## SETS-1 Quiz-14

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. If A={1,2,3}, then the relation R={(1,1),(2,2),(3,1),(1,3)} is
•  Reflexive
•  Symmetric
•  Transitive
•  Equivalence
Solution
(b) Clearly, (3,3)∉R. So, R is not reflexive. Also, (3,1) and (1,3) are in R but (3,3)∉R. So, R is not transitive But, R is symmetric as R=R^(-1)

Q2.Let R be a relation on the set N be defined by {(x,y)│x,y∈N,2 x+y=41}. Then, R is
•  Reflexive
•  Symmetric >
•   Transitive
•  None of these
Solution
(d) We have R={(1,39),(2,37),(3,35),(4,33),(5,31),(6,29), (7,27),(8,25),(9,23),(10,21),(11,19),(12,17), (13,15),(14,13),(15,11),(16,9),(17,7),(18,5), (19,3),(20,1)} Since (1,39)∈R, but (39,1)∉R Therefore, R is not symmetric Clearly, R is not reflexive. Now, (15,11)∈R and (11,19)∈R but (15,19)∉R So, R is not transitive

Q3.  Let A={1,2,3},B={3,4},C={4,5,6}. Then, A∪(B∩C) is
•   {3}
•  {1,2,3,4}
•  {1,2,5,6}
•  {1,2,3,4,5,6}
Solution
B

Q4. If A,B and C are three non-empty sets such that any two of them are disjoint, then (A∪B∪C)∩(A∩B∩C)=
•  A
•  B
•  C
•  Φ
Solution
(d) As A,B,C are pair wise disjoints. Therefore, A∩B=ϕ,B∩C=ϕ and A∩C=ϕ ⇒A∩B∩C=ϕ⇒(A∪B∪C)∩(A∩B∩C)=ϕ

Q5.If A={1,2,3,4}, then the number of subsets of set A containing element 3, is
•  24
•  28
•  8
•   16
Solution
(c) Each subset will contain 3 and any number of elements from the remaining 3 elements 1, 2 and 4 So, required number of elements =2^2=8

Q6. For any three sets A_1,A_2,A_3, let B_1=A_1,B_2=A_2-A_1 and B_3=A_3-(A_1∪A_2 ), then which one of the following statement is always true
•  A_1∪A_2∪A_3⊃B_1∪B_2∪B_3
•  A_1∪A_2∪A_3=B_1∪B_2∪B_3
• A_1∪A_2∪A_3⊂B_1∪B_2∪B_3
•   None of these
Solution
(a) We have, B_1=A_1⇒B_1⊂A_1 B_2=A_2-A_1⇒B_2⊂A_2 B_3=A_3-(A_1∪A_2)⇒B_3⊂A_3 ∴B_1∪B_2∪B_3⊂A_1∪A_2∪A_3

Q7. In an election, two contestants A and B contested x% of the total voters voted for A and (x+20)% for B. If 20% of the voters did not vote, then x=
•  30
•  25
•  40
•  35
Solution
a) Let the total number of voters be n. Then, Number of voters voted for A=nx/100 Number of voters voted for B=(n(x+20))/100 ∴ Number of voters who voted for both =nx/100+(n(x+20))/100 =(n(2x+20))/100 Hence,n-n(2x+20)/100=20n/100⇒x=30

Q8. If X and Y are two sets, then X∩(Y∪X)' equals
•  X
•  Y
•  ϕ
•  None of these
Solution
(c) We have, X∩(Y∪X)^'=X∩(Y^'∩X^' )=(X∩X^' )∩Y^'=ϕ∩Y^'=ϕ

Q9. Given n(U)=20,n(A)=12,n(B)=9,n(A∩B)=4, where U is the universal set, A and B are subsets of U, then n[(A∪B)^c] equals to
•  10
•  9
•  11
•  3
Solution
(d) n(A∪B)=n(A)+n(B)-n(A∩B) =12+9-4=17 Hence, n[(AUB)^c]=n(U)-n(A∪B) =20-17=3

Q10. Let A and B be two sets, then (A∪B)^'∪(A^'∩B)is equal to
•  A'
•  A
•  B'
• None of these
Solution
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