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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: x^{2} – 1 = 0
Adding 1 to both sides of the above equation.
x^{2 } = 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: x^{2} + 1 = 0
Subtracting 1 to both sides of the above equation.
x^{2} = 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.
What is the cardinality of the set of odd positive integers less than 10?
Consider the following relations:
(i) A  B = A  (A ∩ B)
(ii) A = (A ∩ B) ∪ (A  B)
(iii) A  (B ∪ C) = (A  B) ∪ (A  C)
Which of these is/are correct?
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}
In a city, 20 per cent of the population travels by car, 50 per cent travels by bus and 10 per cent travels by both car and bus. Then persons travelling by car or bus is:
► 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
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 5}, B = {6, 7}. Then A ∩ B’ is:
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}
If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12}, C = {4, 5, 6, 12, 14} then (A ∩ B) ∪ (A ∩ C) is equal to:
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}
Let A and B be two sets containing four and two elements respectively.Then the number of subsets of the set A × B,each having atleast three elements is:
⇒ 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) = 2^{8} = 256
⇒ Number of subsets having 0 elements = ^{8}C_{0 }= 1
⇒ Number of subsets having 1 element = ^{8}C_{1} = 8
⇒ Number of subsets having 2 elements = ^{8}C_{2} = 28
∴ Number of subsets having atleast 3 elements = 256  1  8  28 = 219
From VennEuler's Diagram
∴ (A ∪ B)′ ∪ (A′ ∩ B) = A′
The number of elements in the set {(a, b) : 2a^{2} + 3b^{2} = 35, a, b ∈ Z}, where Z is the set of all integers, is:
2a^{2} + 3b^{2} = 35
3b^{2} = 35 – 2a^{2}
Maximum value on RHS is 35.
3b^{2} ≤ 35
b^{2} ≤ 11.66
b = 0, ± 1, ± 2, ± 3….
3b^{2} = 35 – 2a^{2}
Any number multiplied by 2 give even number.
3b^{2} is odd.
If b = 0, then 3b^{2} is not odd.
If b = ± 1, then 3b^{2 }is odd.
If b = ± 2, then 3b^{2 }is even.
If b = ± 3, then 3b^{2 }is odd.
3b^{2} = 35 – 2a^{2}
If b = ± 1, then b^{2} = 1,
3 = 35 – 2a^{2}
2a^{2} = 32
a^{2} = 16
a = ± 4
If b = ± 3, then b^{2} = 9
27 = 35 – 2a^{2}
2a^{2} = 8
a^{2} = 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.
Given n(U) = 20, n(A) = 12, n(B) = 9, n(A ∩ B) = 4, where U is the universal set, A and B are subsets of U, then n((A ∪ B)^{c}) =
In a certain town, 25% families own a cell phone,15% families own a scooter and 65% families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is:
Stepbystep 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.
If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12} and C = {4, 5, 6, 12, 14}, then (A ∪ B) ∩ (A ∪ C) is equal to:
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 nonempty subsets = 2^{n} − 1 = 2^{4} − 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 welldefined collection.
A ∩ (B − A) = ψ,[∵ x ∈ B − A ⇒ x∉ A]
If n(A) = 3, n(B) = 6 and A ⊆ B. Then the number of elements in A ∪ B is equal to:
Since A ⊆ B
∴ A ∪ B = B
So, n(A ∪ B) = n(B) = 6
In a class of 100 students, 55 students have passed in Mathematics and 67 students have passed in Physics. Then the number of students who have passed in Physics only is:
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