Sets

Q1: Multiple Choice Type

i. A set P has 3 elements. The number of proper subsets of set B is:

Step 1:
If a set has \(n\) elements, total subsets = \(2^n\)
Here, \(n = 3 \Rightarrow 2^3 = 8\) subsets
Proper subsets = Total subsets − 1 = \(8 – 1 = 7\)
Answer: d. 7

ii. For sets A and B, where A = {2, 4, 6} and B = {1, 3, 5, 7}, \(A \cap B\) is:

Step 1:
Intersection means common elements
A ∩ B = no common elements
Answer: a. ϕ

iii. If set A = {4, 6, 8} and B = {0}, then \(A \cup B\):

Step 1:
Union means combine all unique elements
A ∪ B = {4, 6, 8, 0}
Answer: b. {4, 6, 8, 0}

iv. If A = students in class 8 of a perticular school and B = all students in school:

Step 1:
All class 8 students are part of the whole school
⇒ A ⊂ B
Answer: c. \(A \subset B\)

v. If universal set ξ = {x : x ∈ W, x < 5} and set A = {1, 3}, then complement o set A is equal to:

Step 1:
Universal Set ξ = {0, 1, 2, 3, 4}
A = {1, 3}
A′ = ξ − A = {0, 2, 4}
Answer: b. {0, 2, 4}

vi. If universal set = N, the set of natural numbers, set A = {multiples of 3 less than or equal to 20} and Set B = {multiples of 4 less then or equal to 20}, then A – B is equal to:

Step 1:
A = {3, 6, 9, 12, 15, 18}
B = {4, 8, 12, 16, 20}
A − B = remove common with B from A
= {3, 6, 9, 15, 18}
Answer: a. {3, 6, 9, 15, 18}

Q2: Find all the subsets of each of the following sets:

i. A = {5, 7}

Step 1:
Set has 2 elements → Total subsets = 2² = 4
Step 2:
Subsets are:
{ }, {5}, {7}, {5, 7}
Answer: Total 4 subsets : { }, {5}, {7}, {5, 7}

ii. B = {a, b, c}

Step 1:
3 elements ⇒ 2³ = 8 subsets
Step 2:
Subsets are:
{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
Answer: Total 8 subsets: { }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}

iii. C = {x : x ∈ W, x ≤ 2}

Step 1:
W = whole numbers = {0, 1, 2} ⇒ Set C = {0, 1, 2}
3 elements ⇒ 2³ = 8 subsets
Step 2:
Subsets are:
{ }, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}
Answer: Total 8 subsets: { }, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}

iv. {p : p is a letter in the word ‘poor’}

Step 1:
Letters in ‘poor’ = {p, o, r} ⇒ only unique letters
3 elements ⇒ 2³ = 8 subsets
Step 2:
Subsets are:
{ }, {p}, {o}, {r}, {p, o}, {p, r}, {o, r}, {p, o, r}
Answer: Total 8 subsets: { }, {p}, {o}, {r}, {p, o}, {p, r}, {o, r}, {p, o, r}

Q3: If C is the set of letters in the word ‘cooler’, find:

i. Set C

Step 1:
Letters in the word ‘cooler’ = c, o, o, l, e, r
Only unique letters are taken in a set
Answer: C = {c, o, l, e, r}

ii. n(C)

Step 2:
n(C) means number of elements in set C
C = {c, o, l, e, r} ⇒ 5 elements
Answer: n(C) = 5

iii. Number of its subsets

Step 3:
Total subsets = \(2^n = 2^5 = 32\)
Answer: 32 subsets

iv. Number of its proper subsets

Step 4:
Proper subsets = Total subsets − 1 (excluding the full set)
= \(32 – 1 = 31\)
Answer: 31 proper subsets

Q4: If T = {x : x is a letter in the word ‘TEETH’}, find all its subsets.

Step 1:
Letters in ‘TEETH’ are: T, E, E, T, H
But in a set, we only list distinct letters
So, T = {T, E, H}
Step 2:
Number of elements in T = 3 ⇒ n(T) = 3
Total subsets = \(2^3 = 8\)
Step 3:
List of all subsets:
{ }, {T}, {E}, {H}, {T, E}, {T, H}, {E, H}, {T, E, H}
Answer: The set T = {T, E, H} has 8 subsets: { }, {T}, {E}, {H}, {T, E}, {T, H}, {E, H}, {T, E, H}

Q5: Given the universal set = {-7, -3, -1, 0, 5, 6, 8, 9}, find:

i. A = {x : x < 2}

Step 1:
From universal set: {-7, -3, -1, 0, 5, 6, 8, 9}
Select elements less than 2
Step 2:
Compare each: -7 ✔, -3 ✔, -1 ✔, 0 ✔, 5 ✘, 6 ✘, 8 ✘, 9 ✘
Answer: A = {-7, -3, -1, 0}

ii. B = {x : -4 < x < 6}

Step 1:
From universal set: {-7, -3, -1, 0, 5, 6, 8, 9}
Select elements between -4 and 6 (excluding -4 and 6)
Step 2:
Valid values: -3 ✔, -1 ✔, 0 ✔, 5 ✔
-7 ✘, 6 ✘, 8 ✘, 9 ✘
Answer: B = {-3, -1, 0, 5}

Q6: Given the universal set = {x : x ∈ N and x < 20}, find:

Universal set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

i. A = {x : x = 3p ; p ∈ N}

Step 1:
x = 3p, where p ∈ N means x is a multiple of 3
Step 2:
List multiples of 3 less than 20:
3×1 = 3
3×2 = 6
3×3 = 9
3×4 = 12
3×5 = 15
3×6 = 18
Answer: A = {3, 6, 9, 12, 15, 18}

ii. B = {y : y = 2n + 3 ; n ∈ N}

Step 1:
Substitute n = 1, 2, 3,… until result is < 20
Step 2:
n = 1 → y = 2×1 + 3 = 5
n = 2 → y = 7
n = 3 → y = 9
n = 4 → y = 11
n = 5 → y = 13
n = 6 → y = 15
n = 7 → y = 17
n = 8 → y = 19
n = 9 → y = 21 ❌ (Not in universal set)
Answer: B = {5, 7, 9, 11, 13, 15, 17, 19}

iii. C = {x : x is divisible by 4}

Step 1:
Find numbers divisible by 4 from the universal set
Step 2:
4, 8, 12, 16
Answer: C = {4, 8, 12, 16}

Q7: Find the proper subset of {x : x² – 9x – 10 = 0}

Step 1: Solve the equation \(x^2 – 9x – 10 = 0\)
Use factorization method:
\(x^2 – 9x – 10 = x^2 – 10x + x – 10 = (x – 10)(x + 1) = 0\)
So, roots are: x = 10 and x = -1
Step 2: The set is {10, -1}
Step 3: Proper subsets are all subsets excluding the full set:
{ }, {10}, {-1}
Answer: The proper subsets of {10, -1} are: { }, {10}, {-1}

Q8: Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}. State whether the following are true or false. Give reasons.

i. \(A \subset B\)

Step 1: A = all triangles (scalene, isosceles, equilateral)
B = only isosceles triangles
Step 2: Not all triangles are isosceles
Answer: False. Reason: A contains all triangles, but B is only a part of A

ii. \(B \subseteq A\)

Step 1: Every isosceles triangle is a triangle
Answer: True.Reason: All elements of B are also in A ⇒ B ⊆ A

iii. \(C \subseteq B\)

Step 1: Equilateral triangles have all three sides equal, so they are also isosceles
Answer: True. Reason: Every equilateral triangle satisfies the condition of an isosceles triangle (at least two sides equal)

iv. \(B \subset A\)

Step 1: All isosceles triangles are triangles, but A has more than just isosceles
Answer: True. Reason: B is a proper subset of A

v. \(C \subset A\)

Step 1: Equilateral triangles are triangles
Answer: True. Reason: All elements of C are in A, but A contains more types

vi. \(C \subseteq B \subseteq A\)

Step 1: C ⊆ B (Equilateral ⊆ Isosceles) ✔
B ⊆ A (Isosceles ⊆ Triangle) ✔
Answer: True. Reason: The subset relationship holds in sequence: C ⊆ B ⊆ A

Q9: Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares} and D = {Rhombuses}. State, giving reasons, whether the following are true or false:

i. \(B \subset C\)

Step 1: B = Rectangles, C = Squares
Every square is a rectangle, but every rectangle is not a square
Answer: False. Reason: B ⊄ C because rectangles can have unequal adjacent sides, while squares have all sides equal

ii. \(D \subset B\)

Step 1: D = Rhombuses, B = Rectangles
Rhombuses have all sides equal, but angles may not be 90°, hence not all rhombuses are rectangles
Answer: False. Reason: D ⊄ B because only squares are both rhombuses and rectangles

iii. \(C \subseteq B \subseteq A\)

Step 1: C = Squares ⊆ Rectangles
B = Rectangles ⊆ Quadrilaterals
Answer: True. Reason: Every square is a rectangle, and every rectangle is a quadrilateral

iv. \(D \subset A\)

Step 1: D = Rhombuses, A = Quadrilaterals
Every rhombus has four sides ⇒ it is a quadrilateral
Answer: True. Reason: All rhombuses are quadrilaterals ⇒ D ⊂ A

v. \(B \supseteq C\)

Step 1: B = Rectangles, C = Squares
Every square is a rectangle, but not every rectangle is a square
Answer: True. Reason: B ⊇ C because all squares are rectangles

vi. \(A \supseteq B \supseteq D\)

Step 1: A = Quadrilaterals ⊇ B = Rectangles ✔
But B ⊇ D is not true (rectangles are not always rhombuses)
Answer: False. Reason: B does not include all rhombuses, so B ⊇ D is incorrect

Q10: Given, universal set = {x : x ∈ ℕ, 10 ≤ x ≤ 35}, A = {x ∈ ℕ, x ≤ 16}, B = {x : x > 29}. Find:

i. Find \(A’\)

Step 1: Universal set ξ = {10, 11, 12, …, 35}
This includes all natural numbers from 10 to 35
Step 2: Set A = {x ∈ ℕ, x ≤ 16} within the universal set
So, A = {10, 11, 12, 13, 14, 15, 16}
Step 3: A′ = ξ − A
A′ = {17, 18, 19, …, 35}
Answer: A′ = {17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35}

ii. Find \(B’\)

Step 1: Set B = {x : x > 29} within universal set ξ
So, B = {30, 31, 32, 33, 34, 35}
Step 2: B′ = ξ − B
B′ = {10 to 29}
Answer: B′ = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29}

Q11: Given, universal set = {x ∈ ℤ : -6 < x ≤ 6}, N = {n : n is a non-negative number}, P = {x : x is a non-positive number}. Find:

Step 1: Universal set = {x ∈ ℤ : -6 < x ≤ 6}
⇒ x ranges from -5 to 6 (inclusive)
Universal Set ξ = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

i. Find \(N′\)

Step 2: N = {n : n is a non-negative number}
Non-negative integers = {0, 1, 2, 3, 4, 5, 6} (from universal set)
Step 3: N′ = ξ − N
N′ = {-5, -4, -3, -2, -1}
Answer: N′ = {-5, -4, -3, -2, -1}

ii. Find \(P′\)

Step 4: P = {x : x is a non-positive number}
Non-positive integers = {-5, -4, -3, -2, -1, 0}
Step 5: P′ = ξ − P
P′ = {1, 2, 3, 4, 5, 6}
Answer: P′ = {1, 2, 3, 4, 5, 6}

Q12: Let M = {letters of the word REAL} and N = {letters of the word LARE}. Write sets M and N in roster form and then state whether:

Step 1: Writing sets M and N in roster form
Word “REAL” has letters: R, E, A, L ⇒
Set M = {R, E, A, L}
Word “LARE” has letters: L, A, R, E ⇒
Set N = {L, A, R, E}
Answer: M = {R, E, A, L}, N = {L, A, R, E}

i. Is \(M \subseteq N\) true?

Step 2: Every element of M is in N
⇒ M ⊆ N is true
Answer: True

ii. Is \(N \subseteq M\) true?

Step 3: Every element of N is in M
⇒ N ⊆ M is true
Answer: True

iii. Is \(M = N\) true?

Step 4: If M ⊆ N and N ⊆ M ⇒ M = N
Answer: True. Reason: Both sets contain exactly the same elements

Q13: Given A = {x ∈ ℕ and 3 < x ≤ 6}, B = {x ∈ 𝕎 and x < 4}. Find:

i. Sets A and B in roster form

Step 1: ℕ = {1, 2, 3, …}
A = {x ∈ ℕ : 3 < x ≤ 6} ⇒ A = {4, 5, 6}
Step 2: 𝕎 = {0, 1, 2, 3, …}
B = {x ∈ 𝕎 : x < 4} ⇒ B = {0, 1, 2, 3}
Answer:
A = {4, 5, 6}, B = {0, 1, 2, 3}

ii. \(A \cup B\)

Step 3: Union includes all elements of A and B without repetition:
A ∪ B = {0, 1, 2, 3, 4, 5, 6}
Answer: A ∪ B = {0, 1, 2, 3, 4, 5, 6}

iii. \(A \cap B\)

Step 4: Common elements between A and B:
A = {4, 5, 6}, B = {0, 1, 2, 3}
No elements are common ⇒ A ∩ B = ∅
Answer: A ∩ B = ∅

iv. \(A – B\)

Step 5: Elements in A but not in B:
A − B = {4, 5, 6} − {0, 1, 2, 3} = {4, 5, 6}
Answer: A − B = {4, 5, 6}

v. \(B – A\)

Step 6: Elements in B but not in A:
B − A = {0, 1, 2, 3} − {4, 5, 6} = {0, 1, 2, 3}
Answer: B − A = {0, 1, 2, 3}

Q14: If P = {x ∈ 𝕎 and 4 ≤ x ≤ 8} and Q = {x ∈ ℕ and x < 6}. Find:

i. Find \(P \cup Q\) and \(P \cap Q\)

Step 1: Define sets P and Q.
ℕ (Natural numbers) = {1, 2, 3, 4, 5, 6, …}
𝕎 (Whole numbers) = {0, 1, 2, 3, 4, 5, …}
P = {x ∈ 𝕎 : 4 ≤ x ≤ 8} = {4, 5, 6, 7, 8}
Q = {x ∈ ℕ : x < 6} = {1, 2, 3, 4, 5}
Step 2: Union (P ∪ Q) = All unique elements from both sets:
P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8}
Step 3: Intersection (P ∩ Q) = Common elements in both sets:
P ∩ Q = {4, 5}
Answer:
P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8}
P ∩ Q = {4, 5}

ii. Is \((P \cup Q) \supset (P \cap Q)\)?

Step 4: Check whether every element of (P ∩ Q) is in (P ∪ Q).
P ∩ Q = {4, 5}, and both 4 and 5 are in P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8}
⇒ P ∩ Q ⊂ P ∪ Q (proper subset)
Answer:Yes, (P ∪ Q) ⊃ (P ∩ Q). Reason: All elements of intersection are contained in the union, but union has more elements than intersection.

Q15: If A = {5, 6, 7, 8, 9}, B = {x : 3 < x < 8 and x ∈ 𝕎}, C = {x : x ≤ 5 and x ∈ ℕ}. Find:

Step 1: Write all sets in roster form
A = {5, 6, 7, 8, 9}
B = {x ∈ 𝕎 : 3 < x < 8} = {4, 5, 6, 7}
C = {x ∈ ℕ : x ≤ 5} = {1, 2, 3, 4, 5}

i. A ∪ B and (A ∪ B) ∪ C

Step 2: A ∪ B = {5, 6, 7, 8, 9} ∪ {4, 5, 6, 7}
⇒ A ∪ B = {4, 5, 6, 7, 8, 9}
Step 3: (A ∪ B) ∪ C = {4, 5, 6, 7, 8, 9} ∪ {1, 2, 3, 4, 5}
⇒ (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Answer:A ∪ B = {4, 5, 6, 7, 8, 9}
(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}

ii. B ∪ C and A ∪ (B ∪ C)

Step 4: B ∪ C = {4, 5, 6, 7} ∪ {1, 2, 3, 4, 5}
⇒ B ∪ C = {1, 2, 3, 4, 5, 6, 7}
Step 5: A ∪ (B ∪ C) = {5, 6, 7, 8, 9} ∪ {1, 2, 3, 4, 5, 6, 7}
⇒ A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Answer:B ∪ C = {1, 2, 3, 4, 5, 6, 7}
A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9}

iii. A ∩ B and (A ∩ B) ∩ C

Step 6: A ∩ B = {5, 6, 7, 8, 9} ∩ {4, 5, 6, 7}
⇒ A ∩ B = {5, 6, 7}
Step 7: (A ∩ B) ∩ C = {5, 6, 7} ∩ {1, 2, 3, 4, 5}
⇒ (A ∩ B) ∩ C = {5}
Answer:A ∩ B = {5, 6, 7}
(A ∩ B) ∩ C = {5}

iv. B ∩ C and A ∩ (B ∩ C)

Step 8: B ∩ C = {4, 5, 6, 7} ∩ {1, 2, 3, 4, 5}
⇒ B ∩ C = {4, 5}
Step 9: A ∩ (B ∩ C) = {5, 6, 7, 8, 9} ∩ {4, 5}
⇒ A ∩ (B ∩ C) = {5}
Answer: B ∩ C = {4, 5}
A ∩ (B ∩ C) = {5}

Is (A ∪ B) ∪ C = A ∪ (B ∪ C)?
Is (A ∩ B) ∩ C = A ∩ (B ∩ C)?

Step 10: Compare (A ∪ B) ∪ C and A ∪ (B ∪ C):
Both are {1, 2, 3, 4, 5, 6, 7, 8, 9}
⇒ (A ∪ B) ∪ C = A ∪ (B ∪ C) ✅
Step 11: Compare (A ∩ B) ∩ C and A ∩ (B ∩ C):
Both are {5}
⇒ (A ∩ B) ∩ C = A ∩ (B ∩ C) ✅
Answer: Yes, Associative Law holds for both union and intersection:
(A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∩ B) ∩ C = A ∩ (B ∩ C)

Q16: Given A = {0, 1, 2, 3, 4, 5}, B = {0, 2, 4, 6, 8} and C = {0, 3, 6, 9}. Show that:

Step 1: Write all the sets
A = {0, 1, 2, 3, 4, 5}
B = {0, 2, 4, 6, 8}
C = {0, 3, 6, 9}

i. \(A\cup\left(B\cup C\right)=(A\cup B)\cup C\) i.e. the union of the sets is associative.

Step 2: Find B ∪ C
B ∪ C = {0, 2, 4, 6, 8} ∪ {0, 3, 6, 9}
= {0, 2, 3, 4, 6, 8, 9}
Step 3: Find A ∪ (B ∪ C)
= {0, 1, 2, 3, 4, 5} ∪ {0, 2, 3, 4, 6, 8, 9}
= {0, 1, 2, 3, 4, 5, 6, 8, 9}
Step 4: Find A ∪ B
= {0, 1, 2, 3, 4, 5} ∪ {0, 2, 4, 6, 8}
= {0, 1, 2, 3, 4, 5, 6, 8}
Step 5: Find (A ∪ B) ∪ C
= {0, 1, 2, 3, 4, 5, 6, 8} ∪ {0, 3, 6, 9}
= {0, 1, 2, 3, 4, 5, 6, 8, 9}
Answer: A ∪ (B ∪ C) = (A ∪ B) ∪ C = {0, 1, 2, 3, 4, 5, 6, 8, 9}
Hence, union is associative.

ii. \(A\cap\left(B\cap C\right)=(A\cap B)\cap C\), i.e. the intersection of sets is associative.

Step 6: Find B ∩ C
= {0, 2, 4, 6, 8} ∩ {0, 3, 6, 9}
= {0, 6}
Step 7: Find A ∩ (B ∩ C)
= {0, 1, 2, 3, 4, 5} ∩ {0, 6}
= {0}
Step 8: Find A ∩ B
= {0, 1, 2, 3, 4, 5} ∩ {0, 2, 4, 6, 8}
= {0, 2, 4}
Step 9: Find (A ∩ B) ∩ C
= {0, 2, 4} ∩ {0, 3, 6, 9}
= {0}
Answer: A ∩ (B ∩ C) = (A ∩ B) ∩ C = {0}
Hence, intersection is associative.

Q17: If A = {x ∈ W : 5 < x < 10}, B = {3, 4, 5, 6, 7} and C = {x = 2n ; n ∈ N and n ≤ 4}, find:

Step 1: Express all sets in roster form
A = {6, 7, 8, 9} (since 5 < x < 10 and x ∈ W)
B = {3, 4, 5, 6, 7}
C = {2, 4, 6, 8} (as x = 2n and n ≤ 4 ⇒ n = 1,2,3,4)

i. A ∩ (B ∪ C)

Step 2: Find B ∪ C
= {3, 4, 5, 6, 7} ∪ {2, 4, 6, 8}
= {2, 3, 4, 5, 6, 7, 8}
Step 3: Find A ∩ (B ∪ C)
= {6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
= {6, 7, 8}
Answer: {6, 7, 8}

ii. (B ∪ A) ∩ (B ∪ C)

Step 4: Find B ∪ A
= {3, 4, 5, 6, 7} ∪ {6, 7, 8, 9}
= {3, 4, 5, 6, 7, 8, 9}
Step 5: Find B ∪ C (already found in step 2)
= {2, 3, 4, 5, 6, 7, 8}
Step 6: Find intersection of both sets
= {3, 4, 5, 6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
= {3, 4, 5, 6, 7, 8}
Answer: {3, 4, 5, 6, 7, 8}

iii. B ∪ (A ∩ C)

Step 7: Find A ∩ C
= {6, 7, 8, 9} ∩ {2, 4, 6, 8}
= {6, 8}
Step 8: Now find B ∪ (A ∩ C)
= {3, 4, 5, 6, 7} ∪ {6, 8}
= {3, 4, 5, 6, 7, 8}
Answer: {3, 4, 5, 6, 7, 8}

iv. (A ∩ B) ∪ (A ∩ C)

Step 9: A ∩ B = {6, 7, 8, 9} ∩ {3, 4, 5, 6, 7} = {6, 7}
A ∩ C = {6, 7, 8, 9} ∩ {2, 4, 6, 8} = {6, 8}
Step 10: Union of both
= {6, 7} ∪ {6, 8} = {6, 7, 8}
Answer: {6, 7, 8}

Compare the sets

From answers above:
– A ∩ (B ∪ C) = {6, 7, 8}
– (A ∩ B) ∪ (A ∩ C) = {6, 7, 8}
⇒ These two sets are equal.
Also:
– (B ∪ A) ∩ (B ∪ C) = {3, 4, 5, 6, 7, 8}
– B ∪ (A ∩ C) = {3, 4, 5, 6, 7, 8}
⇒ These two sets are also equal.
Answer: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(B ∪ A) ∩ (B ∪ C) = B ∪ (A ∩ C)

Q18: If P = (factors of 36) and Q = {factors of 48}; find:

Step 1: Write set P and set Q in roster form
Set P: Factors of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Set Q: Factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

i. \(P \cup Q\)

Step 2: Combine elements of P and Q, remove duplicates
P ∪ Q = {1, 2, 3, 4, 6, 9, 12, 18, 36, 8, 16, 24, 48}
Answer: {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

ii. \(P \cap Q\)

Step 3: Common elements in both sets
P ∩ Q = {1, 2, 3, 4, 6, 12}
Answer: {1, 2, 3, 4, 6, 12}

iii. \(Q – P\)

Step 4: Elements in Q but not in P
Q = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
P = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Q − P = {8, 16, 24, 48}
Answer: {8, 16, 24, 48}

iv. \(P’\cap Q\)

Step 5: Assume universal set U = {1 to 50} (common domain)
Then P′ = All elements from 1 to 50 not in P
P = {1, 2, 3, 4, 6, 9, 12, 18, 36}
So P′ = {5, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, …, 50} excluding above
Step 6: Now, find intersection with Q:
Q = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
From P′, the common elements with Q are {8, 16, 24, 48}
Answer: {8, 16, 24, 48}

Q19: If A = {6, 7, 8, 9}, B = {4, 6, 8, 10} and C = {x : x ∈ N : 2 < x ≤ 7}; find:

Step 1: Write all sets in roster form
A = {6, 7, 8, 9}
B = {4, 6, 8, 10}
C = {x ∈ N | 2 < x ≤ 7} = {3, 4, 5, 6, 7}

i. A – B

Step 2: A – B = elements in A not in B
A = {6, 7, 8, 9}, B = {4, 6, 8, 10}
Common: {6, 8}
A – B = {7, 9}
Answer: {7, 9}

ii. B – C

B = {4, 6, 8, 10}, C = {3, 4, 5, 6, 7}
Common: {4, 6}
B – C = {8, 10}
Answer: {8, 10}

iii. B – (A – C)

Step 3: First find A – C
A = {6, 7, 8, 9}, C = {3, 4, 5, 6, 7}
Common: {6, 7}
A – C = {8, 9}
Now B – (A – C) = B – {8, 9} = {4, 6, 10}
Answer: {4, 6, 10}

iv. A – (B ∪ C)

B ∪ C = {3, 4, 5, 6, 7, 8, 10}
A = {6, 7, 8, 9}
Common: {6, 7, 8}
A – (B ∪ C) = {9}
Answer: {9}

v. B – (A ∩ C)

A ∩ C = elements common in A and C
A = {6, 7, 8, 9}, C = {3, 4, 5, 6, 7}
A ∩ C = {6, 7}
Now B – (A ∩ C) = B – {6, 7} = {4, 8, 10}
Answer: {4, 8, 10}

vi. B – B

B = {4, 6, 8, 10}
B – B = ∅ (empty set)
Answer: ∅

Q20: If A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}; verify:

Step 1: Write all sets in roster form
A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}

i. \(A – (B \cup C) = (A – B) \cap (A – C)\)

Step 2: Find \(B \cup C\)
B ∪ C = {2, 3, 4, 5, 6, 8}
Step 3: Find \(A – (B ∪ C)\)
A = {1, 2, 3, 4, 5}
A – (B ∪ C) = {1}
Step 4: Find \(A – B\) and \(A – C\)
A – B = {1, 3, 5}
A – C = {1, 2}
Step 5: Now find intersection:
(A – B) ∩ (A – C) = {1}
Answer: Verified: LHS = RHS = {1} ✅

ii. \(A – (B \cap C) = (A – B) \cup (A – C)\)

Step 6: Find \(B \cap C\)
B ∩ C = {4, 6}
Step 7: Find \(A – (B ∩ C)\)
A = {1, 2, 3, 4, 5}
A – {4, 6} = {1, 2, 3, 5}
Step 8: Already we had:
A – B = {1, 3, 5}
A – C = {1, 2}
Step 9: Take union:
(A – B) ∪ (A – C) = {1, 2, 3, 5}
Answer: Verified: LHS = RHS = {1, 2, 3, 5} ✅

Q21: Given A = \(\left\{x \in N : x < 6\right\}\), B = {3, 6, 9} and C = \(\left\{x \in N : 2n – 5 \le 8\right\}\). Show that:

Step 1: Write the sets in roster form
A = {1, 2, 3, 4, 5}
B = {3, 6, 9}
We are given C = \({x\in\mathbb{N}:2n – 5 \le 8}\)
⇒ \(2n \le 13\)
⇒ \(n \le 6\), since \(n \in \mathbb{N}\)
Then \(x = n\) (since \(x \in \mathbb{N}\))
So, C = {1, 2, 3, 4, 5, 6}

i. Show \(A\cup(B\cap C) = (A\cup B)\cap(A\cup C)\)

Step 2: Find \(B \cap C\)
B = {3, 6, 9}, C = {1, 2, 3, 4, 5, 6}
Common elements = {3, 6}
So, \(B \cap C = \{3, 6\}\)
Step 3: Find \(A \cup (B \cap C)\)
A = {1, 2, 3, 4, 5}, B ∩ C = {3, 6}
Union = {1, 2, 3, 4, 5, 6}
Step 4: Find \(A \cup B\) and \(A \cup C\)
A ∪ B = {1, 2, 3, 4, 5, 6, 9}
A ∪ C = {1, 2, 3, 4, 5, 6}
Step 5: Find \((A \cup B) \cap (A \cup C)\)
Intersection = common elements = {1, 2, 3, 4, 5, 6}
Answer: Verified: LHS = RHS = {1, 2, 3, 4, 5, 6} ✅

ii. \(A\cap(B\cup C) = (A\cap B)\cup(A\cap C)\)

Step 6: Find \(B \cup C\)
B = {3, 6, 9}, C = {1, 2, 3, 4, 5, 6}
Union = {1, 2, 3, 4, 5, 6, 9}
Step 7: Find \(A \cap (B \cup C)\)
A = {1, 2, 3, 4, 5}
Common with B ∪ C = {1, 2, 3, 4, 5}
Step 8: Find \(A \cap B\) and \(A \cap C\)
A ∩ B = {3}
A ∩ C = {1, 2, 3, 4, 5}
Step 9: Take union:
(A ∩ B) ∪ (A ∩ C) = {1, 2, 3, 4, 5}
Answer: Verified: LHS = RHS = {1, 2, 3, 4, 5} ✅

Q22: If n(A) = 30, n(B) = 20 and \(n\left(A\cup B\right) = 36\), find \(n\left(A\cap B\right)\)

Step 1: Use the formula for union of two sets \[ n(A \cup B) = n(A) + n(B) – n(A \cap B) \]Step 2: Substitute the known values into the formula: \[ 36 = 30 + 20 – n(A \cap B) \]Step 3: Simplify the right-hand side: \[ 36 = 50 – n(A \cap B) \]Step 4: Solve for \(n(A \cap B)\) \[ n(A \cap B) = 50 – 36 = 14 \]Answer: \(n(A \cap B) = 14\)

Q23: If n(A) = 50, n(B) = 30 and \(n\left(A\cap B\right) = 15\), find \(n\left(A\cup B\right)\)

Step 1: Use the formula for union of two sets: \[ n(A \cup B) = n(A) + n(B) – n(A \cap B) \]Step 2: Substitute the given values: \[ n(A \cup B) = 50 + 30 – 15 \]Step 3: Simplify the expression: \[ n(A \cup B) = 80 – 15 = 65 \]Answer: \(n(A \cup B) = 65\)

Q24: If n(A − B) = 30, n(B − A) = 20 and \(n\left(A \cap B\right) = 10\), find:

i. n(A)

Step 1: Use the formula: \[ n(A) = n(A – B) + n(A \cap B) \]Step 2: Substitute the given values: \[ n(A) = 30 + 10 = 40 \]Answer: n(A) = 40

ii. n(B)

Step 1: Use the formula: \[ n(B) = n(B – A) + n(A \cap B) \]Step 2: Substitute the values: \[ n(B) = 20 + 10 = 30 \]Answer: n(B) = 30

iii. \(n\left(A \cup B\right)\)

Step 1: Use the formula: \[ n(A \cup B) = n(A – B) + n(B – A) + n(A \cap B) \]Step 2: Substitute the values: \[ n(A \cup B) = 30 + 20 + 10 = 60 \]Answer: \(n(A \cup B) = 60\)

Q25: If n(A − B) = 30, n(B − A) = 48 and \(n\left(A \cap B\right) = 15\), find \(n\left(A \cup B\right)\).

Step 1: Use the formula for union of two sets: \[ n(A \cup B) = n(A – B) + n(B – A) + n(A \cap B) \]Step 2: Substitute the values: \[ n(A \cup B) = 30 + 48 + 15 \]Step 3: Simplify the expression: \[ n(A \cup B) = 93 \]Answer: \(n(A \cup B) = 93\)