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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