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.**Two points P and Q in a plane are related if OP=OQ, where O is a fixed point. This relation is

Solution

B

B

**Q2.**Let L be the set of all straight lines in the Euclidean plane. Two lines l_1 and l_2 are said to be related by the relation R iff l_1 is parallel to l_2. Then, the relation R is not

Solution

(d)

(d)

**Q3.**If a set has 13 elements and R is a reflexive relation on A with n elements, then

Solution

(c) Since R is a reflexive relation on A. ∴(a,a)∈R for all a∈A ⇒n(A)≤n(R)≤n(A×A)⇒13≤n(R)≤169

(c) Since R is a reflexive relation on A. ∴(a,a)∈R for all a∈A ⇒n(A)≤n(R)≤n(A×A)⇒13≤n(R)≤169

**Q4.**In a class of 45 students, 22 can speak Hindi and 12 can speak English only. The number of students, who can speak both Hindi and English, is

Solution

(b) We have, n(H)-n(H∩E)=22,n(E)-n(H∩E)=12,n(H∪E)=45 ∴n(H∪E)=n(H)+n(E)-n(H∩E) ⇒45=22+12+n(H∪E) ⇒n(H∩E)=11

(b) We have, n(H)-n(H∩E)=22,n(E)-n(H∩E)=12,n(H∪E)=45 ∴n(H∪E)=n(H)+n(E)-n(H∩E) ⇒45=22+12+n(H∪E) ⇒n(H∩E)=11

**Q5.**Let A be the non-void set of the children in a family. The relation 'x is a brother of y' on A is

Solution

(c)

(c)

**Q6.**Let S={1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to

Solution

(d) Required number =(3^4+1)/2=41

(d) Required number =(3^4+1)/2=41

**Q7.**If A and B are two sets such that n(A∩B ̅ )=9,n(A ̅∩B)=10 and n(A∪B)=24, then n(A×B)=

Solution

B

B

Solution

(c) Let A={x∈R:(2x-1)/(x^3+4x^2+3x)} Now, x^3+4x^2+3x=x(x^2+4x+3) =x(x+3)(x+1) ∴ A=R-{0,-1,-3}

(c) Let A={x∈R:(2x-1)/(x^3+4x^2+3x)} Now, x^3+4x^2+3x=x(x^2+4x+3) =x(x+3)(x+1) ∴ A=R-{0,-1,-3}

**Q9.**In a set of ants in a locality, two ants are said to be related iff they walk on a same straight line, then the relation is

Solution

(d) Clearly, R is an equivalence relation

(d) Clearly, R is an equivalence relation

**Q10.**Out of 800 boys in a school 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is

Solution

(a) Let U be the set of all students in the school. Let C,H and B denote the sets of students who played cricket, hockey and basketball respectively. Then, n(U)=800,n(C)=224,n(H)=240,n(B)=336 n(H∩B)=64,n(B∩C)=80,n(H∩C)=40 and,n(H∩B∩C)=24 ∴ Required number =n(C'∩H'∩B') =n(C∪H∪B)' =n(U)-n(C∪H∪B) =n(U)-{n(C)+n(H)+n(B)-n(C∩H)-n(H∩B)-n(B∩C)+n(C∩H∩B)} =800-{224+240+336+336-64-80-40+24} =800-640=160

(a) Let U be the set of all students in the school. Let C,H and B denote the sets of students who played cricket, hockey and basketball respectively. Then, n(U)=800,n(C)=224,n(H)=240,n(B)=336 n(H∩B)=64,n(B∩C)=80,n(H∩C)=40 and,n(H∩B∩C)=24 ∴ Required number =n(C'∩H'∩B') =n(C∪H∪B)' =n(U)-n(C∪H∪B) =n(U)-{n(C)+n(H)+n(B)-n(C∩H)-n(H∩B)-n(B∩C)+n(C∩H∩B)} =800-{224+240+336+336-64-80-40+24} =800-640=160