Test: Sets- 1

As we know that the set of real numbers is made by combining the set of rational numbers and the set of irrational numbers. 
The real numbers include natural numbers or counting numbers, whole numbers, integers, rational numbers(fractions and repeating or terminating decimals), and irrational numbers.

So, now as we can see that the set given in all the above options have real numbers. But there is one additional condition given in each set which defines the elements of the set.

So, we had to solve that additional condition to find the elements of the set.

Option A: x² – 1 = 0
Adding 1 to both sides of the above equation.
x²  = 1
Taking square root to both sides of the above equation. 
 x = 1, -1
So, the set in option A will be {1, -1} because 1 and -1 are also real numbers.
Hence, option A is not correct because this set is not empty.

Option B: x² + 1 = 0
Subtracting 1 to both sides of the above equation.
x²  = -1
Taking square root to both sides of the above equation. 

So, the set in option B will have no elements because i (iota) is not a real number, it is a complex number.

Since there cannot be more than one option correct. So, we need not to check all options.

Hence, option B is correct because this set 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′

2a² + 3b² = 35

3b² = 35 – 2a²

Maximum value on RHS is 35.

3b² ≤ 35

b² ≤ 11.66

b = 0, ± 1, ± 2, ± 3….

3b² = 35 – 2a²

Any number multiplied by 2 give even number. 

3b² is odd.

If b = 0, then 3b² is not odd.

If b = ± 1, then 3b² is odd.

If b = ± 2, then 3b² is even.

If b = ± 3, then 3b² is odd.

3b² = 35 – 2a²

If b = ± 1, then b² = 1,

3 = 35 – 2a²

2a² = 32

a² = 16

a = ± 4

If b = ± 3, then b² = 9 

27 = 35 – 2a²

2a² = 8

a² = 4

a = ± 2

[(4, 1), (4, -1), (-4, 1), (-4, -1), (2, -3), (2, 3), (-2, 3), (-2, -3)]

Therefore there are 8 elements in total.

• 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}