Test: Algebra Of Sets - Commerce MCQ

U = Set of all teachers in a school
B = Set of all Mathematics teachers in a school
B’ = U – B = Set of all Non-Mathematics teachers in a school

B ∩ C = {4}, as only 4 is common between B and C
∴ A ∪ (B ∩ C) = {1, 2, 3, 4}.

No. of proper subsets = 2ⁿ-1

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}

If A = {5, 10, 15}, B = ϕ
B - A will have those elements which are in B but not in A.
B - A = ϕ

A = Set of all triangles
B = Set of all right triangles
A – B = Set of triangles after removing all right triangles from A
= Set of all triangles which do not have right angle

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
A’ = U – A
A’ = {1, 3, 5, 7, 9}

There’s a result in mathematics used for this. It says that a power set B of any set A is a set of all the subsets of A and the number of elements of B will be 2^n where n is the number of elements of A.
So taking your question as an example;
A = {1,2,3}
B : set of all subsets of A
List out all the subsets of A - {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}
Number of elements in A (n) = 3 
so 23 = 8
So, B = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}} 
and the number of elements are 8.

 In set theory, if A and B are sets, then the union of A and B , written A ⋃ B, is {x: x ∈ A or x ∈ B}. By asserting that x ∈ A or x ∈ B, we do not exclude the possibility that x is a member of both A and B. Further, if the same object/element is member of both A and B, then that element is counted only once in the new set formed by the union of A and B.
Given, A = {1,2,3,4} and B = {4,5,6,7,8}.
∴ By definition, A ⋃ B = {1,2,3,4,5,6,7,8}

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 4, 6}
B = {3, 4, 6, 8, 10}
A ∩ B = {4, 6}
(A ∩ B)’ = U – (A ∩ B)
= {1, 2, 3, 5, 7, 8, 9, 10} 

No. of elements in P(S) = 2³ = 8

∴ No. of elements in P(P(S)) = 2⁸ = 256

Set S of odd positive an odd integer less than 10 is {1, 3, 5, 7, 9}.
Then, Cardinality of set S = |S| which is 5.

A = {x : x = 2ⁿ, n ≤ 6}
For (n = 1,2,3,4,5,6) {2,4,8,16,32,64}
B = {x : x = 4ⁿ, n ≤ 2}
For (n = 1,2) {4,8}
A - B = {2, 8, 32, 64}

x² = 16 ⇒ x = ±4 and 2x = 6 ⇒x = 3

There is no value of x which satisfies both the above equations. Thus, A = φ