Q.An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die, C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive?

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