Test: Sets 1

Since x² + 1 = 0,
gives x² = −1 ⇒ x = ±i. ∴x is not real but x is real (given).
∴No value of x is possible.

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.

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

Let A = {1} and B = {1 , 2 } U = {1 , 2 , 3}
so, A U B = {1} U {1 , 2} = {1 , 2}
therefore, (A U B )' = {3}
*option (a):- A' U B' = {2 , 3 } U { 3} = { 2 , 3 }
thus, option (a) is not true
*option (b):- A U B = { 1 , 2}
hence option is not correct
*option (c):- A intersection B = {1} intersection {1,2} = {1}
hence option is not correct
*option (d):- A' intersection B' = {2,3} intersection {3} = {3}
therefore (A U B )' = A' intersection B'

A ∪ {1, 2} = {1, 2, 3, 5, 9}
Now the minimum set of A = {3, 5, 9}
A ∪ 1, 2= 1, 2, 3, 5, 9
Now 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} and the intersection of this set with A will be the common elements in both of these (A and B') i.e = {1, 2, 5} with us 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

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 that 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 gets excluded two times.
So we take B ∪ C to be excluded from A
So 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′

Substitute a=2 and b=3

2x(2)²+3x(3)²=35

Substitute a=4 and b=1

2x(4)²+3x(1)²=35

Substitute a=-2 and b=3

2x(-2)²+3x(3)²=35

Substitute a=-2 and b=-3

2x(-2)²+3x(-3)²=35

Substitute a=2 and b=-3

2x(2)²+3x(-3)²=35

Substitute a=-4 and b=1

2x(-4)²+3x(1)²=35

Substitute a=-4 and b=-1

2x(-4)²+3x(-1)²=35

Substitute a=4 and b=-1

2x(4)²+3x(-1)²=35

{(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,

Hence, the total number of families in the town is 30000.

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}

A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 9)}
n(A × B) = n(A).n(B) = 3 × 3 = 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's 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) = φ,[∵ 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
Thus n(M∩P) = n(M) + n(P) - n(M∪P)
         n(M∩P) = 55 + 67 - 100
         n(M∩P) = 22
Therefore 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}.