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Correct option is (D) 4

Let the number of person is the town be 100. 

Now it is given n(X) = 52, n(Y) = 61, n(Z) = 78, n(X ∩ Y) = 40, n(Y ∩ Z) = 38 

n(X ∩ Z) = 46, n(X ∪ Y ∪ Z) = 100 – 18 = 82 

We have 

n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) – n(X ∩ Y) – n(Y ∩ Z) – n(X ∩ Z) + n(X ∩ Y ∩ Z). 

⇒ 82 = 52 + 61 + 78 – 40 – 38 – 46 + n(X ∩ Y ∩ Z) 

= 191 – 124 + n(X ∩ Y ∩ Z) = 67 + (X ∩ Y ∩ Z) 

∴ n(X ∩ Y ∩ Z) = 82 – 67 = 15 

Here 15% of the people read all the three papers. 

n(A’ ∩ B’) = n(A ∪ B)’ 

= n(U) – n(A ∪ B) = 700 – n(A ∪ B) 

Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 

= 200 + 300 – 100 = 400 

∴ n(A’ ∩ B’) = 700 – 400 = 300

(i) A ∪ B = {1,2, 3, 4} 

∴ (A ∪ B)’ = ∪ – (A ∪ B) = {5, 6, 7} 

(ii) (A ∩ B) = {2,3} 

(A ∩ B)’ = ∪ – (A ∩ B) = {1,4, 5,6} …… (1) 

⇒ A’ = ∪ – A = {4, 5, 6, 7} 

B’ = ∪ – B = {1, 5, 6, 7} 

∴ A’ ∪ B’ = (1,4, 5, 6, 7} ……. (2) 

From (1) and (2) (A ∩ B)’ = A’ ∪ B’

Given, U = {1,2,3,4,5,6,7,8,9,10,12,24} 

A = {2,3,5,7}  

B = {1,2,3,4,5,6,8,12,24}  

Now, A’ = {1,4,6,8,9,10,12,24} 

B’ = {5,7,9,10} 

A ∪ B = {1,2,3,4,5,6,7,8,12,24} 

(A ∪ B)’ = {9,10} 

A ∩ B = {2,3} 

(A ∪ B)’ = {1,4,5,6,7,8,9,10,12,24} 

(i) A – B = A ∩ B’ 

L.H.S = A – B = {2,3,5,7} – {1,2,3,4,6,8,12,24} = {5,7} 

R.H.S = A ∩ B’ = {2,3,5,7} ∩ {5,7,9,10} = {5,7} 

∴ L.H.S = R.H.S,  

(ii) (A ∪ B)’ = A ∩ B’ 

L.H.S = (A ∪ B)’ = {9,10} 

R.H.S = A’∩ B’ = {1,4,6,8,9,10,12,24} ∩ {5,7,9,10} = {9,10} 

∴ L.H.S = R.H.S,  

(iii) (A ∩ B)’ = A’ ∩ B’ 

L.H.S = (A ∩ B)’ = {1,4,5,6,7,8,9,10,12,24} 

R.H.S = A’ ∩ B’ = {1,4,6,8,9,10,12,24} ∩ {5,7,9,10} = {1,4,5,6,7,8,9,10,12,34} 

∴ L.H.S = R.H.S,

A = {5, 7}, B = {6, 7, 8, 9} 

A – B = {5, 6, 7} – {6, 7, 8, 9} 

= {5, 6, 7, 8, 9} = {5} 

A – (A – B) = {5, 6, 7} – {5} = {6, 7} 

A ∩ B – {5, 6, 7} ∩ {6, 7, 8, 9} = {6, 7} 

We observe that A – (A – B) = A ∩ B.

A = {1, 2, 3, 4, ………….} 

B = {2,4, 6, 8, ………….. }; 

C = {1,3,5,7, ………….. } 

For finding, 

A∩B = {1,2,3,4, } ∩ {2,4,6,8, …. } 

= {2, 4, 6, 8…………. } 

= {x/x is an even natural number} 

For finding, 

A ∩ C = {1,2,3,4,…….} ∩ {1,3,5,7, ………. } 

= {1,3,5,7 :…….}

A = {x/x is an odd natural number} 

For finding, 

A – B = {1,2,3,4, ……….. } – {2,4,6,8, …………… } 

= {1,3,5, ………….. } 

= {x/x is an odd natural number} 

For finding 

A – C = {1,2,3,4,…….}-{1,3,5,7, } 

= {2,4,6,8, ,…} 

= {x/x is an even natural number}

A = {Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon} 

B = {Triangles}; 

C = {Quadrilaterals} 

i) A ∩ B = {Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon} ∩ {Triangles} =  {Triangles}

ii) A ∩ C = {Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon} ∩ {Quadrilaterals} = {Quadrilaterals} 

iii) A-B = {Triangles, Quadrilaterals, Pentagon, Hexagon, Heptagon} – {Triangles} = {Quadrilaterals, Pentagon, Hexagon, Heptagon} 

iv) A – C = {Triangles, Quadrilaterals, Pentagon; Hexagon, Heptagon} – {Quadrilaterals} 

= {Triangles, Pentagon, Hexagon, Heptagon}

Correct option is  B) 7

Correct option is B) 0

(a) T is an equivalence relation but S is not an equivalence relation.

(0, 1), (1, 2) it is not an equivalence relation T is an equivalence relation

Answer is (A) A 

Here, x = n² and 2 ≤ n ≤ 5 

∴ n = 2, 3, 4, 5 

[it is given that n is less than equal to 2 and greater than equal to 5] If 

n = 2, x = (2)² = 4

n = 3, x = (3)² = 9 

n = 4, x = (4)² = 16 

n = 5, x = (5)² = 25 

So, D = {4, 9, 16, 25}

Given x ∈ Z and |x| ≤ 2 

Z is a set of integers 

Integers are …-3, -2 , -1, 0, 1, 2, 3, … 

Now, if we take x = -3 then we have to check that it satisfies the given condition |x| ≤ 2 

|-3| = 3 > 2 

So, -3 ∉ H 

If x = -2 then |-2| = 2 [satisfying |x| ≤ 2] 

So, -2 ∈ H 

If x = -1 then |-1| = 1 [satisfying |x| ≤ 2] 

∴ -1 ∈ H 

If x = 0 then |0| = 0 [satisfying |x| ≤ 2] 

∴ 0 ∈ H 

If x = 1 then |1| = 1 [satisfying |x| ≤ 2] 

⇒ 1 ∈ H 

If x = 2 then |2| = 2 [satisfying |x| ≤ 2] 

So, 2 ∈ H 

If x = 3 then |3| = 3 > 2 [satisfying |x| ≤ 2] 

So, 3 ∉ H 

So, H = {-2, -1, 0, 1, 2} 

So, E = {0, 1}

Given: x ∈ Z and x² = x 

Z is a set of integers 

Integers are …-2 , -1, 0, 1, 2, … 

Now, if we take x = -2 then we have to check that it satisfies the given condition x² = x(-2)² = 4 ≠ 2 

So, -2 ∉ E 

If x = -1 then (-1)² = 1 ≠ -1 [not satisfying x² = x] 

So, -1 ∉ E 

If x = 0 then (0)² = 0 [satisfying x² = x] 

∴ 0 ∈ E 

If x = 1 then (1)² = 1 [satisfying x² = x] 

∴ 1 ∈ E 

If x = 2 then (2)² = 4 ≠ 2 [not satisfying x² = x] 

⇒ 2 ∉ E 

So, E = {0, 1}

Given x ∈ Z and

\(-\frac{1}{2}\) <x<\(\frac{13}{2}\)

It can be seen that 

\(-\frac{1}{2}\) = -0.5&\(\frac{13}{2}\) = 6.5

We know that, Z means Set of integers 

∴ F = {0, 1, 2, 3, 4, 5, 6}

A set is a well defined collection of objects.

Sets are usually denoted by capital letters A, B, C, etc. and their elements by small letters a, b, c, etc.

If every element of set A is also an element of set B, then A is a subset of B. 

We write it symbolically as A ⊆ B where ‘⊆’ denotes ‘is a subset of’. 

Ex. A = {4, 8, 12}, B = { 2, 4, 6, 8, 10, 12}, then A ⊆ B. 

⇒ A is contained in B 

⇒ B contains A 

⇒ B is a superset of A 

⇒ B ⊇ A 

Note : (i) Every set is a subset of itself. 

(ii) Null set is a subset of every set.

Two sets A and B containing equal number of elements, which are not necessarily the same are called equivalent sets. 

Ex. A = { letters of the word ‘flower’} 

= { f, l, o, w, e,r } 

B = {2, 4, 6, 8, 10, 12} 

Here, n (A) = 6 and n (B) = 6, then A ∼ B, where ∼ is the symbol for equivalence. 

Note : All equal sets are equivalent but all equivalent sets are not necessarily equal.

There are two main ways of expressing a set: 

(i) Tabular form or Roster form: In this form we list all the members of the set separating them by commas and enclosing them in only brackets. 

Ex. The set of the first ten perfect square numbers is written as 

S = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Note : • The elements of a set can be written in any order. 

• An element of a set is not written more than once

(ii) Set-builder form or Rule method: In this method, instead of listing all elements of a set, we write the set by some special property or properties satisfied by all the elements and write it as: 

A = { \(x\) : P(\(x\))} or A = { \(x\) | \(x\) has the property P(\(x\))} 

Ex. The set of the first ten perfect square numbers is written as: 

S = {\(x\) | \(x\) = n², 1 ≤ n ≤ 10, n ∈ Ν }

There are various types of sets:

(i) Finite set: A set having no elements or a definite number of elements is called a finite set. 

Ex. V = The set of vowels of english alphabet = { a, e, i, o, u} 

(ii) Infinite set: A set having unlimited number of elements is called an infinite set. 

Ex. N = The set of natural numbers = { 1, 2, 3, 4 , ....} 

(iii) Empty set: A set containing no element is called the empty or null or void set. The symbol for empty set is φ. 

φ = { } 

(iv) Singleton set: A set containing only one element is called a singleton set. 

Ex. {2}, {a}, {0}

Note : { } is the empty set whereas {0} is a singleton set.

Two sets P and Q containing the same elements are called equal sets.

Here, P = Q iff a ∈ P ⇒ a ∈ Q and a ∈ Q ⇒ a ∈ P. 

Ex. P = {letters of the word ‘ramp’} 

Q = {letters of the word ‘pram ’} 

⇒ P = Q

The number of distinct elements in a finite set A is called the cardinal number of A and is denoted by n(A) 

Ex. If A = { a, e, i, o, u }, then n (A) = 5.

Given sets are 

A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7} 

A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3} 

B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7} 

∴ No, A – B ≠ B – A.

Given, A = [0,3] and B = [2,6] 

(i) A = (- ∞, 0) ∪ (3, ∞) 

(ii) A ∪ B = [0,3] ∪ [2,6] 

= [0,6] 

(iii) A ∩ B = [0,3] ∩ [2,6] 

= [2,3] 

(iv) A – B = [0,3] − [2,6] 

= (0,2)

i) True (T) 

ii) False (F) 

iii) False (F)

Given sets are 

A = {4, 5, 6} and B = {7, 8} 

A ∪ B = {4, 5, 6} ∪ {7, 8} 

= {4, 5, 6, 7, 8}. 

B ∪ A = {7, 8} ∪ {4, 5, 6} 

= {4, 5, 6, 7, 8} 

∴ A ∪ B = B ∪ A.

(i) (A ∩ B) ∪ (A – B) = A 

L.H.S = (A ∩ B) ∪ (A – B) 

= (A ∩ B) ∪ (A – B’) [∴ (A – B) = (A – B’] 

= A ∩ (B ∪ B’)      [By distributive law] 

= A ∩ (U)      [(B υ B') = U =Universal set] 

= A 

= R.H.S 

(ii) A ∪ (B - A) = A ∪ B 

L.H.S = A ∪ (B - A) 

= A ∪ (B – A’) [∴ (B - A) = (B ∩ A’] 

= (A ∪ B) ∩ (A ∪ A’)      [By distributive law] 

= (A ∪ B) ∩ U 

= A ∪ B      [∴ A υ A' = U =Universal set] 

= R.H.S

Given sets A = (1, 3, 7, 8} and B = {2,4, 7,9} 

A ∩ B = {1, 3, 7, 8} ∩ (2, 4, 7, 9} = {7}

Given sets 

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7} 

A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 

∩ {2, 3, 5, 7} = {2, 3, 5, 7} = B 

∴ A ∩ B = B

Let take three sets A = {1, 2), B = {2, 3} and C = {3, 1}: A ∩ B, B ∩ C and A ∩ C should be non-empty sets. A ∩ B = {2}, B ∩ C = {3} and A ∩ C = {1} 

Therefore, A ∩ B, B ∩ C and A ∩ C are non- empty. Intersection of all three sets is null set, A ∩ B ∩ C = ∅.