**Set Theory Questions For Grade 11 :**

Here we are going to see some practice questions on set theory.

**Question 1 :**

Justify the trueness of the statement:

“An element of a set can never be a subset of itself.”

**Solution :**

The given statement is wrong. Because every element is the subset of itself.

Let us look into the example problem

A = {1, 2, 3}

Subset of A = {{ } {a}, {b}, {c}, {a, b}, {b, c}, {c, a}, {a, b, c} }

Every element is the part of itself. Hence the given statement is wrong.

**Question 2 :**

If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).

**Solution :**

P(A) means power set of A

power set of A, n[P (A)] = 2^{n(A) }

n(P(A)) = 1024 ==> 2^{10}

n(A) = 10

n(P(B)) = 32 ==> 2^{5}

n(B) = 5

n(AnB) = n(A) + n(B) - n(AUB)

= 10 + 5 - 15

n(A n B) = 0

**Question 3 :**

If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(AΔB)).

**Solution :**

(AΔB) = (AUB) - (AnB)

n(AΔB) = n(AUB) - n(AnB)

n(AΔB) = 10 - 3 ==> 7

n(P(AΔB)) = 2 ^{n(AΔB)}

= 2 ^{7}

= 128

Hence the answer is 128.

**Question 4 :**

For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.

**Solution :**

From the given question, we come to know that set A will have 4 elements.

By combining the elements of A and A, we get the elements of A × A. From the given element, we have four different values.

So the elements of A are {0, 1, 2, 3}.

**Question 5 :**

Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.

**Solution :**

We get elements of A × B by combining the elements of set A to set B.

In (x, 1), (y, 2), (z, 1), the first terms will be the elements of set A and second terms will be the elements of set B.

Hence A = {x, y, z} and B = {1, 2}.

**Question 6 :**

If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (−1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.

**Solution :**

From the given two set of elements, we may list out the elements of A.

A = {-1, 0, 1, 2}

A x A = {(-1, -1) (-1, 0) (-1, 1) (-1, 2) (0, -1) (0, 0) (0, 1) (0, 2) (1, -1) (1, 0) (1, 1) (1, 2)(2, -1) (2, 0) (2, 1) (2, 2)}

Hence the remaining elements are {(-1, -1) (-1, 0) (-1, 1) (0, -1) (0, 0) (0, 2) (1, -1) (1, 0) (1, 1) (1, 2)(2, -1) (2, 0) (2, 1) (2, 2)}

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