Q.Let $A = \{2, 4, 6, 8, 10\}$ and $B = \{4, 8, 12, 16\}$. Find $A \cup B$ and $A \cap B$. Then verify that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Showing the 12 most recent of 52 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The shaded region in the given Venn-diagram represents:

   (a) A ∪ B

   (b) A ∩ B

   (c) (A ∪ B)'

   (d) (A ∩ B)'

   ›Reveal solutionSolution

   The shaded region is everything in the universal set except A and B combined, which is exactly $(A \cup B)'$.

   The rectangle is the universal set $U$, and the two overlapping circles are sets $A$ and $B$. The description tells us the shading covers the rectangle except the two circles — i.e. every point that lies outside both $A$ and $B$.

   A point lies in $(A \cup B)'$ exactly when it is not in $A \cup B$, i.e. not in $A$ and not in $B$ (by De Morgan's law, $(A \cup B)' = A' \cap B'$). That is precisely the description of the shaded region.

   ✓Final answer

   The shaded region represents $(A \cup B)'$ — option (d).

2. CBSE 2026Set ANNUAL1 mark
   Q.If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 6, 7}, then (A ∪ B)' = ..............

   ›Reveal solutionSolution

   Find A∪B first, then take its complement in U.

   Given $U=\{1,2,3,4,5,6,7,8,9\}$, $A=\{2,4,6,8\}$, $B=\{2,3,6,7\}$.

   First find $A\cup B$ (all elements in A or B or both):

   $A\cup B = \{2,3,4,6,7,8\}$

   The complement is everything in $U$ not in $A\cup B$:

   $(A\cup B)' = U - (A\cup B) = \{1,5,9\}$

   ✓Final answer

   $(A\cup B)' = \{1,5,9\}$.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $X = \{1, 3, 5\}$ and $Y = \{1, 2, 3\}$ then $X \cap Y = ?$

   (a) $\{1, 2, 3, 4, 5\}$

   (b) $\{1, 2, 3, 5\}$

   (c) $\{1, 3\}$

   (d) $\phi$

   ›Reveal solutionSolution

   $X \cap Y$ consists of elements present in both $X$ and $Y$, which gives $\{1,3\}$.

   Given $X = \{1,3,5\}$ and $Y=\{1,2,3\}$. The intersection $X \cap Y$ contains only those elements that belong to BOTH sets.

   Check each element of $X$: is $1 \in Y$? Yes. Is $3 \in Y$? Yes. Is $5 \in Y$? No.

   So $X \cap Y = \{1,3\}$.

   ✓Final answer

   $X \cap Y = \{1,3\}$, which is option (c).

4. CBSE 2026Set ANNUAL1 mark
   Q.Write True/False: Sets $\{2, 6, 10\}$ and $\{3, 7, 11\}$ are disjoint sets.

   ›Reveal solutionSolution

   Sets are disjoint when their intersection is empty; comparing the elements of $\{2,6,10\}$ and $\{3,7,11\}$ shows no overlap.

   Set $A=\{2,6,10\}$ and set $B=\{3,7,11\}$.

   Comparing every element of A against B: $2\notin B$, $6\notin B$, $10\notin B$. None of A's elements are in B, so $A\cap B=\varnothing$.

   By definition, sets with empty intersection are disjoint sets.

   ✓Final answer

   True.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$A = \{1, 2, 3\}, B = \{2, 3, 7\} \Rightarrow A \cup B =$

   (a) $\{1, 2, 3\}$

   (b) $\{1, 3, 7\}$

   (c) $\{1, 2, 3, 7\}$

   (d) $\{1, 2, 7\}$

   ›Reveal solutionSolution

   $A \cup B = \{1, 2, 3, 7\}$: the union lists every element that is in $A$ or in $B$ (or both), each written once.

   For sets $A$ and $B$, the union is $A \cup B = \{x : x \in A \text{ or } x \in B\}$ — combine both sets and remove duplicate entries.

   Here $A = \{1, 2, 3\}$, $B = \{2, 3, 7\}$. Writing all elements of $A$ then adding any elements of $B$ not already listed: $1, 2, 3$ (from $A$), then $7$ (from $B$, since $2$ and $3$ are already present).

   So $A \cup B = \{1, 2, 3, 7\}$.

   ✓Final answer

   The correct option is (c) $\{1, 2, 3, 7\}$.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{3, 5, 7\}, Y = \{2, 3, 5\} \Rightarrow X \cap Y =$

   (a) $\{3, 2\}$

   (b) $\{3, 7\}$

   (c) $\{5, 7\}$

   (d) $\{3, 5\}$

   ›Reveal solutionSolution

   $X \cap Y = \{3, 5\}$: the intersection keeps only elements that belong to both sets.

   For sets $X$ and $Y$, $X \cap Y = \{x : x \in X \text{ and } x \in Y\}$.

   Here $X = \{3, 5, 7\}$ and $Y = \{2, 3, 5\}$. Checking each element of $X$ against $Y$: $3 \in Y$ (yes), $5 \in Y$ (yes), $7 \in Y$ (no). So $X \cap Y = \{3, 5\}$.

   ✓Final answer

   The correct option is (d) $\{3, 5\}$.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{1, 2\}, Y = \{2, 3, 5\}, Z = \{4, 6\} \Rightarrow X \cup Y \cup Z =$

   (a) $\{1, 2, 3, 5, 6\}$

   (b) $\{2, 3, 4, 5, 6\}$

   (c) $\{1, 2, 3, 4, 5, 6\}$

   (d) $\{1, 2\}$

   ›Reveal solutionSolution

   $X \cup Y \cup Z = \{1, 2, 3, 4, 5, 6\}$: list every element appearing in at least one of the three sets, once each.

   Given $X = \{1, 2\}$, $Y = \{2, 3, 5\}$, $Z = \{4, 6\}$.

   First take $X \cup Y = \{1, 2, 3, 5\}$ (2 is common, written once). Then union with $Z$: $\{1, 2, 3, 5\} \cup \{4, 6\} = \{1, 2, 3, 4, 5, 6\}$, since $Z$ shares no elements with the earlier union.

   ✓Final answer

   The correct option is (c) $\{1, 2, 3, 4, 5, 6\}$.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{1, 2, 3, 6\}, Y = \{4, 5, 6\}, Z = \{4, 2, 3, 6\} \Rightarrow (X \cup Y) \cap Z =$

   (a) $\{2, 3, 4, 6\}$

   (b) $\{1, 5\}$

   (c) $\{1, 2, 5\}$

   (d) $\{1, 2, 3, 5\}$

   ›Reveal solutionSolution

   $(X \cup Y) \cap Z = \{2, 3, 4, 6\}$, found by first taking the union, then intersecting with $Z$.

   Given $X = \{1, 2, 3, 6\}$, $Y = \{4, 5, 6\}$, $Z = \{4, 2, 3, 6\}$.

   Step 1: $X \cup Y = \{1, 2, 3, 4, 5, 6\}$ (combine both, 6 counted once).

   Step 2: $(X \cup Y) \cap Z$ keeps only elements also in $Z = \{2, 3, 4, 6\}$. Checking each element of $X \cup Y$ against $Z$: $1 \notin Z$, $2 \in Z$, $3 \in Z$, $4 \in Z$, $5 \notin Z$, $6 \in Z$. So the result is $\{2, 3, 4, 6\}$.

   ✓Final answer

   The correct option is (a) $\{2, 3, 4, 6\}$.

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.$X = \{a, b, c, d\}, Y = \{c, a, r\}, Z = \{r, o, b\} \Rightarrow (X \cap Y) \cup Z =$

   (a) $\{a, b, c, o, r\}$

   (b) $\{c, a, r, b\}$

   (c) $\{r, o, b, c\}$

   (d) $\phi$

   ›Reveal solutionSolution

   $(X \cap Y) \cup Z = \{a, b, c, o, r\}$, found by first taking the intersection, then the union with $Z$.

   Given $X = \{a, b, c, d\}$, $Y = \{c, a, r\}$, $Z = \{r, o, b\}$.

   Step 1: $X \cap Y$ keeps elements common to both: $a \in Y$, $c \in Y$, so $X \cap Y = \{a, c\}$ ($b$ and $d$ are not in $Y$; $r$ is not in $X$).

   Step 2: $(X \cap Y) \cup Z = \{a, c\} \cup \{r, o, b\} = \{a, b, c, o, r\}$.

   ✓Final answer

   The correct option is (a) $\{a, b, c, o, r\}$.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.$A = \{x : x - 2 = 0\}, B = \{x : 2x = 6\} \Rightarrow A \cup B =$

    (a) $\{2, 6\}$

    (b) $\{-2, 6\}$

    (c) $\{2, 3\}$

    (d) $\{2, -3\}$

    ›Reveal solutionSolution

    $A \cup B = \{2, 3\}$.

    $A = \{x : x - 2 = 0\} = \{2\}$. $B = \{x : 2x = 6\} = \{3\}$.

    $A \cup B = \{2\} \cup \{3\} = \{2, 3\}$.

    ✓Final answer

    The correct option is (c) $\{2, 3\}$.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.$A = \{x : x^2 + 5x + 6 = 0\}, B = \{x : x^2 + 8x + 15 = 0\} \Rightarrow$

    (a) $A \subset B$

    (b) $B \subset A$

    (c) $A = B$

    (d) $A \cap B = \{-3\}$

    ›Reveal solutionSolution

    $A = \{-2, -3\}$, $B = \{-3, -5\}$, and their only common element is $-3$, so $A \cap B = \{-3\}$.

    $A = \{x : x^2 + 5x + 6 = 0\}$: factorising, $(x+2)(x+3) = 0 \Rightarrow x = -2, -3$, so $A = \{-2, -3\}$.

    $B = \{x : x^2 + 8x + 15 = 0\}$: factorising, $(x+3)(x+5) = 0 \Rightarrow x = -3, -5$, so $B = \{-3, -5\}$.

    Neither $A \subset B$ nor $B \subset A$ nor $A = B$ holds (each has an element the other lacks), but both contain $-3$, so $A \cap B = \{-3\}$.

    ✓Final answer

    The correct option is (d) $A \cap B = \{-3\}$.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.Given $U = \{1, 2, \ldots, 15\}$, $A = \{1, 2, 3, 5, 15\}$, $B = \{2, 4, 6, 8, 10, 12, 14\}$, $C = \{2, 3, 5, 7, 11, 13\}$. $A' =$

    (a) $\{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$

    (b) $\{4, 6, 8, 10, 12, 14\}$

    (c) $\{8, 10, 12, 14\}$

    (d) $\phi$

    ›Reveal solutionSolution

    $A' = U - A = \{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$.

    Given $U = \{1, 2, \ldots, 15\}$ and $A = \{1, 2, 3, 5, 15\}$. The complement $A' = U - A$ consists of every element of $U$ not in $A$.

    Removing $1, 2, 3, 5, 15$ from $U$ leaves $\{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$.

    ✓Final answer

    The correct option is (a) $\{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$.