RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY-3

Venn diagrams:
Two sets A and B are shown with the following Venn Diagrams.
A – B is denoted by the blue region.
(A – B) = {2, 4, 8, 10}
A ∩ B is denoted by the common region (orange)
A ∩ B = {6, 12}
B – A is denoted by the yellow region.
B – A = {6, 12}
There is no intersection between these three regions
Hence the three sets are disjoint sets. (Verified)

Exercise 1G

Q. 1. If A and B are two sets such that n(A) = 37, n(B) = 26 and n(A ∪ B) = 51, find n(A ∩ B).

Answer: Given: n(A) = 37; n(B) = 26;
n(A ∪ B) = 51
We know,
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 51 = 37 + 26 – n(A ∩ B)
⇒ n(A ∩ B) = 63 – 51 = 12
∴ n(A ∩ B) = 12  (Ans)

Q. 2. If P and Q are two sets such that n(P ∪ Q) = 75, n(P ∩ Q) = 17 and n(P) = 49, find n(Q).

Answer: Given: n(P ∪ Q) = 75;
n(P ∩ Q) = 17 and n(P) = 39 
We know,
n(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q)
⇒ 75 = 49 + n(Q) – 17
⇒ 75 = 32 + n(Q)
or n(Q) = 43
∴ n(Q) = 43  (Ans)

Q. 3. (i) If A and B are two sets such that n(A) = 24, n(B) = 22 and n(A ∩ B) = 8, find: n(A ∪ B)

Answer: Given: n(A) = 24, n(B) = 22
and n(A ∩ B) = 8
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 24 + 22 – 8 = 38
∴ n(A ∪ B) = 38  (Ans)

Q. 3. (ii) If A and B are two sets such that n(A) = 24, n(B) = 22 and n(A ∩ B) = 8, find: n(A – B)

Answer: Given: n(A) = 24, n(B) = 22
and n(A ∩ B) = 8
n(A – B) = n(A) – n(A ∩ B)
⇒ n(A – B) = 24 – 8 = 16
∴ n(A – B) = 16  (Ans)

Q. 3. (iii) If A and B are two sets such that n(A) = 24, n(B) = 22 and n(A ∩ B) = 8, find: n(B – A)

Answer: Given: n(A) = 24, n(B) = 22
and n(A ∩ B) = 8
n(B – A) = n(B) – n(A ∩ B)
⇒ n(B – A) = 22 – 8 = 14
∴ n(B – A) = 14  (Ans)

Q. 4. (i) If A and B are two sets such that n(A – B) = 24, n(B – A) = 19 and n(A ∩ B) = 11, find: n(A)

Answer: Given: n(A – B) = 24,
n(B – A) =19 and n(A ∩ B) = 11
n(A) = n(A – B) + n(A ∩ B)
⇒ n(A) = 24 + 11 = 35
∴ n(A) = 35  (Ans)

Q. 4. (ii) If A and B are two sets such that n(A – B) = 24, n(B – A) = 19 and n(A ∩ B) = 11, find: n(B)

Answer: Given: n(A – B) = 24,
n(B – A) =19 and n(A ∩ B) = 11
n(B) = n(B – A) + n(A ∩ B)
⇒ n(B) = 19 + 11 = 30
∴ n(B) = 30  (Ans)

Q. 4. (iii) If A and B are two sets such that n(A – B) = 24, n(B – A) = 19 and n(A ∩ B) = 11, find:  n(A ∪ B)

Answer: Given: n(A – B) = 24,
n(B – A) =19 and n(A ∩ B) = 11
∴ n(A) = n(A – B) + n(A ∩ B)
= 24 + 11 = 35 . . . . (i)
∴n(B) = n(B – A) + n(A ∩ B)
= 19 + 11 = 30 . . . . (ii)
We know
n(A ∪ B)  = n(A) + n(B) – n(A ∩ B)
∴ n(A ∪ B)= 35 + 30 – 11 = 54 , , ,[From (i) & (ii2)]
∴ n(A ∪ B) = 54   (Ans)

Q. 5. In a committee, 50 people speak Hindi, 20 speak English and 10 speak both Hindi and English. How many speak at least one of these two languages?

Answer : Let H be the set of people who speak Hindi.
M be the set of people who speak English.
∴ n(H) = 50;
n(E) = 20;
n(H ∩ E) = 10
To find people  who speak at least one of the two languages = n(H ∪ E)
we know that,
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Therefore,
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
∴ n(H ∪ E) = 50 + 20 – 10 = 60
Ans: People who speak at least one of the two languages are 60. 

RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY-3

Q. 6. In a group of 50 persons, 30 like tea, 25 like coffee and 16 like both. How many like
(i) either tea or coffee?
(ii) neither tea nor coffee?

Answer: Let T be the set of people who like Tea.
C be the set of people who like Coffee.
Total number of people = n(X) = 50
∴ n(T) = 30;
n(C) = 25;
n(T ∩ C) = 16
(i) To find: People who like either tea or coffee = n(T ∪ C)
Therefore,
n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
∴ n(T ∪ C) = 30 + 25 – 16 = 39
Ans: People who like either tea or coffee = 39

(ii) People who like neither tea nor coffee
= n(X) – n(T ∪ C)
= 50 – 39 = 11
Ans: People who like neither tea nor coffee = 11 

Q.7. There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C₁, 50 to chemical C₂, and 30 to both the chemicals C₁ and C₂. Find the number of individuals exposed to
(i) Chemical C₁ but not chemical C₂
(ii) Chemical C₂ but not chemical C₁
(iii) Chemical C₁ or chemical C₂

Answer: Given:
Total number of individuals with skin disorder = n(C) = 200
Individuals exposed to chemical C₁ = n(C₁) = 120
Individuals exposed to chemical C₂ = n(C₂) = 50
Exposed to chemicals C₁ and C₂ both = n(C₁ ∩ C₂) = 30
Individuals exposed to Chemical C1 but not C₂ = n(C₁ – C₂)
(i) To Find: Individuals exposed to Chemical C1 but not C₂
= n(C₁ – C₂)
∴ n(C₁ – C₂) = n(C₁) – n(C₁ ∩ C₂)
⇒ n(C₁ – C₂) = 120 – 30 = 90
Number of individuals exposed to chemical C₁ but not C2 = 90  (Ans)

(ii) To Find: Individuals exposed to Chemical C₂ but not C₁
= n(C₂– C₁)
∴ n(C₂ – C₁) = n(C₂) – n(C₁ ∩ C₂)
⇒ n(C₂– C₁) = 50 – 30 = 20
Number of individuals exposed to chemical C₂ but not C₁ = 20  (Ans)

(iii) Individuals exposed to Chemical C₁ or chemical C₂
= n(C₁ ∪ C₂)
∴ n(C₁∪ C₂) = n(C₁) + n(C₂) – n(C₁∩ C₂)
⇒ n(C₁∪ C₂) = 120 + 50 – 30 = 140
Number of individuals exposed to chemical C₁ or C₂ = 140  (Ans)

Q. 8. In a class of a certain school, 50 students offered mathematics, 42 offered biology and 24 offered both the subjects. Find the number of students offering
(i) mathematics only,
(ii) biology only,
(iii) any of the two subjects.

Answer: Let M be the set of students who offered Mathematics.
B be the set of students who offered Biology.
Given: Number of students offered Mathematics  n(M) = 50
Number of students offered Biology n(B) = 42
Number of students offered both Mathematics and Biology = n(M ∩ B) = 24
(i) To Find: Number of students offered Mathematics only = n(M – B)
∴ n(M – B)
= n(M) – n(M ∩ B)
= 50 – 24 = 26
Number of students offered Mathematics only= 26  (Ans)

(ii) To Find: Number of students offered Biology only = n(B – M)
∴ n(B – M)
= n(B) – n(M ∩ B)
= 42 – 24 = 18
Number of students offered Biology only = 18  (Ans)

(iii) To Find:  Number of students offered any of two subjects = n(M ∪ B)
∴ n(M ∪ B)
= n(M) + n(B) – n(M ∩ B)
= 50 + 42 – 24 = 140
Number of students offered any of two subjects = 68  (Ans)

Q. 9. In an examination, 56% of the candidates failed in English and 48% failed in science. If 18% failed in both English and science, find the percentage of those who passed in both the subjects.

Answer: Let E be the set of candidates who failed in English.
S be the set of candidates who failed in Science .
Given: Percentage of candidates who failed in English = n(E) = 56
Percentage of candidates who failed in Science = n(S) = 48
Percentage of candidates who failed in English and Science both = n(E ∩ S) = 18
To Find: Percentage of candidates who failed in English only = n(E – S)
∴  n(E – S)
= n(E) – n(E ∩ S)
= 56 – 18 = 38
Percentage of candidates who failed in Science only = n(S – E)
∴ n(S – E)
= n(S) – n(E ∩ S)
= 48 – 18 = 30
Therefore, Percentage of total candidates who failed
= n(E – S) + n(S – E) + n(E ∩ S)
= 38 + 30 + 18 = 86
The percentage of candidates who passed in both English and Science = 100 – 86 = 14 (Ans)

Q. 10. In a group of 65 people, 40 like cricket and 10 like both cricket and tennis.
How many like tennis only and not cricket?
How many like tennis?

Answer: Let C be the set of people who like cricket.
T be the set of people who like tennis.
Given: Number of people who like cricket n(C) = 40
Number of people who like cricket and tennis both n(C ∩ T) = 10
people who like cricket or tennis n(C ∪ T) = 65
Number of people who like tennis = n(T)
We know,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) – 10
⇒ n(T) = 65 – 40 + 10 = 35
∴ Number of people who like tennis n(T) = 35
Number of people who like tennis only and not cricket= n(T – C)
∴ n(T – C)
= n(T) – n(C ∩ T)
= 35 – 10 = 25
Number of people who like tennis only and not cricket = 25 (Ans)
Number of people who like tennis = 35 (Ans)

Q. 11. A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket. If these medals were bagged by a total of 65 students and only 4 students got medals in all the three sports, how many students received medals in exactly two of the three sports?

Answer: Let H be the set of students who got medals in Hockey.
B be the set of students who got medals in Basketball.
C be the set of students who got medals in Cricket.
Given: Medals awarded in Hockey n(H) = 42
Medals awarded in Basketball n(B) = 18
Medals awarded in Cricket n(C) = 23
Total number of students who bagged medals n(H ∪ B ∪ C) = 65
Total number of students who bagged medals in all the three event’s n(H ∩ B ∩ C) = 4
To Find: Total number of students who bagged medals in exactly two of the three sports = n(H ∪ B) + n(B ∪ C) + n(C ∪ H)
We know,
n(H ∪ B ∪ C) = n(H) + n(B) + n(C) – n(H ∪ B) – n(B ∪ C) – n(C ∪ H) + n(H ∩ B ∩ C)
⇒ 65 = 42 + 18 + 23 – (n(H ∪ B) + n(B ∪ C) + n(C ∪ H)) + 4
⇒ 65 = 87 – [n(H ∪ B) + n(B ∪ C) + n(C ∪ H)]
or, n(H ∪ B) + n(B ∪ C) + n(C ∪ H) = 87 – 65
⇒ n(H ∪ B) + n(B ∪ C) + n(C ∪ H) = 22
22 students received medals in exactly two of the three sports.  (Ans)

Q. 12. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, and 3 read all the three newspapers. Find
(i) The number of people who read at least one of the newspapers,
(ii) The number of people who read exactly one newspaper.

Answer: Given: Number of people who read newspaper H = n(H)= 25
Number of people who read newspaper T = n(T) = 26
Number of people who read newspaper I = n(I) =26
people who read newspaper both H and I = n(H ∩ I ) = 9
Number of people who read newspaper both H and T = n(H ∩ T ) = 11
Number of people who read newspaper both T and I = n(T ∩ I ) = 8
No of people who read all three newspapers = n(H ∩ T ∩ I) = 3
Total number of people = 60

(i) To Find: The number of people who read at least one of the newspapers n(H Ս T Ս I)
Number of people who read at least one of the three newspapers = n(H Ս T Ս I)
We know, n(H Ս T Ս I) = n(H) + n(T) + n(I) – n(H ∩ I) – n(H ∩ T) – n(T ∩ I) + n(H ∩ T ∩ I)
⇒ n(H Ս T Ս I) = 25 + 26 + 26 – 9 – 11 – 8 + 3
⇒ n(H Ս T Ս I) = 52
Ans: 52 people read at least one of the three newspapers.

(ii) Number of people who read newspaper only H and I
= n(H ∩ T ) – n(H ∩ T ∩ I)
= 11 – 3 = 8
Number of people who read newspaper only H and T
= n(H ∩ I ) – n(H ∩ T ∩ I)
= 9 – 3 = 6
Number of people who read newspaper only T and I
= n(T ∩ I ) – n(H ∩ T ∩ I)
= 8 – 3 = 5
∴ Number of people who read exactly two newspapers
= 8 + 6 + 5 = 19
Number of people who read two or more newspapers
= 19 + 3 = 22
The number of people who read exactly one newspaper
= 52 – 22
= 30
Ans: 30 people read exactly one newspaper.

Q. 13. In a survey of 100 students, the number of students studying the various languages is found as English only 18; English but not Hindi 23; English and Sanskrit 8; Sanskrit and Hindi 8; English 26; Sanskrit 48 and no language 24. Find
(i) how many students are studying Hindi?
(ii) how many students are studying English and Hindi both?

Answer: Let E be the set of students who study English.
H be the set of students who study Hindi.
S be the set of students who study Sanskrit.
U be the universal set.
Given: Total number of students n(U) = 100
Number of students who study only English but not Hindi and Sanskrit = n(E ∩ H’ ∩ S’) = 18
Number of students who study English but not Hindi = n(E ∩ H’) = 23
No of students who study both English and Sanskrit = n(E ∩ S) = 8
Number of students who study both Sanskrit and Hindi = n(S ∩ H) = 8
Number of students who study English = n(E) = 26
students who study Sanskrit = n(S) = 48
Number of students who do not study any language = n(E’ ∩ S’ ∩ H’) = 24
Now
n(E ∩ H’ ∩ S’) = 18
⇒ n(E) – n[E ∩ (H Ս S)’] = 18
⇒ 26 – n[(E ∩ H) Ս (E ∩ S)] = 18
or, 26 –  [3 + 8 – n(E ∩ H ∩ S)] = 18
⇒ 15 + n(E ∩ H ∩ S) = 18
⇒ n(E ∩ H ∩ S) = 3
Again
n(E’ ∩ S’ ∩ H’) = 24
⇒ n(U) – n(E Ս H Ս S) = 24
⇒ 100 – n(E Ս H Ս S) = 24
or, n(E Ս H Ս S) = 76
Here
n(E Ս H Ս S) = 76
⇒ n(E) + n(H) + n(S) – n(E Ս H) – n(H Ս S) – n(S Ս E) + n(E ∩ H ∩ S) = 76
⇒ 26 + n(H) + 48 – 3 – 8 – 8 + 3 = 76
or, n(H) + 74 – 16 = 76
⇒ n(H) + 58 = 76
⇒ n(H) = 18
18 students are studying Hindi.  (Ans)
Now
n(E ∩ H’) = 23
⇒ n(E) – n[E ∩ (H) = 23
⇒ 26 – n(E ∩ H) = 23
∴ n(E ∩ H) = 3
3 students who study both English and Hindi.  (Ans) 

Q. 14. In a town of 10,000 families, it was found that 40% of the families buy newspaper A, 20% buy newspaper B, 10% buy newspaper C, 5% buy A and B; 3% buy B and C, and 4% buy A and C. IF 2% buy all the three newspapers, find the number of families which buy
(i) A only,
(ii) B only,
(iii) none of A, B, and C.

Answer: Let U be the universal set of families.
Given: Total number of families = n(U) = 10000
Number of families that buy newspaper A = n(A)
= 10000×40℅ = 4000
Number of families that buy newspaper B = n(B)
= 10000×20℅ = 2000
Number of families that buy newspaper C = n(C)
= 10000×10℅ = 1000
Number of families that buy newspaper A and B = n(A ∩ B)
= 10000×5℅ = 500
Number of families that buy newspaper B and C = n(B ∩ C)
= 10000×3℅ = 300
Number of families that buy newspaper A and C = n(A ∩ C)
= 10000×4℅ = 400
Number of families that buy newspaper A, B and C = n(A ∩ B ∩ C)
= 10000×2℅ = 200
(i) Number of families which buy newspaper A only
= n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)
= 4000 – 500 – 400 + 200
⇒ 4200 – 900
= 3300
3300 families which buy newspaper A only.  (Ans)

(ii) Number of families which buy newspaper B only
= n(B) – n(A ∩ B) – n(B ∩ C) + n(A ∩ B ∩ C)
= 2000 – 500 – 300 + 200
⇒ 2200 – 800
= 1400
1400 families which buy newspaper B only.  (Ans)

(iii) Number of families which buys at least one newspaper
= n(A Ս B Ս C)
= n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
⇒ 4000 + 2000 + 1000 – 500 – 300 – 400 + 200
= 7200 – 1200
= 6000
Number of families which buys none of the newspaper
= N(U) – n(A Ս B Ս C)
= 10000 – 6000 = 4000
4000 families which buys none of the newspaper.  (Ans)

Q. 15. A class has 175 students. The following description gives the number of students studying one or more of the subjects in this class: mathematics 100, physics 70, chemistry 46, mathematics and physics 30; mathematics and chemistry 28; physics and chemistry 23; mathematics, physics and chemistry 18. Find
(i) how many students are enrolled in mathematics alone, physics alone and chemistry alone,
(ii) The number of students who have not offered any of these subjects.

Answer: Let U be the universal set of students.
Given: Total number of students = n(U) = 175
Number of families enrolled in Mathematics = n(M) = 100
Number of students enrolled in Physics = n(P) = 70
No of students enrolled in Chemistry = n(C) = 46
Number of students enrolled in Mathematics and Physics = n(M ∩ P) = 30
Number of students enrolled in Mathematics and Chemistry= n(M ∩ C) = 28
Students enrolled in Physics and Chemistry = n(P ∩ C) = 23
Number of students enrolled in Mathematics, Physics and Chemistry = n(M ∩ P ∩ C) = 18
(i) Number of students enrolled in Mathematics alone
= n(M) – n(M ∩ P) – n(M ∩ C) + n(M ∩ P ∩ C)
= 100 – 30 – 28 + 18
⇒ 118 – 58
= 60
60 students enrolled in Mathematics alone.  (Ans)

Number of students enrolled in Physics alone
= n(P) – n(M ∩ P) – n(P ∩ C) + n(M ∩ P ∩ C)
= 70 – 30 – 23 + 18
⇒ 88 – 53
= 35
35 students enrolled in Physics alone.  (Ans)

Number of students enrolled in Chemistry alone
= n(C) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C)
= 46 – 28 – 23 + 18
⇒ 64 – 51
= 13
60 students enrolled in Chemistry alone.  (Ans)

(ii) Number of students enrolled at least one subjects 
= n(M Ս P Ս C)
= n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C)
⇒ 100 + 70 + 46 – 30 – 28 – 23 + 18
= 234 – 81
= 153
Number of students who have not offered any of these subjects
= N(U) – n(M Ս P Ս C)
= 175 – 153 = 22
22 students enrolled at least one subjects.  (Ans)

Exercise 1H

Very-Short-Answer-Questions

Q. 1. If a set A has n elements then find the number of elements in its power set P(A).

Answer: The power set of set A is a collection of all subsets of A.
∴ Number of subsets of A = 2ⁿ , where n is the number of elements of set A.
∴ Number of elements in power set of A i.e. P(A) is 2ⁿ (Ans)

Q. 2. If A = φ then write P(A).

Answer: Here A = φ
∴ n(A) = 0
∴ P(A) = {Subset of A} = {φ}(Ans)

Q.3. If n(A) = 3 and n(B) = 5, find:
(i) The maximum number of elements in A ∪ B,
(ii) The minimum number of elements in A ∪ B.

Answer: Given: n(A) = 3; n(B) = 5
We know
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(i) n(A ∪ B) to be maximum, if n(A ∩ B) is minimum, which is 0.
If A and B are disjoint sets then n(A ∩ B) = 0
∴ n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 3 + 5 – 0 = 8
Ans: The maximum number of elements in A ∪ B is 8

(ii) n(A ∪ B) to be minimum, if n(A ∩ B) is maximum.
If A is a subset of B then n(A ∩ B) = 3
∴ n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 3 + 5 – 3 = 5
Ans: The minimum number of elements in A ∪ B is 5

Q. 4. If A and B are two sets such than n(A) = 8, n(B) = 11 and n(A ∪ B) = 14 then find n(A ∩ B).

Answer: Given: n(A) = 8, n(B) = 11, n(A ∪ B) = 14
We know
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 14 = 8 + 11 – n(A ∩ B)
⇒ 14 = 19 – n(A ∩ B)
∴ n(A ∩ B) = 19 – 14
⇒ n(A ∩ B) = 5
Ans: n(A ∩ B) = 5

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Q. 5. If A and B are two sets such that n(A) = 23, n(b) = 37 and n(A – B) = 8 then find n(A ∪ B).

Answer: Given: n(A) = 23, n(B) = 37,
n(A – B) = 8
⇒ n(A) – n(A ∩ B) = 8
⇒ 23 – n(A ∩ B) = 8
∴ n(A ∩ B) = 23 – 8
⇒ n(A ∩ B) = 15
We know
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 23 + 37 – 15
⇒ n(A ∪ B) = 60 – 15 = 45
Ans: n(A ∪ B) = 45

Q. 6. If A and B are two sets such than n(A) = 54, n(B) = 39 and n(B – A) = 13 then find n(A ∪ B).

Answer: Given: n(A) = 54, n(B) = 39,
n(B – A) = 13
⇒ n(B) – n(A ∩ B) = 13
= 39 – n(A ∩ B) = 13
⇒ n(A ∩ B) = 26
We know
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 54 + 39 – 26
⇒ n(A ∪ B) = 93 – 26 = 67
∴ n(A ∪ B) = 67
Ans: n(A ∪ B) = 67

Q. 7. If A ⊂ B, prove that B’ ⊂ A’.

Answer:
As A ⊂ B ⇒ A ∪ B = B
Taking compliment
⇒ (A ∪ B)’ = B’
⇒ A’ ∩ B’ = B’ ….. [De-Morgan’s law]
∴ A’ ∩ B’ = B’
⇒ B’ ⊂ A’ (Proved)

Q. 8. If A ⊂ B, show that (B’ – A’) = φ.

Answer:
B’ – A’ = B’ ∩ (A’)’
⇒ B’ – A’ = B’ ∩ A
⇒ B’ – A’ = A ∩ B’….. [ ∵ A ∩ B = B ∩ A]
⇒ B’ – A’ = A – B
⇒ B’ – A’ = φ….. [∵ A ⊂ B]

Q. 9. Let A = {x : x = 6n, n ε N) and B = {x : x = 9n, n ε N}, find A ∩ B.

Answer:
A = {x : x = 6n ∀ n ∈ N)
∴ A = {6, 12, 18, 24, 30, 36…}
B = {x : x = 9n ∀ n ∈ N)
∴ B = {9, 18, 27, 36…}
∴ A ∩ B = {18, 36, 54, …}
= {x: x = 18n ∀ n ∈ N}

Q. 10. If A = {5, 6, 7}, find P(A).
Answer:
A = {5, 6, 7}
∴ P(A) = {φ, {5}, {6}, {7}, {5, 6}, {6, 7}, {7, 5}, {5, 6, 7}}

Q. 11. If A = {2, {2}}, find P(A).
Answer:
A = {2, {2}}
∴ P(A) = {{φ}, {2}, {{2}}, {2, {2}}}

Q. 12. Prove that A ∩ (A ∪ B)’ = φ

Answer:
L.H.S.
= A ∩ (A ∪ B)’
= A ∩ (A’ ∩ B’)….. [Using De-Morgan’s law]
⇒ (A ∩ A’) ∩ (A ∩ B’)
= φ ∩ (A ∩ B’)….[∵ A ∩ A’ = φ]
= φ = R.H.S. (Proved)

Q. 13. Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.

Answer: A = {1, 2, 3}; B = {3, 4, 5}
A Δ B = (A – B) ∪ (B – A)
= {1, 2} ∪ {4, 5}
= {1, 2, 4, 5}
The symmetric difference A Δ B = {1, 2, 4, 5}

Q. 14. Prove that A – B = A ∩ B.’

Answer:
Let x be an arbitrary element in set A – B
∴ x ∈ (A – B)
∴ x ∈ A and ∉ B
⇒ x ∈ A and ∈ B’
⇒ x ∈ (A ∩ B’)
then (A – B) = (A ∩ B’) – – – (i)
Again x ∈ A ∩ B.’ ⇒ x ∈ A – B – – – (ii)
From (i) and (ii),
A – B = A ∩ B. (Proved)

Q. 15. If A = {x : x ε R, x < 5} and B = {x : x ε R, x > 4}, find A ∩ B.

Answer:
A = {x: x ∈ R, x < 5}
⇒ A = ( -∞, 5) B = {x: x ∈ R, x > 4}
⇒ B = (4, ∞)
Hence A ∩ B = (4, 5)
⇒ A ∩ B = {x : x ∈ R, 4 < x < 5}