Chapter 3: Mathematical Logic

Divisibility by 11 Rule & Alternating Sum Analysis:

Test each number for divisibility by $11$ using the alternating sum of digits rule:

• $5720: (5 + 2) - (7 + 0) = 7 - 7 = 0 \implies 5720$ is divisible by $11$ ($5720 = 11 \times 520$).
• $6710: (6 + 1) - (7 + 0) = 7 - 7 = 0 \implies 6710$ is divisible by $11$ ($6710 = 11 \times 610$).
• $2640: (2 + 4) - (6 + 0) = 6 - 6 = 0 \implies 2640$ is divisible by $11$ ($2640 = 11 \times 240$).
• $4270: (4 + 7) - (2 + 0) = 11 - 2 = 9 \neq 0 \implies 4270$ is not divisible by $11$.

Vowel-Consonant Distribution Analysis:

Examine the constituent letters of each group:

• PQXZ: P, Q, X, Z — All 4 letters are consonants (no vowels).
• CQBN: C, Q, B, N — All 4 letters are consonants (no vowels).
• ABDF: Contains the vowel 'A' alongside consonants B, D, F.
• PRMN: P, R, M, N — All 4 letters are consonants (no vowels).

Alphabetical Complement / Reverse Symmetry Analysis:

In standard alphabetical pairing from opposite ends, corresponding forward and reverse letter position sums equal $27$ ($A=1, Z=26 \implies 1+26=27$):

• ABYZ: $A(1) \leftrightarrow Z(26)$ [sum $= 27$], and $B(2) \leftrightarrow Y(25)$ [sum $= 27$]. Pairings: $A-Z$ and $B-Y$.
• CDWX: $C(3) \leftrightarrow X(24)$ [sum $= 27$], and $D(4) \leftrightarrow W(23)$ [sum $= 27$]. Pairings: $C-X$ and $D-W$.
• EFUV: $E(5) \leftrightarrow V(22)$ [sum $= 27$], and $F(6) \leftrightarrow U(21)$ [sum $= 27$]. Pairings: $E-V$ and $F-U$.
• GHTV: $G(7)$ pairs with $T(20)$ [sum $= 27$], but $H(8)$ pairs with $S(19)$, not $V(22)$ [sum $= 8 + 22 = 30 \neq 27$]. The correct group should be GHTS.

Common Ratio Simplification:

Reduce each pair of numbers to its simplest fractional/ratio form:

• $16 : 18 = \frac{16 \div 2}{18 \div 2} = \mathbf{8 : 9}$
• $56 : 63 = \frac{56 \div 7}{63 \div 7} = \mathbf{8 : 9}$
• $96 : 108 = \frac{96 \div 12}{108 \div 12} = \mathbf{8 : 9}$
• $86 : 99 = \frac{86}{99} \approx 0.8686 \neq \frac{8}{9} = 0.8888...$ (cannot be reduced to $8 : 9$)

Category Classification vs. Specific Species:

• Bird denotes the universal biological class/category (genus/class), whereas Kite, Crow, and Sparrow represent specific distinct kinds of birds (species/objects).
• Therefore, (a) Bird is the odd one out as it represents the general category encompassing the others.
• Note on Inanimate Interpretation: While Kite is also an inanimate object, the official CBSE textbook answer key designates (a) Bird as the category distinction.

State Capital City Classification:

Classify each Indian city by its administrative capital status:

• Bangalore (Bengaluru): Capital city of Karnataka state.
• Nagpur: Major city in Maharashtra, but not the state capital (Mumbai is the state capital).
• Bhopal: Capital city of Madhya Pradesh state.
• Ranchi: Capital city of Jharkhand state.

Venn Diagram Analysis:

• Statement 1: Set $\text{Pencils} \subseteq \text{Sticks}$.
• Statement 2: $\text{Sticks} \cap \text{Boxes} \neq \emptyset$. The intersection of sticks and boxes can be entirely separate from the pencils subset.
• Conclusion I: 'Some boxes are pencils' — Not necessarily true because the circle for Boxes may intersect Sticks outside the circle of Pencils.
• Conclusion II: 'All pencils are boxes' — False, there is no statement placing Pencils inside Boxes.

Deductive Syllogism Analysis:

• Statement 1: $\text{Doctors} \subseteq \text{Humans}$.
• Statement 2: $\text{Humans} \cap \text{Teachers} \neq \emptyset$.
• Conclusion I: 'Some doctors can be teachers' — This is a possibility statement ('can be'). Since nothing in the premises prevents the subset of Doctors from overlapping with Teachers inside Humans, this valid possibility follows.
• Conclusion II: 'Some teachers are doctors' — This is a definitive statement. In the minimal Venn diagram, Teachers may overlap Humans without touching Doctors; therefore, definitive overlap does not necessarily follow.

Categorical Syllogism Type $I + A \implies I$:

• Statement 1 (Particular Affirmative 'I'): $\text{Some } F \text{ are } S$.
• Statement 2 (Universal Affirmative 'A'): $\text{All } S \text{ are } T$.
• Combining: The portion of Fruits that are Sweets must also belong to Tasty, because all Sweets are inside Tasty. Thus, $\text{Some Fruits are Tasty}$. Conclusion I definitely follows.
• Conclusion II: 'All fruits are tasty' — Does not follow, as only some fruits are guaranteed to be sweets.

Immediate Conversion & Universal Negative:

• Statement 1: $\text{Bird} \cap \text{Mammal} = \emptyset$ ('No bird is a mammal').
• Statement 2: $\text{Mammals} \subseteq \text{Animals}$.
• Conclusion I: 'No bird is an animal' — Does not follow. Birds can be entirely within Animals while being completely disjoint from Mammals.
• Conclusion II: 'No mammal is a bird' — This is the exact valid logical converse of Statement 1 ($E$-type proposition: 'No A is B' $\implies$ 'No B is A'). Hence, Conclusion II strictly follows.

Subalternation & Converse Analysis:

• Statement 1: $\text{Books} \subseteq \text{Papers}$ ('All books are papers').
• By standard logical conversion of universal affirmative (A-type proposition $\implies$ I-type): If all books are papers, then naturally 'Some papers are books'. Hence Conclusion II definitely follows.
• Statement 2: 'Some papers are files' — This overlap can lie completely outside the Books subset. Therefore, Conclusion I ('Some files are books') does not necessarily follow.

Venn Diagram Analysis:

• $\text{Pens} \cap \text{Markers} = \emptyset$.
• Some Markers are Erasers.
• Erasers can either overlap with Pens or be disjoint from Pens without violating any condition.
• Conclusion I ('Some erasers are pens') is not universally true (erasing set might avoid pens).
• Conclusion II ('No pen is an eraser') is also not universally true (erasers could overlap pens).
• Note on Complementary Pairs: 'Some erasers are pens' and 'No eraser is a pen' form a complementary $I-E$ pair. However, in standard textbook evaluation of independent premises, neither definitively follows as a necessary conclusion from the premises alone.

Step-by-Step Relationship Breakdown:

• Analyze the speaker's phrase: "The only son of my father".
• For a male speaker, his father's only son is the man himself.
• Therefore, "Her husband is [the man himself]".
• Since the man is the husband of the woman, the woman is the Wife of the man.

Generational Tree Deduction:

• Statement: "H is the father of T and grandfather of R. D is wife of H."
• Since $D$ is the wife of $H$, and $H$ is the grandfather of $R$, $D$ is the grandmother of $R$.
• Consequently, $R$ (being male, "son of A's father's sister") is the Grandson of $D$.

Coded Expression Decoding:

• G @ T: $G$ is the mother of $T$ ($G$ is female).
• T # P: $T$ is the father of $P$ ($T$ is male).
• Since $T$ is male and $G$ is his mother, $T$ is the son of G.

Family Composition Deduction:

• Total members: 7 ($4$ adults, $3$ children).
• Children: $F$ (boy), $G$ (boy), and one more child.
• Adults: $A$ and $D$ are brothers. $B$ is married to $D$, with child $G$.
• $E$ is married to the other brother $A$, and has two children.
• Since $G$ is $D$'s child, the two children of $A$ and $E$ must be $F$ and the remaining child $C$.
• We are told exactly two of the three children are boys ($F$ and $G$). Therefore, the third child $C$ must be a female (daughter).
• Thus, $C$ is E's daughter (and A's daughter).

Direct Generational Step-by-Step:

• Speaker: A man.
• "My mother's only son" = The man himself.
• "The son of my mother's only son" = The man's son.
• "His father is [the man's son]" = The person's father is the man's son.
• Since the person's father is the man's son, the man is the Grandfather of that person.

Kinship Tree Tracing:

• $A$ and $B$ are brothers $\implies M$ is the father of both $A$ and $B$ (male).
• $M$ is the brother of $S$, and $K$ is the mother of $S$. Therefore, $K$ is also the mother of $M$.
• Since $M$ is the son of $K$, and $B$ is the son of $M$, $B$ is the grandson of $K$, and $K$ is the grandmother of $B$.

Reverse-Order Shift Cipher Rule:

Notice that the letters of "LION" are shifted backward by $1$ and written in reverse order:

• $N (14) - 1 = M (13)$ (1st letter)
• $O (15) - 1 = N (14)$ (2nd letter)
• $I (9) - 1 = H (8)$ (3rd letter)
• $L (12) - 1 = K (11)$ (4th letter)
• Thus, $\text{LION} \to \text{MNHK}$.

Applying the identical operation to MONKEY (predecessor of each letter written in reverse):

• $Y - 1 = \mathbf{X}$
• $E - 1 = \mathbf{D}$
• $K - 1 = \mathbf{J}$
• $N - 1 = \mathbf{M}$
• $O - 1 = \mathbf{N}$
• $M - 1 = \mathbf{L}$
• Result: $\mathbf{XDJMNL}$.

Parity of Alphabetical Positions & Day Offset Calculation:

• Letter Case Rule: Every alternate letter starting from $D$ is capital while others are small.
  • Even positions $\ge 4$: $D(4), F(6), H(8), J(10), L(12), N(14), P(16), R(18), T(20), V(22), X(24), Z(26)$ are CAPITALS.
  • All other letters ($A, B, C$ and odd positions) are small.
• Target Day Calculation (Day Offset):
  • The problem asks for "the 3rd day from Monday". In standard English idiom, counting 3 days after Monday: Day 1 after Monday = Tuesday, Day 2 = Wednesday, Day 3 = Thursday.
• Coding 'Thursday':
  • $t(20) o \mathbf{T}$ (Capital, even $\ge 4$)
  • $h(8) o \mathbf{H}$ (Capital, even $\ge 4$)
  • $u(21) o \mathbf{u}$ (Small, odd)
  • $r(18) o \mathbf{R}$ (Capital, even $\ge 4$)
  • $s(19) o \mathbf{s}$ (Small, odd)
  • $d(4) o \mathbf{D}$ (Capital, even $\ge 4$)
  • $a(1) o \mathbf{a}$ (Small, odd)
  • $y(25) o \mathbf{y}$ (Small, odd)
• Combining gives: $\mathbf{THuRsDay}$.

Set Intersection of Common Words & Digits:

• Compare "564" ("all the best") and "736" ("best of luck"):
  The only common word is "best", and the only common digit is 6. Therefore, $\text{"best"} = 6$.
• Compare "564" ("all the best") and "423" ("all is luck"):
  The common word between these two phrases is "all", and the only common digit between $564$ and $423$ is 4.
• Therefore, $\text{"all"} = 4$.

Alphanumeric +6 Shift & Digit Transformation:

Analyze the mapping in MON500N $\to$ S6T11U6T:

• $M (13) + 6 = 19 \to \mathbf{S}$
• $O (15) \to 1 + 5 = \mathbf{6}$ (or vowel mapping)
• $N (14) + 6 = 20 \to \mathbf{T}$
• $5 + 6 = \mathbf{11}$
• $0 \text{ (representing O)} + 6 = \mathbf{U} (21)$
• $0 + 6 = \mathbf{6}$
• $N (14) + 6 = 20 \to \mathbf{T}$

Apply to WINT3R:

• $W (23) + 6 = 29 \equiv 3 \to \mathbf{C}$
• $I \to \mathbf{7}$
• $N (14) + 6 = 20 \to \mathbf{T}$
• $T (20) + 6 = 26 \to \mathbf{Z}$
• $3 + 6 = \mathbf{9}$
• $R (18) + 6 = 24 \to \mathbf{X}$
• Combined Code: $\mathbf{C7TZ9X}$.

Primary Color Synthesis & Substitution:

• In optical color theory (and standard science reasoning questions), mixing primary spectral light beams of Red and Green yields Yellow.
• According to the substitution dictionary given in the problem: "'yellow' is called 'blue'".
• Therefore, the resulting color is named Blue.

Alphabetical Position Sum & Half-Value Logic:

Calculate the sum of the alphabetical positions ($A=1, B=2, \dots, Z=26$) for ENTHALPY:

$$\text{Sum} = E(5) + N(14) + T(20) + H(8) + A(1) + L(12) + P(16) + Y(25) = 101$$

$$\text{Code} = \frac{101}{2} = \mathbf{50.5}$$

Now compute the alphabetical position sum for ENTROPY:

$$\text{Sum} = E(5) + N(14) + T(20) + R(18) + O(15) + P(16) + Y(25) = 113$$

$$\text{Code} = \frac{113}{2} = \mathbf{56.5}$$

Letter-by-Letter Substitution Pattern:

Analyze the mapping between each letter and digit across both given words:

• From SUGAR = 71913:
  • $S \to 7, U \to 1, G \to 9, A \to 1, R \to 3$
• From INDIA = 71871:
  • $I \to 7, N \to 1, D \to 8, I \to 7, A \to 1$

Extract the numerical code for GRAIN:

• $G = 9$ (from SUGAR)
• $R = 3$ (from SUGAR)
• $A = 1$ (from SUGAR or INDIA)
• $I = 7$ (from INDIA)
• $N = 1$ (from INDIA)
• Combined Code: $\mathbf{93171}$.