Inequality Questions and Answers

Answer: E

Correct answer: Both conclusions I and II follow

The statements are \( Q = Z \), \( C \ge G \), \( G \ge Q \), \( Q \ge R \) and \( J \ge C \). Lining them up gives one long chain: \[ J \ge C \ge G \ge Q = Z \ge R. \]

Conclusion I claims \( G \ge Z \). Read the chain from G: \( G \ge Q \), and \( Q = Z \), so \( G \ge Z \). Follows.

Conclusion II claims \( C \ge R \). Read from C: \( C \ge G \ge Q \ge R \). Every link points the same way, so \( C \ge R \). Follows.

Both follow. Note that every link here is weak - \(\ge\) or \(=\) - so both conclusions come out weak too; had either conclusion asked for a strict inequality it would have failed, because nothing in the statements forbids all five values being equal.

Answer: D

Correct answer: Neither conclusion I or II follows

The statements are \( A > B > C \), \( D > E > F \) and \( D > C \).

There are two chains here, and the third statement is the only thing joining them - it ties D, the top of the second chain, to C, the bottom of the first. That is a weak join: it tells us D outranks C, but says nothing about where E, F, A or B stand relative to the other chain.

Conclusion I claims \( E > C \). We know \( D > E \) and \( D > C \), so E and C both sit below D with no order between them. Taking \( D = 10 \), \( E = 9, C = 1 \) works and so does \( E = 2, C = 9 \) - provided B and A are raised to suit. Does not follow.

Conclusion II claims \( F > B \). F is the bottom of the second chain and B is in the middle of the first, and no chain of statements runs between them at all. Does not follow.

Neither conclusion follows.

Answer: E

Correct answer: Both conclusions I and II follow

Chain the statements into one line: \[ K \ge M \ge N > O. \] Conclusion I claims \( O < M \). From \( M \ge N \) and \( N > O \) we get \( M > O \), so I follows. Conclusion II claims \( O < K \). Extending the same chain, \( K \ge M > O \), so \( K > O \) and II follows too. A strict inequality anywhere in a chain of \(\ge\) links makes the whole chain strict, which is what carries both conclusions. The symbol between K and M had been lost from this record and is restored above as \(\ge\); had it been the strict \(>\) instead, both conclusions would still follow, so the answer is unaffected either way.

Answer: E

Correct answer: Only IV is true

The chain is \[ L > K \ge X < J = N \le Y < D. \] Its direction reverses at X - everything to the left descends towards X, everything to the right climbs away from it - so X is a low point, and no relation carries across it.

I. \( X \ge Y \): reading right from X gives \( X < J = N \le Y \), so \( X < Y \). The conclusion is definitely false.

II. \( L > N \): L is left of the reversal and N is right of it, so no chain links them. Not determined.

III. \( K > J \): the same break lies between them. Not determined.

IV. \( D > X \): here both elements sit on the same side of the low point, and the chain \( X < J = N \le Y < D \) runs one way throughout with strict links in it. True.

Only IV holds.

Two rules settle every question of this kind. A conclusion follows only when an unbroken chain of links runs between its two elements all pointing the same way; and the chain is strict if it contains at least one strict link, weak if every link is weak.

Answer: D

Correct answer: Only III and IV are true

The chain is \[ P > J \ge V = D > L \le C. \] It descends steadily from P to L and then turns upward again at L, so L is the low point and C is cut off from everything to its left.

I. \( P > C \): C is reached only through the reversal at L, so nothing fixes it against P. Not determined.

II. \( D \ge C \): the same break, one link nearer. Not determined.

III. \( L < P \): the whole run \( P > J \ge V = D > L \) points one way and contains strict links, so \( P > L \). True.

IV. \( J > L \): the same run, started one element later - \( J \ge V = D > L \) - still with a strict link in it. True.

III and IV hold and the two conclusions involving C do not.

Two rules settle every question of this kind. A conclusion follows only when an unbroken chain of links runs between its two elements all pointing the same way; and the chain is strict if it contains at least one strict link, weak if every link is weak.

Answer: B

Correct answer: Only IV is true

Two statements are given: \( B \ge A < K < F \) and \( F < B = D \). The second is what makes the question tractable, because it feeds F back into B - and B into D.

Combining them gives \[ K < F < B = D, \] which settles three of the four conclusions outright.

I. \( B < K \): we have just shown \( K < F < B \), so \( B > K \). Definitely false.

II. \( A > D \): from \( B \ge A \) and \( B = D \) we get \( D \ge A \). Definitely false.

III. \( D < F \): from \( F < B = D \) we get \( D > F \). Definitely false.

IV. \( F \le B \): \( F < B \) is given, and a strict inequality satisfies the weak one. True.

Only IV holds. Conclusion IV is the one candidates most often reject, on the grounds that the statements say less than rather than at most; but \( \le \) asks only that F is not greater than B, which \( F < B \) guarantees.

Answer: A

Correct answer: All I, II, III and IV are true

The statement is a single unbroken chain: \[ M < N < U \le R = T \ge V \ge E. \] It rises from M up to the plateau \( R = T \) and then falls away to E, so T is the high point - and because every conclusion compares T or R against something on one side of it, each one has an unbroken run to work with.

I. \( T > N \): \( N < U \le R = T \), strict because of the first link. True.

II. \( R \ge E \): \( R = T \ge V \ge E \), every link weak, and the conclusion asks only for the weak relation. True.

III. \( M < T \): \( M < N < U \le R = T \), strict twice over. True.

IV. \( T \ge U \): \( U \le R = T \). True.

All four follow. It is worth noticing that no conclusion asks to compare something on the left of T with something on its right - M against E, say - which is exactly the comparison the shape of this chain would refuse.

Answer: C

Correct answer: Only IV and either I or III are true

The statements give \( R > B \ge S \ge D \) and \( S \ge O \ge U \), with \( S \ge D \) repeated.

IV. \( R > O \): the run \( R > B \ge S \ge O \) points one way and opens with a strict link, so \( R > O \). True.

II. \( D \le O \): D and O both hang below S on separate branches, and two elements below a common third have no order between them. Not determined.

Conclusions I and III are the interesting pair. The run \( B \ge S \ge O \ge U \) is weak at every link, so it gives \( B \ge U \) and no more. I claims \( B > U \) and III claims \( U = B \); neither can be established on its own, since all four values could be equal or B could stand well above U. But \( B \ge U \) means B is either greater than U or equal to it, with no third possibility, so one of the two must hold even though we cannot say which. That is precisely the “either…or” case.

So IV is definitely true, and either I or III is true: the keyed answer. The pattern to recognise is a complementary pair - \(>\) against \(=\), or \(<\) against \(\ge\) - arising from a chain of weak links.

Answer: B

Correct answer: Only I is true

The statements give \( D > A \ge M \le N \) and \( M \ge E < L \). M is the pivot: the first chain descends to it from D and then rises to N, and the second descends from it to E and then rises to L. Anything that has to cross M from one branch to another is therefore undecidable.

I. \( D > E \): here the run does not cross a reversal - \( D > A \ge M \ge E \) descends throughout, and the first link is strict. True.

II. \( N \le A \): both N and A stand above M on different branches, so their order is free. With \( M = 1 \), \( A = 2, N = 9 \) and \( A = 9, N = 2 \) both satisfy every statement. Not determined.

III. \( A \ge L \): L stands above E and A stands above M, and the two are joined only through the reversal at E. Not determined.

IV. \( L < D \): the same obstacle - L is reached only by climbing back up from E, so nothing caps it. Not determined.

Only I follows.

Answer: A

Correct answer: Q < P ≥ B > D > F = G < H

This is the only option supplying both chains. Reading it, P ≥ B > D > F = G gives P > G, and F = G with G < H gives F < H. In each of the other options at least one of the two required relations passes through a reversal, so it cannot be established.