Inequality

A direct-chain question gives one line of letters, such as M ≥ N = O > P ≥ Q, and asks which conclusions about it are definitely true. The word to remember is path. To compare two letters, walk along the chain from one to the other. If every link on the walk points the same way, the two letters are related, and the relation comes from combining the links. If the walk changes direction anywhere, the pair is not related (that is Type 2).

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In the picture the walk from M to Q uses ≥, =, >, ≥. All four belong to the greater family and one link is strict, so the result is strict: M > Q. Now match the result with each conclusion, using the allowed-situations idea from the foundation.

The smaller family is the mirror image. Find the path, combine the signs, then read the conclusion against what is left.

Two newer forms use the same logic. The first is the count form: four or five numbered conclusions, and the question "how many are definitely true?". Give every conclusion its own label: definite (it covers every open situation), false (it covers none) or possible (it covers some). Count only at the end, after re-reading the question: "not definitely true" means possible and false together, "may be true" means definite and possible together.

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In the picture, A ≥ B = C > D with E < C is sorted against the conclusions 1. A > D, 2. A = E, 3. B ≥ D, 4. D ≤ E, 5. B > E. Conclusions 1, 3 and 5 are definite, 4 is possible (D and E hang below C) and 2 is false.

The second is the word form: "Radhika scored more than Meena but less than Kavita. Meena scored more than Sunita." Each comparison word is a sign (more, taller, older → >; less, shorter, younger → <; as much as → =) and each name is a letter. Write the first-named person on the left, flip the sign when you turn a sentence around, and join through the common name.

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One difference: two names that hang from the same name in the same direction cannot be ordered, so such a question is "cannot be determined". Questions on exact positions are developed in the Order and Ranking chapter.

This type decides whether a conclusion can be made at all. A chain is a road. As long as you keep going uphill or keep going downhill, you can compare the start with the end. But the moment the road turns, from uphill to downhill (a peak) or from downhill to uphill (a valley), the two ends can no longer be compared.

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Take C < D ≥ E = F > G. Going from C to D is uphill. Going from D to E is downhill. So D is a peak, and C and F lie on different slopes below it. How far down is C, and how far down is F? The statement does not say. So C and F have no relation, and all of greater, equal and smaller stay possible for them.

Find the turning letters by scanning the signs. A turn happens where a "smaller-family" sign meets a "greater-family" sign, in either order. The sign = never makes a turn, and it never cancels one: C < D = E > F is still a peak at the D-E pair.

Now notice that the peak does not spoil every pair. D ≥ E = F > G is a clean downhill walk from D, so D > G, and G ≤ D is definite. Only a pair whose path contains a turn is unrelated; every pair whose path stays on one slope is still comparable.

For the unrelated pair, a conclusion with a single sign (C > F, C = F or C < F) is only possible. Even C ≥ F and C ≤ F are not definite, because each leaves one situation out.

In the paper the two conclusions are often set so that one stays on a slope and the other crosses a turn, so decide each conclusion on its own. A long statement can have many turns. Mark them first (a small ∧ over every peak, a small ∨ under every valley) and read the conclusions only after that.

Sometimes the statement is not one line but two or three short chains separated by semicolons, such as A > B ≥ C; D ≤ C = E. Each chain is a small piece of the puzzle. The pieces are held together by a common letter. Here C appears in both chains, so C is the joint.

The aim is to turn the pieces into one long chain through the common letter. At the joint, look at the two arrows touching it. If one arrow comes into the common letter and the other goes out of it, the direction continues, and you can join. If both arrows point into it, or both point out of it, the joint is a valley or a peak, and the pieces cannot be joined (Type 2 again).

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In the picture, the second chain D ≤ C = E is read from C and becomes C ≥ D and C = E. Reversing a chain flips every sign, so ≤ turns into ≥. Now C feeds two branches, and A > B ≥ C continues through C into both. So A > D and B ≥ E.

Rewrite the second chain so that the common letter stands on the side where the first chain continues, flipping the signs if you turn it around. If there are three statements, join the first two, then join the result with the third.

The paper adds two things. First, some statements are only distractors: one may repeat what an earlier chain already said (T ≥ A after A ≤ T), and another may hang a side branch that no conclusion needs. Use only the statements that lie on the path between the two letters. Second, the question is sometimes worded "Are Statement I and Statement II sufficient to conclude h > i?". The joining is the same: if h > j and j > i fail alone but join at j, the answer is "both together". The full method of this form is in the Data Sufficiency chapter.

Either-or is a question about gaps. After the chain has proved something about a pair of letters, some situations may still be open. Two conclusions are then given about the same pair, and neither is definite on its own. The question is: do they together cover every open situation, with no situation covered by both? If they do, exactly one of them must be true, and the answer is "Either I or II follows".

Take E > F ≥ G = H. For F and H the chain gives F ≥ H. So "greater" and "equal" are open, and "smaller" is closed. Conclusion I (F > H) covers only "greater". Conclusion II (F = H) covers only "equal". Neither is definite. But together they cover both open situations, so the answer is either-or.

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Follow three steps every time. Step 1: list the open situations for the pair. Step 2: tick what each conclusion covers. Step 3: if neither is definite and the ticks together fill every open situation with no situation ticked twice, choose "Either I or II".

You do not need to redraw the table each time. Remember the common pairs:

With no relation, the pair (X > Y, X < Y) is never either-or, because "equal" is left out. Likewise (X ≥ Y, X ≤ Y) is not either-or: both are true when X = Y, so the two overlap. Also keep the order of checks: first see whether either conclusion is already definite. If one is, the answer is about that one, not either-or. Finally, both conclusions must be about the same two letters.

In this type the usual signs are replaced by symbols such as @, $, #, %, © or Greek letters, and a key says what each one means: "A © B means A ≥ B". The logic is exactly the same as before. The only new work is decoding, and the paper sets the key in two ways: in signs ("A © B means A ≥ B") or in words ("A # B means A is not greater than B").

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The routine never changes. Step 1: copy the key as a small table, code on the left and real sign on the right. Step 2: rewrite every statement in real signs, line by line, and do not chain while you decode. A statement may come as one string, such as M Φ N Ω O Δ P, or as separate lines, J @ K and K # L; once decoded they are ordinary chains that join through common letters. Step 3: chain, join and judge as in Types 1 to 4, decoding the conclusions with the same key.

Two habits save marks. The key belongs only to the question in front of you: the same symbol can mean ≥ in one set and < in the next. And letter order matters: "B $ A" is the opposite of "A $ B".

When the key is in words, use the cross-out method from the foundation. Start from the three situations: greater, equal, smaller. "Not" removes one; "neither … nor" removes two; what is left is the sign.

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So "not greater than" removes > and leaves ≤. "Not smaller than" leaves ≥. "Neither greater than nor equal to" removes > and = and leaves <. "Neither smaller than nor equal to" leaves >. A plain phrase such as "smaller than" removes nothing and is just <. The two common slips are reading "not greater than" as <, and reading "neither greater nor equal" as ≠. Decode first, chain second, judge last.

Here no conclusions I and II are given. The options themselves are relations, and you must find the one that is definitely true (or the one that is not true). The same logic also comes as "What is the relation between A and D?" with sign options, and with one statement or with two.

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Only the path between the two letters matters. Everything outside it, however long, is a distraction. Walk from the first letter to the second and check three things: do all links point the same way, is any link strict, and what does the combination give?

In the picture the statement is A < B ≤ C = D ≥ E. The path from A to D uses <, ≤, =: all smaller family, with one strict link, so A < D. But A and E lie on different sides of D. The path turns at D, so nothing can be said about A and E.

Now choose the option with care.

• A strict result (> or <) gives the strict option. Do not pick a weaker sign such as ≤ if the strict option exists.
• A non-strict result (≥ or ≤) is the answer only if that sign is among the options. If the options are only >, <, = and "cannot be determined", choose "cannot be determined".
• If the path has a turn, that option is out. If every relation is out in a "relation between A and D" question, the answer is "cannot be determined".
• In a "which is NOT true" question, three options are definite and one is not. Check each option on its own path.

Two statements. Sometimes two statements are given, such as P ≥ Q ≥ R ≥ S < T and J ≤ S ≤ K ≤ R. First join them through the common letters (here S and R, as in Type 3), and then test the options one by one. Look only at the path; the rest of the chain is there to distract you.