Q.In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

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 $B$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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):

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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.

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $B=\{3,7,11\}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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.

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $Y$, $X \cap Y = \{x : x \in X \text{ and } x \in Y\}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $X = \{1, 2\}$, $Y = \{2, 3, 5\}$, $Z = \{4, 6\}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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).

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $X = \{a, b, c, d\}$, $Y = \{c, a, r\}$, $Z = \{r, o, b\}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
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

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

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

    …

    🔒 Start your 14-day free trial to see the full solution →
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 $B = \{-3, -5\}$.

    …

    🔒 Start your 14-day free trial to see the full solution →
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 $A$.

    …

    🔒 Start your 14-day free trial to see the full solution →