Statements: P > Q ≥ R = S Conclusions:I. P > SII. R < T

Answer: Both I and II follow. Exam Attack: For I: Path P to S has >. (P > Q ≥ R = S). True.For II: Path R to T is R = S < T. The supreme sign < is present. True. Both follow.

Statements: A ≥ B = C ≤ D Conclusions:I. A ≥ CII. C < E

Answer: Both I and II follow. Exam Attack: Path A to C: A ≥ B = C. Highest priority is ≥. True.Path C to E: C ≤ D < E. Highest priority is <. True. Both follow.

Statements: M O = P ≥ QConclusions:I. M > OII. N > Q

Answer: Only Conclusion II follows. Exam Attack: I is False (Blocked at N). II is True (N > O = P ≥ Q leads to N > Q).

Statements: W ≥ X = Y ≥ ZConclusions:I. W > ZII. W = Z

Answer: Either I or II follows. Exam Attack: Result between W and Z is W ≥ Z. Individually, > and = are false. Together they form an Either/Or pair.

Statements: L = M O ≤ PConclusions:I. L II. O < M

Answer: Only Conclusion I follows. Exam Attack: L = M O is a blocked path, so O < M is False.

Statements: J ≥ K > L = M Conclusions:I. J > MII. K < N

Answer: Only Conclusion I follows. Exam Attack: Path J to M: J ≥ K > L = M. Supreme sign is >. J > M is True. Path K to N is blocked (K > L = M < N), so II is False.

Statements: S WConclusions:I. T II. T = V

Answer: Either I or II follows. Exam Attack: T ≤ U = V gives T ≤ V. Neither < nor = is completely true alone. They form an Either/Or case.

Statements: D = E > F ≥ G Conclusions:I. D > GII. E > H

Answer: Only Conclusion I follows. Exam Attack: D = E > F ≥ G results in D > G (True). For E and H, the path E > F ≥ G < H is blocked. False.

Which of the following expressions will be true if the expression H ≥ I = J > K ≤ L is definitely true?

Answer: H > K. Exam Attack: From H to K: H ≥ I = J > K. The supreme sign is >. Therefore, H > K is definitely true.

In which of the following expressions does the expression 'O < M' and 'R ≥ P' definitely hold true?

Answer: M = N > O ≥ P = Q ≤ R. Exam Attack: In option B: M = N > O gives M > O (True). Also P = Q ≤ R gives P ≤ R which means R ≥ P (True).

Statements: W ≤ X A ≥ BConclusions:I. W II. Z > B

Answer: Both I and II follow. Exam Attack: W ≤ X B. Both are true.

Statements: H ≥ I = J > K ; L Conclusions:I. H ≥ LII. M > K

Answer: Only Conclusion II follows. Exam Attack: H ≥ I = J > L -> H > L (Not ≥). So I is False. K K M > K (True). Thus, only II follows.

Statements: C Conclusions:I. G II. H > D

Answer: Only Conclusion II follows. Exam Attack: C C D H > D. True.

Statements: I = J ≥ K ≥ L ; K Conclusions:I. I > MII. I ≤ M

Answer: Either I or II follows. Exam Attack: Path from I to M is I = J ≥ K , <, =). This is a textbook 'Either-Or' condition.

Which of the following symbols should replace the question mark (?) in the given expression in order to make the expressions 'A > D' and 'F ≥ C' definitely true?A > B ≥ C ? D ≤ E = F

Answer: =. Exam Attack: To make A>D true, ? can be >, ≥, or =. To make F≥C true (which means C≤F, path: C ? D ≤ E = F), ? MUST be ≤ or =. The common valid symbol that satisfies both directions is '='.

Statement: M ≤ N Which of the following conclusions is definitely FALSE?

Answer: N > P. Exam Attack: N to P path is N P is absolutely 100% FALSE. (Note: O > R is possible but not certain, but N > P directly conflicts the given statement).

Reverse Inequality: Which of the following expressions shows that 'D > B' is definitely true?

Answer: E ≤ D > C = B Exam Attack: Check paths from D to B. In A: D > C = B, which means D > B is True. (In B it is D ≥ B, in C it is D ≥ B, in D it is D > B but via C=D and B C>B -> D>B). Wait! Let's re-eval D. A > B D = C > B -> D > B. Ah, option A also has D > C = B. Both A and D make it true? Let's check A again: D > C = B (D>B). Option D: B B. Let's assume the most direct one (A).

Which of the following expressions will be FALSE if the expression 'J = K N = O' is definitely true?

Answer: L ≤ O. Exam Attack: Path between L and O is L ≤ M > N = O. The gates oppose, meaning no relation. Hence L ≤ O is a false conclusion. (A, B, C, E are all definitely true).

In which of the following expressions does 'T > Q' as well as 'Q < W' hold definitely true?

Answer: T > R = S ≥ Q Exam Attack: In C: T > R = S ≥ Q gives T > Q (True). Also Q < P ≤ W gives Q < W (True).

Statements: A > B ; C Which of the following is true?

Answer: C Exam Attack: Chain them: A > B ≥ C C < F. True.

Fill the blank with appropriate symbols to make 'P > S' and 'T ≤ R' true.P > Q _ R _ S _ T

Answer: ≥ , = , ≥. Exam Attack: Try Option A (≥, =, ≥): P > Q ≥ R = S ≥ T. P > S is True. R = S ≥ T implies R ≥ T or T ≤ R. True. Option A satisfies both.

Statements: X > Y ≥ Z = A ; B F ≥ XConclusions:I. D > CII. E > A

Answer: Both I and II. Exam Attack: Combine D to C: D = E > F ≥ X > Y ≥ Z ≥ C. Clearly D > C (True). Combine E to A: E > F ≥ X > Y ≥ Z = A. Clearly E > A (True). Both Follow.

If 'A + B' means A>B, 'A - B' means A≥B, 'A * B' means A=B, 'A / B' means AWhich of the following means P>S is False?

Answer: P ^ Q + R - S. Exam Attack: In Option D: P ^ Q + R - S decodes to P ≤ Q > R ≥ S. The path is blocked. Therefore, P > S is False.

Statement: M > N ≥ O = P Conclusions:I. M > PII. P < S

Answer: Both Conclusion I and Conclusion II follow. Exam Attack: From M > N ≥ O = P, we have M > P. From P < Q ≤ R < S, we have P < S. Both follow.

Statement: A Conclusions:I. E > AII. F > D

Answer: Both Conclusion I and Conclusion II follow. Exam Attack: I: E ≥ C ≥ B > A => E > A (True). II: F > C = D => F > D (True). Both follow.

Statement: U > V = W ≥ X ; Y Conclusions:I. U > YII. Z ≥ X

Answer: Both Conclusion I and Conclusion II follow. Exam Attack: U > V > Y => U > Y. Also Z ≥ V = W ≥ X => Z ≥ X. Both conclusions are definitively correct.

Statement: J ≥ K > L ; M ≤ K Conclusions:I. J ≥ MII. N > L

Answer: Both Conclusion I and Conclusion II follow. Exam Attack: J ≥ K ≥ M => J ≥ M (True). N > K > L => N > L (True). Both are definitively correct.

Statement: M > N ≥ O = P Conclusions:I. M > PII. P < S

Both Conclusion I and Conclusion II follow

Either Conclusion I or Conclusion II follows

Neither Conclusion I nor Conclusion II follows

Both Conclusion I and Conclusion II follow

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Inequality (Coded & Reverse)

Q: Statements: P > Q ≥ R = S Conclusions:I. P > SII. R < T

Answer: Both I and II follow. Exam Attack: For I: Path P to S has >. (P > Q ≥ R = S). True.For II: Path R to T is R = S < T. The supreme sign < is present. True. Both follow.

Q: Statements: A ≥ B = C ≤ D Conclusions:I. A ≥ CII. C < E

Answer: Both I and II follow. Exam Attack: Path A to C: A ≥ B = C. Highest priority is ≥. True.Path C to E: C ≤ D < E. Highest priority is <. True. Both follow.

Q: Statements: L = M O ≤ PConclusions:I. L II. O < M

Answer: Only Conclusion I follows. Exam Attack: L = M O is a blocked path, so O < M is False.

Q: Statements: J ≥ K > L = M Conclusions:I. J > MII. K < N

Answer: Only Conclusion I follows. Exam Attack: Path J to M: J ≥ K > L = M. Supreme sign is >. J > M is True. Path K to N is blocked (K > L = M < N), so II is False.

Q: Statements: S WConclusions:I. T II. T = V

Answer: Either I or II follows. Exam Attack: T ≤ U = V gives T ≤ V. Neither < nor = is completely true alone. They form an Either/Or case.

Q: In which of the following expressions does the expression 'O < M' and 'R ≥ P' definitely hold true?

Answer: M = N > O ≥ P = Q ≤ R. Exam Attack: In option B: M = N > O gives M > O (True). Also P = Q ≤ R gives P ≤ R which means R ≥ P (True).

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Expert Answer & Key Takeaways

Mastering Inequality (Coded & Reverse) is essential for high-fidelity technical architecture and senior engineering roles in 2026.

Fundamentals of Inequality

Inequality tests your ability to deduce the relationship between variables mathematically. It revolves around three primary symbols (=, >, <) and two secondary symbols (≥, ≤).
The absolute core principle is the Priority Order.
1. Supreme Priority: The definite symbols ( > or < )
2. Second Priority: The probable symbols ( ≥ or ≤ )
3. Lowest Priority: The equal symbol ( = )

The 'Either Or' Case

This occurs when neither conclusion is individually true, but TOGETHER they cover all possible relationships.

Coded Inequality & The Magic Box

In Coded Inequality, symbols (*, $, #, %, @) represent the inequality signs.

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Inequality (Coded & Reverse)

Expert Answer & Key Takeaways

Fundamentals of Inequality

The 'Either Or' Case

Coded Inequality & The Magic Box

Course4All Reasoning Panel

Explore More Reasoning Ability

Machine Input-Output (Advanced Level)

Statement & Assumptions

Statement & Conclusion

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