In mathematics a set is a collection of distinct
elements. The elements that make up a set can be any kind of things: people, letters of the alphabet,
numbers, points in space, lines, other geometrical shapes, variables, or even other sets. Two sets are equal if
and only if they have precisely the same elements.
Sets are ubiquitous in modern mathematics. Indeed, set theory, more specifically Zermelo–Fraenkel set theory, has
been the standard way to provide rigorous foundations for all branches of mathematics since the first half of the
20th century..

**Q1.**In a battle 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. The minimum value of x is

Solution

(a) 10

(a) 10

**Q2.**If n(A) denotes the number of elements in the set A and if n(A)=4,n(B)=5 and n(A∩B)=3, then n[(A×B)∩(B×A)] is equal to

Solution

(b) Given, n(A)=4, n(B)=5 and n(A∩B)=3 ∴n[(A×B)∩(B×A)]=3^2=9

(b) Given, n(A)=4, n(B)=5 and n(A∩B)=3 ∴n[(A×B)∩(B×A)]=3^2=9

**Q3.**If R={(a,b):|a+b|=a+b} is a relation defined on a set {-1,0,1}, then R is⊂

Solution

(b) Let (a,b)∈R. Then, |a+b|=a+b⇒|b+a|=b+a⇒(b,a)∈R ⇒R is symmetric

(b) Let (a,b)∈R. Then, |a+b|=a+b⇒|b+a|=b+a⇒(b,a)∈R ⇒R is symmetric

**Q4.**Let n(U)=700,n(A)=200,n(B)=300 and n(A∩B)=100. Then, n(A^c∩B^c )=

Solution

(c) We have, n(A^c∩B^c) =n{(A∪B)^c } =n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B)} =700-(200+300-100)=300

(c) We have, n(A^c∩B^c) =n{(A∪B)^c } =n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B)} =700-(200+300-100)=300

**Q5.**Consider the following statements: (i) Every reflexive relation is antisymmetric (ii) Every symmetric relation is antisymmetric Which one among (i) and (ii) is true?

Solution

D

D

**Q6.**For real numbers x and y, we write x Ry⇔x-y+√2 is an irrational number. Then, the relation R is

Solution

(a) For any x∈R, we have x-x+√2=√2 an irrational number ⇒x R x for all x So, R is reflexive R is not symmetric, because √2 R 1 but 1 √2 R is not transitive also because √2 R 1 and 1 R 2 √2 but √2 2√2

(a) For any x∈R, we have x-x+√2=√2 an irrational number ⇒x R x for all x So, R is reflexive R is not symmetric, because √2 R 1 but 1 √2 R is not transitive also because √2 R 1 and 1 R 2 √2 but √2 2√2

**Q7.**If A,B and C are three non-empty sets such that A and B are disjoint and the number of elements contained in A is equal to those contained in the set of elements common to the sets A and C, then n(A∪B∪C) is necessarily equal to

Solution

(a) We have, A∩B=Ï• and A⊂C ⇒A∩B=Ï• and A∪C=C ∴n(A∪B∪C)=n(A∪C∪B)=n(C∪B)=n(B∪C)

(a) We have, A∩B=Ï• and A⊂C ⇒A∩B=Ï• and A∪C=C ∴n(A∪B∪C)=n(A∪C∪B)=n(C∪B)=n(B∪C)

**Q8.**If A={Ï•,{Ï•}}, then the power set of A is

Solution

C

C

**Q9.**If R is a relation from a set A to a set B and S is a relation from B to a set C, then the relation SoR

Solution

(a)

(a)

**Q10.**If A_1⊂A_2⊂A_3⊂⋯⊂A_50 and n(A_i )=i-1, then n(⋂_(i=11)^50▒A_i )=

Solution

(d) It is given A_1⊂A_2⊂A_3⊂A_4…⊂A_50 ∴⋃_(i=11)^50▒〖A_i=A_11 〗 ⇒n(⋃_(i=11)^50▒A_i )=n(A_11 )=11-1=10

(d) It is given A_1⊂A_2⊂A_3⊂A_4…⊂A_50 ∴⋃_(i=11)^50▒〖A_i=A_11 〗 ⇒n(⋃_(i=11)^50▒A_i )=n(A_11 )=11-1=10