Inequalities

Study Materials › Reasoning › Inequalities

You are given a chain such as A > B ≥ C = D and asked which conclusions hold. The whole topic rests on one idea: a conclusion is valid only when an unbroken chain pointing in a single direction connects the two terms. Both 2022 papers carried one of these, and the SI version wrapped it in a code.

Which Combinations Give a Conclusion

ChainConclusion
A > B > CA > C  ✓
A > B ≥ CA > C  ✓
A ≥ B ≥ CA ≥ C  ✓
A = B > CA > C  ✓
A > B < Cnothing — directions oppose
A ≠ Bnothing about which is larger
The weakest link decides the answer. A chain containing one ≥ and several > can only conclude > if the strict sign genuinely lies on the path. Mixing > with < anywhere in the path yields nothing at all.

The ≠ Trap

A ≠ B tells you they differ, not which is larger.

It therefore breaks a chain completely for > and < conclusions — but it can still support a conclusion of the form "A ≠ C" if equalities connect the rest.

Either-Or Cases

If two conclusions concern the same pair of terms and between them cover all possibilities — typically A ≥ B and A < B, or A > B and A = B — and neither holds alone, the answer is "either I or II follows".

Worked Examples from Real Papers

TS Police Constable 2022 — Q115

Let A = B ≠ C = D ≤ E < F. Which of the following is correct?
(1) A > C   (2) D > B   (3) A > C ⇒ D > B   (4) D > B ⇒ A < C

From the chain: A = B and C = D, with B ≠ C. So A and C differ, but nothing tells us which is larger.
(1) A > C — undetermined, so not correct.
(2) D > B — equally undetermined.
(3) Suppose A > C. Since A = B and C = D, that means B > D, so D > B is false. The implication fails.
(4) Suppose D > B. Since D = C and B = A, that means C > A, i.e. A < C. The implication holds.
Answer: D > B ⇒ A < C.
Method: options (1) and (2) are plain claims, which the ≠ makes undecidable. Options (3) and (4) are conditionals — and a conditional can be true even when neither side is separately known. Substituting the equalities is all that is needed.

TS Police SI 2022 — Q84

The symbols ≠̲, ≠, ≠̅, >, < and = are represented respectively by A, B, C, D, E and F. Given the statements PAQ, QFR and SER, which implications are correct?
(1) PDS, RCP   (2) PES, RBF   (3) RBQ, PFS   (4) PAS, PCR

Decode the letters first: A means "not less than" (≥), B means ≠, C means "not greater than" (≤), D means >, E means <, F means =.
PAQ → P ≥ Q  ·  QFR → Q = R  ·  SER → S < R.
Combining: P ≥ Q = R, so P ≥ R. And S < R ≤ P, so P > S.
Now test option (1): PDS is P > S — established above. RCP is R ≤ P — also established. Both hold.
Answer: PDS, RCP.
Method: write the decoded chain out in ordinary symbols before touching the options. Working directly in the coded letters is where this question is lost.

Watch for negated symbols. "Not less than" is ≥, not >, and "not greater than" is ≤, not <. Both appear in the SI question above, and reading either as a strict inequality changes the answer.

Also see: Syllogism · Coding & Decoding · Non-Verbal Reasoning

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