Logical Reasoning

Logical reasoning tests your analytical thinking, pattern recognition, and problem-solving abilities. This section covers the most common types found in placement aptitude tests.

Syllogisms

What is a Syllogism?

A syllogism is a logical argument with two premises and a conclusion. You must determine if the conclusion follows from the premises.

Structure

Premise 1 (Major): All A are B.
Premise 2 (Minor): All B are C.
Conclusion: Therefore, All A are C. ✓

Types of Statements

Venn Diagram Method

Draw overlapping circles for each category and check if the conclusion is always true.

Example:

Premise 1: All dogs are animals.
Premise 2: All animals are living beings.
Conclusion: All dogs are living beings.

Draw: Dogs circle inside Animals circle, which is inside Living Beings circle. ✓

Rules for Syllogisms

1. “All A are B” does NOT mean “All B are A”
2. “Some A are B” does NOT mean “Some A are not B”
3. Two negative premises → No valid conclusion
4. Two particular premises (Some…) → No valid conclusion
5. If one premise is negative, conclusion must be negative

Quick Check Method

For each conclusion, draw the most restrictive Venn diagram consistent with the premises. If the conclusion holds in ALL possible diagrams, it’s valid.

Coding-Decoding

Letter Coding

Each letter is shifted by a fixed number in the alphabet.

Example: If COMPUTER = DPNQVUFS (each letter +1):

PROGRAM → QSPHSBN

Number Coding

Letters are assigned numbers: A=1, B=2, …, Z=26.

Example: If CAB = 3+1+2 = 6, then BAD = 2+1+4 = 7

Reverse Coding

A=Z, B=Y, C=X, … (mirror of alphabet)

A B C D E F G H I J K L M
Z Y X W V U T S R Q P O N

Example: HELLO → SVOOL

Pattern-Based Coding

Example: If TIGER = 20-9-7-5-18, then LION = 12-9-15-14

Conditional Coding

Rules change based on position, vowel/consonant, etc.

Example:

• Vowels → next letter
• Consonants → previous letter
• APPLE → BQOKF

Blood Relations

Common Relations

Solving Strategy

1. Draw a family tree
2. Use symbols: Male = □, Female = ○, Marriage = =
3. Start from the given relationship and work outward

Example: “A is the son of B. C is the daughter of B. D is the brother of A. How is C related to D?”

B ─── (spouse)
├── A (son) ── D (brother of A, so also B's child)
└── C (daughter)
C is D's sister.

Common Tricks

• “A is B’s son” → A is male, B is parent
• “A is B’s brother” → A is male, same parents as B
• “A is B’s wife” → A is female, married to B
• “Only son” → No other male siblings

Direction Sense

Cardinal Directions

        North
          |
West ----+---- East
          |
        South

Key Rules

1. Right turn from North → East
2. Left turn from North → West
3. Right turn from East → South
4. Left turn from East → North

Solving Steps

1. Draw the starting point
2. Follow each movement step by step
3. Track direction and distance
4. Calculate final position and distance from start

Example: “A walks 5 km North, turns right, walks 3 km, turns right, walks 5 km. How far from start?”

Start → 5 km N → turn right (E) → 3 km E → turn right (S) → 5 km S

Net: 0 km N/S, 3 km E
Distance = 3 km

Pythagorean Theorem for Distance

When the path forms a right triangle:

Distance = √(x² + y²)

Seating Arrangements

Linear Arrangement

People sitting in a row. Key clues:

• “A sits to the left of B” → A is on B’s left (from B’s perspective)
• “A is second from the left” → A is at position 2 from left end

Circular Arrangement

People sitting around a table. Key clues:

• “A sits opposite B” → directly across
• “A sits to the right of B” → clockwise from B (usually)

Solving Strategy

1. Fix one person’s position (reference point)
2. Use definite clues first
3. Fill in possibilities for ambiguous clues
4. Eliminate contradictions

Example: 5 people (A, B, C, D, E) sit in a row.

• C is at one end
• A sits next to C
• B does not sit next to D
• E sits in the middle

E is at position 3.
C is at position 1 or 5.
If C is at 1, A is at 2: C A E _ _
B and D fill 4 and 5. B not next to D is impossible (4 and 5 are adjacent).
So C is at 5, A is at 4: _ _ E A C
B and D fill 1 and 2. B not next to D → B at 1, D at 2 (or vice versa, they're adjacent)
Both work: B D E A C or D B E A C

Puzzles

Types of Puzzles

1. Ordering/Ranking — Who is tallest, who scored highest
2. Scheduling — Days, months, time slots
3. Floor/Building — Who lives on which floor
4. Comparison — More/less than relationships

Strategy

1. List all given information
2. Create a table/grid
3. Fill definite information first
4. Use process of elimination
5. Verify all constraints are met

Clock Problems

Key Facts

• Minute hand: 360° in 60 minutes → 6°/minute
• Hour hand: 360° in 12 hours → 0.5°/minute
• Relative speed: 5.5°/minute (minute hand gains on hour hand)

Angle Between Hands

Angle = |30H - 5.5M|

Where H = hours, M = minutes.

Example: Angle at 3:30:

= |30×3 - 5.5×30| = |90 - 165| = 75°

When Do Hands Coincide?

Hands coincide when:

M = 60H/11

Example: Between 3 and 4:

M = 60×3/11 = 180/11 = 16.36 minutes
At 3:16:22 (approximately)

Coincidence Times (12-hour clock)

Hands coincide 11 times in 12 hours (not 12, because between 11 and 12, they coincide at 12:00 which is counted with the next cycle).

Clock Gains/Loses

Problem: A clock gains 5 minutes every hour. It shows 12:00 noon. What is the actual time after 3 hours?

Solution:

In 3 real hours, clock advances: 3 × 65 = 195 minutes = 3 hours 15 minutes
Clock shows: 12:00 + 3:15 = 3:15 PM
Actual time: 3:00 PM

Calendar Problems

Key Facts

• Ordinary year: 365 days = 52 weeks + 1 odd day
• Leap year: 366 days = 52 weeks + 2 odd days
• Century years divisible by 400 are leap years (1600, 2000, 2400)
• Other century years are NOT leap years (1700, 1800, 1900)

Odd Days

Odd days = Total days mod 7

Finding the Day

Method — Odd Days. Count the total number of “odd days” (days beyond complete weeks) up to the target date, then add that count to the known day-of-week of the starting reference.

Reference: Jan 1, 1 AD (proleptic Gregorian) was a Monday.

Odd-days table:

(100 years = 24 leap + 76 ordinary → 24·2 + 76·1 = 124 → 124 mod 7 = 5. Every 400-year cycle is exactly 0 odd days, which is why the Gregorian calendar repeats every 400 years.)

Worked example: What day of the week was January 1, 2000?

Years completed before Jan 1, 2000: 1999
Split: 1999 = 1600 + 399

1600 years → 0 odd days  (multiple of 400)
399 years  → 399 + ⌊399/4⌋ − ⌊399/100⌋ + ⌊399/400⌋
            = 399 + 99 − 3 + 0
            = 495  →  495 mod 7 = 5  odd days

Total odd days = 0 + 5 = 5
Jan 1, 1 AD was Monday (day 1).
Monday + 5 = Saturday  ✓  (historically, Jan 1, 2000 was a Saturday)

Cross-check using the 400-year rule: 2000 is a multiple of 400, so Jan 1, 2000 has the same day-of-week as Jan 1, 2000 − 1600 = Jan 1, 400, which by the 400-year cycle is also a Saturday.

Example (leap year): January 1, 2024 was a Monday. What day is December 31, 2024?

2024 is a leap year → 366 days
Jan 1 to Dec 31 = 365 days after Jan 1
365 mod 7 = 1
Monday + 1 = Tuesday  →  Dec 31, 2024 = Tuesday

Series & Patterns

Number Series

Common patterns:

• Arithmetic: +3, +3, +3, …
• Geometric: ×2, ×2, ×2, …
• Alternating: +1, ×2, +1, ×2, …
• Squares/Cubes: 1, 4, 9, 16, 25, …
• Fibonacci: 1, 1, 2, 3, 5, 8, 13, …
• Prime: 2, 3, 5, 7, 11, 13, …

Example: Find the next term: 2, 6, 12, 20, 30, ?

Differences: 4, 6, 8, 10, ?
Second differences: 2, 2, 2, 2
Next difference: 12
Next term: 30 + 12 = 42

Pattern: n(n+1) → 1×2, 2×3, 3×4, 4×5, 5×6, 6×7 = 42

Letter Series

Example: A, C, F, J, O, ?

A(+2)C(+3)F(+4)J(+5)O(+6)U
Answer: U

Figure Series

Look for rotation, addition, deletion, or color change patterns.

Mirror & Water Images

Mirror Image

Left and right are reversed; top and bottom remain same.

Mirror of "AMBULANCE" → ƎƆИA⅃U8MA

For clock times:

Mirror of 3:00 → 9:00
Mirror of 4:30 → 7:30

Formula: Mirror of H:M = 11:60 − H:M (clock-arithmetic subtraction; equivalently (11 − H) mod 12 hours and (60 − M) minutes, treating the result modulo 12 hours).

Verification:

• 3:00 → (11−3):(60−0) = 8:60 = 9:00 ✓
• 4:30 → (11−4):(60−30) = 7:30 ✓
• 6:00 → (11−6):(60−0) = 5:60 = 6:00 ✓ (6:00 is symmetric)
• 12:00 → (11−0):(60−0) = 11:60 = 12:00 ✓ (12:00 is symmetric)
• 5:30 → (11−5):(60−30) = 6:30 ✓

⚠️ Common mistake: the formula (12 − H):(60 − M) is incorrect — for 4:30 it gives 8:30, but the correct mirror time is 7:30. Always use 11:60 − H:M (or equivalently subtract one extra hour to account for the hour hand’s continuous motion through the next hour).

Water Image

Top and bottom are reversed; left and right remain same.

Practice Questions

Q1: Syllogism

Premises: All cats are dogs. Some dogs are birds. Conclusion: Some cats are birds.

Solution:

Draw Venn: Cats inside Dogs. Dogs partially overlaps Birds.
The overlap of Dogs and Birds might not include the Cats part.
Conclusion does NOT follow.

Q2: Coding

If TIGER = UJHFS, then PROGRAM = ?

Solution:

Each letter is shifted forward by +1:
T(+1)=U, I(+1)=J, G(+1)=H, E(+1)=F, R(+1)=S  →  UJHFS  ✓

Apply the same rule to PROGRAM:
P→Q, R→S, O→P, G→H, R→S, A→B, M→N
PROGRAM = QSPHSBN

Why a constant shift works: the rule is a simple Caesar cipher with shift = +1. Each letter’s alphabet index (A=1, …, Z=26, wrapping) is incremented by the same amount. The same rule applied to any input word produces a deterministic, verifiable output.

Q3: Blood Relation

A’s mother is B’s daughter. C is B’s son. How is A related to C?

Solution:

B's daughter = A's mother
B's son = C
So A's mother and C are siblings.
A is C's nephew/niece.

Q4: Direction

A walks 3 km North, turns left, walks 4 km. How far from start?

Solution:

North 3 km, then left (West) 4 km
Distance = √(3²+4²) = √(9+16) = √25 = 5 km
Direction: North-West

Q5: Clock

What is the angle between hands at 4:20?

Solution:

= |30×4 - 5.5×20| = |120 - 110| = 10°

Q6: Calendar

If March 1 is Wednesday, what day is April 15?

Solution:

March has 31 days. March 1 to April 15 = 31 + 14 = 45 days
45 mod 7 = 3
Wednesday + 3 = Saturday

Q7: Series

Find the missing number: 3, 8, 18, 38, ?

Solution:

Pattern: ×2 + 2
3×2+2=8, 8×2+2=18, 18×2+2=38
Next: 38×2+2=78

Q8: Seating Arrangement

5 people sit in a row. A is not at either end. B sits to the right of C. D sits at one end. E sits next to A.

Solution:

A is at position 2, 3, or 4.
D is at position 1 or 5.
B is to the right of C (not necessarily adjacent).
E is next to A.

Let's try D at position 1: D _ _ _ _
A at 2: D A _ _ _ → E at 3 (next to A): D A E _ _
C must be left of B: C at 4, B at 5: D A E C B ✓
Or C at 5, B at 4: D A E B C → B not right of C ✗

Solution: D A E C B

Summary Table