Q.If the set $A$ contains 5 elements, then the number of elements in the power set $P(A)$ is

Showing the 12 most recent of 26 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The number of all non-empty subsets of the set $\{1, 2, 3\}$ is:

   (a) 8

   (b) 7

   (c) 3

   (d) 9

   ›Reveal solutionSolution

   A set of $n$ elements has $2^n$ subsets in total; removing the empty subset gives $2^n-1$ non-empty subsets.

   For the set $\{1,2,3\}$, $n=3$.

   Total subsets $= 2^n = 2^3 = 8$ (this includes $\varnothing$ and the full set itself).

   Non-empty subsets $= 2^n - 1 = 8 - 1 = 7$.

   …

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2. CBSE 2026Set ANNUAL1 mark
   Q.Define subset.

   ›Reveal solutionSolution

   $A\subseteq B$ means every element of A belongs to B.

   Formally, A is a subset of B if, for every element $x$, $x\in A \Rightarrow x\in B$.

   …

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3. CBSE 2026Set 1A1 mark
   Q.List all the elements of the set $C = \{x : x \text{ is an integer}, x^2 \le 4\}$.

   ›Reveal solutionSolution

   Integers with $x^2\le 4$ are $-2,-1,0,1,2$.

   We need integers $x$ with $x^2\le 4$, i.e. $-2\le x\le 2$. The integers in this r …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.All possible subsets of set $A = \{2, 3, 4\}$ are

   (a) $P(A) = \{2\}, \{3\}, \{4\}$

   (b) $P(A) = \{\{2\}, \{3\}, \{4\}, \{2, 3, 4\}\}$

   (c) $P(A) = \{\phi, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{3, 4\}, \{4, 2\}, \{2, 3, 4\}\}$

   (d) None of these

   ›Reveal solutionSolution

   A set with $n$ elements has $2^n$ subsets. For $A=\{2,3,4\}$, $n=3$, so there are $8$ subsets total, exactly the list in option (c).

   The power set $P(A)$ is the collection of ALL subsets of $A$ -- including the empty set $\phi$ and $A$ itself.

   Here $n(A) = 3$, so $n(P(A)) = 2^3 = 8$.

   Listing every subset by size:

   • Size 0: $\phi$
   • Size 1: $\{2\}, \{3\}, \{4\}$
   • Size 2: $\{2,3\}, \{3,4\}, \{2,4\}$
   • Size 3: $\{2,3,4\}$ …

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5. CBSE 2025Set ANNUAL1 mark
   Q.If set A has 4 elements, then number of subsets is .............

   ›Reveal solutionSolution

   The number of subsets of a set with $n$ elements is $2^n$.

   Set $A$ has $4$ elements, so the number of subsets is $2^4 = 16$ (this counts the empt …

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6. CBSE 2025Set sz1 markMCQ
   Q.If $X = \{0, -1\}$ then $P(A)$ has :

   (a) 4 elements

   (b) 3 elements

   (c) 2 elements

   (d) 6 elements

   ›Reveal solutionSolution

   Reading the paper's $P(A)$ as the power set of the given set $X$, a 2-element set has $2^2 = 4$ subsets.

   Honest note on the printed stem: the question states "If $X = \{0, -1\}$ then $P(A)$ has :" — the set introduced is $X$, but the power-set notation printed is $P(A)$, not $P(X)$. This is almost certainly a printing slip in the original paper (there is no set $A$ defined anywhere in this question), and the only way the question makes sense is if it means the power set of $X$, i.e. $P(X)$.

   Solving for $P(X)$: $X = \{0, -1\}$ has $n = 2$ elements.

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Match the column: Column A entry 'Number of subsets of the set $\{2, 3, 5\}$' — find the matching value from Column B.

   (a) 2

   (b) 8

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

   (d) $\dfrac{2\tan x}{1 - \tan^2 x}$

   (e) $\sin 2x$

   (f) 10

   (g) 20

   (h) $\dfrac{2\tan x}{1 + \tan^2 x}$

   (i) 4

   ›Reveal solutionSolution

   A set with $n$ elements has exactly $2^n$ subsets; here $n = 3$ gives $2^3 = 8$.

   The set $\{2, 3, 5\}$ has $n = 3$ elements. The number of subsets of a set with $n$ elements is $2^n$ (this includes the empty set $\varnothing$ and the set itself).

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $A = \{2, 4, 6\}$, then number of total subsets of $A$ is

   (a) 8

   (b) 16

   (c) 15

   (d) None of these

   ›Reveal solutionSolution

   The number of subsets of a set with $n$ elements is $2^n$.

   $A = \{2, 4, 6\}$ has $n(A) = 3$ elements. The total number of subsets of a set (i.e. the number of elements in its power set) is $2^n$, because each element independently is either included or excluded from a subset.

   …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The number of subsets of a set having $n$ elements is:

   (a) $2n$

   (b) $n^2$

   (c) $2^n$

   (d) $2^n - 1$.

   ›Reveal solutionSolution

   A set with $n$ elements has exactly $2^n$ subsets.

   For a set $A=\{a_1,a_2,\ldots,a_n\}$ with $n$ elements, to build any subset we decide, for each element, whether to include it or not — that is $2$ independent choices per element. By the multiplication principle, the total number of subsets is …

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10. CBSE 2024Set sz1 markMCQ
    Q.Subsets of set $\{-1, 1\}$ are:

    (a) $\phi, \{-1\}$

    (b) $\phi$ only

    (c) $\phi, \{-1\}, \{1\}, \{-1, 1\}$

    (d) $0$

    ›Reveal solutionSolution

    A set with $n$ elements has $2^n$ subsets; the set $\{-1,1\}$ has 2 elements so it has $2^2=4$ subsets: the empty set, the two singletons, and the set itself.

    A subset of a set $S$ is any set formed by taking zero or more elements of $S$ (including none, giving the empty set $\phi$, and all of them, giving $S$ itself).

    For $S = \{-1, 1\}$, which has $n = 2$ elements, the number of subsets is $2^n = 2^2 = 4$.

    Listing them out:

    • Taking 0 elements: $\phi$
    • Taking 1 element: $\{-1\}$, $\{1\}$
    • Taking 2 elements: $\{-1, 1\}$ …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The number of subsets of the set $\{1, 2\}$ is:

    (a) 1

    (b) 2

    (c) 3

    (d) 4

    ›Reveal solutionSolution

    A set with $n$ elements has $2^n$ subsets, so $\{1,2\}$ (2 elements) has $2^2=4$ subsets.

    Step 1. The number of subsets of a set with $n$ elements is $2^n$, since each element is independently either included or excluded.

    …

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12. CBSE 2024Set ANNUAL1 mark
    Q.Write true or false: $A = \{1, 2, 3\}$ is a proper subset of $B = \{1, 2, 3, 4\}$.

    ›Reveal solutionSolution

    $A$ is a proper subset of $B$ because every element of $A$ lies in $B$, but $B$ has an extra element $A$ lacks.

    Step 1. Check $A \subseteq B$: every element of $A=\{1,2,3\}$ — namely $1,2,3$ — is present in $B=\{1,2,3,4\}$. So $A \subseteq B$.

    Step 2. Check $A\ne B$: $B$ contains $4$, which is not in $A$. So $A\ne B$.

    …

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