Top 5 Reasoning Topics that Always Appear in RRB Clerk Exam
Top 5 Reasoning Topics that Always Appear in RRB Clerk Exam
Home » IBPS RRB Clerk » Top 5 Reasoning Topics that Always Appear in RRB Clerk Exam
The Reasoning Ability section in the IBPS RRB Clerk exam is truly a score booster if approached well. With 40 questions in Prelims and 40 in Mains, it tests not just your logical thinking but also your ability to solve under time pressure. While the exact difficulty level may vary, certain topics always appear in the exam year after year. By mastering these high‑weightage areas, you can secure a strong base score and maximize your chances of clearing both sectional and overall cutoffs. Let’s dive into the Top 5 Reasoning Topics that you must prepare thoroughly.
Most Important Reasoning Topics for IBPS RRB Clerk Exam
The Reasoning Ability section of the IBPS RRB Clerk exam consistently revolves around a few high‑weightage topics that aspirants must master to maximize their score. Among these, Puzzles & Seating Arrangement dominate the paper, followed by Syllogism, Inequality, Blood Relation, and Direction Sense. These five areas together account for nearly 70–75% of the reasoning section, making them the most reliable scoring opportunities. By prioritizing these topics in practice sessions, candidates can build both speed and accuracy, ensuring they clear the sectional cutoff with confidence.
Top 5 Reasoning Topics & Their Weightage in IBPS RRB Clerk Exam
There are a total of 40 questions of 40 marks in Reasoning Section of IBPS RRB Clerk Prelims. And in mains, there are 40 questions of 50 marks. In the Prelims exam, the questions revolve around the 5 topics given below. Check expected weightage based on previous year trends and their importance level in the exam. Mastering these topics may not fully ensure it, but it will most probably help you to clear the cut-offs.
Reasoning Topic
Expected Weightage (Prelims)
Importance Level
Puzzles & Seating Arrangement
15–20 marks
Very High
Syllogism
4–5 marks
High
Inequality
4–5 marks
High
Blood Relation
2–3 marks
Moderate
Direction Sense
2–3 marks
Moderate
1️. Puzzles & Seating Arrangement
Puzzles and seating arrangement questions form the backbone of the reasoning section, often carrying the highest weightage. They test your ability to organize information systematically, whether it’s arranging people in a line, around a circle, across floors, or within boxes. The key to solving them quickly is to start with definite clues, use structured tables or diagrams, and avoid solving mentally. In Prelims, aim to attempt the simpler puzzles first and skip overly complex ones if they consume more than four minutes, while in Mains, practice multi‑variable puzzles to build stamina and accuracy.
Why it’s important:
Accounts for 15–20 marks in Prelims and even more in Mains.
Covers linear, circular, floor‑based, box‑based, and variable puzzles.
Shortcut Strategy:
Always start with definite clues (e.g., “A sits third to the left of B”).
Use tables or diagrams instead of solving mentally.
Skip overly complex puzzles in Prelims if they consume more than 4 minutes.
Example: Seven people A, B, C, D, E, F, and G are sitting in a straight line facing North. The following information is known:
C sits third to the left of E.
A sits second to the right of C.
D sits at one of the extreme ends.
B sits immediately to the left of G.
F is not an immediate neighbor of E.
Exactly one person sits between A and B.
Find the correct seating order from left to right.
Step-by-step solution
From C and E: C is third to the left of E, so E = C + 3.
From A and C: A is second to the right of C, so A = C + 2.
Direct inference: E = A + 1, so E sits immediately to the right of A.
This means the pattern “C — A — E” occupies consecutive positions with gaps of +2 and +3 from C.
Try valid anchors for C
C can be at positions 1–4 (so that C+3 stays within 7). Test C = 1.
If C = 1 → A = 3 → E = 4. This fits the line length.
Place D at an extreme
D must be at 1 or 7. Since 1 is occupied by C, D = 7.
Place B and G together
B is immediately left of G and exactly one person sits between A and B.
With A at 3, B can be at 1 or 5.
B ≠ 1 (C already at 1), so B = 5 → G = 6.
Place F with the restriction near E
F is not an immediate neighbor of E (neighbors of E at 4 are 3 and 5, which are A and B).
Available positions left: 2 (free) and 7 (D already).
F = 2 satisfies all constraints.
Final arrangement (left to right)
Position
1
2
3
4
5
6
7
Person
C
F
A
E
B
G
D
Direct answer: C, F, A, E, B, G, D.
Quick tips for puzzles
Eliminate neighbors: Non-neighbor clauses (F not near E) help finalize leftover positions.
Anchor chains first: Lock “C–A–E” together before placing others.
Use extremes early: If someone must sit at an end (like D), place them once conflicts appear.
Pair constraints: Combine “B left of G” with “one person between A and B” to narrow options fast.
2️. Syllogism
Syllogism questions check your logical reasoning skills by asking you to derive valid conclusions from given statements. They usually involve terms like “All,” “Some,” “No,” and “Only a few,” which can be tricky if misread. The fastest way to solve them is by using Venn diagrams or memorizing standard conclusion rules. Pay special attention to “Possibility” and “Either/Or” cases, as these are common traps in the exam. With consistent practice, syllogism becomes one of the most scoring areas, often solvable in under 30 seconds per question.
Why it’s important:
Consistently contributes 4–5 questions.
Tests your ability to interpret statements and draw logical conclusions.
Shortcut Strategy:
Use Venn diagrams for clarity.
Memorize standard rules for “All,” “Some,” “No,” and “Only a few.”
Practice “Possibility” and “Either/Neither” cases, as they are common traps.
Example: Example 1: Standard set (All / Some / No)
Statements
S1: All pens are blue.
S2: Some blues are markers.
S3: No marker is metal.
Conclusions to test
C1: Some pens are markers.
C2: No pen is metal.
C3: Some blues are not metal.
C4: Some metals are not pens.
Step-by-step Solution
From S1 (All pens ⊆ blue): Pens lie entirely within blue.
From S2 (Some blue ∩ markers): There is at least one common area between blue and markers. Pens may or may not overlap with that specific “some” part—no compulsion.
From S3 (markers ∩ metal = ∅): Markers are completely outside metal; hence all markers are non-metal.
Evaluate each conclusion:
C1: Some pens are markers. Not forced. The “some blue are markers” could be a blue region disjoint from pens. So C1 is not necessarily true (it’s a possibility, not a must).
C2: No pen is metal. We only know markers are not metal. Pens could be blue areas outside markers, and blue could overlap metal unless restricted. No statement forbids blue-metal overlap. So C2 is not necessarily true.
C3: Some blues are not metal. Since all markers are not metal (S3) and some blues are markers (S2), those “some blues” (which are markers) are certainly not metal. So C3 is definitely true.
C4: Some metals are not pens. Pens are inside blue. It’s possible metal lies entirely outside blue, or partly overlaps blue. Even if metal overlapped blue, it might still sit outside pens. However, none of the statements force metal to exist or to be outside pens. Without existence or placement guarantees, C4 is not necessarily true.
Direct answer
Only Conclusion C3 follows.
Example 2: Possibility-based set
Statements
S1: All books are papers.
S2: Some papers are files.
S3: No file is plastic.
Conclusions to test
C1: Some books are files.
C2: Some papers are not plastic.
C3 (Possibility): It is possible that all books are not plastic.
C4: No book is plastic.
Step-by-step Solution
From S1: Books lie inside papers.
From S2: Papers overlap with files in at least one region.
From S3: Files are entirely outside plastic.
Evaluate each:
C1: Some books are files. Not forced. The “some papers are files” portion might lie outside the books subset of papers. So C1 is not necessarily true.
C2: Some papers are not plastic. Since “some papers are files” and no file is plastic, those papers (that are files) are definitely not plastic. So C2 is definitely true.
C3 (Possibility): All books are not plastic. Could this be arranged? Yes. Place the “books” subset inside the “papers” region that is disjoint from plastic (e.g., within the “files” or any other non-plastic paper area). The statements don’t force any book into plastic. So C3 is true as a possibility.
C4: No book is plastic. This is stronger than C3. While it’s possible, it’s not compelled—books could be in a papers region that overlaps plastic. So C4 is not necessarily true.
Direct answer
Conclusion C2 follows.
Conclusion C3 (possibility) is valid.
C1 and C4 do not follow.
Quick rules for Syllogism
Possibility questions (“can be”): Check if a valid Venn arrangement exists without violating any statement, even if the conclusion isn’t necessary.
All A are B: A lies completely inside B. Doesn’t imply “Some A are B” unless A’s existence is certain; IBPS generally assumes class existence unless contradicted.
Some A are B: A and B overlap in at least one element; never forces neighbors like “Some A are not B.”
No A is B: A and B are disjoint; any subset of A is also disjoint from B.
3️. Inequality
Inequality questions are quick and straightforward, making them a must‑attempt in the exam. They involve comparing variables using symbols like >, <, ≥, ≤, and =, or decoding symbol‑based inequalities. The trick is to combine statements step by step and check whether the given conclusions follow logically. Always watch out for “Either/Or” cases, which appear frequently. Since these questions can be solved in seconds with accuracy, they are excellent confidence boosters at the start of the reasoning section.
Why it’s important:
Usually 5 questions appear in Prelims.
Quick and scoring if practiced well.
Shortcut Strategy:
Learn symbol combinations (>, <, ≥, ≤, =) thoroughly.
For coded inequalities, decode the symbols first, then apply logic.
Watch out for “Either/Or” cases.
Example:
Direct comparison set (multiple statements, conclude relation)
Example 1
Statements:
(A > B)
(B >= C)
(C < D)
(E <= A)
Question: Find the relation between (A) and (D).
Solution:
Chain build: From (B >= C) and (C < D), we get (B > D) or (B >= D).
Propagate to (A): Since (A > B), it must be (A > D).
Direct answer: (A > D).
Example 2
Statements:
(P >= Q)
(Q > R)
(R >= S)
(S > T)
Question: Determine the relation between (P) and (T).
Solution:
Upward chain: (Q > R >= S > T -> Q > T).
Include (P): (P >= Q) and (Q > T -> P > T) or (P >= T). Since (Q > T), minimum (P) is equal to (Q), so (P >= T) with strictness not guaranteed.
Direct answer: (P >= T).
Conclusion-based set (I, II type)
Example 3
Statements:
(X >= Y)
(Y > Z)
(Z <= W)
Conclusions:
I: (X > Z)
II: (W < X)
Solution:
For I: (X >= Y > Z -> X > Z). Conclusion I is true.
For II: From (Z <= W) and (Y > Z), we only know (W) is at least (Z). With (X >= Y), we cannot force (W < X) because (W) could be large (e.g., (W = X) or (W > X)). Conclusion II is not necessarily true.
Direct answer: Only Conclusion I follows.
Example 4
Statements:
(M < N)
(N <= O)
(O > P)
Conclusions:
I: (M < O)
II: (P < N)
Solution:
For I: (M < N \le O -> M < O). True.
For II: (O > P) and (N <= O -> N) could be equal to (O) or less, but we still get (O > P & N >= P). This does not guarantee (P < N) strictly (if (N = P), II fails). So II is not necessarily true.
Direct answer: Only Conclusion I follows.
Coded inequality set (symbol decoding)
Example 5
Code rules:
(A @ B -> A > B)
(A # B -> A \ge B)
(A $ B -> A < B)
(A % B -> A \le B)
Given: (H @ I,, I % J,, J # K).
Question: Find the relation between (H) and (K).
Solution:
Decode chain: (H > I,, I <= J,, J >= K).
Consolidate middle: From (I <= J) and (J >= K), we get (I <= J) and (K <= J), but no fixed relation between (I) and (K).
Propagate from (H > I): Two minimal cases:
If (I = J) and (J = K), then (H > K).
If (I < J) and (K < J), (K) could be below or equal to (I). Still, (H > I) and if (K <= I), (H > K). If (K = J) and (I < J), we still get (H > I) but cannot force (H > K) unless we fix (K) relative to (I).
Check counterexample for strictness: Let (I = 5, J = 6, K = 6, H = 7) gives (H > K). Try to break strictness: Let (I = 5, J = 6, K = 7) violates (J \ge K). Let (I = 5, J = 5, K = 4, H = 6) gives (H > K). Because (K <= J) and (I <= J) with (H > I), it’s not possible to make (K > H) or (K = H) under these constraints. Thus (H > K) holds.
Direct answer: (H > K).
Quick tipsfor Inequality
Coded sets: Decode first, then apply the same chaining logic; watch for equality traps.
Chain carefully: Convert all statements to a linear chain; push strict “>” wherever possible and keep “≥” when strictness isn’t guaranteed.
Test edge cases: For “≥” and “≤”, check equality scenarios to avoid over-asserting “>” or “<”.
Use contradiction checks: If you suspect a relation, try constructing a counterexample. If none fits the constraints, the relation likely holds.
4. Blood Relation
Blood relation questions test your ability to decode family connections, either directly or through coded statements. They require careful attention to generational differences (parent, child, grandparent) and gender identification. The best approach is to draw a family tree or diagram instead of trying to solve mentally, which reduces errors. In coded relations, always identify the reference person first before mapping others. With practice, these questions become easy marks, usually taking less than a minute to solve.
Why it’s important:
Appears in both Prelims and Mains (2–3 questions).
Tests clarity of family tree logic.
Shortcut Strategy:
Always identify the reference person first.
Draw a family tree for coded relations.
Remember: Generations differ by ±25 years (approx.) to avoid confusion.
Example: Example 1: Mixed generational puzzle
Eight family members: A, B, C, D, E, F, G, H. The following is known:
Fact 1: A is the mother of C.
Fact 2: B is the father of D.
Fact 3: C is the sister of D.
Fact 4: E is the wife of B.
Fact 5: F is the only son of C.
Fact 6: G is the brother of A.
Fact 7: H is the mother-in-law of E.
Questions:
What is the relation of H with B?
How is G related to F?
How is E related to F?
Step-by-step solution
Build the core parents of D and C:
From Facts 1 and 3: A is mother of C; C is sister of D, so A must also be mother of D or at least in the same parent set with D.
From Fact 2 and 4: B is father of D; E is wife of B → E is mother of D.
Align parents for C and D:
Since C and D are siblings (Fact 3), they share parents. We already have A as mother of C, and E as mother of D. To keep siblings consistent, A = E (contradiction) or A is the mother and E is the step-mother? Avoid contradictions by reading carefully:
Fact 4 says E is wife of B (present generation). Fact 2 says B is father of D. Fact 1 says A is mother of C. The consistent way is: A and B are parents of C and D, and E is wife of B (implies A is not B’s wife). That creates conflict unless E and A are the same person. But names differ, so we need another interpretation:
Resolve using Fact 7: H is mother-in-law of E → E is married to B, so H is B’s mother.
Therefore, E is B’s wife and H is B’s mother. A must be the mother of C, but C and D are siblings. The only way: A is also mother of D, and A is B’s wife. That contradicts E being B’s wife. Hence, we deduce C and D are siblings from one parent only? IBPS never does that.
Fix the overlap using timing/generation:
The puzzle’s standard resolution is: E is wife of B; H is mother-in-law of E → H is B’s mother.
C is sister of D and A is mother of C → A is mother of C (and D) too.
Therefore A must be B’s wife to be mother of D, but E is given as B’s wife. This is inconsistent unless E is B’s current wife and A is former spouse, which IBPS typically avoids.
Reframe with minimal conflict: Replace Fact 4 to a workable exam variant:
Use: “E is the sister-in-law of A” instead of “E is the wife of B.” Here’s a corrected, consistent set:
Revised statements:
A is the mother of C.
B is the father of D.
C is the sister of D.
E is the sister-in-law of A.
F is the only son of C.
G is the brother of A.
H is the mother of B.
Now solve:
Parents of C and D: A is mother of C; C is sister of D → A is also mother of D. With B father of D, B is father of C as well. So A (mother) + B (father) are parents of C and D.
H as B’s mother: H is B’s mother → H is grandmother of C and D, and great-grandmother of F.
E as sister-in-law of A: A’s sister-in-law can be the wife of A’s brother (G) or the sister of A’s husband (B). Without specifics, both are valid options.
F as only son of C: F is male child, grandson of A and B.
G as brother of A: G is maternal uncle of C and D, grand-uncle of F.
Answer the questions:
Relation of H with B: H is B’s mother → Mother.
Relation of G with F: G is maternal uncle of C and D, so to F (child of C), G is grand-uncle (mother’s uncle).
Relation of E with F: If E is wife of G, E is grand-aunt to F; if E is B’s sister, E is great-aunt to F. Since “sister-in-law of A” has two valid paths, the safest exam answer is aunt (elder generation, not direct). If the question expects a single answer, additional info is usually provided.
Direct answers:
Mother.
Grand-uncle (mother’s uncle).
Aunt (grand-aunt/great-aunt depending on which sister-in-law path applies).
Example 2: Point-based relation identification
Statements:
Fact 1: P is the son of Q.
Fact 2: Q is the daughter of R.
Fact 3: R is married to S.
Fact 4: T is the brother of P.
Fact 5: U is the mother of T.
Questions:
How is U related to Q?
How is S related to T?
If V is the wife of T, how is V related to R?
Step-by-step solution
Build generations:
P (male) is child of Q.
Q (female) is child of R. So R is grandparent of P.
R married S → S is the other grandparent of P.
T (male) is brother of P → T is also child of Q.
U is mother of T → U is mother of P too.
Resolve parental roles:
Since Q is female and parent of P and T, and U is the mother of T, U must be Q herself (single consistent mother). Therefore, U = Q.
Answer the questions:
U related to Q: U = Q → Self (or “U and Q are the same person”).
S related to T: S is spouse of R (grandparent of P/T) → Grandmother or Grandfather depending on S’s gender. Since gender of S not given, use Grandparent.
V related to R: V is wife of T (grandchild of R) → Granddaughter-in-law to R.
Direct answers:
Self (U is Q).
Grandparent.
Granddaughter-in-law.
Quick exam tips
Beware sister-in-law cases: Sister-in-law means the wife of your brother or sister, or the sister of your husband or wife, or the wife of the brother or sister of your husband or wife; check which path the puzzle implies.
Anchor parents first: Identify parent-child links before siblings; it prevents loops and contradictions.
Track genders explicitly: Son/daughter/brother/sister instantly restrict directions of relations.
Use concise trees: Draw three tiers (grandparents → parents → children) and drop each name where it fits.
5. Direction Sense
Direction sense questions assess your ability to track movements and orientations on a plane. Candidates are asked to determine final positions, distances, or directions after a series of moves. The most effective strategy is to sketch rough diagrams, mark the four cardinal directions (N, S, E, W), and carefully follow each step. For left/right turns, always visualize the person’s current facing direction before applying the turn. These questions are generally straightforward and can be solved quickly with accuracy if you rely on diagrams rather than mental calculation.
Why it’s important:
2–3 questions in Prelims, sometimes clubbed with puzzles.
Easy marks if solved with diagrams.
Shortcut Strategy:
Always mark N, S, E, W on rough sheet.
Track each movement step by step.
For left/right turns, visualize facing direction first.
Example: Example 1: Final direction and shortest distance
Aman starts from point P, walks 6 km east, then turns left and walks 4 km. He turns left again and walks 2 km, then turns right and walks 3 km. Finally, he turns right and walks 5 km to reach point Q.
Questions:
In which direction is Q with respect to P?
What is the shortest distance between P and Q?
Step-by-step solution
Layout path:
Start: P at origin; face unspecified, but first move is absolute: 6 km east.
Turn 1 (left from east): Now facing north; move 4 km.
Turn 2 (left from north): Now facing west; move 2 km.
Turn 3 (right from west): Now facing north; move 3 km.
Turn 4 (right from north): Now facing east; move 5 km → reach Q.
Net displacement:
East–West: (6 – 2 + 5 = 9) km east.
[ 6 – 2 + 5 = 9 ]
North–South: (4 + 3 = 7) km north.
[ 4 + 3 = 7 ]
Answers:
Direction of Q from P: North-East (east-positive, north-positive).
Shortest distance (PQ): Use Pythagoras: (\sqrt{9^2 + 7^2} = \sqrt{81 + 49} = \sqrt{130} \approx 11.40) km.
Example 2: Relative position after multiple turns
Neha starts from point A, walks 8 km south, turns right and walks 6 km, turns right and walks 10 km, turns left and walks 4 km, and finally turns left and walks 6 km to reach point B.
Questions:
In which direction is B with respect to A?
What is the shortest distance between A and B?
Step-by-step solution
Trace moves:
Start: From A, 8 km south.
Turn 1 (right from south): Face west; 6 km.
Turn 2 (right from west): Face north; 10 km.
Turn 3 (left from north): Face west; 4 km.
Turn 4 (left from west): Face south; 6 km → reach B.
Net displacement:
North–South: (10 – 8 – 6 = -4) km → 4 km south overall.
Avoid drawing delays: A quick T-chart of x and y updates is enough under time pressure.
Keep axes mental: East–West on the x-axis, North–South on the y-axis. Update signs after each turn.
Turn logic anchors: From East: left→North, right→South. From North: left→West, right→East. From West: left→South, right→North. From South: left→East, right→West.
Compress sums early: Combine total East–West and North–South separately, then apply Pythagoras only once.
Sanity checks: If both nets are positive, answer will be North-East; positive x and negative y → South-East; etc.
You should prepare the complete syllabus; just focus a bit more on these topics, as these have been asked repeatedly.
Conclusion
The Reasoning section in RRB Clerk is not about solving everything, it’s about solving the right questions quickly. By focusing on these Top 5 topics includes Puzzles, Syllogism, Inequality, Blood Relation, and Direction Sense, you can secure 30+ marks with accuracy. The key is consistent practice, error analysis, and smart time management. Remember: reasoning is logic‑based, not memory‑based, so clarity of thought is your biggest weapon.
Know What Our Successful Aspirant Says…
Success Story of Rajneet Kumar Yadav who cleared IBPS RRB Clerk Exam
Our Banking Preparation Package includes topic tests, sectional tests, rank boosters for prelims, previous year paper tests, e-books, CA tests, Quizzes, live tests, PDF Course, and more. Discover our banking and insurance packages in detail from the link provided below!
Join our exclusive Telegram group, where our experts are ready to answer all your queries, guide you in banking exam preparation, and give personalised tips to boost your success. Get access to real-time solutions, expert advice, and valuable resources to improve your study journey.
Yes, puzzles and seating arrangements are the backbone of the reasoning section, contributing the highest weightage.
Can I clear the cutoff by skipping puzzles?
In Prelims, it’s possible if you maximize accuracy in other topics, but in Mains, puzzles are unavoidable.
Which topic should I attempt first in Reasoning?
Start with quick‑scoring topics like inequality, syllogism, and blood relation before moving to puzzles.
How much time should I spend on one puzzle in Prelims?
Ideally, no more than 4 minutes. If stuck, skip and revisit later.
Sandhya
Hi, I'm Sandhya Sadhvi (B.E. in ECE from GTU 2017-2021). Over the years, I've been a dedicated government job aspirant, having attempted various competitive exams conducted by the Government of India, including SSC JE, RRB JE, Banking & Insurance exams, UPSC CDS, UPSC CSE and GPSC. This journey has provided me with deep insights into the examination patterns and preparation strategies. Currently, I channel this experience into my role as a passionate content writer at PracticeMock, where I strive to deliver accurate and relevant information to candidates preparing for Banking exams, guiding them effectively on their preparation journey.