Sets- 2 - Free MCQ Practice Test with solutions, Commerce Maths Class 11

n(A) = 3, n(B) = 6
It is given that A is a subset of B which means that all elements of A are present in B.
n(A∪B) = n(A) + n(B) - n(A∩B) ----------- [ As A⊂B ]
= 3 + 6 - 3 = 6

Since  n(A) = 3   n(B) = 6

A⊆B,A⊂B=B

So, n (A⋂B) =n(A) =3

A - B means part of A that does not contain B,
So, A - B = A−(A∩B) 
= A ∩ B^c

According to De'Morgan's Law;
A-(BUC) = (A-B)∩(A-C)

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.

In the problem statement we are taking union of B and C and then taking its intersection with A.
This means A∩(B∪C)will contain elements that are in A and are in either B or C.
∴ A∩(B∪C) is equivalent to taking intersection of A,B and A,C and then taking there union i.e. (A∩B)∪(A∩C)

According to the given notation, 
3N = {3x:x∈R} = {3,6,9,12.......}
and 7N = {7x:x∈R} = {7,14,21,28,35,42,.......}
Hence 3N∩7N = {21,42,63}
= {21x:x∈N}
= 21N

Let A be the smallest set.
A ∪ {1,2} = {1,2,3,5,9}
A = {3,5,9} or {1,2,3,5,9}
As A is smallest set,we take A = {3,5,9} only
Check:-
{3,5,9} ∪ {1,2} = {1,2,3,5,9}

As we know that A-B = A
that implies x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x ∈ to A-B and x ∈ B
⇒ (x ∈ A and x doesnt belongs to B) and x ∈ to B
So, our assumption is wrong
⇒ A∩B = ϕ

The question should be: “Which set is the subset of all the sets?”
Φ = { } (Empty set) is a subset of all the sets.

We know that

n(A∪B)=n(A)+n(B)−n(A∩B)......(i)

n(A∪B)=n(A)+n(B)-n(A∩B)......(i)

Case 1 From (i) , it is clear that n(A∪B)

n(A∪B) will be maximum when n(A∩B)=0

In that case, 

n(A∪B)=n(A)+n(B)=(3+6)=9

∴ Maximum number of elements in 

(A∪B)=9

Case 2 From (i) , it is clear that n(A∪B)

n(A∪B) will be minimum when n(A∩B)=0 maximum ,i.e, when 

n(A∩B)=3

In this case, 

n(A∪B)=n(A)+n(B)−n(A∩B)=(3+6−3)=6

∴ minimum number of elements in 

A∪B=6

Clearly, A = {2, 3}, B = {2, 4}, C = {4,5} B ∩ C  = {4}

∴ A × (B ∩ C) = {(2, 4), (3, 4)}.

A = {x : x ∈ R,−1 < x < 1}

B = {x : x ∈ R : x − 1 ≤ −1 or x − 1 ≥ 1]

  = {x : x ∈ R : x ≤ 0 or x ≥ 2}

∴A ∪ B = R − D , where D = [x : x ∈ R,1 ≤ x < 2]

Use venn diagram approach

Since the curly brackets denote a set, so the other three options are a subset of the set {1, 2, 3} and and only 3 belongs to the set.

Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C           [ ∴(AUC)∩C = C ]
⇒ (A∩B) U (B∩C) = C     ..........(1)  [ ∴ (A∩C) = A∩B ]
Again AUB = AUC
(AUB) ∩ B = (AUC) ∩ B
B = (A∩B) U (C∩B)
= (A∩B) U (B∩C)   ...........(2)
From 1 & 2 we get
B = C

Total number of subsets of the first set = 2^m
Total number of subsets of the second set = 2ⁿ
2^m = 2ⁿ + 48
By hit and trial,
m = 6, n = 4

A = {2, 5, 3} and B = {7, 8, 5} and A ∪ B = {2, 3, 5, 7, 8}