Sets-1-Quiz-8

SETS-1 Quiz-8

Dear Readers,

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 certain town 25% families own a cell phone, 15% families own a scooter and 65% families own neither a cell phone nor a scooter. If 1500 families own both a cell phone and a scooter, then the total number of families in the town is

•  10000
•  20000
•  30000
•  40000

Solution
C

Q2.Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B^'. Then, R is

•  Reflexive
•  Symmetric >
•   Antisymmetric
•  Transitive

Solution
(b) Clearly, the relation is symmetric but it is neither reflexive nor transitive

Q3.  Given the relation R={(1,2),(2,3)} on the set A={1,2,3}, the minimum number of ordered pairs which when added to R make it an equivalence relation is

•   5
•  6
•  7
•  8

Solution
(c) R is reflexive if it contains (1,1),(2,2),(3,3) ∵(1,2)∈R,(2,3)∈R ∵R is symmetric, if (2,1),(3,2)∈R Now, R={(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2)} R will be transitive, if (3,1),(1,3)∈R Thus, R becomes an equivalence relation by adding (1,1) (2,2) (3,3),(2,1) (3,2),(1,3),(3,1). Hence, the total number of ordered pairs is 7

Q4. Let R_1 be a relation defined by R_1={(a,b)│a≥b,a,b∈R}. Then, R_1 is

•  An equivalence relation on R
•  Reflexive, transitive but not symmetric
•  Symmetric, transitive but not reflexive
•  Neither transitive not reflexive but symmetric

Solution
(b) For any a∈R, we have a≥a Therefore, the relation R is reflexive. R is not symmetric as (2,1)∈R but (1,2)∉R. The relation R is transitive also, because (a,b)∈R,(b,c)∈R imply that a≥b and b≥c which in turn imply that a≥c

Q5. An integer m is said to be related to another integer n if m is a multiple of n. Then, the relation is

•  Reflexive and symmetric
•  Reflexive and transitive
•  Symmetric and transitive
•  Equivalence relation

Solution
 (b) For any integer n, we have n|n⇒n R n So, n R n for all n ∈Z ⇒R is reflexive Now, 2|6 but 6 does not divide 2 ⇒(2,6)∈R but (6,2)∉R So, R is not symmetric Let (m,n)∈R and (n,p)∈R. Then, ├ █((m,n)∈R⇒m|n@(n,p)∈R⇒n|p)}⇒m|p⇒(m,p)∈R So, R is transitive Hence, R is reflexive and transitive but it is not symmetric

Q6. Let X and Y be the sets of all positive divisors of 400 and 1000 respectively (including 1 and the number). Then, n(X∩Y) is equal to

•  4
•  6
• 8
•  12

Solution
(d) X∩Y={1,2,4,5,8,10,20,25,40,50,100,200} ∴ n(X∩Y)=12

Q7. If R⊂A×B and S⊂B×C be relations, then (SoR)^(-1)=

•  S^(-1) o R^(-1)
•  R^(-1) o S^(-1)
•  SoR
•  RoS

Solution
B

Q8. Let R={(a,a)} be a relation on a set A. Then, R is

•  Symmetric
•  TAntisymmetric
•  Symmetric and antisymmetric
•  Neither symmetric nor antisymmetric

Solution
(c)

Q9. In order that a relation R defined on a non-empty set A is an equivalence relation, it is sufficient, if R

•  Is reflective
•  Is symmetric
•  Is transitive
•  Possesses all the above three properties

Solution
D

Q10. Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then,

•  m≥n
•  m≤n
•  m=n
• None of these

Solution
(a) Since R is reflexive relation on A ∴(a,a)∈R for all a∈A ⇒ The minimum number of ordered pairs in R is n Hence, m≥n