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={a,b,c,l,m,n}, then the maximum number of elements in
any relation on A is
Solution
(d) Any relation on A is a subset of A×A which contains 36 elements. Hence, maximum number of elements in a relation on A can be 36
(d) Any relation on A is a subset of A×A which contains 36 elements. Hence, maximum number of elements in a relation on A can be 36
Q2.Let A be the set of all animals.
A relation R is defined as "aRb iff a and b are in different zoological parks”. Then R is
Solution
(b) Clearly, (a,a)∈R for any a∈A Also, (a,b)∈R ⇒a and b are in different zoological parks ⇒b and a are in different zoological parks ⇒(b,a)∈R Now, (a,b)∈R and (b,a)∈R but (a,a)∉R So, R is not transitive
(b) Clearly, (a,a)∈R for any a∈A Also, (a,b)∈R ⇒a and b are in different zoological parks ⇒b and a are in different zoological parks ⇒(b,a)∈R Now, (a,b)∈R and (b,a)∈R but (a,a)∉R So, R is not transitive
Q3. In a class of 35 students, 17 have taken Mathematics, 10
have taken Mathematics but not Economics. If each student has taken either Mathematics or Economics or
both, then the number of students who have taken Economics but not Mathematics is
Solution
(c) Let M and E denote the sets of students who have taken Mathematics and Economics respectively. Then, we have n(M∪E)=35,n(M)=17 and n(M∩E^' )=10 Now, n(M∩E^' )=n(M)-n(M∩E) ⇒10=17-n(M∩E)⇒n(M∩E)=7 Now, n(M∪E)=n(M)+n(E)-n(M∩E) ⇒35=17+n(E)-7⇒n(E)=25 ∴n(E∩M^' )=n(E)-n(E∩M)=25-7=18
(c) Let M and E denote the sets of students who have taken Mathematics and Economics respectively. Then, we have n(M∪E)=35,n(M)=17 and n(M∩E^' )=10 Now, n(M∩E^' )=n(M)-n(M∩E) ⇒10=17-n(M∩E)⇒n(M∩E)=7 Now, n(M∪E)=n(M)+n(E)-n(M∩E) ⇒35=17+n(E)-7⇒n(E)=25 ∴n(E∩M^' )=n(E)-n(E∩M)=25-7=18
Q4. A and B are any two non-empty sets and A is proper subset of B. If n(A)=5, then find the minimum possible value of n(A∆B)
Solution
(a) It is given that A is a proper subset of B ∴A-B=Ï•⇒n(A-B)=0 We have, n(A)=5. So, minimum number of elements in B is 6 Hence, the minimum possible value of n(A Δ B) is n(B)-n(A)=6-5=1
(a) It is given that A is a proper subset of B ∴A-B=Ï•⇒n(A-B)=0 We have, n(A)=5. So, minimum number of elements in B is 6 Hence, the minimum possible value of n(A Δ B) is n(B)-n(A)=6-5=1
Q5.
If n(A_i )=i+1 and A_1⊂A_2⊂A_3⊂⋯⊂A_99, then n(⋃_(i=1)^99▒A_i )=
Solution
(c) It is given that A_1⊂A_2⊂A_3…⊂A_99 ⋃_(i=1)^999▒〖A_i=A_99 〗 ⇒n(⋃_(i=1)^99▒A_i )=n(A_99 )=99+1=100
(c) It is given that A_1⊂A_2⊂A_3…⊂A_99 ⋃_(i=1)^999▒〖A_i=A_99 〗 ⇒n(⋃_(i=1)^99▒A_i )=n(A_99 )=99+1=100
Q6. If S is a set with 10 elements and A={(x,y):x,y∈S,x≠y},
then the number of elements in A is
Solution
(b) Number of element is S=10 And A={(x,y);x,y∈S,x≠y} ∴ Number of element in A=10×9=90
(b) Number of element is S=10 And A={(x,y);x,y∈S,x≠y} ∴ Number of element in A=10×9=90
Q7.
Let X={1,2,3,4,5} and Y={1,3,5,7,9}. Which of the following is/are not relations from X to
Y?
Solution
(d) R_4 is not a relation from A to B, because (7,9)∈R_4 but (7,9)∉A×B
(d) R_4 is not a relation from A to B, because (7,9)∈R_4 but (7,9)∉A×B
Q8.
If A is a finite set having n elements, then P(A) has
Solution
B
B
Q9.
In the above question, the number of families which buy none of A,B and C is
Solution
(a) We have, Required number of families =n(A'∩B'∩C') =n(A∪B∪C)' =N-n(A∪B∪C) =10000-{n(A)+n(B)+n(C)-n(A∩B)} -n(B∩C)-n(A∩C)+n(A∩B∩C)} =10000-4000-2000-1000+500+300+400-200 =4000
(a) We have, Required number of families =n(A'∩B'∩C') =n(A∪B∪C)' =N-n(A∪B∪C) =10000-{n(A)+n(B)+n(C)-n(A∩B)} -n(B∩C)-n(A∩C)+n(A∩B∩C)} =10000-4000-2000-1000+500+300+400-200 =4000
Q10. If A⊆B, then B∪A is equal
to
Solution
(c) ∵ A⊆B ∴ B∪A=B
(c) ∵ A⊆B ∴ B∪A=B