Coded Inequality Questions for Bank Exams: Quick Answer
This set has 20 original coded inequality questions with full step-by-step answers, written at bank Prelims and Mains level. Every question gives a statement in code — symbols such as @, #, %, $ and * standing in for <, ≤, =, ≥ and > — followed by two conclusions, and every answer has been checked by a brute-force program that tried every possible ordering of the letters. As of 6 November 2026.
Coded inequality is one of the fastest marks in bank Reasoning — and one of the most thrown away. It has appeared in IBPS and SBI Clerk, IBPS and SBI PO and IBPS RRB papers, often as a set of three to five questions. SBI PO Mains runs Reasoning and Computer Aptitude as one combined section of 40 questions for 60 marks in 50 minutes; IBPS PO Mains runs Reasoning separately at 40 questions for 60 marks in 45 minutes. IBPS RRB PO and Clerk Mains run Reasoning at 40 questions for 50 marks in 30 minutes — barely 45 seconds a question. IBPS Clerk Mains runs the same combined section at 40 questions for 60 marks in 35 minutes, SBI Clerk Mains at 50 questions for 60 marks in 45 minutes. A coded question is worth the same one mark as a five-minute puzzle, so the fastest marks in Reasoning are the coded questions you can solve in under half a minute. That is what this set trains: decode once, build one chain, read both conclusions off it.
Key takeaway: Never solve a coded question in symbols. Decode every statement into real signs first, join them into one chain through the common letters, and then test each conclusion against that chain. The two checks that settle almost every question are: do all the signs between the two letters point the same way, and is at least one of them strict? If the signs change direction anywhere on the path, the chain is broken and the conclusion does not follow.
What This Set Covers
| Block | Questions | Topics |
|---|---|---|
| Direct-coded | 1–5 | Straight chains, decoding, both-follows and either-or |
| Reverse-coded | 6–10 | Statements written with the larger letter first, chains read backwards |
| Mixed with ≥ and ≤ | 11–15 | Non-strict chains where "greater than" alone does not follow |
| Either-or cases | 16–20 | Genuine either-or pairs and the traps that look like them |
| Total | 20 | Prelims and Mains level, 5 options each |
The answer key spreads evenly — A 4, B 4, C 4, D 4, E 4 — with no option repeating in a row.
How to Use This Set
Give yourself 15 minutes for the 20 questions — 45 seconds a question. Write the decoded chain on paper before looking at the options; the options reward the chain and punish the guess. Mark every question you guessed, and rework the guessed ones before reading the explanations. Note the exact step where you went wrong — wrong decoding, a broken chain, a wrong either-or — because the same three slips cause most losses. Do the direct-coded block first, then the mixed block, and save reverse-coded and either-or for last.
How Coded Inequality Works: Decode, Chain, Test
The code used in every question in this set is fixed, so learn it once: @ means "less than" (<), # means "less than or equal to" (≤), % means "equal to" (=), $ means "greater than or equal to" (≥), and * means "greater than" (>). So "M @ N" reads "M is less than N" and "R * S" reads "R is greater than S". The code changes from paper to paper, so never carry one over from another paper — decode from the definition in front of you, every time.
Step 1: Decode. Rewrite every statement with real signs on your rough sheet — once, not once per conclusion. In Q1, "M @ N, N # P, P % Q" becomes M < N, N ≤ P, P = Q. If you skip this step and try to reason in symbols, you will mix up # and $ — the two that look most alike and mean opposite things.
Step 2: Join through common letters. If the statements are M < N, N ≤ O and O = P, write them as one line: M < N ≤ O = P. Flip a statement if needed so that the common letter sits in the middle: "W # T" can be written "T $ W", because W ≤ T is the same as T ≥ W.
Step 3: Read each conclusion off the chain. Find both letters of the conclusion in the chain and look at every sign between them:
| Signs between the two letters | Result |
|---|---|
| All signs point the same way and at least one is strict (> or <) | The strict relation (> or <) is definitely true |
| All signs point the same way, none strict, at least one is ≥ or ≤ | ≥ or ≤ is definitely true; > alone and = alone are NOT definite |
| Only "=" signs | "=" is definitely true |
| Signs point in opposite directions (like > and <, or ≥ and ≤) | No definite relation: the chain is broken |
Remember: the strongest sign on the path wins, as long as the path does not change direction. From A ≥ B ≥ C you get A ≥ C, not A > C, because all three letters can be equal; for A > C you need a strict sign on the path, as in Q1.
The either-or rule. Mark "Either conclusion I or II follows" only when all three hold: neither conclusion is definitely true on its own; both talk about the same two letters; and together they cover every possible case. The pairs that pass it are ≥ with <, and ≤ with >, whenever the relation is not definite; > with = only when the chain gives ≥; < with = only when the chain gives ≤. Two traps follow. First, "> and =" on a broken chain is not either-or — < is also possible, so the pair does not cover every case, and the answer is "Neither". Second, two conclusions that say the same thing are never either-or: "E < H" and "H > E" are one relation written two ways. Q4, Q10, Q13 and Q19 are built around these traps.
Direct-Coded Inequality Questions
In this block the statements are written in chain order, so the only skill tested is clean decoding. The code is: @ means <, # means ≤, % means =, $ means ≥, * means >. Decode every statement before you read the conclusions.
Q1. Statements: M @ N, N # P, P % Q
Conclusions: I. M @ Q II. Q $ N
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: C — Both conclusions I and II follow.
Explanation: Decode first. "M @ N" means M < N, "N # P" means N ≤ P, and "P % Q" means P = Q. Join them through the common letters into one chain: M < N ≤ P = Q. Conclusion I, "M @ Q", means M < Q. From M to Q the signs are <, ≤ and =, all pointing the same way, with one strict sign, so M < Q is definitely true. Conclusion II, "Q $ N", means Q ≥ N. From N to Q the signs are ≤ and =, so N ≤ Q, which is the same as Q ≥ N — also definitely true. Both follow.
Q2. Statements: R * S, S $ T, T @ U
Conclusions: I. R * T II. U * S
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: A — Only conclusion I follows.
Explanation: Decode first. "R * S" means R > S, "S $ T" means S ≥ T, and "T @ U" means T < U. The chain is R > S ≥ T < U. Conclusion I, "R * T", means R > T. From R to T the signs are > and ≥, same direction with a strict sign, so R > T is definitely true. Conclusion II, "U * S", means U > S. Between S and U the signs are ≥ and <, pointing in opposite directions, so the chain is broken there and S and U have no definite relation. Only conclusion I follows.
Q3. Statements: J @ K, K $ L, L % M
Conclusions: I. J * M II. K $ M
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: B — Only conclusion II follows.
Explanation: Decode first. "J @ K" means J < K, "K $ L" means K ≥ L, and "L % M" means L = M. The chain is J < K ≥ L = M. Conclusion I, "J * M", means J > M. Both J and M sit below K, and nothing in the statements fixes which of the two is bigger — J could be above M, below M, or equal to M — so conclusion I does not follow. Conclusion II, "K $ M", means K ≥ M. From K to M the signs are ≥ and =, so K ≥ M is definitely true. Only conclusion II follows.
Q4. Statements: P * Q, R @ Q, R # S
Conclusions: I. P * S II. Q # S
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: D — Neither conclusion I nor II follows.
Explanation: Decode first. "P * Q" means P > Q, "R @ Q" means R < Q, and "R # S" means R ≤ S. The chain is P > Q > R ≤ S. Conclusion I, "P * S", means P > S. From P to S the signs are >, > and then ≤, so the direction changes at R and the chain is broken — S can sit anywhere at or above R, above P or below it, so P > S is not definite. Conclusion II, "Q # S", means Q ≤ S. Between Q and S the signs are > and ≤, again in opposite directions, so Q and S have no definite relation either. The two conclusions also compare different pairs of letters, P with S and Q with S, so either-or cannot apply. Neither follows.
Q5. Statements: E $ F, G # F
Conclusions: I. E % G II. E * G
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: E — Either conclusion I or II follows.
Explanation: Decode first. "E $ F" means E ≥ F, and "G # F" means G ≤ F, which is the same as F ≥ G. Join them into one chain: E ≥ F ≥ G, so E ≥ G is definitely true. Conclusion I, "E % G", means E = G. That is possible — all three letters can be equal — but it is not definite, so conclusion I does not follow on its own. Conclusion II, "E * G", means E > G. That is also possible, but not definite, so conclusion II does not follow on its own either. The two conclusions compare the same two letters, E and G, and together "=" and ">" make exactly "≥", which is what the chain gives. So one of them must be true, and we cannot tell which. Either-or.
Reverse-Coded Inequality Questions
In this block the statements are written with the larger letter first, so the chain has to be read backwards. The code is the same: @ means <, # means ≤, % means =, $ means ≥, * means >. Flip each statement so the common letter sits in the middle, then read the chain the usual way — forgetting to flip is the most common error here.
Q6. Statements: V * U, U $ T, W # T
Conclusions: I. V @ W II. V * T
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: B — Only conclusion II follows.
Explanation: Decode first. "V * U" means V > U, "U $ T" means U ≥ T, and "W # T" means W ≤ T, which is the same as T ≥ W. Read backwards, the chain is W ≤ T ≤ U < V. Conclusion I, "V @ W", means V < W. But the chain puts V above U, U above T, and T above W, so V is always greater than W — V < W is impossible, so conclusion I does not follow. Conclusion II, "V * T", means V > T. From V to T the signs are > and ≥, same direction with a strict sign, so V > T is definitely true. Only conclusion II follows.
Q7. Statements: J $ H, H * G, F @ G
Conclusions: I. H * F II. J @ F
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: A — Only conclusion I follows.
Explanation: Decode first. "J $ H" means J ≥ H, "H * G" means H > G, and "F @ G" means F < G, which is the same as G > F. Read backwards, the chain is F < G < H ≤ J. Conclusion I, "H * F", means H > F. From H to F the signs are > and >, so H > F is definitely true. Conclusion II, "J @ F", means J < F. But the chain puts J above H, H above G, and G above F, so J is always greater than F — J < F is impossible, so conclusion II does not follow. Only conclusion I follows.
Q8. Statements: K @ L, L # M, M % N
Conclusions: I. K @ N II. N $ L
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: C — Both conclusions I and II follow.
Explanation: Decode first. "K @ L" means K < L, "L # M" means L ≤ M, and "M % N" means M = N. The chain is K < L ≤ M = N. Conclusion I, "K @ N", means K < N. From K to N the signs are <, ≤ and =, all pointing the same way with one strict sign, so K < N is definitely true. Conclusion II, "N $ L", means N ≥ L. From L to N the signs are ≤ and =, so L ≤ N, which is the same as N ≥ L — also definitely true. Both follow.
Q9. Statements: S $ T, U # T, V * U
Conclusions: I. S % U II. S * U
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: E — Either conclusion I or II follows.
Explanation: Decode first. "S $ T" means S ≥ T, "U # T" means U ≤ T, which is the same as T ≥ U, and "V * U" means V > U. The chain for S and U is S ≥ T ≥ U, so S ≥ U is definitely true. The third statement, V > U, does no work for these conclusions — it only tells you about V, which neither conclusion mentions, so ignore it once you have checked that. Conclusion I, "S % U", means S = U. That is possible — S, T and U can all be equal — but it is not definite, so conclusion I does not follow on its own. Conclusion II, "S * U", means S > U. That is also possible, but not definite, so conclusion II does not follow on its own either. The two conclusions compare the same two letters, S and U, and together "=" and ">" make exactly "≥", which is what the chain gives. Either-or.
Q10. Statements: E * G, H @ G, H $ K
Conclusions: I. K * E II. G # K
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: D — Neither conclusion I nor II follows.
Explanation: Decode first. "E * G" means E > G, "H @ G" means H < G, which is the same as G > H, and "H $ K" means H ≥ K. Read backwards, the chain is K ≤ H < G < E. Conclusion I, "K * E", means K > E. But the chain puts K below H, H below G, and G below E, so K is always smaller than E — K > E is impossible, so conclusion I does not follow. Conclusion II, "G # K", means G ≤ K. But the chain gives G > H ≥ K, so G is always greater than K — G ≤ K is impossible too, so conclusion II does not follow. Neither follows.
Mixed Inequality Questions with ≥ and ≤
In this block every chain is built from non-strict signs, and that is the trap: a chain of ≥ and ≤ proves ≥ or ≤, never > or < on its own. The code is the same: @ means <, # means ≤, % means =, $ means ≥, * means >. When a conclusion claims a strict sign, check whether the path has one — if it does not, the conclusion fails.
Q11. Statements: M $ L, L # N, N # P
Conclusions: I. P $ L II. M * P
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: A — Only conclusion I follows.
Explanation: Decode first. "M $ L" means M ≥ L, "L # N" means L ≤ N, and "N # P" means N ≤ P. The chain is M ≥ L ≤ N ≤ P. Conclusion I, "P $ L", means P ≥ L. From L to P the signs are ≤ and ≤, so L ≤ P, which is the same as P ≥ L — definitely true. Conclusion II, "M * P", means M > P. Between M and P the signs are ≥ and then ≤, pointing in opposite directions, so the chain is broken and M and P have no definite relation. Only conclusion I follows.
Q12. Statements: U % S, S # R, R # Q
Conclusions: I. U # Q II. Q $ S
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: C — Both conclusions I and II follow.
Explanation: Decode first. "U % S" means U = S, "S # R" means S ≤ R, and "R # Q" means R ≤ Q. The chain is U = S ≤ R ≤ Q. Conclusion I, "U # Q", means U ≤ Q. From U to Q the signs are =, ≤ and ≤, so U ≤ Q is definitely true. Conclusion II, "Q $ S", means Q ≥ S. From S to Q the signs are ≤ and ≤, so S ≤ Q, which is the same as Q ≥ S — also definitely true. Both follow. Notice that neither conclusion claims a strict sign: with only = and ≤ on the path, that is all the chain can prove, and both conclusions stay inside what it proves.
Q13. Statements: F $ H, J # H, J * K
Conclusions: I. K $ F II. H % K
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: D — Neither conclusion I nor II follows.
Explanation: Decode first. "F $ H" means F ≥ H, "J # H" means J ≤ H, which is the same as H ≥ J, and "J * K" means J > K. The chain is F ≥ H ≥ J > K. Conclusion I, "K $ F", means K ≥ F. But the chain puts K below J, J below H, and H below F, so K is always smaller than F — K ≥ F is impossible, so conclusion I does not follow. Conclusion II, "H % K", means H = K. But the chain gives H ≥ J > K, so H is always greater than K — H = K is impossible too, so conclusion II does not follow. Neither follows.
Q14. Statements: P $ N, N # H, H # G
Conclusions: I. P * G II. G $ N
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: B — Only conclusion II follows.
Explanation: Decode first. "P $ N" means P ≥ N, "N # H" means N ≤ H, and "H # G" means H ≤ G. The chain is P ≥ N ≤ H ≤ G. Conclusion I, "P * G", means P > G. Between P and G the signs are ≥ and then ≤, pointing in opposite directions, so the chain is broken and P and G have no definite relation. Conclusion II, "G $ N", means G ≥ N. From N to G the signs are ≤ and ≤, so N ≤ G, which is the same as G ≥ N — definitely true. Only conclusion II follows.
Q15. Statements: T $ V, W # V
Conclusions: I. T % W II. T * W
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: E — Either conclusion I or II follows.
Explanation: Decode first. "T $ V" means T ≥ V, and "W # V" means W ≤ V, which is the same as V ≥ W. Join them into one chain: T ≥ V ≥ W, so T ≥ W is definitely true. Conclusion I, "T % W", means T = W. That is possible — all three letters can be equal — but it is not definite, so conclusion I does not follow on its own. Conclusion II, "T * W", means T > W. That is also possible, but not definite, so conclusion II does not follow on its own either. The two conclusions compare the same two letters, T and W, and together "=" and ">" make exactly "≥", which is what the chain gives. Either-or.
Either-Or Case Questions
This block is where coded inequality questions are won and lost, because the either-or option is the most guessed and the most misused. The code is the same: @ means <, # means ≤, % means =, $ means ≥, * means >. Before you mark E, run the three checks: neither conclusion definite alone, both about the same two letters, together covering every case. If any check fails, the answer is not E.
Q16. Statements: K * E, E $ M, M % S
Conclusions: I. K * S II. E $ S
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: C — Both conclusions I and II follow.
Explanation: Decode first. "K * E" means K > E, "E $ M" means E ≥ M, and "M % S" means M = S. The chain is K > E ≥ M = S. Conclusion I, "K * S", means K > S. From K to S the signs are >, ≥ and =, all pointing the same way with a strict sign, so K > S is definitely true. Conclusion II, "E $ S", means E ≥ S. From E to S the signs are ≥ and =, so E ≥ S is definitely true. Both follow. This question is in the either-or block as a control: both conclusions are definite, so the answer is C, not E — either-or only applies when neither conclusion follows on its own.
Q17. Statements: R # L, L * J, J $ F
Conclusions: I. L * F II. R * F
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: A — Only conclusion I follows.
Explanation: Decode first. "R # L" means R ≤ L, "L * J" means L > J, and "J $ F" means J ≥ F. The chain is R ≤ L > J ≥ F. Conclusion I, "L * F", means L > F. From L to F the signs are > and ≥, same direction with a strict sign, so L > F is definitely true. Conclusion II, "R * F", means R > F. Between R and F the signs are ≤ and then >, pointing in opposite directions, so the chain is broken and R and F have no definite relation. Only conclusion I follows.
Q18. Statements: W @ T, T # N, G $ N
Conclusions: I. W * G II. N * W
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: B — Only conclusion II follows.
Explanation: Decode first. "W @ T" means W < T, "T # N" means T ≤ N, and "G $ N" means G ≥ N, which is the same as N ≤ G. The chain is W < T ≤ N ≤ G. Conclusion I, "W * G", means W > G. But the chain puts W below T, T below N, and N below G, so W is always smaller than G — W > G is impossible, so conclusion I does not follow. Conclusion II, "N * W", means N > W. From W to N the signs are < and ≤, so W < N, which is the same as N > W — definitely true. Only conclusion II follows.
Q19. Statements: H $ L, P # L, P * U
Conclusions: I. U $ H II. L % U
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: D — Neither conclusion I nor II follows.
Explanation: Decode first. "H $ L" means H ≥ L, "P # L" means P ≤ L, which is the same as L ≥ P, and "P * U" means P > U. The chain is H ≥ L ≥ P > U. Conclusion I, "U $ H", means U ≥ H. But the chain puts U below P, P below L, and L below H, so U is always smaller than H — U ≥ H is impossible, so conclusion I does not follow. Conclusion II, "L % U", means L = U. But the chain gives L ≥ P > U, so L is always greater than U — L = U is impossible too, so conclusion II does not follow. Neither follows. Note that the two conclusions compare different pairs of letters, U with H and L with U, so either-or could not apply here even if both had merely been unproven.
Q20. Statements: J $ N, Q # N, V * Q
Conclusions: I. J % Q II. J * Q
A. Only conclusion I follows B. Only conclusion II follows C. Both conclusions I and II follow D. Neither conclusion I nor II follows E. Either conclusion I or II follows
Answer: E — Either conclusion I or II follows.
Explanation: Decode first. "J $ N" means J ≥ N, "Q # N" means Q ≤ N, which is the same as N ≥ Q, and "V * Q" means V > Q. The chain for J and Q is J ≥ N ≥ Q, so J ≥ Q is definitely true. The third statement, V > Q, does no work for these conclusions — it only tells you about V, which neither conclusion mentions. Conclusion I, "J % Q", means J = Q. That is possible — J, N and Q can all be equal — but it is not definite, so conclusion I does not follow on its own. Conclusion II, "J * Q", means J > Q. That is also possible, but not definite, so conclusion II does not follow on its own either. The two conclusions compare the same two letters, J and Q, and together "=" and ">" make exactly "≥", which is what the chain gives. Either-or.
Answer Key
| Question | Answer | Question | Answer |
|---|---|---|---|
| Q1 | C | Q11 | A |
| Q2 | A | Q12 | C |
| Q3 | B | Q13 | D |
| Q4 | D | Q14 | B |
| Q5 | E | Q15 | E |
| Q6 | B | Q16 | C |
| Q7 | A | Q17 | A |
| Q8 | C | Q18 | B |
| Q9 | E | Q19 | D |
| Q10 | D | Q20 | E |
Key spread: A 4, B 4, C 4, D 4, E 4. No option is correct more than four times out of twenty, and no option repeats in a row anywhere in the key.
Common Mistakes
The first and most expensive mistake in coded inequality is decoding from memory instead of from the definition in the question. The code changes from paper to paper: in this set $ means ≥, but another paper may use $ for a different sign altogether. Every explanation in this set decodes from it. If you assume a symbol means what it meant in the last paper you solved, the whole question collapses before you have drawn a chain.
The second mistake is reading a ≥ chain as a > chain. A ≥ B ≥ C only proves A ≥ C, because all three letters can be equal. Q5, Q9, Q15 and Q20 all turn on this: the "=" and ">" conclusions are each possible but neither is definite, which is what makes them either-or. If you mark ">" as following from a ≥ chain, you will choose B or C where the answer is E, and you will do it four times in one set.
The third mistake is skipping the direction check. A > B < C tells you nothing about A and C, and a chain that changes direction anywhere between the two letters you care about is broken for that pair. Q2, Q4, Q11, Q14 and Q17 all have a broken chain for at least one conclusion. Always trace the whole path between the two letters before you decide.
The fourth mistake is applying either-or to a pair that is not complementary. Either-or needs the two conclusions to cover every possibility between the same two letters. In Q4 and Q19 the conclusions compare different pairs, so either-or cannot apply. Before choosing E, ask: if both conclusions were false, would that be possible? If yes, the answer is D, not E.
Memory Tricks
"Decode once, chain once." Write the code at the top of your rough sheet — @ <, # ≤, % =, $ ≥, * > — and decode all the statements into one chain before you look at either conclusion.
"Strongest sign wins, if the road stays straight." On a path that never changes direction, the strongest sign decides the relation: > beats ≥, ≥ beats =. In Q16, K > E ≥ M = S gives K > S; if the road bends, there is no relation at all.
"Either-or needs the same two letters and a full cover." Before choosing E, check three things: neither conclusion follows alone, both name the same two letters, and together they cover every case. The safe pairs are "≥ with <", "≤ with >", "> with =" (only when the chain gives ≥) and "< with =" (only when the chain gives ≤).
"Flip before you read." In reverse-coded questions, rewrite every statement so the smaller letter comes first — "W # T" becomes "T $ W" — and only then join the chain.
Frequently Asked Questions
What do the symbols @, #, %, $ and * mean in this set?
In every question in this set, @ means less than (<), # means less than or equal to (≤), % means equal to (=), $ means greater than or equal to (≥), and * means greater than (>). In the real exam the code is defined in the question and changes from paper to paper, so decode from the definition in front of you, never from memory.
When is the answer "either I or II" in coded inequality?
Only when neither conclusion is definitely true on its own, both conclusions talk about the same two letters, and together they cover every possible case for those two letters. Q5, Q9, Q15 and Q20 are genuine either-or questions; Q4 and Q19 show when the answer is "neither follows" instead.
Is A ≥ B ≥ C enough to say A > C?
No. A ≥ B ≥ C only proves A ≥ C, because all three letters can be equal. To get A > C you need at least one strict sign (> or <) on the path between A and C, as in Q1 and Q16.
What does a broken chain mean in coded inequality?
The signs between two letters point in opposite directions, like A > B < C, so A and C have no definite relation and any conclusion about them alone is false. Q2, Q4, Q11, Q14 and Q17 each have a broken chain for at least one conclusion.
Does coded inequality come in bank Prelims as well as Mains?
Yes. It has appeared in IBPS and SBI Clerk, IBPS and SBI PO and IBPS RRB papers, often as a set of three to five questions in the Prelims. The method on this page solves both, because a direct inequality is a coded question with the code already removed.
How much time should one coded inequality question take?
With practice, about 20 to 30 seconds: around 10 seconds to decode and build the chain, and a few seconds for each conclusion.
Practice Next
- Coded inequality tricks — the full method behind this set.
- Syllogism questions for bank exams — 20 solved syllogism questions, including coded syllogism.
- Either-or case rules — every complementary pair, and when "either" is wrong.
- Syllogism questions with answers — 20 direct, either-or and possibility questions.
- Blood relation questions with answers — the other big reasoning topic, solved the same way.
- Direction sense questions for bank exams — fast one-mark reasoning questions to pair with this set.
- Alphanumeric series tricks for bank exams — the other coded format in bank Reasoning.
- Input output reasoning for SBI PO — Mains-level reasoning practice.
- IBPS PO inequality mock test 01 — timed inequality practice at IBPS PO level.
- SBI Clerk inequality mock test 01 — timed inequality practice at SBI Clerk level.
- IBPS RRB PO inequality mock test 03 — timed inequality practice at RRB PO level.
Sources: SBI PO, IBPS PO, IBPS Clerk, SBI Clerk and IBPS RRB Mains exam patterns as published in their official notifications. All 20 questions are original, written for this page, and every conclusion was checked by a brute-force program. As of 6 November 2026.