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.**Which one of the following relations on R is an equivalence relation?

Solution

A

A

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**Q2.**Three sets A,B,Care such that A=B∩C and B=C∩A, then

Solution

(c) Since, A=B∩C and B=C∩A, Then A≡B

(c) Since, A=B∩C and B=C∩A, Then A≡B

**Q3.**Suppose A_1,A_2,…,A_30 are thirty sets, each having 5 elements and B_1,B_2,…,B_n are n sets each with 3 elements, let ⋃_(i=1)^30▒〖A_i=〗 ⋃_(j=1)^n▒〖B_j=〗 S and each element of S belongs to exactly 10 of the A_i's and exactly 9 of the B_j's. Then, n is equal to

Solution

(c) Given, A's are 30 sets with five elements each, so ∑_(i=1)^30▒〖n(A_i )=5×30=150〗 ...(i) If the m distinct elements in S and each elements of S belongs to exactly 10 of the A_i's, then ∑_(i=1)^30▒〖n(A_i )=10m〗 ...(ii) From Eqs. (i) and (ii), m=15 Similarly, ∑_(j=1)^n▒〖n(B_j )=3n〗 and ∑_(j=1)^n▒〖n(B_j )=9m〗 ∴ 3n=9m ⇒ n=9m/3=3×15=45

(c) Given, A's are 30 sets with five elements each, so ∑_(i=1)^30▒〖n(A_i )=5×30=150〗 ...(i) If the m distinct elements in S and each elements of S belongs to exactly 10 of the A_i's, then ∑_(i=1)^30▒〖n(A_i )=10m〗 ...(ii) From Eqs. (i) and (ii), m=15 Similarly, ∑_(j=1)^n▒〖n(B_j )=3n〗 and ∑_(j=1)^n▒〖n(B_j )=9m〗 ∴ 3n=9m ⇒ n=9m/3=3×15=45

**Q4.**Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of the set of all determinants with value -1. Then

Solution

(b) Since the value of a determinant charges by minus sign by interchanging any two rows or columns. Therefore, corresponding to every element ∆ of B there is an element ∆' in C obtained by interchanging two adjacent rows (or columns) in ∆. It follows from this that n(B)≤n(C) Similarly, we have n(C)≤n(B) Hence, n(B)=n(C)

(b) Since the value of a determinant charges by minus sign by interchanging any two rows or columns. Therefore, corresponding to every element ∆ of B there is an element ∆' in C obtained by interchanging two adjacent rows (or columns) in ∆. It follows from this that n(B)≤n(C) Similarly, we have n(C)≤n(B) Hence, n(B)=n(C)

**Q5.**If A={1,2,3},B{3,4},C{4,5,6}. Then, A∪(B∩C) is

Solution

(d) B∩C={4}. ∴A∪(B∩C)={1,2,3,4}

(d) B∩C={4}. ∴A∪(B∩C)={1,2,3,4}

**Q6.**If A={1,3,5,7,9,11,13,15,17},B={2,4…,18} and N is the universal set, then A'∪((A∪B)∩B') is

Solution

(b) We have, (A∪B)∩B^'=A ∴((A∪B)∩B^' )∪A^'=A∪A^'=N

(b) We have, (A∪B)∩B^'=A ∴((A∪B)∩B^' )∪A^'=A∪A^'=N

**Q7.**In a set of teachers of a school, two teachers are said to be related if they “teach the same subject”, then the relation is

Solution

(d) Clearly, given relation is an equivalence relation

(d) Clearly, given relation is an equivalence relation

**Q8.**In a town of 10,000 families it was found that 40% families buy newspaper A,20% families buy newspaper B and 10% families buy newspaper C,5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then the number of families which buy A only is

Solution

(b) We have, N=10,000,n(A)=40% of 10,000=4000, n(B)=2000,n(C)=1000,n(A∩B)=500, n(B∩C)=300,n(C∩A)=400,n(A∩B∩C)=200 Now, Required number of families = n(A∩B ̅∩C ̅ )=n(A∩(B∪C)^') =n(A)-n(A∩(B∪C)) =n(A)-n((A∩B)∪(A∩C)) =n(A)-{n(A∩B)+n(A∩C)-n(A∩B∩C)} =4000-(500+400-200)=3300

(b) We have, N=10,000,n(A)=40% of 10,000=4000, n(B)=2000,n(C)=1000,n(A∩B)=500, n(B∩C)=300,n(C∩A)=400,n(A∩B∩C)=200 Now, Required number of families = n(A∩B ̅∩C ̅ )=n(A∩(B∪C)^') =n(A)-n(A∩(B∪C)) =n(A)-n((A∩B)∪(A∩C)) =n(A)-{n(A∩B)+n(A∩C)-n(A∩B∩C)} =4000-(500+400-200)=3300

**Q9.**In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?

Solution

(c) We have, c+e+f+g=100 a+d+e+g=70 b+d+f+g=40 e+g=30 g+f=28 d+g=23 g=18 ∴g=18,f=10,e=12,d=15,a=35,b=7,c=60

(c) We have, c+e+f+g=100 a+d+e+g=70 b+d+f+g=40 e+g=30 g+f=28 d+g=23 g=18 ∴g=18,f=10,e=12,d=15,a=35,b=7,c=60

**Q10.**Let R be a relation defined on S, the set of squares on a chess board such that xRy iff x and y share a common side. Then, which of the following is false for R?

Solution

(c) Clearly, R is reflexive and symmetric but it is not transitive

(c) Clearly, R is reflexive and symmetric but it is not transitive

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