Test: Sets- 1

⇒ x² + 1 = 0
⇒ x² = −1
⇒ x = ± √-1
i.e., x is an imaginary number.
Hence {x : x is a real number and x² + 1 = 0} is an empty set.

• Set S of odd positive integer less than 10 is {1, 3, 5, 7, 9}.
• Then, the cardinality of set S, i.e., the number of elements in the set, is 5.

Using the Venn Diagram,
(i) A - B = A - (A ∩ B) is true
(ii) A = (A ∩ B) ∪ (A - B) is true
(iii) A - (B ∪ C) = (A - B) ∪ (A - C) is false

By applying De Morgan's Law,
(A U B)' = A' ∩ B'

A ∪ {1, 2} = {1, 2, 3, 5, 9}
If A = {3, 5, 9}
Then, A ∪ {1, 2} = {1, 2, 3, 5, 9}
So, the minimum set of A = {3, 5, 9}

n(Persons travelling by car) = 20 = n(C)
n(Persons travelling by bus) = 50 = n(B)
n(Persons travelling by both car and bus) = 10 = n(C ∩ B)
∴ n(C ∪ B) = n(C) + n(B) - n(C ∩ B)
= 20 + 50 - 10
= 60 

• B' gives us all the elements in U other than 6 and 7 i.e., B' = {1, 2, 3, 4, 5, 8, 9, 10}
• The intersection of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is set A itself.

• This method involves specifying a rule or condition which can be used to decide whether an object can belong to the set.
• This rule is written inside a pair of curly braces and can be written either as a statement or expressed symbolically or written using a combination of statements and symbols.
• {x :  x ≠ x}
  This implies a null set.

If A ∪ B = A ∩ B then A = B.
Example: Let A = {1, 2, 3, 4} & B = {1, 2, 3, 4}
(AUB) = (A⋂B)
⇒{1,2,3,4} = {1,2,3,4}

A = {2, 3, 4, 8, 10}
B = {3, 4, 5, 10, 12}
C = {4, 5, 6, 12, 14}
► (A ∩ B) = {3,4,10}
► (A ∩ C) = {4}
► (A ∩ B) U (A ∩ C) = {3,4,10}

• (A - B) is A but not B which means A excluding the intersection of A and B.
• Similarly, in the figure, the shaded region is A but not C and B which means A excluding B and C. But if B and C both are excluded from A then the intersection of A, B and C get excluded two times.
• So, we take B ∪ C to be excluded from A. Therefore the answer is A - (B ∪ C).

⇒ Number of elements in Set A = 4
⇒ Number of elements in Set B = 2
∴ Number of elements in set (A × B) = 8
∴ Total number of subsets of (A×B) = 2⁸ = 256
⇒ Number of subsets having 0 elements = ⁸C₀ = 1
⇒ Number of subsets having 1 element = ⁸C₁ = 8
⇒ Number of subsets having 2 elements = ⁸C₂ = 28
∴ Number of subsets having atleast 3 elements = 256 - 1 - 8 - 28 = 219

From Venn-Euler's Diagram

∴ (A ∪ B)′ ∪ (A′ ∩ B) = A′

• Since 3b² = 35 - 2a² is odd and at most 35, so b must be odd and b² < 12. Thus, b= ± 1 or ± 3.
• If  b= ± 1, then  a= ± 4.
  If  b= ± 3, then  a= ± 2.
• This will give us 8 solutions:
  {(2,3),(-2,-3),(-2,3),(2,-3),(4,1),(4,-1),(-4,-1),(-4,1)}
• Therefore, total elements in the set = 8.

• n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 12 + 9 − 4 = 17
• Now, n((A ∪ B)^c) = n(U) − n(A ∪ B) = 20 − 17 = 3

Step-by-step Explanation:
Since we have given that,
Percentage of families own a cell phone = 25%
Percentage of families own a scooter = 15%
Percentage of families neither own = 65%
So, it becomes,
P(C ∪ S)^' = 1 - P(C ∪ S)
0.65 = 1 - P(C ∪ S)
1 - 0.65 = P(C ∪ S)
P(C ∪ S) = 0.35
So, it becomes, 
P(C ∪ S) = P(C) + P(S) - P(C ∩ S)
0.35 = 0.25 + 0.15 - x
0.35 = 0.40 - x
x = 0.05
Thus, 0.05 x Total number of families in town = 1500
Total number of families in the town =
1500 / 0.05 = 30,000
Hence, the total number of families in the town is 30,000.

A = {2,3,4,8,10}, B = {3,4,5,10,12} and C = {4,5,6,12,14}
⇒ (AUB) = {2,3,4,5,8,10,12}
⇒ (AUC) = {2,3,4,5,6,8,10,12,14}
⇒ (AUB) ⋂ (AUC) = {2,3,4,5,8,10,12}

n(A × B) = n(A).n(B)
⇒ 3 × 3 = 9
A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 9)}

As n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
But in case of disjoint sets, n(A ∩ B) = 0
∴ n(A ∪ B) = n(A) + n(B)

The number of non-empty subsets = 2ⁿ − 1 = 2⁴ − 1 = 16 − 1 = 15
The subsets are:
[{1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,4}, {1,2,3}, {1,3,4} ,{2,3,4}, {1,2,3,4}]

As the opinions of different person is different about the intelligent student. So, we can't exactly have the same student's name in a set, hence it is not a well-defined collection.

A ∩ (B − A) = ψ,[∵ x ∈ B − A ⇒ x∉ A]

Since A ⊆ B
∴ A ∪ B = B
So, n(A ∪ B) = n(B) = 6

• Let M denote the sets of students who have passed in Mathematics and P denote the sets of students who have passed in Physics.
• Thus from the question we get:
  ► n(M∪P) = 100
  ► n(M) = 55
  ► n(P) = 67
• n(M∩P) = n(M) + n(P) - n(M∪P)
  ► n(M∩P) = 55 + 67 - 100
  ► n(M∩P) = 22
• ∴ The no. of who passed in physics only is given by:
  = n(P) - n(M∩P) = 67 - 22 = 45

• (A - B) is nothing but the elements which are present in set A but not present in set B.
• In other words, it is elements of set A excluding the elements which are in set B.
• Here A = {1, 2, 3, 4} and B = {4, 5, 6, 7}
  So, (A - B) = {1, 2, 3}