Statement I : Let X and Y be two sets then, X ∩ Y ∪ X ′ = ϕ Statement II : If X ∪ Y has m elements and X ∩ Y has n elements, then the symmetrical difference X Δ Y has (m - n) elements.

Statement I is true, Statement II is also true and Statement II is not the correct explanation of the Statement I.

Statement I is true, Statement II is also true and Statement II is the correct explanation of the Statement I

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Statement I : Let X and Y be two sets then, X ∩ Y ∪ X ′ = ϕ Statement II : If X ∪ Y has m elements and X ∩ Y has n elements, then the symmetrical difference X Δ Y has (m - n) elements.

Statement I is true, Statement II is also true and Statement II is not the correct explanation of the Statement I.

Let A = θ ∈ R : 1 3 sin θ + 2 3 cos θ 2 = 1 3 sin 2 θ + 2 3 cos 2 θ . Then

A ∩ 0 , π has exactly one point

A ∩ 0 , π has exactly two points

A ∩ 0 , π has more than two points

HARD

Earn 100

Statement I : Let X and Y be two sets then, $X \cap {(Y \cup X)}^{'} = ϕ$

Statement II : If $X \cup Y$ has m elements and $X \cap Y$ has n elements, then the symmetrical difference $X Δ Y$ has (m - n) elements.

50% studentsanswered this correctly

Important Questions on Set Theory and Relations

HARD

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let

A

and

B

be finite sets such that

n (A - B) = 18, n (A \cap B) = 25

and

n (A \cup B) = 70

. Then

n (B)

is equal to

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

U

is the universal set with

100

elements;

A

and

B

are two sets such that

n (A) = 50, n (B) = 60

n (A \cap B) = 20

then

n (A^{'} \cap B^{'}) =

HARD

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

In a group of

100

persons,

80

people can speak Malayalam and

60

can speak English. Then the number of people who speak English only is

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

A survey shows that

73 %

of the persons working in an office like coffee, whereas

65 %

like tea. If

x

denotes the percentage of them, who like both coffee and tea, then

x

cannot be:

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let

X = \{n \in N : 1 \leq n \leq 50\}

. If

A = \{n \in X : n is a multiple of 2\}

and

B = \{n \in X : n is a multiple of 7\}

, then the number of elements in the smallest subset of

X

, containing both

A

and

B

, is.

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

A survey shows that

63 %

of the people in a city read newspaper

A

whereas

76 %

read news paper

B

. If

x %

of the people read both the newspapers, then a possible value of

x

can be:

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

A \subset B,

then

A Δ B

is equal to

HARD

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

In a certain town,

25 %

families own a phone,

15 %

families own a car,

65 %

families own neither a phone nor a car and

2000

families own both a car and a phone. Consider the following Statements

(S) :

(S_{1}) :

35 %

families own at least one of a car or a phone.

(S_{2}) :

40, 000

families live in the town.
Then:

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

A = {x | x \in N, x

is a prime number less than

12}

and

B = {x |

x \in N

x

is factor of

10}

, then

A \cap B

=

…

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

A = \{x \in R : |x| < 2\}

and

B = \{x \in R : |x - 2| \geq 3\};

then

HARD

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let $\cup_{i = 1}^{50} X_{i} = \cup_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains $10$ elements and each $Y_{i}$ contains $5$ elements. If each element of the set $T$ is an element of exactly $20$ of sets $X_{i}$ 's and exactly $6$ of sets $Y_{i}$ 's then $n$ is equal to :

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

In a certain town,

25 %

of the families own a phone and

15 %

own a car;

65 %

families own neither a phone nor a car and

2000

families own both a car and a phone. Consider the following three statements:

(i)

5 %

families own both a car and a phone.

(i i)

35 %

families own either a car or a phone.

(i i i)

40000

families live in the town.

Then,

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Consider the two sets:

A = {m \in R :

both the roots of

x^{2} - (m + 1) x + m + 4 = 0

are real

}

and

B = [- 3, 5)

Which of the following is not true?

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

If $A = {1, 2, 3, 4, 5}$ and $B = {2, 4, 6}$ , then $A - B =$

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

n (A) = 3, n (B) = 6

and

A \subseteq B,

then the number of elements in

(A \cup B)

is equal to

HARD

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let

A = \{θ \in R : {(\frac{1}{3} \sin θ + \frac{2}{3} \cos θ)}^{2} = \frac{1}{3} \sin^{2} θ + \frac{2}{3} \cos^{2} θ\}

. Then

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let

A

and

B

be two sets such that

A \cap X = B \cap X = ϕ

and

A \cup X = B \cup X

for same set

X

. Then,

EASY

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Let

A, B, C

and

D

be four non-empty sets. The contrapositive statement of “If

A \subseteq B

and

B \subseteq D

, then

A \subseteq C

” is

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

Two newspapers

A

and

B

are published in a city. It is known that

25 %

of the city population reads

A

and

20 %

reads

B

while

8 %

reads both

A

and

B .

Further,

30 %

of those who read

A

but not

B

look into advertisements and

40 %

of those who read

B

but not

A

also look into advertisements, while

50 %

of those who read both

A

and

B

look into advertisements. Then the percentage of the population who look into advertisements is:

MEDIUM

Mathematics>Algebra>Set Theory and Relations>Algebra of Sets

In a class of

140

students numbered

1

140

, all even numbered students opted Mathematics course, those whose number is divisible by

3

opted Physics course and those whose number is divisible by

5

opted Chemistry course. Then the number of students who did not opt for any of the three courses is:

Statement I : Let X and Y be two sets then, X ∩ Y ∪ X ′ = ϕ Statement II : If X ∪ Y has m elements and X ∩ Y has n elements, then the symmetrical difference X Δ Y has (m - n) elements.

Important Questions on Set Theory and Relations

Learn and Prepare for any exam you want !

Statement I : Let X and Y be two sets then, $X \cap {(Y \cup X)}^{'} = ϕ$

Statement II : If $X \cup Y$ has m elements and $X \cap Y$ has n elements, then the symmetrical difference $X Δ Y$ has (m - n) elements.