Syllogism for IBPS Clerk — Statements, Conclusions, and Venn Diagram Method

intermediate 18 min read

Concept

Syllogism is a form of deductive reasoning where you are given two or more statements and asked to determine which conclusions logically follow from them — and nothing more than what the statements allow.

Think of it this way: the statements are a locked room. Your job is not to imagine what might be outside the room. Your job is to describe only what you can see inside it with certainty.

The statements use four quantifiers:

Here is the analogy that makes this click. Imagine two sets as circles drawn on a whiteboard. "All A are B" means the circle for A sits completely inside the circle for B. "Some A are B" means the two circles overlap partially. "No A are B" means the circles do not touch at all. "Some A are not B" means part of circle A is hanging outside circle B.

Your entire job in syllogism is to combine these circles correctly when there are multiple statements, and then check whether each conclusion must be true given every possible valid diagram — not just one diagram.

This last point is the one most test-takers miss. A conclusion follows only if it is true in every possible Venn diagram arrangement consistent with the statements. If you can draw even one valid diagram where the conclusion is false, it does not follow.

IBPS Clerk syllogism questions typically give you four to six statements and three to four conclusions to evaluate. The answer choices combine which conclusions follow. Once you understand the core rules, these questions become mechanical — a 90-second problem, not a 3-minute puzzle.


Deep Dive

The Four Statement Types and Their Symbols

Before combining statements, label them:

| Statement | Type | Symbol | |-----------|------|--------| | All A are B | Universal Positive | A+ | | No A are B | Universal Negative | A− | | Some A are B | Particular Positive | I+ | | Some A are not B | Particular Negative | I− |

The Combination Rules (Syllogism Laws)

These rules tell you what conclusion you can draw when two statements share a common term (the middle term).

Rule 1: All + All → All All A are B. All B are C.All A are C.

Rule 2: All + Some → Some All A are B. Some B are C. → No definite conclusion about A and C. But: Some A are B. All B are C.Some A are C. Direction matters. The "Some" must be on the subject side flowing forward.

Rule 3: All + No → No All A are B. No B is C.No A is C.

Rule 4: Some + No → Some Not Some A are B. No B is C.Some A are not C.

Rule 5: No + Some → Some Not (reversed) No A is B. Some B are C.Some C are not A.

Rule 6: Two Particular Statements → No Conclusion Some A are B. Some B are C. → No conclusion about A and C.

Rule 7: Two Negative Statements → No Conclusion No A is B. No B is C. → Nothing about A and C.

Rule 8: Universal Negative is Reversible No A is BNo B is A (always valid)

Rule 9: Particular Positive is Reversible Some A are BSome B are A (always valid)

The Chain Logic — How to Connect Multiple Statements

Most IBPS Clerk questions chain four or five statements. Work through them two at a time, linking via the shared middle term.

Example: Some A are B → All B are C → No C is D → Some D are E

Step 1: Some A are B + All B are C → Some A are C (Rule 2, forward direction) Step 2: All B are C + No C is D → No B is D (Rule 3, and therefore No D is B) Step 3: Some A are C + No C is D → Some A are not D (Rule 4) Step 4: No C is D + Some D are E → Some E are not C (Rule 5, reversed)

Now check each given conclusion against this derived set.

The "Some + All" Trap — The Most Common Error

Look carefully: Some A are B and All B are C gives Some A are C. But All A are B and Some B are C does NOT give Some A are C. Why? Because the "some B" that are C might be the B's that are not A. The direction of the "All" determines which way the chain can travel.

Concrete test: All managers are leaders. Some leaders are innovators. Can you conclude "Some managers are innovators"? No. The innovators among leaders might be the ones who are not managers at all. This exact trap appears repeatedly in the PYQs below.

Complementary Pair — The "Either-Or" Escape Hatch

When neither Some A are B nor No A is B follows individually, but together they form a complementary pair, the answer is "Either I or II follows." This applies specifically when:

This shows up occasionally in IBPS Clerk. Don't confuse it with the standard "does not follow" situation.

Venn Diagram Approach vs. Rules Approach

Both work. Here is when to use which:


Memory Tricks & Shortcuts

patternPUNS: Particular-Universal, No-Some

Remember which statement combinations yield conclusions with this pattern: P-U gives P (Particular + Universal gives Particular, but only when the Universal is the second statement and the Particular is feeding into it). U-U gives U (All+All→All, All+No→No). Two Particulars or Two Negatives give Nothing. This collapses 9 rules into 3 memory slots. Standard rule-memorization: 4 minutes to recall. This pattern: 20 seconds.

patternThe Forward Arrow Test

When you see a chain like Some A→B and All B→C, ask: is the "All" pointing forward in the chain? If yes, the Some travels through. If you see All A→B then Some B→C, the "All" is pointing forward but the "Some" is the second link — the Some might be the B's that aren't A. Conclusion does NOT follow. Drawing this mental arrow takes 5 seconds vs. drawing a full Venn diagram (45 seconds).

patternNo+Some = Some Not (Reversed Subject)

No X is Y. Some Y are Z.Some Z are not X. The conclusion flips to start from Z, not from X. Many students write "Some Y are not X" — wrong subject. The rule: start the conclusion from the term that appears only in the second statement (Z), end it at the term from the first statement (X). This single flip-check eliminates a 2-step Venn diagram, saving 30 seconds.

eliminationAll+No Chain Elimination

All A are B. No B is C.No A is C. This is the cleanest chain — when you see All feeding into a No, the conclusion is always a No about the outer terms. And No A is C immediately means No C is A by reversal. If a conclusion says "Some A are C" or "Some C are A," eliminate it instantly. This takes 3 seconds vs. drawing circles (40 seconds).

estimationThe Boundary Diagram for Some+No

When Some A are B and No B is C, draw a quick mental picture: the overlapping part of A and B is completely outside C. So the A's that are also B cannot be C. Therefore "Some A are not C" must follow. Mentally: shade the overlap, check it cannot touch C's circle. If you practice this 3-step mental image, you resolve these in 8 seconds vs. full written Venn (50 seconds).


Fast-Solving Framework

Here is the decision tree to use in the exam hall:

Step 1 — Identify the chain. Write out each statement in the form quantifier [A] → [B]. Spot which terms are shared between consecutive statements.

Step 2 — Apply combination rules in order. Start from statement 1 + statement 2. Generate an intermediate conclusion. Then take that + statement 3, and so on. If two statements share no common term, skip and try combining non-adjacent ones.

Step 3 — Check each conclusion. For each given conclusion, ask: does this match any derived conclusion or its valid reversal? If yes, it follows. If not, it does not follow.

Step 4 — Watch for the three classic traps: (a) "All + Some" with wrong direction, (b) "No + Some" with wrong subject in conclusion, (c) Two-particular chains that yield nothing.

Step 5 — Eliminate answer choices. Once you confirm Conclusion I and II follow, check if any option includes them both. Usually you can eliminate 2-3 options before evaluating the remaining conclusions.

Target time per question: 60-90 seconds.


Solved PYQs

Why this question: This is the most structurally common IBPS Clerk syllogism format — four statements, four conclusions, one "All+Some wrong direction" trap and one correct "Some+No" chain.

Previous Year Questionपिछले वर्ष का प्रश्न
Statements: Some books are novels. All novels are interesting. No interesting thing is boring. Some boring things are old. Conclusions: I. Some books are interesting. II. Some old things are not interesting. III. No novel is old. IV. Some books are not boring.
कथन: कुछ किताबें उपन्यास हैं। सभी उपन्यास रोचक हैं। कोई भी रोचक चीज़ उबाऊ नहीं है। कुछ उबाऊ चीजें पुरानी हैं। निष्कर्ष: I. कुछ किताबें रोचक हैं। II. कुछ पुरानी चीजें रोचक नहीं हैं। III. कोई उपन्यास पुराना नहीं है। IV. कुछ किताबें उबाऊ नहीं हैं।
  1. Only I, II and IV follow
  2. Only I and II follow
  3. Only I and IV follow
  4. Only II and IV follow
  1. केवल I, II और IV अनुसरण करते हैं
  2. केवल I और II अनुसरण करते हैं
  3. केवल I और IV अनुसरण करते हैं
  4. केवल II और IV अनुसरण करते हैं
Solutionसमाधान
Since some books are novels and all novels are interesting, some books are definitely interesting (I follows). Since some boring things are old and no interesting thing is boring, some old things are definitely not interesting (II follows). We cannot conclude no novel is old (III is false). Since some books are interesting and no interesting thing is boring, some books are definitely not boring (IV follows).
चूंकि कुछ किताबें उपन्यास हैं और सभी उपन्यास दिलचस्प हैं, कुछ किताबें निश्चित रूप से दिलचस्प हैं (I सही है)। चूंकि कुछ उबाऊ चीजें पुरानी हैं और कोई दिलचस्प चीज उबाऊ नहीं है, कुछ पुरानी चीजें निश्चित रूप से दिलचस्प नहीं हैं (II सही है)। हम यह निष्कर्ष नहीं निकाल सकते कि कोई उपन्यास पुराना नहीं है (III गलत है)। चूंकि कुछ किताबें दिलचस्प हैं और कोई दिलचस्प चीज उबाऊ नहीं है, कुछ किताबें निश्चित रूप से उबाऊ नहीं हैं (IV सही है)।

Solving path:

Chain Stmt 1+2: Some books are novels + All novels are interesting → Some books are interesting (Rule 2, forward). Conclusion I: follows.

Chain Stmt 3+4 (reverse): No interesting is boring → No boring is interesting. Some boring are old + No boring is interesting → Some old are not interesting (Rule 4, reversed). Conclusion II: follows.

Conclusion III — "No novel is old": Novels are interesting (All), No interesting is boring, Some boring are old. The old things are among boring, which is separate from interesting. But can novels be old? A novel is interesting, and no interesting thing is boring, and old things here came from boring. So novels cannot be old via this chain... actually, wait — old things are a subset of boring things in this direction, and novels are not boring. So can some novels be old through another path? No other path exists. But can we definitively say NO novel is old? The "some old are boring" doesn't mean "all old are boring." Old things can exist outside boring too. So old things outside boring could overlap with novels. We cannot conclude "No novel is old." Conclusion III: does not follow.

Chain Stmt 1+2+3: Some books are interesting (derived) + No interesting is boring → Some books are not boring (Rule 4). Conclusion IV: follows.

Answer: Only I, II, and IV follow.


Why this question: Structurally identical to the previous — confirms the pattern. Recognize it immediately and apply the same chain logic.

Previous Year Questionपिछले वर्ष का प्रश्न
Statements: Some cars are red. All red things are beautiful. No beautiful thing is cheap. Some cheap things are useful. Conclusions: I. Some cars are beautiful. II. Some useful things are not beautiful. III. No red thing is useful. IV. Some cars are not cheap.
कथन: कुछ कारें लाल हैं। सभी लाल चीजें सुंदर हैं। कोई भी सुंदर चीज सस्ती नहीं है। कुछ सस्ती चीजें उपयोगी हैं। निष्कर्ष: I. कुछ कारें सुंदर हैं। II. कुछ उपयोगी चीजें सुंदर नहीं हैं। III. कोई भी लाल चीज उपयोगी नहीं है। IV. कुछ कारें सस्ती नहीं हैं।
  1. Only I, II and IV follow
  2. Only I and IV follow
  3. Only II and IV follow
  4. Only I and II follow
  1. केवल I, II और IV अनुसरण करते हैं
  2. केवल I और IV अनुसरण करते हैं
  3. केवल II और IV अनुसरण करते हैं
  4. केवल I और II अनुसरण करते हैं
Solutionसमाधान
Since some cars are red and all red things are beautiful, some cars are definitely beautiful (I follows). Since some cheap things are useful and no beautiful thing is cheap, some useful things are definitely not beautiful (II follows). We cannot conclude no red thing is useful (III is false). Since some cars are beautiful and no beautiful thing is cheap, some cars are definitely not cheap (IV follows).
चूंकि कुछ कारें लाल हैं और सभी लाल चीजें सुंदर हैं, कुछ कारें निश्चित रूप से सुंदर हैं (I सही है)। चूंकि कुछ सस्ती चीजें उपयोगी हैं और कोई सुंदर चीज सस्ती नहीं है, कुछ उपयोगी चीजें निश्चित रूप से सुंदर नहीं हैं (II सही है)। हम यह निष्कर्ष नहीं निकाल सकते कि कोई लाल चीज उपयोगी नहीं है (III गलत है)। चूंकि कुछ कारें सुंदर हैं और कोई सुंदर चीज सस्ती नहीं है, कुछ कारें निश्चित रूप से सस्ती नहीं हैं (IV सही है)।

Solving path:

Stmt 1+2: Some cars are red + All red are beautiful → Some cars are beautiful (I: follows). Stmt 3+4 reversed: No beautiful is cheap → No cheap is beautiful. Some cheap are useful + No cheap is beautiful → Some useful are not beautiful (II: follows, Rule 5). Conclusion III — "No red thing is useful": Red → beautiful → not cheap. Useful things come from cheap. No beautiful thing is cheap, so no red (beautiful) thing is cheap, so no red thing is useful via this chain. But can a red thing be useful through a path not involving cheap? The statements only link useful to cheap. So yes, III holds logically... but let's check: is "no red thing is useful" a definitive conclusion? Red things are all beautiful. No beautiful thing is cheap. Useful things are a subset of cheap things. So red things cannot be useful (since useful ⊆ cheap, and red ⊆ beautiful, and beautiful ∩ cheap = empty). This seems valid — but the explanation says III is false. The reason: "Some cheap things are useful" does not mean "All useful things are cheap." Useful things might exist outside cheap. So a red thing could be useful through a path not captured by these statements. We cannot definitively say no red thing is useful. III does not follow. Stmt 2+3: All red are beautiful + No beautiful is cheap → No red is cheap. Some cars are beautiful (derived) + No beautiful is cheap → Some cars are not cheap (IV: follows).

Answer: Only I, II, and IV follow.


Why this question: This is the classic "broken chain" question — the trap is assuming All+Some travels in any direction. The only valid conclusion is the No+Some chain at the end.

Previous Year Questionपिछले वर्ष का प्रश्न
Statements: All smartphones are gadgets. Some gadgets are useful. All useful things are necessary. No necessary thing is wasteful. Some wasteful things are luxuries. Conclusions: I. Some smartphones are necessary. II. No gadget is wasteful. III. Some luxuries are not necessary.
कथन: सभी smartphones गैजेट हैं। कुछ गैजेट उपयोगी हैं। सभी उपयोगी चीजें जरूरी हैं। कोई भी जरूरी चीज फिजूल नहीं है। कुछ फिजूल चीजें विलासिता (luxuries) हैं। निष्कर्ष: I. कुछ smartphones जरूरी हैं। II. कोई भी गैजेट फिजूल नहीं है। III. कुछ luxuries जरूरी नहीं हैं।
  1. Only I follows
  2. Only III follows
  3. Only I and III follow
  4. None follows
  1. केवल I सही है
  2. केवल III सही है
  3. केवल I और III सही हैं
  4. कोई भी सही नहीं है
Solutionसमाधान
We cannot conclude that some smartphones are necessary as we don't know if smartphones fall in the useful category among gadgets. We cannot conclude that no gadget is wasteful as only some gadgets are useful, and only useful things are necessary (not wasteful). Since some wasteful things are luxuries and no necessary thing is wasteful, some luxuries are definitely not necessary (III follows).
हम यह निष्कर्ष नहीं निकाल सकते कि कुछ स्मार्टफोन आवश्यक हैं क्योंकि हम नहीं जानते कि स्मार्टफोन गैजेट्स में उपयोगी श्रेणी में आते हैं या नहीं। हम यह निष्कर्ष नहीं निकाल सकते कि कोई गैजेट बर्बादी नहीं है। चूंकि कुछ बर्बादी चीज़ें विलासिता हैं और कोई आवश्यक चीज़ बर्बादी नहीं है, कुछ विलासिता निश्चित रूप से आवश्यक नहीं हैं (III सत्य है)।

Solving path:

Conclusion I — "Some smartphones are necessary": All smartphones are gadgets. Some gadgets are useful. Can we say some smartphones are useful? No — the useful gadgets might be entirely non-smartphone gadgets. Chain breaks here. I does not follow.

Conclusion II — "No gadget is wasteful": Some gadgets are useful → those useful gadgets are necessary → those specific gadgets are not wasteful. But the other gadgets (not useful) have no constraint preventing them from being wasteful. II does not follow.

Conclusion III — "Some luxuries are not necessary": Some wasteful things are luxuries. No necessary thing is wasteful, so no wasteful thing is necessary. Therefore, those luxuries (which are wasteful) are definitely not necessary. Some luxuries are not necessary. III follows.

Answer: Only III follows.


Why this question: Identical structure to the previous question — All+Some broken chain pattern repeated. Recognizing this template saves 60 seconds.

Previous Year Questionपिछले वर्ष का प्रश्न
Statements: All managers are leaders. Some leaders are innovators. No innovator is a follower. Some followers are employees. Conclusions: I. Some managers are innovators. II. No leader is a follower. III. Some employees are not innovators.
कथन: सभी मैनेजर लीडर हैं। कुछ लीडर इनोवेटर हैं। कोई भी इनोवेटर फॉलोअर नहीं है। कुछ फॉलोअर कर्मचारी हैं। निष्कर्ष: I. कुछ मैनेजर इनोवेटर हैं। II. कोई भी लीडर फॉलोअर नहीं है। III. कुछ कर्मचारी इनोवेटर नहीं हैं।
  1. Only I follows
  2. Only II follows
  3. Only III follows
  4. Only I and III follow
  1. केवल I सही है
  2. केवल II सही है
  3. केवल III सही है
  4. केवल I और III सही हैं
Solutionसमाधान
From the given statements, we cannot conclude that some managers are innovators (I is false). We cannot conclude that no leader is a follower as only some leaders are innovators and no innovator is follower (II is false). Since some followers are employees and no innovator is follower, some employees are definitely not innovators (III is true).
दिए गए कथनों से, हम यह निष्कर्ष नहीं निकाल सकते कि कुछ प्रबंधक नवाचारी हैं (I गलत है)। हम यह निष्कर्ष नहीं निकाल सकते कि कोई नेता अनुयायी नहीं है क्योंकि केवल कुछ नेता नवाचारी हैं और कोई नवाचारी अनुयायी नहीं है (II गलत है)। चूंकि कुछ अनुयायी कर्मचारी हैं और कोई नवाचारी अनुयायी नहीं है, कुछ कर्मचारी निश्चित रूप से नवाचारी नहीं हैं (III सत्य है)।

Solving path:

Conclusion I — "Some managers are innovators": All managers are leaders. Some leaders are innovators. The innovative leaders might not include any manager. I does not follow.

Conclusion II — "No leader is a follower": Some leaders are innovators. Innovators are not followers. But only some leaders are innovators — the non-innovative leaders have no restriction against being followers. II does not follow.

Conclusion III — "Some employees are not innovators": Some followers are employees. No innovator is a follower → no follower is an innovator. Those employees (being followers) cannot be innovators. Some employees are not innovators. III follows.

Answer: Only III follows.


Why this question: Tests whether you correctly handle a five-statement chain and identify the broken link precisely.

Previous Year Questionपिछले वर्ष का प्रश्न
Statements: All teachers are educators. Some educators are researchers. All researchers are scholars. No scholar is ignorant. Some ignorant people are students. Conclusions: I. Some teachers are scholars. II. No educator is ignorant. III. Some students are not scholars.
कथन: सभी शिक्षक, educators हैं। कुछ educators, researchers हैं। सभी researchers, scholars हैं। कोई भी scholar अज्ञानी नहीं है। कुछ अज्ञानी लोग students हैं। निष्कर्ष: I. कुछ शिक्षक scholars हैं। II. कोई भी educator अज्ञानी नहीं है। III. कुछ students scholars नहीं हैं।
  1. Only I follows
  2. Only III follows
  3. Only I and III follow
  4. Only II and III follow
  1. केवल I सही है
  2. केवल III सही है
  3. केवल I और III सही हैं
  4. केवल II और III सही हैं
Solutionसमाधान
We cannot conclude that some teachers are scholars as we don't know if teachers fall in the researcher category. We cannot conclude that no educator is ignorant as only some educators are researchers. Since some ignorant people are students and no scholar is ignorant, some students are definitely not scholars (III follows).
हम यह निष्कर्ष नहीं निकाल सकते कि कुछ शिक्षक विद्वान हैं क्योंकि हम नहीं जानते कि शिक्षक शोधकर्ता श्रेणी में आते हैं या नहीं। हम यह निष्कर्ष नहीं निकाल सकते कि कोई शिक्षक अज्ञानी नहीं है क्योंकि केवल कुछ शिक्षक शोधकर्ता हैं। चूंकि कुछ अज्ञानी लोग छात्र हैं और कोई विद्वान अज्ञानी नहीं है, कुछ छात्र निश्चित रूप से विद्वान नहीं हैं (III सत्य है)।

Solving path:

Conclusion I — "Some teachers are scholars": All teachers are educators. Some educators are researchers. The researcher-educators might not include any teacher. Chain breaks. I does not follow.

Conclusion II — "No educator is ignorant": Some educators are researchers → those researchers are scholars → not ignorant. The non-researcher educators have no constraint. II does not follow.

Conclusion III — "Some students are not scholars": Some ignorant people are students. No scholar is ignorant → no ignorant person is a scholar. Those students (being ignorant) cannot be scholars. Some students are not scholars. III follows.

Answer: Only III follows.


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