Class 9 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 9 (FBISE / NBF)

Mastery Guide: Sets and Relations — Set Operations, Venn Diagrams, Survey Inclusion-Exclusion, Cartesian Products & Binary Relations

📖 Chapter 3: Sets and Relations 📅 Updated: Sep 18, 2026

National Curriculum (SNC) & FBISE

Chapter 3: Sets and Relations

Mastery Guide • Set Operations • Venn Diagrams • Inclusion-Exclusion Principle • Cartesian Products • Binary Relations

Unit 03 Class 9 Mathematics

1 Unit Overview & Learning Outcomes

In this unit, students explore mathematics as the foundational language of patterns, structures, and their relationships. From classifying observable universe galaxies to solving complex multi-variable demographic surveys and defining binary mappings, students master:

🎯 Set Definitions & Forms Descriptive, Tabular (Roster), and Set-Builder (Rule) notations; standard number systems ($\mathbb{N}, \mathbb{W}, \mathbb{Z}, \mathbb{E}, \mathbb{O}, \mathbb{P}, \mathbb{Q}, \mathbb{Q}', \mathbb{R}$).

🎯 Three-Set Operations & Venn Diagrams Union, Intersection, Difference, and Complement under Disjoint, Overlapping, and Nested Subset configurations.

🎯 Set Laws & Algebraic Proofs Rigorous set-theoretic proofs and Venn diagram verifications of Associative, Distributive, and De Morgan's Laws.

🎯 Real-World Inclusion-Exclusion Solving 2-set and 3-set survey models, cataloging systems, school enrollment, and demographic polls.

🎯 Cartesian Products & Ordered Pairs Ordered pairs equality, Cartesian products $A \times B$, coordinate grid graphs, and arrow mapping diagrams.

🎯 Binary Relations & Inverses Total possible relations ($2^{m \times n}$), predicate filters, Domain $\text{Dom}(R)$, Range $\text{Range}(R)$, and Inverse Relations $R^{-1}$.

2 Conceptual Foundations of Sets

2.1 Rigorous Definition of a Set

A Set is a well-defined collection of distinct objects.

Essential Criterion	Detailed Explanation	Comparative Examples
1. Well-Defined	There exists an unambiguous, objective mathematical rule or criterion to decide conclusively whether any given element belongs ($\in$) or does not belong ($\notin$) to the set.	✓ Set: "Stones weighing less than $1\text{ kg}$" ✗ Not a Set: "Lighter stones" (subjective)
2. Distinct Objects	Each object in the set must be listed exactly once. Repetition of elements is strictly not allowed and does not alter the set.	Letters in the word `"sets"` $\implies \{s, e, t\}$. Natural numbers among $3, -9, 2, 3, 4, 2 \implies \{2, 3, 4\}$.

2.2 Mathematics as Patterns, Structures, and Relationships

Patterns allow us to predict unknown terms through functional rules. In structural geometry, square numbers represent the area of regular grid squares:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

Sequence of square numbers: $\{1^2, 2^2, 3^2, 4^2, \dots\} = \{1, 4, 9, 16, 25, \dots\}$, where $T_n = n^2$.

2.3 Standard Mathematical Sets of Numbers

Symbol	Name of Number Set	Tabular / Set-Builder Representation
$\mathbb{N}$	Natural Numbers	$\{1, 2, 3, 4, 5, \dots\}$
$\mathbb{W}$	Whole Numbers	$\{0, 1, 2, 3, 4, \dots\}$
$\mathbb{Z}$	Integers	$\{0, \pm 1, \pm 2, \pm 3, \dots\}$
$\mathbb{E}$	Even Integers	$\{0, \pm 2, \pm 4, \pm 6, \dots\}$
$\mathbb{O}$	Odd Integers	$\{\pm 1, \pm 3, \pm 5, \dots\}$
$\mathbb{P}$	Prime Numbers	$\{2, 3, 5, 7, 11, 13, 17, 19, \dots\}$
$\mathbb{Q}$	Rational Numbers	$\{ \frac{p}{q} \mid p, q \in \mathbb{Z} \land q \ne 0 \}$
$\mathbb{Q}'$	Irrational Numbers	$\{ x \mid x \ne \frac{p}{q}, p, q \in \mathbb{Z} \land q \ne 0 \}$
$\mathbb{R}$	Real Numbers	$\mathbb{Q} \cup \mathbb{Q}' = \{ x \mid x \in \mathbb{Q} \lor x \in \mathbb{Q}' \}$

💡 Critical Checkpoint: Why can the set of Real Numbers $\mathbb{R}$ never be written in Tabular (Roster) Form?
Reason: The real numbers are uncountably infinite and dense (between any two distinct real numbers, there are infinitely many real numbers). Thus, elements cannot be enumerated sequentially one after another.

3 Fundamental Set Operations & Venn Diagram Configurations

Disjoint Sets ($A \cap B = \emptyset$)

No common elements: $A \cap B = \emptyset$. Circles do not touch.

Overlapping Sets ($A \cap B \ne \emptyset$)

Common region exists, neither is a subset of the other.

Nested Subset ($A \subseteq B$)

All elements of $A$ are contained in $B$: $A \subseteq B$.

3.1 Set Difference & Complements

Set Difference ($A - B$ or $A \setminus B$)

Collection of all elements that belong to $A$ but do not belong to $B$:


        A - B = { x | x ∈ A ∧ x ∉ B }

• Note: In general, $A - B \ne B - A$ (Non-commutative).

Complement of a Set ($A^c$ or $A'$)

Collection of all elements in universal set $U$ that do not belong to $A$:


        A' = U - A = { x | x ∈ U ∧ x ∉ A }

• Note: $A \cup A' = U$, $A \cap A' = \emptyset$, and $(A')' = A$.

4 Rigorous Laws of Set Operations

4.1 Associative Laws

1. Union: $(A \cup B) \cup C = A \cup (B \cup C)$
2. Intersection: $(A \cap B) \cap C = A \cap (B \cap C)$

4.2 Distributive Laws

1. Union over Intersection: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
2. Intersection over Union: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

4.3 De Morgan's Laws

1. $(A \cup B)' = A' \cap B'$ • "The complement of a union is the intersection of the complements."
2. $(A \cap B)' = A' \cup B'$ • "The complement of an intersection is the union of the complements."

General 3-Set Venn Diagram (8 Mutually Disjoint Regions)

Every 3-set problem partitions the universe $U$ into exactly $2^3 = 8$ mutually disjoint regions.

5 Inclusion-Exclusion Principle (Survey Word Problems)

In real-world demographic and survey analysis, cardinalities $n(S)$ satisfy fundamental counting equations:

Two-Set Cardinality Formulas:

Union Formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Intersection Formula: $n(A \cap B) = n(A) + n(B) - n(A \cup B)$
Only $A$ (Difference): $n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B)$
Disjoint Partition: $n(A \cup B) = n(A - B) + n(A \cap B) + n(B - A)$
Neither $A$ nor $B$: $n((A \cup B)') = n(U) - n(A \cup B)$

Three-Set Cardinality Formula:

$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$$

• Neither of the three: $n((A \cup B \cup C)') = n(U) - n(A \cup B \cup C)$.

6 Cartesian Products, Binary Relations & Mappings

6.1 Ordered Pairs & Equality

An Ordered Pair $(a, b)$ consists of two elements where order is non-invertible ($(a, b) \ne (b, a)$ if $a \ne b$).

Fundamental Equality Axiom: $$(a, b) = (c, d) \iff a = c \quad \text{and} \quad b = d$$

6.2 Cartesian Product $A \times B$

The Cartesian Product of two sets $A$ and $B$, denoted $A \times B$, is the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$: $$A \times B = \{(x, y) \mid x \in A \land y \in B\}$$

Cardinality & Subsets

$n(A \times B) = n(A) \times n(B)$
Total subsets of $A \times B = 2^{n(A \times B)}$
$A \times B = \emptyset$ if $A = \emptyset$ or $B = \emptyset$

Commutativity & Distributivity

$A \times B \ne B \times A$ (unless $A = B$ or one is $\emptyset$)
$A \times (B \cup C) = (A \times B) \cup (A \times C)$
$A \times (B \cap C) = (A \times B) \cap (A \times C)$

6.3 Binary Relation, Domain, Range & Inverse

Binary Relation: Any subset $R \subseteq A \times B$ is called a binary relation from $A$ to $B$.
Total Relations: If $n(A) = p$ and $n(B) = q$, the number of binary relations is $2^{p \times q}$.
Domain $\text{Dom}(R)$: Set of all first elements of the ordered pairs in $R$: $$\text{Dom}(R) = \{ x \in A \mid (x, y) \in R \}$$
Range $\text{Range}(R)$: Set of all second elements of the ordered pairs in $R$: $$\text{Range}(R) = \{ y \in B \mid (x, y) \in R \}$$
Inverse Relation $R^{-1}$: Obtained by interchanging the coordinates of each ordered pair in $R$: $$R^{-1} = \{(y, x) \mid (x, y) \in R\} \subseteq B \times A$$ $$\text{Dom}(R^{-1}) = \text{Range}(R) \quad \text{and} \quad \text{Range}(R^{-1}) = \text{Dom}(R)$$

Arrow Diagram Representation of Relation $R: A \to B$

Relation $R = \{(a, b) \in A \times B \mid a > b\} = \{(1, 0), (2, 0), (2, 1), (3, 0), (3, 1)\}$.

3.1 Exercise 3.1 — Operations on Sets & Venn Diagram Verifications

7 Major Questions • Complete Visual Proofs

Question 1: Shade Set Operations Across ALL 4 Textbook Configurations

Shade each of the four operations: (a) $A \cup (B \cap C)$, (b) $A \cap (B \cup C)$, (c) $(A \cup B) \cup C$, and (d) $A \cap (B \cap C)$ across all four distinct Venn diagram arrangements.

Configuration (i): Overlapping Sets $A$ and $B$, with $C$ Intersecting Both

(a) $A \cup (B \cap C)$ Entire circle $A$ + lens $B \cap C$

(b) $A \cap (B \cup C)$ $(A \cap B) \cup (A \cap C)$

(c) $(A \cup B) \cup C$ All three full sets shaded

(d) $A \cap (B \cap C)$ Central triple intersection

Configuration (ii): Disjoint Sets $A$ and $B$, with $C$ Overlapping Both ($A \cap B = \emptyset$)

(a) $A \cup (B \cap C)$ Entire $A$ + lens $B \cap C$

(b) $A \cap (B \cup C)$ Lens $A \cap C$ only (since $A \cap B = \emptyset$)

(c) $(A \cup B) \cup C$ Entire region of $A, B$, and $C$

(d) $A \cap (B \cap C)$ Empty ($\emptyset$) • No common region

Configuration (iii): Three Mutually Overlapping Sets (Symmetric Triad)

(a) $A \cup (B \cap C)$ Entire $A$ + lens $B \cap C$

(b) $A \cap (B \cup C)$ $(A \cap B) \cup (A \cap C)$

(c) $(A \cup B) \cup C$ All three full sets

(d) $A \cap (B \cap C)$ Central triple-intersection

Configuration (iv): Nested $B \subseteq A$ with Disjoint $C$ ($A \cap C = \emptyset, B \cap C = \emptyset$)

(a) $A \cup (B \cap C)$ Entire circle $A$ ($B \cap C = \emptyset$)

(b) $A \cap (B \cup C)$ Inner circle $B$ only ($A \cap C = \emptyset$)

(c) $(A \cup B) \cup C$ Both circles $A$ and $C$ fully shaded

(d) $A \cap (B \cap C)$ Empty ($\emptyset$) • No intersection

Question 2: Evaluations & Venn Diagram for $X = \{a, b, c, d, e\}$, $Y = \{a, c, e\}$, $Z = \{g, h, i, j\}$

Given Sets:
• $X = \{a, b, c, d, e\}$
• $Y = \{a, c, e\}$ &implies; Note: $Y \subset X$ (Proper Subset)
• $Z = \{g, h, i, j\}$ &implies; Note: $Z \cap X = \emptyset$ (Disjoint)

Step-by-Step Evaluations:
(i) $(X \cup Y) \cup Z = \{a, b, c, d, e\} \cup \{g, h, i, j\} = \mathbf{\{a, b, c, d, e, g, h, i, j\}}$
(ii) $X \cup (Y \cup Z) = \{a, b, c, d, e\} \cup \{a, c, e, g, h, i, j\} = \mathbf{\{a, b, c, d, e, g, h, i, j\}}$
(iii) $(X \cap Y) \cap Z = \{a, c, e\} \cap \{g, h, i, j\} = \mathbf{\emptyset}$
(iv) $X \cap (Y \cap Z) = X \cap \emptyset = \mathbf{\emptyset}$
(v) $(X \cup Y) \cap Z = \{a, b, c, d, e\} \cap \{g, h, i, j\} = \mathbf{\emptyset}$
(vi) $(X \cap Y) \cup Z = \{a, c, e\} \cup \{g, h, i, j\} = \mathbf{\{a, c, e, g, h, i, j\}}$

Venn Diagram for Question 2 ($Y \subset X$ and $Z$ Disjoint)

Questions 3 & 4: Verification of Associative & Distributive Laws via Diagrams

Question 3: Associative Laws

• Union: Shading $(A \cup B) \cup C$ combines all 3 regions, which is visually identical to $A \cup (B \cup C)$.
• Intersection: Shading $(A \cap B) \cap C$ isolates the mutual common region, identically matching $A \cap (B \cap C)$.

Question 4: Distributive Laws

• Union over Intersection: $A \cup (B \cap C)$ shades all of $A$ and lens $B \cap C$, identically equal to $(A \cup B) \cap (A \cup C)$.
• Intersection over Union: $A \cap (B \cup C)$ shades the sections of $A$ inside $B$ or $C$, matching $(A \cap B) \cup (A \cap C)$.

Question 5: Proof of Associative Laws with Specific Element Sets & Diagrams

Part (i): $P = \{0, 1, 2, 3\}$, $Q = \{2, 3, 4, 5, 6\}$, $R = \{5, 6, 7, 8, 9\}$

(a) Associative Law of Union: $(P \cup Q) \cup R = P \cup (Q \cup R)$
LHS: $(P \cup Q) \cup R = \{0, 1, 2, 3, 4, 5, 6\} \cup \{5, 6, 7, 8, 9\} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$
RHS: $P \cup (Q \cup R) = \{0, 1, 2, 3\} \cup \{2, 3, 4, 5, 6, 7, 8, 9\} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$
$\implies \text{LHS} = \text{RHS}$.

(b) Associative Law of Intersection: $(P \cap Q) \cap R = P \cap (Q \cap R)$
LHS: $(P \cap Q) \cap R = \{2, 3\} \cap \{5, 6, 7, 8, 9\} = \emptyset$
RHS: $P \cap (Q \cap R) = \{0, 1, 2, 3\} \cap \{5, 6\} = \emptyset$
$\implies \text{LHS} = \text{RHS} = \emptyset$.

Part (ii): $P = \{m, n, o, p, q\}$, $Q = \{r, s, t, u\}$, $R = \{t, u, v, w\}$

• Union: $(P \cup Q) \cup R = P \cup (Q \cup R) = \{m, n, o, p, q, r, s, t, u, v, w\}$.
• Intersection: $(P \cap Q) \cap R = \emptyset \cap R = \emptyset$; $P \cap (Q \cap R) = P \cap \{t, u\} = \emptyset$. $\implies \text{LHS} = \text{RHS} = \emptyset$.

Questions 6 & 7: Distributive Laws Proofs with Exact Element Venn Placement

Question 6: Verify $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$

Given: $X = \{-3, -2, -1\}$, $Y = \{0, 1, 2, 3\}$, $Z = \{-3, -2, -1, 0, 1, 2, 3\}$.
Note: $X \subset Z$ and $Y \subset Z$, with $X \cap Y = \emptyset$.

LHS: $Y \cap Z = \{0, 1, 2, 3\}$. $X \cup (Y \cap Z) = \{-3, -2, -1, 0, 1, 2, 3\} = Z$.
RHS: $X \cup Y = \{-3, -2, -1, 0, 1, 2, 3\} = Z$; $X \cup Z = Z$. $Z \cap Z = Z$.
$\implies \text{LHS} = \text{RHS} = Z$. (Verified)

Question 7: Verify $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$

Given: $X = \{a, e, i\}$, $Y = \{e, n, r, g, y\}$, $Z = \{a, l, g, e, b, r\}$.

LHS: $Y \cup Z = \{a, b, e, g, l, n, r, y\}$. $X \cap (Y \cup Z) = \{a, e, i\} \cap \{a, b, e, g, l, n, r, y\} = \mathbf{\{a, e\}}$.
RHS: $X \cap Y = \{e\}$, $X \cap Z = \{a, e\}$. $(X \cap Y) \cup (X \cap Z) = \{e\} \cup \{a, e\} = \mathbf{\{a, e\}}$.
$\implies \text{LHS} = \text{RHS} = \{a, e\}$. (Verified)

3.2 Exercise 3.2 — Survey Word Problems & Inclusion-Exclusion

14 Complete Problems • Rich Diagrams

Question 1: Cardinality of Intersection

Given: $n(A) = 24, n(B) = 18, n(A \cup B) = 31$. Find $n(A \cap B)$.

Formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$$31 = 24 + 18 - n(A \cap B)$$ $$31 = 42 - n(A \cap B)$$ $$n(A \cap B) = 42 - 31 = 11$$

✓ Result: $n(A \cap B) = 11$

Question 2: Finding $n(B - A)$ and $n(B)$

Given: $n(A - B) = 23, n(A \cup B) = 44, n(A \cap B) = 2$.

Step 1: Finding $n(B - A)$
$$n(A \cup B) = n(A - B) + n(A \cap B) + n(B - A)$$ $$44 = 23 + 2 + n(B - A) \implies 44 = 25 + n(B - A)$$ $$n(B - A) = 44 - 25 = 19$$ Step 2: Finding $n(B)$
$$n(B) = n(A \cap B) + n(B - A) = 2 + 19 = 21$$

✓ Result: $n(B - A) = 19$, $n(B) = 21$

Question 3: Algebra & Geometry Students Survey

In a group of 30 Mathematics students, 20 like Algebra and 15 like both Geometry and Algebra. Show the data by Venn diagram and find how many students like Geometry.

Mathematical Formulation:
Let $A =$ Students who like Algebra, $G =$ Students who like Geometry.
Total $n(A \cup G) = 30$, $n(A) = 20$, $n(A \cap G) = 15$.

$$n(A \cup G) = n(A) + n(G) - n(A \cap G)$$ $$30 = 20 + n(G) - 15 \implies 30 = 5 + n(G)$$ $$n(G) = 30 - 5 = 25$$ Disjoint Regions:
• Algebra only $= n(A) - n(A \cap G) = 20 - 15 = 5$
• Both $= 15$
• Geometry only $= n(G) - n(A \cap G) = 25 - 15 = 10$

✓ Students who like Geometry $= 25$

Question 4: Street Houses Survey (Lawns & Porches)

In a street of 50 houses, 25 houses have lawns, 32 houses have car porches, and 15 houses have both lawn and car porch. Show the data by Venn diagram. Find how many houses have neither lawn nor porch.

Data: Total $n(U) = 50$, $n(L) = 25$, $n(P) = 32$, $n(L \cap P) = 15$.

Union:
$$n(L \cup P) = n(L) + n(P) - n(L \cap P) = 25 + 32 - 15 = 42$$ Neither Lawn nor Porch:
$$n((L \cup P)') = n(U) - n(L \cup P) = 50 - 42 = 8$$ Disjoint breakdown: Lawn only $= 25-15=10$, Both $= 15$, Porch only $= 32-15=17$, Neither $= 8$.

✓ Houses with neither lawn nor porch $= 8$

Question 5: Survey of 940 Children (Disjoint Categories)

Total children $n(U) = 940$. Primary level $= 400$, Elementary level $= 240$, Secondary level $= 175$. Since educational levels are mutually exclusive (disjoint): $$n(P \cup E \cup S) = 400 + 240 + 175 = 815$$ $$\text{Out of school children} = n(U) - n(P \cup E \cup S) = 940 - 815 = 125$$

✓ Result: $125$ children are out of school.

Question 6: ABC Dairy Ice Cream Poll

$n(M) = 100, n(V) = 90, n(C) = 40, n(M \cap V) = 20, n(C \cap V) = 14$ (with $M \cap C = \emptyset$).

• (i) Only Mango: $n(M) - n(M \cap V) = 100 - 20 = \mathbf{80}$
• (ii) Only Vanilla: $n(V) - n(M \cap V) - n(C \cap V) = 90 - 20 - 14 = \mathbf{56}$
• (iii) Only Chocolate: $n(C) - n(C \cap V) = 40 - 14 = \mathbf{26}$

Question 7: University 200 Students (Laptops, Cell Phones, iPods)

Total $n(U) = 200$. $n(L) = 42, n(C) = 80, n(I) = 100$. Pairwise: $n(L \cap C) = 23, n(L \cap I) = 10, n(C \cap I) = 14$. All three: $n(L \cap C \cap I) = 8$.

Partitioning all regions:
• All three $= 8$
• $L \cap C$ only $= 23 - 8 = 15$
• $L \cap I$ only $= 10 - 8 = 2$
• $C \cap I$ only $= 14 - 8 = 6$
• Laptop only $= 42 - (15 + 2 + 8) = 17$
• Cell phone only $= 80 - (15 + 6 + 8) = 51$
• iPod only $= 100 - (2 + 6 + 8) = 84$

Answers:
(a) Only cell phone: $\mathbf{51}$
(b) None of the three: Total union $= 17 + 51 + 84 + 15 + 2 + 6 + 8 = 183$. None $= 200 - 183 = \mathbf{17}$.
(c) iPod and Laptop but not Cellphone: $n(L \cap I) - n(L \cap C \cap I) = 10 - 8 = \mathbf{2}$.

Questions 8 to 14: Direct Algebraic Resolutions

Q8: Badminton & Table Tennis $n(B) = 350, n(T) = 280, n(B \cap T) = 150$.
Total students $= n(B \cup T) = 350 + 280 - 150 = \mathbf{480}$.

Q9: English & Chinese Learners $n(E \cup C) = 50, n(E \cap C) = 8, n(E) = 26$.
$50 = 26 + n(C) - 8 \implies n(C) = 50 - 18 = \mathbf{32}$.

Q10: Tea & Coffee Preferences $n(T \cup C) = 70, n(T) = 48, n(C) = 40$.
Both $= n(T \cap C) = 48 + 40 - 70 = \mathbf{18}$.

Q11: Science & Arts Groups (Disjoint) $n(S) = 46, n(A) = 50, S \cap A = \emptyset$.
Either group $= n(S \cup A) = 46 + 50 = \mathbf{96}$.

Q12: Arabic & French Speakers $n(A) = 52, n(F) = 112, n(A \cap F) = 12$.
• Arabic only $= 52 - 12 = \mathbf{40}$
• French only $= 112 - 12 = \mathbf{100}$
• Total group $= 40 + 12 + 100 = \mathbf{152}$.

Q13: Reading & Practical Activities $n(S) = 360, n(P) = 170, n(S \cap P) = 150$.
(i) Story books only $= 360 - 150 = \mathbf{210}$
(ii) Activities only $= 170 - 150 = \mathbf{20}$
(iii) Total students $= 210 + 150 + 20 = \mathbf{380}$.

Q14: 3 TV Channels Survey (60 People) $n(A)=25, n(B)=16, n(C)=13$, $n(A \cap B)=4, n(B \cap C)=7, n(A \cap C)=8$, $n(A \cap B \cap C)=3$.
• Watch at least one $= 25+16+13-(4+7+8)+3 = \mathbf{38}$
• Watch none $= 60 - 38 = \mathbf{22}$.

3.3 Exercise 3.3 — Cartesian Products & Ordered Pairs

8 Major Questions • Complete Mathematical Derivations

Question 1: Find Values of Unknowns from Equality of Ordered Pairs

Fundamental Axiom of Ordered Pairs: Two ordered pairs $(u, v)$ and $(w, z)$ are equal if and only if their corresponding coordinates are identically equal: $$(u, v) = (w, z) \iff u = w \quad \text{and} \quad v = z$$

Part (i): $(a, -b) = (7, 1)$ Equating first coordinates: $$a = 7$$ Equating second coordinates: $$-b = 1 \implies b = -1$$

Result: $a = 7, \quad b = -1$

Part (ii): $(2a, 2b+3) = (-10, -b)$ Equating first coordinates: $$2a = -10 \implies a = \frac{-10}{2} = -5$$ Equating second coordinates: $$2b + 3 = -b \implies 2b + b = -3 \implies 3b = -3 \implies b = -1$$

Result: $a = -5, \quad b = -1$

Part (iii): $(2a - 4, 6) = (8, -b + 1)$ Equating first coordinates: $$2a - 4 = 8 \implies 2a = 12 \implies a = 6$$ Equating second coordinates: $$6 = -b + 1 \implies b = 1 - 6 \implies b = -5$$

Result: $a = 6, \quad b = -5$

Part (iv): $(x + 2y, y - 3) = (2, 5)$ Equating second coordinates first: $$y - 3 = 5 \implies y = 5 + 3 = 8$$ Equating first coordinates and substituting $y = 8$: $$x + 2(8) = 2 \implies x + 16 = 2 \implies x = 2 - 16 = -14$$

Result: $x = -14, \quad y = 8$

Part (v): $(2x - y, y - 3x) = (4, 2)$ System of linear equations: $$\begin{aligned} 2x - y &= 4 \quad \text{--- (1)} \\ -3x + y &= 2 \quad \text{--- (2)} \end{aligned}$$ Adding equations (1) and (2) eliminates $y$: $$(2x - 3x) + (-y + y) = 4 + 2 \implies -x = 6 \implies x = -6$$ Substituting $x = -6$ into equation (1): $$2(-6) - y = 4 \implies -12 - y = 4 \implies y = -16$$

Result: $x = -6, \quad y = -16$

Part (vi): $(4x + 6y, x - 12y) = (6, -3)$ System of linear equations: $$\begin{aligned} 4x + 6y &= 6 \implies 2x + 3y = 3 \quad \text{--- (1)} \\ x - 12y &= -3 \implies x = 12y - 3 \quad \text{--- (2)} \end{aligned}$$ Substituting (2) into (1): $$2(12y - 3) + 3y = 3 \implies 24y - 6 + 3y = 3 \implies 27y = 9 \implies y = \frac{9}{27} = \frac{1}{3}$$ Substitute $y = \frac{1}{3}$ into (2): $$x = 12\left(\frac{1}{3}\right) - 3 = 4 - 3 = 1$$

Result: $x = 1, \quad y = \frac{1}{3}$

Part (vii): $(5x + y, -x + y) = (6, 1)$ System of linear equations: $$\begin{aligned} 5x + y &= 6 \quad \text{--- (1)} \\ -x + y &= 1 \quad \text{--- (2)} \end{aligned}$$ Subtracting (2) from (1) eliminates $y$: $$(5x - (-x)) + (y - y) = 6 - 1 \implies 6x = 5 \implies x = \frac{5}{6}$$ From equation (2): $$y = 1 + x = 1 + \frac{5}{6} = \frac{11}{6}$$

Result: $x = \frac{5}{6}, \quad y = \frac{11}{6}$

Question 2: Cartesian Products & Cardinality

Given: $A = \{1, 4, 8\}$ ($n(A) = 3$), $B = \{1, 0\}$ ($n(B) = 2$).

(i) $A \times B$:
$\{(1,1), (1,0), (4,1), (4,0), (8,1), (8,0)\}$ • Cardinality: $n(A \times B) = 3 \times 2 = \mathbf{6}$

(ii) $B \times A$:
$\{(1,1), (1,4), (1,8), (0,1), (0,4), (0,8)\}$ • Cardinality: $n(B \times A) = 2 \times 3 = \mathbf{6}$

(iii) $A \times A$:
$\{(1,1), (1,4), (1,8), (4,1), (4,4), (4,8), (8,1), (8,4), (8,8)\}$ • Cardinality: $n(A \times A) = 3 \times 3 = \mathbf{9}$

(iv) $B \times B$:
$\{(1,1), (1,0), (0,1), (0,0)\}$ • Cardinality: $n(B \times B) = 2 \times 2 = \mathbf{4}$

Question 4: Inverse Extraction of Sets from Product Set

Given: $L \times M = \{(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)\}$.

• Set $L$ (Set of First Coordinates / Projections):
$L = \{0, 1\}$ • $n(L) = 2$

• Set $M$ (Set of Second Coordinates / Projections):
$M = \{2, 3, 4\}$ • $n(M) = 3$

• Reversed Product $M \times L$:
$M \times L = \{(2, 0), (2, 1), (3, 0), (3, 1), (4, 0), (4, 1)\}$
Note: $L \times M \ne M \times L$ (Non-commutativity of Cartesian Product).

Question 3: Graphical Representation of Cartesian Products on Coordinate Grid

Given Sets: $E = \{1, 3\}$, $F = \{4, 6, 8\}$. Plot the Cartesian products on 2D coordinate planes:

(i) $E \times F = \{(1,4), (1,6), (1,8), (3,4), (3,6), (3,8)\}$ 6 Points plotted

(ii) $F \times E = \{(4,1), (4,3), (6,1), (6,3), (8,1), (8,3)\}$ 6 Points plotted

(iii) $E \times E = \{(1,1), (1,3), (3,1), (3,3)\}$ 4 Points plotted

(iv) $F \times F$ (9 Ordered Pairs) 9 Points plotted

Questions 5 to 8: Distributive Laws of Cartesian Product & Arrow Diagrams

Question 5: $A \times (B \cup C) = (A \times B) \cup (A \times C)$ Given: $A = \{1, 3, 5\}$, $B = \{2, 4\}$, $C = \{6, 7\}$.

LHS: $B \cup C = \{2, 4, 6, 7\}$
$A \times (B \cup C) = \{(1,2), (1,4), (1,6), (1,7), (3,2), (3,4), (3,6), (3,7), (5,2), (5,4), (5,6), (5,7)\}$ (12 elements)

RHS: $A \times B = \{(1,2), (1,4), (3,2), (3,4), (5,2), (5,4)\}$
$A \times C = \{(1,6), (1,7), (3,6), (3,7), (5,6), (5,7)\}$
$(A \times B) \cup (A \times C) = \text{Union of all 12 pairs above}$.

✓ LHS = RHS (Verified)

Question 6: $D \times (E \cap F) = (D \times E) \cap (D \times F)$ Given: $D = \{a, e, i\}$, $E = \{a, c\}$, $F = \{b, c\}$.

LHS: $E \cap F = \{c\}$
$D \times (E \cap F) = \{(a, c), (e, c), (i, c)\}$ (3 elements)

RHS: $D \times E = \{(a,a), (a,c), (e,a), (e,c), (i,a), (i,c)\}$
$D \times F = \{(a,b), (a,c), (e,b), (e,c), (i,b), (i,c)\}$
$(D \times E) \cap (D \times F) = \{(a, c), (e, c), (i, c)\}$.

✓ LHS = RHS (Verified)

Question 7: Distributivity of Cartesian Product over Difference Given: $A = \{1, 2\}$, $B = \{0, 1\}$, $C = \{0, 2, 4\}$.

Part (i): $A \times (B - C) = (A \times B) - (A \times C)$
LHS: $B - C = \{1\} \implies A \times (B - C) = \{(1, 1), (2, 1)\}$.
RHS: $A \times B = \{(1,0), (1,1), (2,0), (2,1)\}$, $A \times C = \{(1,0), (1,2), (1,4), (2,0), (2,2), (2,4)\}$.
$(A \times B) - (A \times C) = \{(1, 1), (2, 1)\}$. ✓ LHS = RHS

Part (ii): $(A - B) \times C = (A \times C) - (B \times C)$
LHS: $A - B = \{2\} \implies (A - B) \times C = \{(2, 0), (2, 2), (2, 4)\}$.
RHS: $A \times C = \{(1,0), (1,2), (1,4), (2,0), (2,2), (2,4)\}$, $B \times C = \{(0,0), (0,2), (0,4), (1,0), (1,2), (1,4)\}$.
$(A \times C) - (B \times C) = \{(2, 0), (2, 2), (2, 4)\}$. ✓ LHS = RHS

Question 8: Vector Arrow Diagrams for $X = \{0, 1, 2\}$ and $Y = \{-1, -2, -3\}$

(i) Arrow Diagram of $X \times Y$ ($9$ directed mappings)

(ii) Arrow Diagram of $Y \times X$ ($9$ directed mappings)

3.4 Exercise 3.4 — Binary Relations, Domain, Range & Inverses

9 Complete Questions • Vector Arrow Visualizations

Question 1: Total Number of Possible Binary Relations

Formula: Total Relations from $A$ to $B$ $= 2^{n(A \times B)} = 2^{n(A) \times n(B)}$.

(i) $A = \{1, 3\}$, $B = \{0, 2, 4\}$:
$n(A) = 2, n(B) = 3 \implies n(A \times B) = 2 \times 3 = 6$
$\text{Total relations} = 2^6 = \mathbf{64}$.

(ii) $n(C) = 7$ (Relations on $C$):
$n(C \times C) = 7 \times 7 = 49$
$\text{Total relations} = \mathbf{2^{49}}$.

(iii) $D = \{1, 3, 5\}$ (Relations on $D$):
$n(D) = 3 \implies n(D \times D) = 3 \times 3 = 9$
$\text{Total relations} = 2^9 = \mathbf{512}$.

Question 2: Complete Enumeration of All Binary Relations

(i) $A = \{\sqrt{2}, \sqrt{3}, \sqrt{5}\}, B = \{\sqrt[3]{5}\}$ ($2^3 = 8$ relations):
$R_1 = \emptyset$,
$R_2 = \{(\sqrt{2}, \sqrt[3]{5})\}$, $R_3 = \{(\sqrt{3}, \sqrt[3]{5})\}$, $R_4 = \{(\sqrt{5}, \sqrt[3]{5})\}$,
$R_5 = \{(\sqrt{2}, \sqrt[3]{5}), (\sqrt{3}, \sqrt[3]{5})\}$, $R_6 = \{(\sqrt{2}, \sqrt[3]{5}), (\sqrt{5}, \sqrt[3]{5})\}$,
$R_7 = \{(\sqrt{3}, \sqrt[3]{5}), (\sqrt{5}, \sqrt[3]{5})\}$,
$R_8 = A \times B$.

(ii) $C = \{a, b\}, D = \{x\}$ ($2^2 = 4$ relations):
$R_1 = \emptyset$, $R_2 = \{(a, x)\}$, $R_3 = \{(b, x)\}$, $R_4 = \{(a, x), (b, x)\}$.

(iii) $D = \{5\}, E = \{1, 10\}$ ($2^2 = 4$ relations):
$R_1 = \emptyset$, $R_2 = \{(5, 1)\}$, $R_3 = \{(5, 10)\}$, $R_4 = \{(5, 1), (5, 10)\}$.

Question 3: Successor Relation on $A = \{1, 2, 3, 4, 5, 6\}$

Condition: $R = \{(x, y) \mid y = x + 1, \quad x, y \in A\}$

• Tabular Form:
$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$
(For $x = 6$, $y = 7 \notin A$, so $(6, 7) \notin R$)

• Domain: $\text{Dom}(R) = \{1, 2, 3, 4, 5\}$
• Range: $\text{Range}(R) = \{2, 3, 4, 5, 6\}$
• Inverse: $R^{-1} = \{(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)\}$

Question 4: Binary Relations between $H$ and $G$

Given: $H = \{5, 6, 7, 8, 9\}$ and $G = \{5, 7, 9, 11\}$.

(i) $R_1 = \{(x, y) \mid x = y\}$:
$R_1 = \{(5, 5), (7, 7), (9, 9)\}$
$\text{Dom} = \{5, 7, 9\}, \quad \text{Range} = \{5, 7, 9\}$

(ii) $R_2 = \{(x, y) \mid x < y\}$:
$R_2 = \{(5,7), (5,9), (5,11), (6,7), (6,9), (6,11), (7,9), (7,11), (8,9), (8,11), (9,11)\}$

(iii) $R_3 = \{(x, y) \mid x > y\}$:
$R_3 = \{(6,5), (7,5), (8,5), (8,7), (9,5), (9,7)\}$

(iv) $R_4 = \{(x, y) \mid y - x = 2\}$:
$R_4 = \{(5, 7), (7, 9), (9, 11)\}$

Question 5: Factor Relation & Arrow Diagram

$A = \{2, 3, 4, 5\}$, $B = \{4, 6, 8, 10\}$, $R = \{(x, y) \mid x \text{ divides } y\}$

• $R = \{(2,4), (2,6), (2,8), (2,10), (3,6), (4,4), (4,8), (5,10)\}$
• $\text{Dom}(R) = \{2, 3, 4, 5\}$, $\text{Range}(R) = \{4, 6, 8, 10\}$
• $R^{-1} = \{(4,2), (6,2), (8,2), (10,2), (6,3), (4,4), (8,4), (10,5)\}$

Question 6: Linear Form $y = 2x + 1$ on $A = \{1,2,3,4,5,6,7,8\}$

$R = \{(x, y) \mid y = 2x + 1, \quad x, y \in A\}$

• $x = 1 \implies y = 3 \in A$
• $x = 2 \implies y = 5 \in A$
• $x = 3 \implies y = 7 \in A$
• $x \ge 4 \implies y \ge 9 \notin A$
Relation: $R = \{(1, 3), (2, 5), (3, 7)\}$
• $\text{Dom}(R) = \{1, 2, 3\}$, $\text{Range}(R) = \{3, 5, 7\}$
• $R^{-1} = \{(3, 1), (5, 2), (7, 3)\}$

Questions 7, 8 & 9: Inverse Relations and Symmetric Properties

Question 7: Constant Sum $x + y = 6$ on $X = \{1, 2, 3, 4, 5\}$ Pairs summing to 6:
$R = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$

• $\text{Dom}(R) = \{1, 2, 3, 4, 5\} = X$
• $\text{Range}(R) = \{1, 2, 3, 4, 5\} = X$
• $R^{-1} = \{(5, 1), (4, 2), (3, 3), (2, 4), (1, 5)\} = R$
$R = R^{-1}$ (Self-Inverse / Symmetric Relation)

Question 8: Strict Inequality $x < y$ on $A = \{1, 2, 3, 4\}$ Pairs where $x < y$:
$R = \{(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)\}$

• $\text{Dom}(R) = \{1, 2, 3\}$
• $\text{Range}(R) = \{2, 3, 4\}$
• $R^{-1} = \{(2,1), (3,1), (4,1), (3,2), (4,2), (4,3)\} = \{(x,y) \mid x > y\}$
• $\text{Dom}(R^{-1}) = \{2, 3, 4\} = \text{Range}(R)$
• $\text{Range}(R^{-1}) = \{1, 2, 3\} = \text{Dom}(R)$

Question 9: General Inverses & Domain-Range Interchange Given: $R = \{(1, 2), (2, 4), (3, 6), (4, 8)\}$.

• $R^{-1} = \{(2, 1), (4, 2), (6, 3), (8, 4)\}$
• $\text{Dom}(R) = \{1, 2, 3, 4\}, \quad \text{Range}(R) = \{2, 4, 6, 8\}$
• $\text{Dom}(R^{-1}) = \{2, 4, 6, 8\} = \text{Range}(R)$
• $\text{Range}(R^{-1}) = \{1, 2, 3, 4\} = \text{Dom}(R)$

✓ Fundamental Duality Identity Verified

3.M Miscellaneous Exercise 3 — Comprehensive Chapter Review Solutions

11 MCQs + 8 Theoretical & Applied Problems

Question 1: Multiple Choice Questions (With Rigorous Mathematical Rationale)

i. Set builder form of $A - B$ is:
(a) $\{x \mid x \in A\}$ (b) $\{x \mid x \in A \land x \notin B\}$ (c) $\{x \mid x \in A \land x \in B\}$ (d) $\{x \mid x \in B\}$
✓ Correct Option: (b) • Rationale: By definition of relative complement/set difference, $A - B$ consists of elements in $A$ that do not belong to $B$.

ii. If $A \cup B = A$ and $A \cap B = B$, then:
(a) $B \subseteq A$ (b) $A \not\supseteq B$ (c) $A \supseteq B$ (d) $A \ne B$
✓ Correct Option: (a) / (c) • Rationale: $A \cup B = A \iff B \subseteq A \iff A \supseteq B$.

iii. If $A \subseteq B$ then $A - B =$:
(a) $A$ (b) $B$ (c) $\emptyset$ (d) $B - A$
✓ Correct Option: (c) $\emptyset$ • Rationale: Every element of $A$ is in $B$, so no element remains in $A$ after removing elements of $B$.

iv. If $A - B = B - A = \emptyset$, then:
(a) $A = B$ (b) $B \subseteq A$ (c) $A \subseteq B$ (d) all a, b & c
✓ Correct Option: (d) all a, b & c • Rationale: $A - B = \emptyset \implies A \subseteq B$, and $B - A = \emptyset \implies B \subseteq A$, which together imply $A = B$.

v. Set of common elements of $A$ and $A'$ is ________ set:
(a) infinite (b) null (c) universal (d) singleton
✓ Correct Option: (b) null set • Rationale: By law of complementation, $A \cap A' = \emptyset$.

vi. If $A = \{1, 2, 3\}$ and $B = \{2, 3, 4\}$, then $A \Delta B$ is:
(a) $\{1, 4\}$ (b) $\{2, 3\}$ (c) $\{1, 2, 3, 4\}$ (d) $\emptyset$
✓ Correct Option: (a) $\{1, 4\}$ • Rationale: Symmetric difference $A \Delta B = (A - B) \cup (B - A) = \{1\} \cup \{4\} = \{1, 4\}$.

vii. If $n(A) = 3$ and $n(B) = 2$, then $n(A \times B)$ is:
(a) 5 (b) 6 (c) 8 (d) 9
✓ Correct Option: (b) 6 • Rationale: Cardinality product $n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6$.

viii. If $A$ is a set with $m$ elements, the total number of binary relations on $A$ is:
(a) $2^m$ (b) $m^2$ (c) $2^{m^2}$ (d) $2^{2m}$
✓ Correct Option: (c) $2^{m^2}$ • Rationale: $n(A \times A) = m \times m = m^2$, so power set has $2^{m^2}$ subsets.

ix. The domain of $R = \{(1, 2), (2, 3), (3, 4)\}$ is:
(a) $\{1, 2, 3\}$ (b) $\{2, 3, 4\}$ (c) $\{1, 2, 3, 4\}$ (d) $\{1, 4\}$
✓ Correct Option: (a) $\{1, 2, 3\}$ • Rationale: Domain is the set of all first coordinates: $\{1, 2, 3\}$.

x. The range of $R = \{(a, 1), (b, 2), (c, 3)\}$ is:
(a) $\{a, b, c\}$ (b) $\{1, 2, 3\}$ (c) $\{a, 1, b, 2, c, 3\}$ (d) $\{1, 3\}$
✓ Correct Option: (b) $\{1, 2, 3\}$ • Rationale: Range is the set of all second coordinates: $\{1, 2, 3\}$.

xi. If $R = \{(x, y) \mid y = 2x, x \in \{1, 2\}\}$, then $R^{-1}$ is:
(a) $\{(1, 2), (2, 4)\}$ (b) $\{(2, 1), (4, 2)\}$ (c) $\{(1, 1), (2, 2)\}$ (d) $\{(2, 4), (1, 2)\}$
✓ Correct Option: (b) $\{(2, 1), (4, 2)\}$ • Rationale: $R = \{(1, 2), (2, 4)\} \implies R^{-1} = \{(2, 1), (4, 2)\}$.

Question 2: Fundamental Set Theory & Relation Definitions

1. Null Set ($\emptyset$): A set containing no elements, denoted $\emptyset$ or $\{\}$. Cardinality $n(\emptyset) = 0$.

2. Disjoint Sets: Two sets $A$ and $B$ having no elements in common, i.e., $A \cap B = \emptyset$.

3. Overlapping Sets: Two sets having at least one common element, but neither is a subset of the other.

4. Universal Set ($U$): The super-set containing all objects under consideration in a particular context.

5. Power Set ($\mathcal{P}(A)$): The set of all subsets of $A$. If $n(A) = k$, then $n(\mathcal{P}(A)) = 2^k$.

6. Binary Relation: Any subset $R \subseteq A \times B$ of ordered pairs.

Question 3: Venn Diagram Shading for 3-Set Operations

Visual verification of algebraic set operations using shaded region geometry:

(i) $A \cup (B \cap C)$ Entire $A$ + region $B \cap C$

(ii) $A \cap (B \cup C)$ $(A \cap B) \cup (A \cap C)$

(iii) $(A \cup B) - C$ $(A \cup B)$ excluding region $C$

(iv) $(A \cap B)'$ Entire rectangle shaded EXCEPT lens $A \cap B$

Question 4: De Morgan's Laws Verification

$U = \{1, 2, 3, \dots, 10\}$, $A = \{2, 4, 6, 8, 10\}$, $B = \{1, 2, 3, 4, 5, 6\}$.

Part (i): $(A \cup B)' = A' \cap B'$
$A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$
$\text{LHS} = U - (A \cup B) = \{7, 9\}$
$A' = U - A = \{1, 3, 5, 7, 9\}$
$B' = U - B = \{7, 8, 9, 10\}$
$\text{RHS} = A' \cap B' = \{7, 9\}$
✓ LHS = RHS = $\{7, 9\}$

Part (ii): $(A \cap B)' = A' \cup B'$
$A \cap B = \{2, 4, 6\}$
$\text{LHS} = U - \{2, 4, 6\} = \{1, 3, 5, 7, 8, 9, 10\}$
$\text{RHS} = A' \cup B' = \{1, 3, 5, 7, 8, 9, 10\}$
✓ LHS = RHS (Verified)

Question 5: Distributive Laws Verification

$A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$, $C = \{3, 4, 5, 6\}$.

Part (i): $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
$B \cap C = \{4, 6\}$
$\text{LHS} = A \cup \{4, 6\} = \{1, 2, 3, 4, 6\}$
$A \cup B = \{1, 2, 3, 4, 6, 8\}$, $A \cup C = \{1, 2, 3, 4, 5, 6\}$
$\text{RHS} = (A \cup B) \cap (A \cup C) = \{1, 2, 3, 4, 6\}$
✓ LHS = RHS

Part (ii): $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
$B \cup C = \{2, 3, 4, 5, 6, 8\}$
$\text{LHS} = A \cap (B \cup C) = \{2, 3, 4\}$
$A \cap B = \{2, 4\}$, $A \cap C = \{3, 4\}$
$\text{RHS} = (A \cap B) \cup (A \cap C) = \{2, 3, 4\}$
✓ LHS = RHS

Questions 6 to 9: Applied Survey Demographics & Cartesian Relations

Question 6: Cartesian Relations for $P = \{a, b\}$ and $Q = \{1, 2, 3\}$ $P \times Q = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\}$ ($n = 6$).
Total possible relations $= 2^6 = 64$.
Four specific relations:
• $R_1 = \{(a, 1), (b, 2)\} \implies \text{Dom}=\{a, b\}, \text{Range}=\{1, 2\}$
• $R_2 = \{(a, 3)\} \implies \text{Dom}=\{a\}, \text{Range}=\{3\}$
• $R_3 = \{(a, 2), (b, 3)\} \implies \text{Dom}=\{a, b\}, \text{Range}=\{2, 3\}$
• $R_4 = P \times Q \implies \text{Dom}=P, \text{Range}=Q$

Question 7: Town Newspaper Readership Survey $n(U) = 100$, $n(E) = 65$, $n(U_r) = 45$, $n(E \cap U_r) = 25$.

(i) Read at least one newspaper:
$$n(E \cup U_r) = n(E) + n(U_r) - n(E \cap U_r) = 65 + 45 - 25 = \mathbf{85}$$ (ii) Read neither newspaper:
$$n((E \cup U_r)') = n(U) - n(E \cup U_r) = 100 - 85 = \mathbf{15}$$

Question 8: Qiraat Recitation Styles (10 Boys) $n(U) = 10$, $n(H) = 6$ (Hafs), $n(W) = 5$ (Warsh), $n(H \cap W) = 3$.

Recite in at least one style:
$$n(H \cup W) = 6 + 5 - 3 = 8$$ Recite in neither style:
$$n((H \cup W)') = 10 - 8 = \mathbf{2}$$

Question 9: 3-Set Comprehensive Examination (100 Students) $n(U)=100, n(M)=50, n(P)=40, n(C)=30$, $n(M \cap P)=20, n(P \cap C)=15, n(M \cap C)=10, n(M \cap P \cap C)=5$.

(i) Passed at least one subject:
$$n(M \cup P \cup C) = (50+40+30) - (20+15+10) + 5 = 120 - 45 + 5 = \mathbf{80}$$ (ii) Failed all three subjects:
$$n((M \cup P \cup C)') = 100 - 80 = \mathbf{20}$$ (iii) Passed Mathematics only:
$$n(M_{\text{only}}) = 50 - (15 + 5 + 5) = \mathbf{25}$$