Subsets & Supersets - Commerce Free MCQ Test with solutions

Null set is the subset of every set so Y ⊂ X and Y ⊂ Z.
Since set X is equal to set Z so, Z = X and X = Z.

If set A is equal to set B, it means both sets contain exactly the same elements.
Therefore, every element of A is also in B → A ⊆ B, and every element of B is also in A → B ⊆ A.
Hence, A ⊆ B and B ⊆ A.

Set of rational numbers { x: x=p/q where p and q are integers and q≠0}.
Set of real numbers is not a subset of X. Set of natural numbers, whole numbers, integers are subset of X.

Number of elements in a power set = 2ⁿ, where n = number of elements in the set P.
Hence, 2⁴ = 16.

• A subset is a set whose elements are all members of another set.
• Since all the elements of set A are the members of set E. So A is a subset of E.  

(2, 4) written in set builder form is:

{x : 2 < x < 4}

If A = Ф that means A does not contain any element i.e., n = 0.
Now, number of elements in a power set is 2ⁿ.
∴, n[P(A)] = 2⁰ = 1
Therefore P(A) contains 1 element.

• The set containing all objects or elements and of which all other sets are subsets is a Universal set.
• Here, option A has all the elements of set P, Q and R. So {1, 2, 3, 4, 5, 6, 7, 9} may be considered as universal set for all the three sets P, Q and R.

Square brackets [a, b] implies that values of a and b should be included in the range of x i.e., a ≤ x ≤ b.

• We have, A^c in X = The set of elements in X which are not in A = 0,1
• {0} ∈ A^c w.r.t. X is false, because {0} is not an element of A^c in X.
• ϕ ∈ A^c in X is false because ϕ is not an element of A^c in X.
• {0} ⊂ A^c in X is correct because the only element of {0} namely 0 also belong to A^c in X.
• 0 ⊂ A^c in X is false because 0 is not a set.

All the elements in set-A are presented in set-B. So "A" is a subset of "B".

A singleton set is a set that contains only one element, so statement A is correct.
The set of all even prime numbers is {2}, and {2} is a subset of the set of natural numbers, so statement B is correct.
Every set is a subset of itself, so statement D is also correct.
However, statement C is incorrect because the empty set Φ is a subset of every set.
Therefore, the incorrect statement is C.

• P contains {0, 1, 2, 3, 4} and Q contain natural numbers which start with 1 and P = {0, 1, 2, 3, 4} Q = {1, 2, 3}
• Here P isn't a subset of Q because all the elements of P are not in Q.

|x| < 1 ⇒ -1 < x < 1
∴ No natural number exists between (-1, 1).

L.C.M of 4 and 6 is 12.

Given, A = {x:x is a multiple of 4}
              = {4,8,12,16,20,…}

and     B = {x:x is a multiple of 6}
              = {6,12,18,24,…}

∴A ∩ B = {12,24,…}
             = {x:x is amultiple of 12}

If a set has n elements, then the total number of subsets of that set is given by the formula 2ⁿ.
Here, the set has 6 elements, so:

2⁶ = 64

Therefore, the total number of subsets of the set is 64.

• Number of proper subsets of a given set = 2^m - 1, where m is the number of elements.
• Here the number of elements is 3. So the number of proper subsets of A = 2³ - 1 = 7.

A universal set is a set that contains all elements under consideration.
For the set of all natural numbers, the universal set must contain every natural number, and it may also include additional elements.
Among the given options, the set of all integers includes every natural number along with 0 and negative numbers.
Hence, it is the only option that contains all natural numbers, so it can serve as the universal set for natural numbers.

Therefore, the correct answer is C.

The number of elements in the power set of a set with n elements is given by the formula:

2ⁿ

In this case, we have:

• n = 7
• Therefore, the number of elements in the power set is:

2⁷ = 128

A = {1, 2, {3, 4}, 5}

• 1 ∈ A is correct because 1 is an element of the set.
• {3, 4} ∈ A is also correct because {3, 4} appears as a single element inside the set.
• {3, 4} ⊂ A is incorrect because to be a subset, the elements 3 and 4 must be present individually in A — but they are not.
• {{3, 4}} ⊂ A is correct because the set {{3, 4}} contains a single element {3, 4}, and {3, 4} is an element of A.

Therefore, the incorrect statement is A, i.e., {3, 4} ⊂ A.