CBSE Class 11 Mathematics Set Theory Worksheet

Multiple Choice Questions

Question. Total number of elements in the power set of A containing 15 elements is
(a) 2¹⁵
(b) 15²
(c) 2¹⁵⁻¹
(d) 2¹⁵−1
Answer : A

Question. If A = P ({1, 2}), where P denotes the power set, then which one of the following is correct?
(a) {1,2} ⊂ A
(b) 1 ∈ A
(c) Φ ∉ A
(d) {1,2} ∈ A
Answer : D

Question. If A = {1, 3, 5, 7}, then what is the cardinality of the power set P (A)?
(a) 8
(b) 15
(c) 16
(d) 17
Answer : C

Question. If A and B are two sets, then A ∩ (A ∪ B ) equals to
(a) A
(b) B
(c) f
(d) A ∩ B
Answer : A

Question. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of people who speak atleast one of these two languages, is
(a) 40
(b) 60
(c) 20
(d) 80
Answer : B

Question. If A ={1, 3, 9} and B ={2, 4, 5, 6}, then A ∪ B is
(a) {1, 3, 2, 4, 9}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4, 5, 9}
(d) {1, 2, 3, 4, 5, 6, 9}
Answer : D

Question. Write the {x : x ∈ R, − 5 < x ≤ 6} as interval, then the length of the interval is
(a) 9
(b) 10
(c) 11
(d) 12
Answer : C

Question. If A = {(x, y ) : x² + y² = 25} and B = {(x, y ):x² + 9y² = 144}, then A ∩ B contains
(a) one point
(b) three points
(c) two points
(d) four points
Answer : D

Question. From the following sets given below, pair the equivalent sets. A = {1, 2, 3}, B = {t , p,q, r, s},C = {a, b, γ} and D = {a, e, i, o, u}
(a) Sets A, C and A, D
(b) Sets A, B andB, D
(c) Sets A, C andB, D
(d) Sets A, C andB, C
Answer : C

Question. A set B = {5} is called
(a) singleton set
(b) empty set
(c) infinite set
(d) None of these'
Answer : A

Question. If S = {x | x is a positive multiple of 3 less than 100} and P = {x | x is a prime number less than 20}. Then, n(S ) + n(P ) is equal to
(a) 34
(b) 31
(c) 33
(d) 41
Answer : D

Question. Are the given sets disjoint?
A = {x : x is the boys of your school}
B = {x : x is the girls of your school}
(a) Yes
(b) No
(c) Can’t say
(d) Insufficient data
Answer : A

Question. Let A and B be two sets such that n (A) = 0.16, n (B) = 0.14 and n (A υ B ) = 0.25. Then, n (A ∩ B ) is equal to
(a) 0.3
(b) 0.5
(c) 0.05
(d) None of these
Answer : C

Question. If A = {2, 4, 6}, B = {1, 3, 5} and C = {0, 7},then the universal set will be
(a) {0, 7}
(b) {1, 2, 3, 4, 5, 6}
(c) {0,1, 3, 5, 7}
(d) {0, 1, 2, 3, 4, 5, 6, 7}
Answer : D

Question. Let A = {a, e, i, o, u} and B = {a, i, u}.  Then, A ∪ B is …X… . Here, X refers to
(a) A
(b) B
(c) A and B
(d) None of these
Answer : A

Question. In a town with a population of 5000, 3200 people are egg-eaters, 2500 meat-eaters and 1500 eat both egg and meat. Then, the number of pure vegetarians is
(a) 800
(b) 1000
(c) 1600
(d) 2000
Answer : A

Question. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. The number of students in the group is
(a) 50
(b) 125
(c) 75
(d) 175
Answer : B

Question. In a school sports event, 50 students participate for football, 30 students participate for cricket and 15 students participate for both. Then, the number of students who participated for either football or for cricket, is
(a) 60
(b) 55
(c) 65
(d) 75
Answer : C

Assertion-Reasoning MCQs

Directions Each of these questions contains two statements 
Assertion (A) and Reason (R). Each of the questions has four alternative choices, any one of the which is the correct answer. You have to select one of the codes (a), (b), (c) and
(d) given below.
(a) A is true, R is true; R is a correct explanation of A.
(b) A is true, R is true; R is not a correct explanation of A.
(c) A is true; R is false.
(d) A is false; R is true.

Question. Assertion (A) If A = set of prime numbers less than 6 and B = set of prime factors of 30, then A = B.
Reason (R) If P = {1, 2, 3} andQ = {2, 2, 1, 3, 3}, then P and Q are not equal.
Answer : C

Question. Assertion (A) Let A = {1, 2, 3} and B = {1, 2, 3, 4}. Then, A Ì B.
Reason (R) If every element of X is also an element of Y, then X is a subset of Y.
Answer : A

Question. Assertion (A) The interval {x : x ∈ R, −4 < x ≤ 6} is represented by (−4, 6].
Reason (R) The interval {x : x ÎR, −12 < x < −10} is represented by [−12, −10].
Answer : C

Question. Assertion (A) Set of English alphabets is the universal set for the set of vowels in English alphabets.
Reason (R) The set of vowels is the subset of set of consonants in the English alphabets.
Answer : C

Question. Assertion (A) The power set of the set {1, 2} is the set {f, {1}, {2}, {1,2}}.
Reason (R) The power set is set of all subsets of the set.
Answer : A

Question. Assertion (A) Let A = {a, b} and B = {a, b, c }. Then, A Ë B.
Reason (R) If A ⊂ B, then A ∪ B = B.
Answer : D

Question. Assertion (A) If n(A) = 3, n(B ) = 6 and A ⊂ B, then the number of elements in A ∪ B is 9.
Reason (R) If A and B are disjoint, thenn(A ∪ B ) is n(A) + n(B).
Answer : D

Case Based MCQs

The school organised a farewell party for 100 students and school management decided three types of drinks will be distributed in farewell party i.e. Milk (M), Coffee (C) and Tea (T). Organiser reported that 10 students had all the three drinks M,C,T. 20 students had M and C; 30 students had C and T; 25 students had M and T. 12 students had M only; 5 students had C only; 8 students had T only.

Based on the above information, answer the following questions.

Question. The number of students who did not take any drink, is
(a) 20
(b) 30
(c) 10
(d) 25
Answer : A

Question. The number of students who prefer Milk is
(a) 47
(b) 45
(c) 53
(d) 50
Answer : A

Question. The number of students who prefer Coffee is
(a) 47
(b) 53
(c) 45
(d) 50
Answer : C

Question. The number of students who prefer Tea is
(a) 51
(b) 53
(c) 50
(d) 47
Answer : B

Question. The number of students who prefer Milk and Coffee but not tea is
(a) 12
(b) 10
(c) 15
(d) 20
Answer : B

The school organised a cultural event for 100 students. In the event, 15 students participated in dance, drama and singing. 25 students participated in dance and drama; 20 students participated in drama and singing; 30 students participated in dance and singing. 8 students participated in dance only; 5 students in drama only and 12 students in singing only.

Based on the above information, answer the following questions.

Question. The number of students who participated in dance, is
(a) 18
(b) 30
(c) 40
(d) 48
Answer : A

Question. The number of students who participated in drama, is
(a) 35
(b) 30
(c) 25
(d) 20
Answer : A

Question. The number of students who participated in singing, is
(a) 42
(b) 45
(c) 47
(d) 37
Answer : C

Question. The number of students who participated in dance and drama but not in singing, is
(a) 20
(b) 5
(c) 10
(d) 15
Answer : C

Question. The number of students who did not participate in any of the events, is
(a) 20
(b) 30
(c) 25
(d) 35
Answer : B

Q1. If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {7, 8, 9, 10, 11} and D = {10, 11, 12, 13, 14}. Find (i) A U B
(ii) B U C (iii) A ΠC (iv) A Π D (v) A Π B
Verify the following
(i) A U (B Π C) = (A U B) Π (A U C)
(ii) A Π (B UC) = (A Π B) U (A Π C)
(iii) A Π (B - C) = (A Π B) - (A Π C)

Q2. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9}. A = {2, 4, 6, 8} and B = {2, 3, 5, 7, 8}. Find (i) A′ (ii) (A′) ′ (iii) (A U B) ′
(iv) (A Π B) ′
Verify the following
(i) (A U B) ′ = A′ Π B′
(ii) (A Π B) ′ = A′ U B′
(III) B – A = B Π A′

Q3. Let A and B be two sets such that n(A) = 24, n(A U B) = 46 and n (A Π B) = 8. Find (i) n(B) (ii) n (A – B)
(iii) n (B – A)

Q4. What is the number of subsets and proper sub sets of a set containing n-elements.

Q5. In a survey of 800 students in a school 200 were listed as taking apple juice, 250 taking orange juice and 125 were taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

Q6. There 40 students in a chemistry class and 60 students in physics class. Find the number of students which are either in Physics class or Chemistry class in the cases.
(i) the two classes meet at the same hour.
(ii) the two classes meet at different hours and 20 students are enrolled in both the subjects.

Q7. In a class of 35 students, 17 have taken mathematics 10 have taken mathematics but not economics. Find the number of students who have taken both mathematics and economics and the number of students who have taken economics but not mathematics, if it is given that each student has taken either mathematics or economics or both.

Q8. If A {x : x = 2n + 1, n ≤ 4, n Є N} and B = { y : 2 < y < 7, y Є N}, find (i) A Π B (ii) A U B

Q9. Using laws of algebra of sets, show that (i) (A U B) Π (A U B’) = A (ii) AU ( B – A) = A U B

Q10. Of the members of three athletic teams in a certain school, 21 are in the basket ball team, 26 in hockey team and 29 in the football team. 14 play hockey and basket ball, 15 play hockey and football, 12 play football and basket ball and 8 play all the three games. How many members are there in all?

Q11. If A {a, b, c}. write subsets of set A. Also mention the proper subsets of A.

Q12. Describe the following sets in set builder form :-
(i) { 1, 1, 1, 1, 1 }
(ii) {3, 6, 9, - - -}
2 3 4 5

Q13. Write the following intervals in set-builder form :-
(i) (-6, 0) (ii) (2, 5]
(iii) [-20, 3) (iv) [5, 10]

Q14. Draw venu diagram of
(i) (A U B) ′ Π C
(ii) (A – B) ′ Π (B – A) ′
(iii) A – (A U B)

Q15. How many element has P(A) , if A = { }

Q16. If A is a subset of { }. Prove that A = { }.

Q17. Find the smallest set A such that
A U {3, 5} = {1, 2, 3, 5, 4}

Q.1 If U = {1,2,3,4,……..,10} is the universal set for the sets A = {2,3,4,5} and B = {1,2,3,4,5,6}, then verify that (A∪B)'' = A''∩B''.

Q.2 If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 8}, C = {2, 5, 7, 8}, verify that A– (B U C) = (A–B) ∩ (A –C).

Q.3 Which type of set is the set of odd natural numbers divisible by 2?

Q.4 Out of 20 members in a family, 11 like to take tea and 14 like coffee. Assume that each one likes at least one of two drinks. how many like, only tea and not coffee?

Q.5 Decide, among the following sets are subsets of one and another : A={x : x ∈ R and x satisfy : x² - 4x + 3 = 0} B = {1,3}, C = {1,3,5}, D = {4,5,6}.

Q.6 A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B. What is the leastnumber that must have liked both products?

Q.7 Let A and B be two finite sets such that n(A – B) = 30, n (A U B) = 180, n(A ∩ B) = 60, find n(B).

Q.8 Write the set A = {x : x ∈N and x² < 25} in roster form.

Q.9 In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked (i) product C only (ii) product A and C but not product B (iii) at least one of three products.

Q.10 If A × B = {(p,q),(p,r),(m,q),(m,r)}, find A and B.

Q.11 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, 3 read all 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.

Q.12 In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Q.13 In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Q.14 If A = {x : x is a prime number ∀x ∈N}, then find A^c.

Q.15 If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(XUY) = 38, find n(X∩Y).

Q.16 From the sets given below, select equal sets :
A = { 2,4,8,12}, B = {1,2,3,4}, C ={ 4,8,12,14}, D = {3,1,4,2}, E = {-1,1}, F = {0,a}, G = {1,-1}, H={0,1}.

Q.17 Draw appropriate Venn diagram for each of the following:
(i) (A U B)
(ii) A ∩ B
(iii) (A ∩ B)
(iv) (A U B)

Q.18 Show that A ∩ B = A ∩ C need not imply B = C. 

Q.19 Let U = {1,2,3,4,5,6,7,8,9,10} and A = {1,3,5,7,9}. Find A''''.

Q.20 In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Find the number of persons who read neither.

Question. There are 200 individuals with skin disorder, 120 has exposed to chemical 𝐶₁, 50 to 𝐶₂ and 30 to both 𝐶₁ 𝑎𝑛𝑑 𝐶₂. Find the number of individuals exposed to (i) 𝐶₁ but not 𝐶₂ (ii) 𝐶₂ but not 𝐶₁ (iii) 𝐶1 or 𝐶₂ (iv) neither 𝐶₁ nor 𝐶₂.
Answer : Given, n(𝐶₁) = 120
= n(𝐶₂) = 50
= n(𝐶₁ ∩ 𝐶₂) = 30
and n(∪) = 200
(i)n(𝐶₁ ∩ 𝐶₂) = 𝑛(𝐶₁) − n(𝐶₁ ∩ 𝐶₂)
= 120 − 30 = 90
∴ 90 persons exposed to chemical 𝐶₁ but not 𝐶₂
(ii) n(𝐶₂ ∩ 𝐶₁) = n(𝐶₂) − n(𝐶₁ ∩ 𝐶₂)
= 50 − 30 = 120
∴ 20 persons exposed to chemical 𝐶2 but not 𝐶₁
(iii) n(𝐶₁ ∪ 𝐶₂) = 𝑛(𝐶₁) + 𝑛(𝐶₂) − n(𝐶1 ∩ 𝐶₂)
= 120 + 50 − 30 = 140
∴ 140 persons exposed to chemical 𝐶₁ or 𝐶₂
(iv) n(𝐶₁ ∩ 𝐶₂) = n(∪) − 𝑛 n(𝐶₁ ∪ 𝐶₂)
= 200 − 140 = 60
∴ 60 persons exposed to chemical 𝐶₁ nor 𝐶₂.

Question. In a class of 35 students 17 have taken mathematics, 10 have taken mathematics but not economics. Find the number of students who have taken economics but not n mathematics, if it is given that each student has taken either maths or economics or both.
Answer : Let A → set of students taken mathematics
B→ set of students taken economics
Given, 𝑛(𝐴 ∪ 𝐵) = 35
𝑛(𝐴) = 17
𝑛(𝐴 ∩ 𝐵¹) = 10
Now, 𝑛(𝐴 ∩ 𝐵¹) = 𝑛(𝐴) − 𝑛(𝐴 ∩ 𝐵)
⇒ 10 = 17 − 𝑛(𝐴 ∩ 𝐵)
⇒ 𝑛(𝐴 ∩ 𝐵) = 7
Now, 𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)
⇒ 35 = 17 + 𝑛(𝐵) − 7
⇒ 35 = 10 + 𝑛(𝐵)
⇒ 𝑛(𝐵) = 25
Now, 𝑛(𝐵 ∩ 𝐴) = 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵)
𝑛(𝐵 ∩ 𝐴1) = 25 − 7 = 18
∴ 18 students have taken economics but not mathematics. Ans.

Question. A market research group conducted a survey of 2000 consumers and reported that 1720 consumers liked product 𝑃1 and 1450 consumers liked product 𝑃2. What is the least number that must have liked both the products?
Answer : Let A → set of consumers who liked product 𝑃1
B→ set of consumers who liked product 𝑃2
Given, 𝑛(∪) = 2000
𝑛(𝐴) = 1720
𝑛(𝐵) = 1450
We have, 𝑛(𝐴 ∪ 𝐵) ≤ 𝑛(∪)
⇒ 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) ≤ 𝑛(∪)
⇒ 1720 + 1450 − 𝑛(𝐴 ∩ 𝐵) ≤ 2000
⇒ 3170 − 2000 ≤ 𝑛(𝐴 ∩ 𝐵)
⇒ 1170 ≤ 𝑛(𝐴 ∩ 𝐵)
OR 𝑛(𝐴 ∩ 𝐵) ≥ 1170
∴ the least number of 𝑛(𝐴 ∩ 𝐵) is 1170 ans.

Question. A survey shows that 63% of the Americans like cheese whereas 76% like apples. If 𝑥% of the Americans like both cheese & apples, find the value of 𝑥.
Answer : Let A → set of Americans who like cheese
B→ set of Americans who like Apples
Let the population of America is 100
i.e., 𝑛(∪) = 100
∴ 𝑛(𝐴) = 63 , 𝑛(𝐵) = 76 and 𝑛(𝐴 ∩ 𝐵) = 𝑥
We have, 𝑛(𝐴 ∪ 𝐵) ≤ 𝑛(∪)
⇒ 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) ≤ 𝑛(∪)
⇒ 63 + 76 − 𝑥 ≤ 100
⇒ 139 − 𝑥 ≤ 100
⇒ −𝑥 ≤ −39
⇒ 𝑥 ≥ 39 ……… (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 − 𝑣𝑒 & 𝑠𝑖𝑔𝑛 𝑐ℎ𝑎𝑛𝑔𝑒)
∴ 𝑥 ≥ 39 ……….. (1)
Now, 𝑛(𝐴 ∩ 𝐵) ≤ 𝑛(𝐴) also 𝑛(𝐴 ∩ 𝐵) ≤ 𝑛(𝐵)
⇒ 𝑥 ≤ 63 and 𝑥 ≤ 76
⇒ 𝑥 ≤ 63 …….. (2) ……….. {∵ a number less than 63 also less than 76}
From (1) and (2), 39 ≤ 𝑥 ≤ 63 ans.

Question. In a town of 10,000 families, it was found that 40% families buy newspaper H, 20% families buy newspaper T, 10% families buy newspaper I, 5% buy H & T, 3% buy T & I and 4% buy H & I. If 2% buy all the here newspapers, find the numbers of families which buy:
(i) Exactly 1 newspaper, (ii) Exactly 2 newspapers, (iii) At least newspaper, (iv) Only newspaper H, (v) Only newspaper I, (vi) H & T but not I, (vii) None of the newspaper.
Answer : Given, 𝑛(∪) = 10000
𝑛(𝐻) = 40% 𝑜𝑓 10000 = 4000 = 𝑎 + 𝑏 + 𝑒 + 𝑑
𝑛(𝑇) = 20% 𝑜𝑓 10000 = 2000 = 𝑏 + 𝑐 + 𝑒 + 𝑓
𝑛(𝐼) = 10% 𝑜𝑓 10000 = 1000 = 𝑑 + 𝑒 + 𝑓 + 𝑔
𝑛(𝐻 ∩ 𝑇) = 5% 𝑜𝑓 10000 = 500 = 𝑏 + 𝑒
𝑛(𝑇 ∩ 𝐼) = 3% 𝑜𝑓 10000 = 300 = 𝑒 + 𝑓 venn diagram
𝑛(𝐻 ∩ 𝐼) = 4% 𝑜𝑓 10000 = 400 = 𝑑 + 𝑒
𝑛(𝐻 ∩ 𝑇 ∩ 𝐼) = 2% 𝑜𝑓 10000 = 200 = 𝑒
Solving above seven equation we get,
𝑒 = 200, 𝑑 = 200, 𝑓 = 100, 𝑏 = 300, 𝑔 = 500, 𝑐 = 1400, 𝑎 = 3300
(i) Exactly one newspaper = 𝑎 + 𝑐 + 𝑔
= 3300 + 1400 + 500 = 600
(ii) Exactly two newspaper = 𝑏 + 𝑑 + 𝑓
= 300 + 200 + 100 = 600
(iii) At least one newspaper = 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 + 𝑓 + 𝑔
= 3300 + 300 + 1400 + 200 + 200 + 100 + 500 = 6000
(iv) Only newspaper 𝐻 = 𝑎 = 3300
(v) Only newspaper 𝑇 = 𝑐 = 1400
vi) 𝐻 and 𝑇 but not 𝐼 = (𝑏 + 𝑒) − 𝑒 = 𝑏
= 300
(vii) None of the newspaper = 10000 − (𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒)
= 10000 − 6000 = 4000 ans.

Question. A college awarded 38 medals in football, 15 in Basketball and 20 in Cricket. If these medals went to a total of a 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports?
Answer : Given, 𝑛(𝐹 ∪ 𝐵 ∪ 𝐶) = 58
𝑛(𝐹) = 38
𝑛(𝐵) = 15
𝑛(𝐶) = 20
𝑛(𝐹 ∩ 𝐵 ∩ 𝐶) = 3(𝑎𝑙𝑠𝑜 𝑒)
We have,
𝑛(𝐹 ∪ 𝐵 ∪ 𝐶) = 𝑛(𝐹) + 𝑛(𝐵) + 𝑛(𝐶) − 𝑛(𝐹 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − 𝑛(𝐹 ∩ 𝐶) +
𝑛(𝐹 ∩ 𝐵 ∩ 𝐶)
⇒ 58 = 38 + 15 + 20 − 𝑛(𝐹 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − 𝑛(𝐹 ∩ 𝐶) + 3
⇒ 58 = 76 − 𝑛(𝐹 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − 𝑛(𝐹 ∩ 𝐶)
⇒ 𝑛(𝐹 ∩ 𝐵) + 𝑛(𝐵 ∩ 𝐶) + − 𝑛(𝐹 ∩ 𝐶) = 18
⇒ (𝑏 + 𝑒) + (𝑒 + 𝑓) + (𝑑 + 𝑒) = 18
⇒ 𝑏 + 𝑑 + 𝑓 + 3𝑒 = 18
⇒ 𝑏 + 𝑑 + 𝑓 + 3(3) = 18 ………… {∵ e = 𝑛(𝐹 ∩ 𝐵 ∩ 𝐶) = 3}
⇒ 𝑏 + 𝑑 + 𝑟 = 18 − 9 = 9
∴ 9 man received medals in exactly two of the three sports.

Question. A survey of 500 television viewers produced the following information: 285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football & basketball, 70 watch football & hockey, 50 watch hockey & basketball, 50 do not watch any of the three games. How many watch all the three games? How many watch exactly one of the three games? How many watch exactly two the three games?
Answer : Given, 𝑛(∪) = 500 universal
𝑛(𝐹) = 285 = 𝑎 + 𝑏 + 𝑒 + 𝑑
𝑛(𝐻) = 195 = 𝑏 + 𝑐 + 𝑒 + 𝑓
𝑛(𝐵) = 115 = 𝑑 + 𝑒 + 𝑓 + 𝑔
𝑛(𝐹 ∩ 𝐵) = 45 = 𝑑 + 𝑒
𝑛(𝐹 ∩ 𝐻) = 70 = 𝑏 + 𝑒
𝑛(𝐻 ∩ 𝐵) = 50 = 𝑒 + 𝑓
𝑛(𝐹1 ∩ 𝐵¹ ∩ 𝐻1) = 50
We have, 𝑛(𝐹1 ∩ 𝐵¹ ∩ 𝐻1) = 𝑛(∪) − 𝑛(𝐹 ∪ 𝐵 ∪ 𝐻)
(𝑛𝑜𝑛𝑒) = (𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙) – (𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒)
⇒ 50 = 500 − 𝑛(𝐹 ∪ 𝐵 ∪ 𝐻)
⇒ 𝑛(𝐹 ∪ 𝐵 ∪ 𝐻) = 450
Now, 𝑛(𝐹 ∪ 𝐵 ∪ 𝐻) = 𝑛(𝐹) + 𝑛(𝐵) + 𝑛(𝐻) − 𝑛(𝐹 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐻) − 𝑛(𝐹 ∩ 𝐻) +
𝑛(𝐹 ∩ 𝐵 ∩ 𝐻)
⇒ 450 = 285 + 195 + 115 − 45 − 50 − 70 + 𝑛(𝐹 ∩ 𝐵 ∩ 𝐻)
⇒ 450 = 430 + 𝑛(𝐹 ∩ 𝐵 ∩ 𝐻)
⇒ 𝑛(𝐹 ∩ 𝐵 ∩ 𝐻) = 20
i.e., 𝑒 = 20
∴ 𝑓 = 30, 𝑏 = 50, 𝑑 = 25, 𝑔 = 40, 𝑐 = 95, 𝑎 = 190
20 percent watch all the three games
Exactly one of the games = 𝑎 + 𝑐 + 𝑔
= 190 + 95 + 40 = 325
Exactly two of the games = 𝑏 + 𝑑 + 𝑓
= 50 + 25 + 30 = 105 ans.

Question. Show that 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵 ⇒ 𝐴 = 𝐵.
Answer : Given, 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵
To prove: 𝐴 = 𝐵
Let 𝑥 ∈ 𝐴
⇒ 𝑥 ∈ (𝐴 ∪ 𝐵)
⇒ 𝑥 ∈ (𝐴 ∩ 𝐵) …………. {𝑔𝑖𝑣𝑒𝑛 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵}
⇒ 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵
⇒ 𝐴∁𝐵 …….. (1)
Let 𝑦 ∈ 𝐵
⇒ 𝑦 ∈ (𝐴 ∪ 𝐵)
⇒ 𝑦 ∈ (𝐴 ∩ 𝐵) …………. {𝑔𝑖𝑣𝑒𝑛 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵}
⇒ 𝑦 ∈ 𝐴 𝑎𝑛𝑑 𝑦 ∈ 𝐵
⇒ 𝐵∁𝐴 …….. (2)
From (1) and (2), 𝐴 = 𝐵 (proved)

Question. Show that 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ∩ 𝑃(𝐵)
Answer : Let 𝑋 ∈ 𝑃(𝐴 ∩ 𝐵)
⇒ 𝑋 ∁ (𝐴 ∪ 𝐵)
⇒ 𝑋 ∁ 𝐴 𝑎𝑛𝑑 𝑋 ∁ 𝐵
⇒ 𝑋 ∈ 𝑃(𝐴) 𝑎𝑛𝑑 𝑋 ∈ 𝑃(𝐵)
⇒ 𝑋 ∈ 𝑃(𝐴) ∩ 𝑃(𝐵)
⇒ 𝑃(𝐴 ∩ 𝐵) ∁ 𝑃(𝐴) ∩ 𝑃(𝐵) ………….. (1)
Let 𝑌 ∈ 𝑃(𝐴) ∩ 𝑃(𝐵)
⇒ 𝑌 ∈ 𝑃(𝐴) 𝑎𝑛𝑑 𝑌 ∈ 𝑃(𝐵)
⇒ 𝑌 ∁ 𝐴 𝑎𝑛𝑑 𝑌 ∁ 𝐵
⇒ 𝑌 ∁ (𝐴 ∩ 𝐵)
⇒ 𝑌 ∁ 𝑃(𝐴 ∩ 𝐵)
⇒ 𝑃(𝐴) ∩ 𝑃(𝐵) ∁ 𝑃(𝐴 ∩ 𝐵) ………. (2)
From (1) and (2), 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) ∩ 𝑃(𝐵) (proved)

Question. Assume that, 𝑃(𝐴) = 𝑃(𝐵). Show that, 𝐴 = 𝐵.
Answer : Given, 𝑃(𝐴) = 𝑃(𝐵)
To prove: 𝐴 = 𝐵
Let 𝑥 ∈ 𝐴
⇒ 𝑥 ∁ 𝐴
⇒ 𝑥 ∈ 𝑃(𝐴)
⇒ 𝑥 ∈ 𝑃(𝐵) ………. {𝑔𝑖𝑣𝑒𝑛 𝑃(𝐴) = 𝑃(𝐵)}
⇒ 𝑥 ∁ 𝐵
⇒ 𝑥 ∈ 𝐵
⇒ 𝐴 ∁ 𝐵 …….. (1)
Let, 𝑦 ∈ 𝐵
⇒ 𝑦 ∁ 𝐵
⇒ 𝑦 ∈ 𝑃(𝐵)
⇒ 𝑦 ∈ 𝑃(𝐴) ………. {𝑔𝑖𝑣𝑒𝑛 𝑃(𝐴) = 𝑃(𝐵)}
⇒ 𝑦 ∁ 𝐴
⇒ 𝑦 ∈ 𝐴
⇒ 𝐵 ∁ 𝐴 …….. (2)
From (1) and (2), 𝐴 = 𝐵 (proved)

Q.11) Show that, (𝐴 ∪ 𝐵 ∪ 𝐶) ∩ (𝐴 ∩ 𝐵¹ ∩ 𝐶¹) ∩ 𝐶¹ = 𝐵 ∩ 𝐶¹
Sol.11) L.H.S. (𝐴 ∪ 𝐵 ∪ 𝐶) ∩ (𝐴 ∩ 𝐵¹ ∩ 𝐶¹) ∩ 𝐶¹
= (𝐴 ∪ 𝐵 ∪ 𝐶) ∩ (𝐴 ∩ (𝐵1 ∩ 𝐶¹)) ∩ 𝐶¹ …….. {𝐷𝑒 − 𝑚𝑜𝑟𝑔𝑎𝑛′𝑠 𝑙𝑎𝑤}
= (𝐴 ∪ 𝐵 ∪ 𝐶) ∩ (𝐴¹ ∪ (𝐵 ∩ 𝐶)) ∩ 𝐶¹…….. {𝐷𝑒 − 𝑚𝑜𝑟𝑔𝑎𝑛′𝑠 𝑙𝑎𝑤}
⇒ ((𝐵 ∪ 𝐶) ∪ 𝐴) ∩ ((𝐵 ∪ 𝐶) ∪ 𝐴¹) ∩ 𝐶¹
= (𝐵 ∪ 𝐶) ∪ (𝐴 ∩ 𝐴1) ∩ 𝐶……… {𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦}
= ((𝐵 ∪ 𝐶) ∪⊄) ∩ 𝐶¹ ………. {𝐴 ∩ 𝐴¹ =⊄}
= (𝐵 ∪ 𝐶) ∩ 𝐶¹ ………. {𝐴 ∪⊄= A}
= (𝐵 ∩𝐶¹) ∪ (𝐶 ∩ 𝐶¹)……… {𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
= (𝐵 ∩ 𝐶¹) ∪⊄
= 𝐵 ∩ 𝐶¹ R.H.S. ans.

Q.12) If A and B are two sets containing and 6 elements respectively, what can be the minimum number of elements in 𝐴 ∪ 𝐵. Find also maximum number of elements in 𝐴 ∪ 𝐵 and 𝐴 ∩ 𝐵.
Sol.12) 𝑛(𝐴) = 3 ; 𝑛(𝐵) = 6
Minimum no. of elements in 𝐴 ∪ 𝐵 = 6
Maximum no. of elements in 𝐴 ∪ 𝐵 = 9
Maximum no. of elements in 𝐴 ∩ 𝐵 = 3

Q.13) From 50students taking examination in Maths, physics and chemistry, each of the students has passed in at least one of the subject, 37 passed maths, 24 passed physics and 43 passed chemistry. At most 19 passed maths & physics, at most 29 passed maths & chemistry and at most 20 passed physics & chemistry.
What is the largest possible number that could have passed all the three subjects?
Sol.13) Given, 𝑛(𝑀 ∪ 𝑃 ∪ 𝐶) = 50
𝑛(𝑀) = 37, 𝑛(𝐶) = 43; 𝑛(𝑃) = 24
Since, at most if given
∴ 𝑛(𝑀 ∩ 𝑃) ≤ 19
𝑛(𝑀 ∩ 𝐶) ≤ 29
𝑛(𝑃 ∩ 𝐶) ≤ 20
𝑛(𝑀 ∪ 𝑃 ∪ 𝐶) = 𝑛(𝑀) + 𝑛(𝑃) + 𝑛(𝐶) − 𝑛(𝑀 ∩ 𝑃) − 𝑛(𝑃 ∩ 𝐶) − 𝑛(𝑀 ∩ 𝐶) + 𝑛(𝑀 ∩ 𝑃 ∩ 𝐶)
⇒ 37 + 24 + 43 − 19 − 29 − 20 + 𝑛(𝑀 ∩ 𝑃 ∩ 𝐶) ≤ 50
⇒ 𝑛(𝑀 ∩ 𝑃 ∩ 𝐶) ≤ 50 − 36
⇒ 𝑛(𝑀 ∩ 𝑃 ∩ 𝐶) ≤ 14
∴ largest possible number that could have passed all the three exams is 14 ans.

Q.14) Suppose 𝐴₁, 𝐴₂ … … … 𝐴30 are thirty sets each having 5 elements and 𝐵1, 𝐵2, … … … … 𝐵𝑛 are 𝑛 sets with each 3 elements. Let ⋃³⁰_𝑖=1 𝐴_𝑖 = ⋃^𝑛_{𝑖 = 1}𝐵_𝑖 = 𝑆 and each element of 𝑆 belongs to exactly 10 of the 𝑆 and exactly 9 of the 𝐵𝑖 𝑆 then find the value of 𝑛.
Sol.14) n(A₁), n(A₂), n(A3) … … … (A₃₀) = 5
⋃³⁰_𝑖=1 𝐴_𝑖 = S ∴ we get n(S) = 150
But each element of 𝑆 belongs to exactly 10 of 𝐴_𝑖 𝑆
∴ 150/10 = 15 are the number of distinct elements in 𝑆
Also each element of 𝑆 belong to exactly 9 of the 𝐵𝑖 𝑆 and each 𝐵𝑖 contains 3 elements
∴ and ⋃^𝑛_{𝑖 = 1} 𝐵_𝑖 =1 = 𝑆
⇒ 3𝑛/9 = 15
⇒ 𝑛 = 45
∴ 𝑛(𝐵) = 45 ans.

Q.15) Using properties show that, (𝐴 ∪ 𝐵) − (𝐴 ∩ 𝐵) = (𝐴 − 𝐵) ∪ (𝐵 − 𝐴)
Sol.15) (𝐴 ∪ 𝐵) − (𝐴 ∩ 𝐵)
= (𝐴 ∪ 𝐵) ∩ (𝐴 ∩ 𝐵)¹ ………. {𝐴 − 𝐵 = 𝐴 ∩ 𝐵¹}
= (𝐴 ∪ 𝐵) ∩ (𝐴¹ ∪ 𝐵1)…….. {𝐷𝑒 − 𝑚𝑜𝑟𝑔𝑎𝑛′𝑠 𝑙𝑎𝑤}
= [(𝐴 ∪ 𝐵) ∩ 𝐴¹] ∪ [(𝐴 ∪ 𝐵) ∩ 𝐵¹]……… {𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
= [(𝐴 ∩ 𝐴¹) ∪ (𝐵 ∩ 𝐴¹)] ∪ [(𝐴 ∩ 𝐵1) ∪ (𝐵 ∩ 𝐵¹)]
= [∅ ∪ (𝐵 − 𝐴)] ∪ [(𝐴 − 𝐵) ∪⊄]
= (𝐵 − 𝐴) ∪ (𝐴 − 𝐵) ………….. {∅ ∪ 𝐴 = 𝐴}
= (𝐴 − 𝐵) ∪ (𝐵 − 𝐴) ……….{𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴} R.H.S. (proved)

Q.16) For any three sets show that 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶)
Sol.16) Let (𝑎, 𝑏) ∈ 𝐴 × (𝐵 ∪ 𝐶)
⇒ 𝑎 ∈ 𝐴 and 𝑏 ∈ (𝐵 ∪ 𝐶)
⇒ 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵 𝑜𝑟 𝑏 ∈ 𝐶
⇒ (𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵 )𝑜𝑟 (𝑎 ∈ 𝐴 𝑎𝑛𝑑 𝑏 ∈ 𝐶)
⇒ (𝑎, 𝑏 ) ∈ 𝐴 × 𝐵 𝑜𝑟 (𝑎, 𝑏) ∈ 𝐴 × 𝐶
⇒ (𝑎, 𝑏 ) ∈ (𝐴 × 𝐵) ∪ (𝐴 × 𝐶)
∴ 𝐴 × (𝐵 ∪ 𝐶) ⊆ (𝐴 × 𝐵) ∪ (𝐴 × 𝐶) ………… (1)
Now, let (𝑥, 𝑦) ∈ (𝐴 × 𝐵) ∪ (𝐴 × 𝐶)
⇒ (𝑥, 𝑦) ∈ 𝐴 × 𝐵 𝑜𝑟 (𝑥, 𝑦) ∈ 𝐴 × 𝐶
⇒ (𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑦 ∈ 𝐵) 𝑜𝑟 (𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐶)
⇒ 𝑥 ∈ 𝐴 𝑎𝑛𝑑 (𝑦 ∈ 𝐵 𝑜𝑟 𝑦 ∈ 𝐶)
⇒ 𝑥 ∈ 𝐴 and 𝑦 ∈ (𝐵 ∪ 𝐶)
⇒(𝑥, 𝑦) ∈ 𝐴 × (𝐵 ∪ 𝐶)
∴ (𝐴 × 𝐵) ∪ (𝐴 × 𝐶) ⊆ A × (𝐵 ∪ 𝐶 )………… (2)
From (1) and (2), 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶)

Question. Show that if 𝐴 ∁ 𝐵 then, (𝐶 − 𝐵)∁(𝐶 − 𝐴).
Answer : Given: 𝐴 ∁ 𝐵
To prove, (𝐶 − 𝐵)∁(𝐶 − 𝐴)
Let 𝑥 ∈ (𝐶 − 𝐵)
⇒ 𝑥 ∈ (𝐶 ∩ 𝐵¹) ………… {∵ A − B = A ∩ B1}
⇒ 𝑥 ∈ 𝐶 𝑎𝑛𝑑 𝑥 ∈ B¹
⇒ 𝑥 ∈ 𝐶 𝑎𝑛𝑑 𝑥 ∉ B
⇒ 𝑥 ∈ 𝐶 𝑎𝑛𝑑 𝑥 ∉ A …………. {∵ A ∁ B if 𝑥 ∉ B then 𝑥 ∉ A}
⇒ 𝑥 ∈ 𝐶 𝑎𝑛𝑑 𝑥 ∈ A¹
⇒ 𝑥 ∈ (𝐶 ∩ A¹)
⇒ 𝑥 ∈ (𝐶 − A)
⇒ (𝐶 − 𝐵) ∁ (𝐶 − A) (proved)

Question. Let A,B and C be the sets such that 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐶 . Show that B = C.
Answer : Given: (𝐴 ∪ 𝐵) = (𝐴 ∩ 𝐶) and
= 𝐴 ∩ 𝐵 = 𝐴 ∪ 𝐶
To prove, B = C
We have, 𝐴 ∩ 𝐵 = 𝐴 ∪ 𝐶
⇒ 𝐵 ∩ (𝐴 ∪ 𝐵) = 𝐵 ∩ (𝐴 ∩ 𝐶)
⇒ (𝐵 ∩ 𝐴) ∪ (𝐵 ∪ 𝐵) = (𝐵 ∩ 𝐴) ∪ (𝐵 ∩ 𝐶) ………… {𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
⇒ (𝐴 ∩ 𝐵) ∪ (𝐵) = (𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶)
⇒ 𝐵 = (𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶) …. (1)….. {𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝐴 ∩ 𝐵 𝑎𝑙𝑤𝑎𝑦𝑠 𝑏𝑒𝑙𝑜𝑛𝑔𝑠 𝑡𝑜 𝐵}
Again, 𝐴 ∪ 𝐵 = 𝐴 ∪ 𝐶 ……. (taking same equation again)
⇒ 𝐶 ∩ (𝐴 ∪ 𝐵) = 𝐶 ∩ (𝐴 ∪ 𝐶)
⇒ (𝐶 ∩ 𝐴) ∪ (𝐶 ∩ 𝐵) = (𝐶 ∩ 𝐴) ∪ (𝐶 ∩ 𝐶)………… {𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
⇒ (𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶) = (𝐴 ∩ 𝐶) ∪ 𝐶
⇒ (𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶) = 𝐶
⇒ (𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶) = 𝐶 ……(2)…… (Given: (𝐴 ∪ 𝐵) = (𝐴 ∩ 𝐶))
From (1) and (2), B = C (proved)

Question. If 𝐴 ∩ 𝑋 = 𝐵 ∩ 𝑋 = ∅, show that A = B.
Answer : Given: 𝐴 ∪ 𝑋 = 𝐵 ∪ 𝑋
𝐴 ∩ 𝑋 = 𝐵 ∩ 𝑋 = ∅
To prove: A = B
We have, 𝐴 ∪ 𝑋 = 𝐵 ∪ 𝑋
⇒ 𝐴 ∩ (𝐴 ∪ 𝑋) = 𝐴 ∩ (𝐵 ∪ 𝑋)
⇒ (𝐴 ∩ 𝐴) ∪ (𝐴 ∪ 𝑋) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝑋)………… {𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
⇒ 𝐴 ∪⊄= (𝐴 ∩ 𝐵) ∪ ∅ ………….. (𝑔𝑖𝑣𝑒𝑛 𝐴 ∩ 𝑋 = ∅)
⇒ 𝐴 = (𝐴 ∩ 𝐵) ………… (𝑔𝑖𝑣𝑒𝑛, 𝐴 ∪⊄= A)
Again, A ∪ X = B ∪ X
⇒ 𝐵 ∩ (𝐴 ∪ 𝑋) = 𝐵 ∩ (𝐵 ∪ 𝑋)
⇒ (𝐵 ∩ 𝐴) ∪ (𝐵 ∩ 𝑋) = (𝐵 ∩ 𝐵) ∪ (𝐵 ∩ 𝑋)
⇒ (𝐴 ∩ 𝐵) ∪⊄= B ∪⊄ ………….. (𝑔𝑖𝑣𝑒𝑛 𝐴 ∩ 𝑋 = ∅)
⇒ 𝐴 = (𝐴 ∩ 𝐵) ………… (𝑔𝑖𝑣𝑒𝑛, 𝐵 ∩ 𝑋 = ∅)
⇒ (𝐴 ∩ 𝐵) = B …….. (2)
From (1) and (2), A = B (proved)

Question. Show that (i) 𝐴 = (𝐴 ∩ 𝐵) ∪ (A − B), (ii) 𝐴 ∪ = (𝐵 − 𝐴) = 𝐴 ∪ 𝐵.
Answer : (i) Taking R.H.S. = (𝐴 ∩ 𝐵) ∪ (A − B)
= (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐵1)
= 𝐴 ∪ (𝐴 ∩ 𝐵¹)………… {𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
= 𝐴 ∩∪ ………….. {∵ 𝐴 ∪ 𝐴¹ =∪}
= A = L.H.S. (proved)
(ii) Taking L.H.S. = ∪ = (𝐵 − 𝐴)
= 𝐴 ∪ (𝐵 ∩ 𝐴)
= (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐴¹)………… {𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑙𝑎𝑤}
= (𝐴 ∪ 𝐵) ∩∪
= 𝐴 ∪ 𝐵 = R.H.S. (proved)

Question. Find sets A,B and C such that 𝐴 ∩ 𝐵, 𝐵 ∪ 𝐶, 𝐴 ∪ 𝐶 are non-empty sets and 𝐴 ∩ 𝐵 ∩ 𝐶 is an empty set.
Answer : Let 𝐴 = {1,2}
𝐵 = {2,3}
𝐶 = {1,3}
𝐴 ∩ 𝐵 = {2}: 𝐵 ∩ 𝐶 = {3}: 𝐶 ∩ 𝐴 = {1} and 𝐴 ∩ 𝐵 ∩ 𝐶 =⊄ ans.

Question. Is it true for any sets A and B ; 𝑃(𝐴) ∪ 𝑃(𝐵) = 𝑃(𝐴 ∪ 𝐵). Justify your answer.
Answer : To check: 𝑃(𝐴) ∪ 𝑃(𝐵) = 𝑃(𝐴 ∪ 𝐵)
Let 𝐴 = {1,2} and
𝐵 = {2,3}
𝑃(𝐴) = {⟨1⟩, ⟨2⟩, ⟨1,2⟩, ⊄} ; 𝑃(𝐵) = {⟨2⟩, ⟨3⟩, ⟨2,3⟩, ∅}
𝐴 ∪ 𝐵 = {1,2,3}
𝑃(𝐴 ∪ 𝐵) = {⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨1,2⟩, ⟨2,3⟩, ⟨3,1⟩, ⟨1,2,3⟩, ∅}
𝑃(𝐴) ∪ 𝑃(𝐵) = {⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨1,2⟩, ⟨2,3⟩, ⊄}
Clearly, 𝑃(𝐴) ∪ 𝑃(𝐵) ≠ 𝑃(𝐴 ∪ 𝐵) 𝑎𝑛𝑠.

Question. State and prove 𝐷𝐸 − 𝑀𝑂𝑅𝐺𝐴𝑁′𝑆 𝐿𝐴𝑊.
Answer : De-Morgan’s Law
(i) (𝐴 ∪ 𝐵)¹ = 𝐴¹ ∩ 𝐵¹
Let 𝑥 ∈ (𝐴 ∪ 𝐵)¹
⇒𝑥 ∉ (𝐴 ∪ 𝐵)
⇒𝑥 ∉ A and 𝑥 ∉ B
⇒𝑥 ∈ A¹ and 𝑥 ∈ B¹
⇒𝑥 ∈ A¹ ∪ B¹
∴ (𝐴 ∪ 𝐵)¹ ∁ 𝐴¹ ∩ 𝐵¹ ……. (1)
Now let 𝑦 ∈ 𝐴1 ∩ 𝐵1
⇒𝑦 ∉ (𝐴 ∪ 𝐵)
⇒𝑦 ∈ A¹ and 𝑦 ∈ B¹
⇒𝑦 ∉ A and 𝑦 ∉ B
⇒𝑦 ∉ (𝐴 ∪ 𝐵)
⇒𝑦 ∈ (𝐴 ∪ 𝐵)¹
∴ 𝐴¹ ∩ 𝐵¹∁(𝐴 ∪ 𝐵)¹ ……. (2)
From (1) and (2), (𝐴 ∪ 𝐵)¹ = 𝐴¹ ∩ 𝐵1(proved)
(ii) (𝐴 ∩ 𝐵)¹ = 𝐴1 ∪ 𝐵¹
Let 𝑥 ∈ (𝐴 ∩ 𝐵)¹
⇒𝑥 ∉ (𝐴 ∩ 𝐵)
⇒𝑥 ∉ A or 𝑥 ∉ B
⇒𝑥 ∈ A¹ or 𝑥 ∈ B¹
⇒𝑥 ∈ (A¹ ∪ B¹)
∴ (𝐴 ∩ 𝐵)¹ ∁ 𝐴¹ ∪ 𝐵¹ ……. (1)
Now let 𝑦 ∈ 𝐴1 ∪ 𝐵¹
⇒𝑦 ∈ A¹ or 𝑥 ∈ B¹
⇒𝑦 ∉ A and 𝑦 ∉ B
⇒𝑦 ∉ 𝐴 ∩ 𝐵
⇒𝑦 ∈ (𝐴 ∩ 𝐵)¹
∴ 𝐴¹ ∪ 𝐵¹∁(𝐴 ∩ 𝐵)¹ ……. (2) (proved)
From (1) and (2), (𝐴 ∩ 𝐵)1 = 𝐴¹ ∪ 𝐵¹ (proved)

Question. State and prove DISTRIBUTIVE LAW.
Answer : (i) 𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
Let 𝑥 ∈ 𝐴 ∪ (𝐵 ∩ 𝐶)
⇒𝑥 ∈ A and 𝑥 ∈ (𝐵 ∩ 𝐶)
⇒𝑥 ∈ A or (𝑥 ∈ 𝐵 𝑎𝑛𝑑 𝑥 ∈ 𝐶)
⇒(𝑥 ∈ A or 𝑥 ∈ B) and (𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐶)
⇒𝑥 ∈ (A ∪ B) and 𝑥 ∈ (A ∪ C)
⇒𝑥 ∈ [(A ∪ B) ∩ (A ∪ C)]
∴ 𝐴 ∪ (𝐵 ∩ 𝐶)∁(𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶) ……. (1)
Now let 𝑦 ∈ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
⇒𝑦 ∈ (𝐴 ∪ 𝐵) 𝑎𝑛𝑑 (𝐴 ∪ 𝐶)
⇒(𝑦 ∈ A or 𝑦 ∈ 𝐵) and (𝑦 ∈ A or 𝑦 ∈ 𝐶)
⇒𝑦 ∈ A or y ∈ (𝐵 ∩ 𝐶)
⇒𝑦 ∈ 𝐴 ∪ (𝐵 ∩ 𝐶)
∴ (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)∁ 𝐴 ∪ (𝐵 ∩ 𝐶) ……. (2)
From (1) and (2), 𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶) (proved)
(ii) 𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)

Question. Two finite sets have 𝑚 𝑎𝑛𝑑 𝑛 elements. The number of elements in the power set of first set is 48 more than the number of elements in power set of the second set. find the values of 𝑚 𝑎𝑛𝑑 𝑛.
Answer : Let two sets are A and B
Given 𝑛(𝐴) = 𝑛 and 𝑛(𝐵) = 𝑚
No. of subsets of Set A = 2𝑛
(𝑂𝑅) No. of elements in P(𝐴)
No. of subsets of Set B = 2^𝑚
(𝑂𝑅) No. of elements in P(𝐵)
Given that, 2^𝑛 − 2𝑚 = 48
⇒ 2^𝑛 − 2^𝑚 = 48
⇒ 2^𝑛 − 2^𝑚 = 64 − 16
⇒ 2^𝑛 − 2^𝑚 = 26 − 24
Compare both sides, 𝑛 = 6 𝑎𝑛𝑑 𝑚 = 4 ans.

Question. if A=⊄, fond 𝑃(𝐴), 𝑃(𝑃(𝐴)) and 𝑃 (𝑃(𝑃(𝐴)))
Answer : A=⊄
⇒ 𝑛(𝐴) = 0
No. of subsets of 𝐴 = 2 0 = 1
Subsets =⊄
∴ 𝑃(𝐴) = {⊄}
Here, n(P(A)) = 1
No. of subsets of P(A) = 21 = 2
Subsets = {⊄}, ⊄
∴ P(P(A)) = {⊄}, ⊄
Here, n (𝑃(𝑃(𝐴))) = 2
No. of subsets of (𝑃(𝑃(𝐴))) = 22 = 4
Subsets = {{⊄}}, {⊄}, {{⊄}, ⊄}, ⊄
∴ (𝑃(𝑃(𝐴))) = {{⊄}}, {⊄}, {{⊄}, ⊄}, ⊄ ans.

Question. Show that 𝐴 − (𝐵 − 𝐶) = (𝐴 − 𝐵) ∪ (𝐴 ∩ 𝐶)
Answer : L.H.S. 𝐴 − (𝐵 − 𝐶)
= 𝐴 − (𝐵 ∩ 𝐶¹) ……… {𝐴 − 𝐵 = 𝐴 ∩ 𝐵¹}
= 𝐴 ∩ (𝐵 ∩ 𝐶¹)1 ………… {𝐴 − 𝐵 = 𝐴 ∩ 𝐵¹}
= 𝐴 ∩ (𝐵¹ ∪ 𝐶) …….. {𝐷𝑒 − 𝑚𝑜𝑟𝑔𝑎𝑛′𝑠 𝑙𝑎𝑤}
= (𝐴 ∩ 𝐵¹) ∪ (𝐴 ∩ 𝐶) …….. {𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦}
= (𝐴 − 𝐵) ∪ (𝐴 ∩ 𝐶) R.H.S. ans.

Question. Draw Venn diagram of 𝐴Δ𝐵
Answer : Δ → is called “Symmetric Difference”
𝐴Δ𝐵 = (𝐴 − 𝐵) ∪ (𝐵 − 𝐴)

Question. Write in Roster Form 𝐴 = {𝑥: 𝑥 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 10 𝑎𝑛𝑑 2^𝑥−1 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟}
Answer : 𝐴 = {1,2,3,4,5,6,7,8,9}
Since, 2^𝑥−1 is always an odd no. for all 𝑥 < 10

Question. Show that (𝐴 − 𝐵) ∩ (𝐶 − 𝐵) = (𝐴 ∩ 𝐶) − 𝐵
Answer : L.H.S. (𝐴 − 𝐵) ∩ (𝐶 − 𝐵)
= (𝐴 ∩ 𝐵1) ∩ (𝐶 ∩ 𝐵¹) …….. {𝐴 − 𝐵 = 𝐴 ∩ 𝐵¹}
= (𝐴 ∩ 𝐶) ∩ 𝐵¹ ……… {𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦}
= (𝐴 ∩ 𝐶) − 𝐵 R.H.S. (proved)

Question. If 𝑦 = {𝑡: 𝑡3 = 𝑡; 𝑡 ∈ 𝑅}
Answer : 𝑡³ = 𝑡
⇒ 𝑡³ − 𝑡 = 0
⇒ 𝑡(𝑡² − 1) = 0
⇒ 𝑡 = 0, 𝑡 = ±1
∴ 𝑦 = {0, −1,1} ans.