 7 \mathcal{N} \text { is }$

    (A) $10 \mathcal{N}$ (B) $21 \mathcal{N}$ (C) $4 \mathcal{N}$ (D) $3 \mathcal{N}$

    ›Reveal solutionSolution

    The intersection of the sets of multiples of 3 and multiples of 7 is exactly the set of multiples of their least common multiple, which is 21. So the answer is $21\mathcal{N}$.

    We are asked: if $a\mathcal{N} = \{a x : x \in \mathcal{N}\}$ (where $\mathcal{N}$ is the set of natural numbers, usually $\{1,2,3,\dots\}$), then what is $3\mathcal{N} \cap 7\mathcal{N}$?

    Concept and intuition:

    The set $a\mathcal{N}$ is all positive multiples of $a$. The intersection of two such sets consists of numbers that are multiples of both $a$ and $b$. A number is a multiple of both if and only if it is a multiple of the least common multiple (LCM) of $a$ and $b$. Why? Because being a multiple of both means the number contains all prime factors of $a$ and all prime factors of $b$; the smallest number that does this is the LCM, and any common multiple is a multiple of that LCM. So the intersection is exactly the set of multiples of $\operatorname{lcm}(a,b)$.

    Step-by-step:

    1. Interpret the sets:

       $3\mathcal{N} = \{3, 6, 9, 12, 15, 18, 21, 24, \dots\}$ — all multiples of 3.

       $7\mathcal{N} = \{7, 14, 21, 28, 35, 42, \dots\}$ — all multiples of 7.

    2. Find the common elements:

       A number $n$ belongs to both sets exactly when $n$ is divisible by 3 and by 7. Since 3 and 7 are coprime (they share no common prime factors), the smallest positive number divisible by both is $3 \times 7 = 21$. Every common multiple must be a multiple of 21.

    3. General principle:

       For any positive integers $a$ and $b$,

       $a\mathcal{N} \cap b\mathcal{N} = \operatorname{lcm}(a,b)\,\mathcal{N}.$ …

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