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 class of 30 pupils 12 take needls work, 16 take physics and 18 take history. If all the 30 students take at least one subject and no one takes all three, then the number of pupils taking 2 subjects is

Solution

(a) Given, n(N)=12,n(P)=16,n(H)=18, n(N∪P∪H)=30 And n(N∩P∩H)=0 Now, n(N∪P∪H)=n(N)+n(P)+n(H) -n(N∩P)-n(P∩H)-n(H∩N) +n(N∩P∩H) ⇒n(N∩P)+n(P∩H)+n(H∩N)=(12+16+18)-30 =46-30=16

(a) Given, n(N)=12,n(P)=16,n(H)=18, n(N∪P∪H)=30 And n(N∩P∩H)=0 Now, n(N∪P∪H)=n(N)+n(P)+n(H) -n(N∩P)-n(P∩H)-n(H∩N) +n(N∩P∩H) ⇒n(N∩P)+n(P∩H)+n(H∩N)=(12+16+18)-30 =46-30=16

**Q2.**If A={Î¸∶cosÎ¸>-1/2,0≤Î¸≤Ï€} and B={Î¸∶sin〖Î¸>1/2,Ï€/3≤Î¸≤Ï€〗 }, then

Solution

(a) We have, cos〖Î¸>-1/2 and 0≤Î¸≤Ï€〗 ⇒0≤Î¸≤2Ï€/3 and 0≤Î¸≤Ï€ ⇒0≤Î¸≤2Ï€/3 ⇒A={Î¸:0≤Î¸≤2Ï€/3} Also, sin〖Î¸>1/2 and Ï€/3≤Î¸≤Ï€〗 ⇒Ï€/3≤Î¸≤(5 Ï€)/6⇒B={Î¸:Ï€/3≤Î¸≤(5 Ï€)/6} ∴A∩B={Î¸:Ï€/3≤Î¸≤(2 Ï€)/3} and A∪B={Î¸:0≤Î¸≤5Ï€/6}

(a) We have, cos〖Î¸>-1/2 and 0≤Î¸≤Ï€〗 ⇒0≤Î¸≤2Ï€/3 and 0≤Î¸≤Ï€ ⇒0≤Î¸≤2Ï€/3 ⇒A={Î¸:0≤Î¸≤2Ï€/3} Also, sin〖Î¸>1/2 and Ï€/3≤Î¸≤Ï€〗 ⇒Ï€/3≤Î¸≤(5 Ï€)/6⇒B={Î¸:Ï€/3≤Î¸≤(5 Ï€)/6} ∴A∩B={Î¸:Ï€/3≤Î¸≤(2 Ï€)/3} and A∪B={Î¸:0≤Î¸≤5Ï€/6}

**Q3.**Let A={1,2,3,4},B={2,4,6}. Then, the number of sets C such that A∩B⊆C⊆A ∪B is

Solution

(c) A∩B={2,4} {A∩B}⊆{1,2,4},{3,2,4},{6,2,4},{1,3,2,4}, {1,6,2,4},{6,3,2,4},{2,4},{1,3,6,2,4}⊆A∪B ⇒ n(C)=8

(c) A∩B={2,4} {A∩B}⊆{1,2,4},{3,2,4},{6,2,4},{1,3,2,4}, {1,6,2,4},{6,3,2,4},{2,4},{1,3,6,2,4}⊆A∪B ⇒ n(C)=8

**Q4.**If A={1,2,3,4,5},B={2,4,6},C={3,4,6}, then (A∪B)∩C is

Solution

(a) ∵ A∪B={1,2,3,4,5,6} ∴ (A∪B)∩C={1,2,3,4,5,6}∩{3,4,6} ={3,4,6}

(a) ∵ A∪B={1,2,3,4,5,6} ∴ (A∪B)∩C={1,2,3,4,5,6}∩{3,4,6} ={3,4,6}

**Q5.**From 50 students taking examinations in Mathematics, Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 passed Mathematics and Chemistry and at most 20 passed Physics and Chemistry. The largest possible number that could have passed all three examinations is

Solution

D

D

**Q6.**The relation R={(1,3),(3,5)} is defined on the set with minimum number of elements of natural numbers. The minimum number of elements to be included in R so that R is an equivalence relation, is

Solution

(a) To make R a reflexive relation, we must have (1,1),(3,3) and (5,5) in it. In order to make R a symmetric relation, we must inside (3,1) and (5,3) in it. Now, (1,3)∈R and (3,5)∈R. So, to make R a transitive relation, we must have, (1,5)∈R. But, R must be symmetric also. So, it should also contain (5,1). Thus, we have R={(1,1),(3,3),(5,5),(1,3),(3,5),(3,1),(5,3),(1,5),(5,1)} Clearly, it is an equivalence relation on A{1,3,5}

(a) To make R a reflexive relation, we must have (1,1),(3,3) and (5,5) in it. In order to make R a symmetric relation, we must inside (3,1) and (5,3) in it. Now, (1,3)∈R and (3,5)∈R. So, to make R a transitive relation, we must have, (1,5)∈R. But, R must be symmetric also. So, it should also contain (5,1). Thus, we have R={(1,1),(3,3),(5,5),(1,3),(3,5),(3,1),(5,3),(1,5),(5,1)} Clearly, it is an equivalence relation on A{1,3,5}

**Q7.**Let R be a relation on a set A such that R=R^(-1), then R is

Solution

(b) Let (a,b)∈R. Then, (a,b)∈R⇒(b,a)∈R^(-1) [By def. of R^(-1)] ⇒(b,a)∈R [∵R=R^(-1)] So, R is symmetric

(b) Let (a,b)∈R. Then, (a,b)∈R⇒(b,a)∈R^(-1) [By def. of R^(-1)] ⇒(b,a)∈R [∵R=R^(-1)] So, R is symmetric

**Q8.**Let A={x:x is a multiple of 3} and B={x:x is a multiple of 5}. Then, A∩Bis given by

Solution

(c) A∩B={x:x a multiple of 3}and {x:x is a multiple of 5} ={x:x is a multiple of 15} ={15,30,45,……….}

(c) A∩B={x:x a multiple of 3}and {x:x is a multiple of 5} ={x:x is a multiple of 15} ={15,30,45,……….}

**Q9.**If R is a relation on a finite set having n elements, then the number of relations on A is

Solution

B

B

**Q10.**If A⊆B, then B∪A is equal to

Solution

(c) ∵ A⊆B ∴ B∪A=B

(c) ∵ A⊆B ∴ B∪A=B