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 rehabilitation programme, a group of 50 families were assured new houses and compensation by the government. Number of families who got both is equal to the number of families who got neither of the two. The number of families who got new houses is 6 greater than the number of families who got compensation. How many families got houses?

Solution

(b) Let A and B denote respectively the sets of families who got new houses and compensation It is given that n(A∩B)=n((A∪B) ̅) ⇒n(A∩B)=50-n(A∪B) ⇒n(A)+n(B)=50 ⇒n(B)+6+n(B)=50 [∵n(A)=n(B)+6 (given)] ⇒n(B)=22⇒n(A)=28

(b) Let A and B denote respectively the sets of families who got new houses and compensation It is given that n(A∩B)=n((A∪B) ̅) ⇒n(A∩B)=50-n(A∪B) ⇒n(A)+n(B)=50 ⇒n(B)+6+n(B)=50 [∵n(A)=n(B)+6 (given)] ⇒n(B)=22⇒n(A)=28

**Q2.**If A and B are two sets, then A-(A-B) is equal to

Solution

(C)

(C)

**Q3.**Let A and B be two non-empty subsets of a set X such that A is not a subset of B. Then,

Solution

(D)

(D)

**Q4.**If A={a,b,c},B={b,c,d} and C={a,d,c}, then (A-B)×(B∩C) is equal to

Solution

(a) Given, A={a,b,c}, B={b,c,d} and C={a,d,c} Now, A-B={a,b,c}-{b,c,d}={a} And B∩C={b,c,d}∩{a,d,c}={c,d} ∴(A-B)×(B∩C)={a}×{c,d} ={(a,c),(a,d)}

(a) Given, A={a,b,c}, B={b,c,d} and C={a,d,c} Now, A-B={a,b,c}-{b,c,d}={a} And B∩C={b,c,d}∩{a,d,c}={c,d} ∴(A-B)×(B∩C)={a}×{c,d} ={(a,c),(a,d)}

**Q5.**If A and B are two sets such that n(A)=7,n(B)=6 and (A∩B)≠Ï•. The least possible value of n(A Î” B),

Solution

(a) We have, A ∆ B=(A∪B)-(A∪B) ⇒n(A ∆ B)=n(A)+n(B)-2 n(A∩B) So, n(A ∆ B) is greatest when n(A∩B) is least It is given that A∩B≠Ï•. So, least number of elements in A ∩B can be one ∴ Greatest possible value of n(A ∆ B) is 7+6-2×1=11

(a) We have, A ∆ B=(A∪B)-(A∪B) ⇒n(A ∆ B)=n(A)+n(B)-2 n(A∩B) So, n(A ∆ B) is greatest when n(A∩B) is least It is given that A∩B≠Ï•. So, least number of elements in A ∩B can be one ∴ Greatest possible value of n(A ∆ B) is 7+6-2×1=11

**Q6.**If A={1,2,3,4}, then the number of subsets of A that contain the element 2 but not 3, is

Solution

(b) Required number of subsets is equal to the number of subsets containing 2 and any number of elements from the remaining elements 1 and 4 So, required number of elements =2^2=4

(b) Required number of subsets is equal to the number of subsets containing 2 and any number of elements from the remaining elements 1 and 4 So, required number of elements =2^2=4

**Q7.**If A={p∈N:p is a prime and p=(7n^2+3n+3)/n for some n∈N}, then the number of elements in the set A, is

Solution

(a) We have, p=(7n^2+3n+3)/n⇒p=7n+3+3/n It is given that n∈N and p is prime. Therefore, n=1 ∴n(A)=1

(a) We have, p=(7n^2+3n+3)/n⇒p=7n+3+3/n It is given that n∈N and p is prime. Therefore, n=1 ∴n(A)=1

**Q8.**Let Y={1,2,3,4,5},A{1,2},B={3,4,5} and Ï• denotes null set. If (A×B) denotes cartesian product of the sets A and B; then (Y×A)∩(Y×B) is

Solution

(d) (Y×A)={(1,1),(1,2),(2,1),(2,2), (3,1),(3,2),(4,1),(4,2),(5,1),(5,2)} And(Y×B)={(1,3),(1,4),(1,5),(2,3), (2,4),(2,5),(3,3),(3,4),(3,5),(4,3), (4,4),(4,5),(5,3),(5,4),(5,5)} ∴(Y×A)∩(Y×B)=Ï•

(d) (Y×A)={(1,1),(1,2),(2,1),(2,2), (3,1),(3,2),(4,1),(4,2),(5,1),(5,2)} And(Y×B)={(1,3),(1,4),(1,5),(2,3), (2,4),(2,5),(3,3),(3,4),(3,5),(4,3), (4,4),(4,5),(5,3),(5,4),(5,5)} ∴(Y×A)∩(Y×B)=Ï•

**Q9.**In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is

Solution

(c) Clearly, Required percent =20+50-10=60% [∵n(A∪B)=n(A)+n(B)-n(A∩B)]

(c) Clearly, Required percent =20+50-10=60% [∵n(A∪B)=n(A)+n(B)-n(A∩B)]

**Q10.**If A={x∶x is a multiple of 4} and, B={x∶x is a multiple of 6}, then A∩B consists of multiples of

Solution

(b) Let x∈A∩B. Then, x∈A and x∈B ⇒x is a multiple of 4 and x is a multiple of 6 ⇒x is a multiple of 4 and 6 both ⇒x is a multiple of 12

(b) Let x∈A∩B. Then, x∈A and x∈B ⇒x is a multiple of 4 and x is a multiple of 6 ⇒x is a multiple of 4 and 6 both ⇒x is a multiple of 12