Chapter 3 of 11

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IBPS PO Reasoning Ability★ Prelims: 3–5 marks · Mains: 0–3 marks (coded) — the most mechanical, highest-accuracy reasoning block

Inequalities — Direct & Coded

3–5 of the most mechanical marks in the reasoning section — pure symbol-chaining with one golden rule. Learn how relations combine, when a conclusion is definite vs an "either–or", and how to decode the coded (symbol-substitution) version, and inequalities become sub-20-second gifts you never get wrong.

⏱ 18 min readEasy difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

IBPS PO PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Combine relations using the common-sign, unbroken-chain rule
2Decide when a result is strict (>) versus equal (≥)
3Recognise a broken chain (opposing signs) and mark 'does not follow'
4Identify complementary pairs and apply the either–or rule
5Decode coded inequalities into real symbols before solving

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Why this chapter matters in IBPS PO

Inequalities are the most mechanical topic in the reasoning section — no diagram, no imagined cases, just symbol-chaining with one rule. At 3–5 Prelims marks solvable in under 20 seconds each with near-perfect accuracy, they are pure sectional-score insurance, and the coded variant carries the same method into Mains. Locking this topic frees time and confidence for the puzzle sets that actually decide the section.

Before you start — revise these

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Syllogism (incl. New Pattern)

The same either–or (complementary + exhaustive) logic recurs here.

Inequalities — IBPS PO Reasoning

Inequalities are the most mechanical topic in the whole reasoning paper: no diagram, no cases to imagine, just a chain of symbols and one rule about when a relation carries through. IBPS asks 3–5 of them in Prelims and folds the coded version into Mains. Once you internalise the "common sign, unbroken chain" rule and the either–or pattern, these are 20-second marks with near-perfect accuracy — the kind of block that quietly lifts your sectional score.

1. What IBPS actually asks

Direct inequalities: a statement like A > B ≥ C = D < E and two conclusions (e.g. A > D, B ≥ E) — mark which definitely follow.
Coded inequalities: symbols are redefined ("P © Q means P ≥ Q; P % Q means P < Q…"). You decode to real symbols first, then it's a direct inequality.
The either–or pair: two conclusions that are individually uncertain but together exhaust the possibilities.

The five relations: > (greater), < (less), ≥ (greater or equal), ≤ (less or equal), = (equal).

2. The one golden rule: common sign, unbroken chain

A conclusion between two variables is definite only if you can trace an unbroken chain between them where all the inequality signs point the same way (all "greater" or all "less"), treating = as transparent.

A > B > C ⇒ A > C ✓ (all >).
A > B < C ⇒ nothing between A and C — the chain breaks at B (signs oppose). This is the core trap.
A ≥ B ≥ C ⇒ A ≥ C ✓. But A ≥ B = C ⇒ A ≥ C ✓ (= passes through).

The strict-vs-equal subtlety: the result is strict (>) only if at least one strict > appears in the chain; if every link is ≥/=, the result is ≥.

A > B ≥ C ⇒ A > C (one strict link makes it strict).
A ≥ B ≥ C ⇒ A ≥ C (no strict link ⇒ ≥, not >).

$X □ Y is definite ⟺ \exists an unbroken same-direction chain X \to Y$

3. The either–or (complementary) case

When a single pair cannot be fixed as > or = alone, but the two given conclusions are complementary (e.g. A > D and A = D, or A ≥ D and A < D) and together cover all cases, the answer is "either follows".

The classic setup: A ≥ B gives, between A and B, either A > B or A = B. So if the two conclusions are exactly A > B and A = B, neither alone is definite but either–or holds. Requirements: same pair of variables, complementary signs, and together exhaustive.

The either–or only applies to a single, ambiguous pair produced by a ≥/≤. If the chain is broken (opposite signs), you get plain "does not follow", not either–or.

4. Coded inequalities (Mains)

The question redefines symbols: "P @ Q = P > Q; P # Q = P < Q; P $ Q = P ≥ Q; P & Q = P ≤ Q; P * Q = P = Q." Your only first step:

Rewrite the whole statement in real symbols. P @ Q # R $ S → P > Q < R ≥ S.
Then apply the common-sign rule exactly as for direct inequalities.

Never try to reason on the letters/symbols directly — decode, then solve. The coding is just a translation layer.

Solved examples

Question 1 of 4

Q1. Statement: A > B ≥ C = D < E. Conclusions: (I) A > C (II) A > E.

Show explanation

Solution. (I) A > B ≥ C is an unbroken >/≥ chain with a strict link ⇒ A > C follows. (II) A … < E? Chain C = D < E reverses direction relative to A — broken ⇒ A > E does not follow. Only I.

Question 2 of 4

Q2 (either–or). Statement: P ≥ Q = R. Conclusions: (I) P > R (II) P = R.

Show explanation

Solution. P ≥ Q = R ⇒ P ≥ R, i.e. either P > R or P = R. Neither alone is definite; together they are complementary and exhaustive. Either I or II follows.

Question 3 of 4

Q3 (broken chain). Statement: M < N > O ≥ P. Conclusion: M < P.

Show explanation

Solution. M < N then N > O — signs oppose at N ⇒ no relation between M and O/P. Does not follow.

Question 4 of 4

Q4 (coded). Code: X @ Y = X ≥ Y; X © Y = X < Y. Statement: A @ B @ C © D. Conclusion: A ≥ C.

Show explanation

Solution. Decode: A ≥ B ≥ C < D. A ≥ B ≥ C ⇒ A ≥ C follows (no strict link, so ≥).

6. The 20-second protocol

For every inequality question:

(Coded only) rewrite in real symbols first.
Locate the two variables in the conclusion and trace the chain between them.
Same direction, unbroken? Yes ⇒ follows. Broken (opposite signs) ⇒ does not follow.
Strict or equal? A > anywhere in the chain ⇒ strict; all ≥/= ⇒ ≥.
If a single pair is ambiguous, check the other conclusion for a complementary partner → either–or.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Definite conclusion

X □ Y definite ⇔ unbroken same-direction chain X→Y

'=' is transparent and passes the chain through.

Broken chain

A > B < C ⇒ nothing between A and C

Opposite signs at a node break the relation — the core trap.

Strict vs equal

One '>' in the chain ⇒ strict '>'; all '≥/=' ⇒ '≥'

A ≥ B ≥ C ⇒ A ≥ C (not >); A > B ≥ C ⇒ A > C.

Either–or

A ≥ B ⇒ either A > B or A = B

If the two conclusions are exactly this complementary pair, answer 'either follows'.

Coded step 1

Rewrite every coded symbol as >, <, ≥, ≤, = BEFORE solving

Never reason on the code letters directly.

IBPS PO test series

Ready to be tested on this?

Inequalities — Direct & Coded shows up in every IBPS PO paper — sit a timed one and find out where you actually stand.

3–5Qs in Prelims

Highpriority

Full IBPS PO mocks, real clock
Reasoning Ability-only sectionals & chapter tests
All-India rank + topic-level analysis

Free forever · solver-verified questions

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Traps IBPS PO sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Concluding a relation across a broken chain (A > B < C ⇒ A ? C)

✓ Signs must point the same way along the whole path. At any node where '>' meets '<', the chain breaks and no conclusion between the far ends is definite.

WATCH OUT

✗ Writing '>' when the chain only supports '≥'

✓ The result is strict only if at least one strict '>' link appears. A ≥ B ≥ C gives A ≥ C, not A > C. This exact distinction is IBPS's favourite either–or trap.

WATCH OUT

✗ Missing the either–or pair

✓ When a single pair is ambiguous (from a ≥/≤), check the other conclusion: if it's the complementary sign on the same pair and together they're exhaustive, the answer is 'either I or II follows'.

WATCH OUT

✗ Applying either–or to a broken chain

✓ Either–or only arises from a ≥/≤ ambiguity on a SINGLE valid pair. A broken chain (opposite signs) is plain 'does not follow', never either–or.

WATCH OUT

✗ Reasoning on coded symbols without decoding

✓ Translate the entire coded statement to real inequality symbols first, then solve as a direct question. The code is only a translation layer.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Inequalities — Direct & Coded?

6 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

6 questions~4 min worth ~5 marks in IBPS PO exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Inequalities are pure symbol-chaining — no diagrams, no cases.
•A conclusion is definite only along an unbroken, same-direction chain ('=' is transparent).
•Opposite signs at a node break the chain ⇒ no relation between the far ends.
•Result is strict '>' only if one '>' link exists; all '≥/=' ⇒ '≥'.
•A single ambiguous ≥/≤ pair with complementary, exhaustive conclusions ⇒ either–or.
•Coded: rewrite in real symbols first, then apply the same rule.
•Target under 20 seconds — this is a time source, not a sink.

IBPS PO question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: Prelims: 3–5 marks (of 40) · Mains: 0–3 marks (of 60)

Question style	Marks each	Typical count	What it tests
Direct inequality	1 each	3–5	Chain-tracing, strict-vs-equal, broken chains, either–or
Coded inequality (Mains)	1.5 each	0–3	Decode symbols then chain

Prep strategy

Day 1: memorise the common-sign rule and the strict-vs-equal condition.
Day 2–3: 20 direct questions daily; aim for zero errors before adding speed.
Day 4+: coded sets and mixed 'either–or' spotting under a 20-second timer.

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Decode any coded statement to real symbols before doing anything else.
Trace the chain only between the two variables in the conclusion.
Confirm same-direction and unbroken; then decide strict vs equal.
For an ambiguous ≥/≤ pair, look for the complementary either–or partner.
Cap each question at ~20 seconds — inequalities should never eat your clock.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Threshold logic in banking

Ranking loan risk, comparing rates and applying 'greater-than' thresholds to approvals is exactly this relation-chaining.

Fast comparison

Chaining 'A cheaper than B, B pricier than C' to a definite conclusion is the everyday version of the common-sign rule.

Where else this topic is tested

Prepare once, score in every exam that asks it.

IBPS Clerk / SBI PO & ClerkVery high — direct + coded inequalities every paper

RRB PO & ClerkVery high — 3–5 questions per paper

RBI Grade BMedium — usually the coded variant

📝 My notes

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Typically 3–5 in Prelims (direct) and, in Mains, a coded set of up to 3. They're the most mechanical and highest-accuracy questions in reasoning — you should almost never lose one.

Only when a single pair of variables is left ambiguous by a ≥ or ≤ (so it's 'either > or ='), and the two given conclusions are precisely that complementary pair covering all cases. A broken chain is never either–or — it's plain 'does not follow'.

Almost always the strict-vs-equal rule: A ≥ B ≥ C only gives A ≥ C, not A > C, because no strict '>' link is present. You need at least one '>' somewhere in the path for a strict conclusion.

No — they're identical after one translation step. Rewrite every coded symbol as its real inequality, then solve normally. The difficulty is entirely removed by decoding first.