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..

**Q2.**Let n be a fixed positive integer. Define a relation R on the set Z of integers by, a R b⇔n | a-b. Then, R is not

Solution

D

D

**Q3.**Let R and S be two equivalence relations on a set A. Then,

Solution

B

B

**Q4.**Let A be the set of all students in a school. A relation R is defined on A as follows: "aRb iff a and b have the same teacher”

Solution

(d) We have, n(A∩B ̅ )=9,n(A ̅∩B)=10 and n(A∪B)=24 ⇒n(A)-n(A∩B)=9,n(B)-n(A∩B)=10 and, n(A)+n(B)-n(A∩B)=24 ⇒n(A)+n(B)-2n(A∩B)=19 and n(A)+n(B)-n(A∩B)=24 ⇒n(A∩B)=5 ∴n(A)=14 and n(B)=15 Hence, n(A×B)=14×15=210

(d) We have, n(A∩B ̅ )=9,n(A ̅∩B)=10 and n(A∪B)=24 ⇒n(A)-n(A∩B)=9,n(B)-n(A∩B)=10 and, n(A)+n(B)-n(A∩B)=24 ⇒n(A)+n(B)-2n(A∩B)=19 and n(A)+n(B)-n(A∩B)=24 ⇒n(A∩B)=5 ∴n(A)=14 and n(B)=15 Hence, n(A×B)=14×15=210

**Q5.**Let R be an equivalence relation on a finite set A having n elements. Then, the number of ordered pairs in R is

Solution

(b) Since R is an equivalence relation on set A. Therefore (a,a)∈R for all a∈A. Hence, R has at least n ordered pairs

(b) Since R is an equivalence relation on set A. Therefore (a,a)∈R for all a∈A. Hence, R has at least n ordered pairs

**Q6.**Consider the following statements: p∶ Every reflexive relation is symmetric relation q∶ Every anti-symmetric relation is reflexive Which of the following is/are true?

Solution

(d) Clearly, none of the statements is true

(d) Clearly, none of the statements is true

**Q7.**If P is the set of all parallelograms, and T is the set of all trapeziums, then P∩T is

Solution

(a) Clearly, P⊂T ∴P∩T=P

(a) Clearly, P⊂T ∴P∩T=P

**Q8.**The relation R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)} on set A={1,2,3} is

Solution

(a) Since (1,1),(2,2),(3,3)∈R. Therefore, R is reflexive. We observe that (1,2)∈R but (2,1)∉R, therefore R is not symmetric. It can be easily seen that R is transitive

(a) Since (1,1),(2,2),(3,3)∈R. Therefore, R is reflexive. We observe that (1,2)∈R but (2,1)∉R, therefore R is not symmetric. It can be easily seen that R is transitive

**Q9.**If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A×B and B×A are

Solution

(b) Number of elements common to each set is 99×99=99^2.

(b) Number of elements common to each set is 99×99=99^2.

**Q10.**A survey shows that 63% of the Americans like cheese whereas 76% like apples. If x% of the Americans like both cheese and apples, then

Solution

(c) Given, n(C)=63,n(A)=76 and n(C∩A)=x We know that, n(C∪A)=n(C)+n(A)-n(C∩A) ⇒ 100=63+76-x⇒x=139-100=39 And n(C∩A)≤n(C) ⇒ x≤63 ∴39≤x≤63

(c) Given, n(C)=63,n(A)=76 and n(C∩A)=x We know that, n(C∪A)=n(C)+n(A)-n(C∩A) ⇒ 100=63+76-x⇒x=139-100=39 And n(C∩A)≤n(C) ⇒ x≤63 ∴39≤x≤63