Inequality Questions and Answers

Answer: B

Correct answer: Only Conclusion II is true.

The statements are \( P < A \le C \ge K \), \( C \le O < L \) and \( R \le N \le C \).

II. \( R < L \): assemble the run through C. \( R \le N \le C \le O < L \) points one way throughout, and the final link is strict, so \( R < L \). True.

I. \( A = K \): A and K both hang below C - \( A \le C \) and \( K \le C \) - on separate branches. That permits them to be equal but does not require it: with \( C = 10 \), \( A = 10, K = 3 \) satisfies every statement, and so does \( A = 3, K = 10 \). Not determined.

Only II follows. Conclusion I is worth pausing on, because an equality conclusion can only be established by an unbroken run of \(=\) or \(\le\)-and-\(\ge\) links that pin the two values together; two elements merely capped by the same third are never forced into equality.

Answer: E

Correct answer: Both conclusions I and II are true.

The statements are \( S \le T \le K \le E \) and \( N \ge E \le P \). E is the junction: the first chain climbs to it from S, and both N and P stand at or above it.

I. \( S \le N \): the run \( S \le T \le K \le E \le N \) points one way at every link, so \( S \le N \). True.

II. \( S \le P \): the same run with P in place of N - \( S \le T \le K \le E \le P \). True.

Both follow. What makes this one work where so many others fail is that the two branches leaving E both point upwards, in the same direction as the chain arriving at it, so E is not a pivot and the relations carry straight through. The comparison this arrangement would still refuse is N against P - those two are the pair on separate branches above a common element.

Answer: D

Correct answer: Neither conclusion I nor II is true.

The statements are \( S > O \ge U \ge P \), \( O \le T < R \) and \( T \ge E > A \). T is the hub: O climbs to it, and both R and E hang off it, R above and E below.

I. \( R < A \): follow the chain from R downwards. \( R > T \ge E > A \) gives \( R > A \), so the conclusion is definitely false - the comparison can be made, and it comes out the other way.

II. \( U = A \): U sits at or below O, which sits at or below T; A sits below E, which sits at or below T. Both are therefore capped by T from different directions, and nothing pins them together. With \( T = 10 \), \( U = 9, A = 1 \) and \( U = 1, A = 9 \) both satisfy every statement. Not determined.

So one conclusion is false and the other undecidable, and neither follows.

Answer: A

Correct answer: Both conclusions I and II are true.

The statements are \( P < A \le C \ge K \), \( C \le O < L \) and \( R \le N \le C \). C is the junction, and both conclusions happen to route through it in a single direction.

I. \( O \ge R \): the run \( R \le N \le C \le O \) points one way at every link, giving \( R \le O \), which is what the conclusion asks. True.

II. \( P \le O \): the run \( P < A \le C \le O \) also points one way, and its opening link is strict, so in fact \( P < O \). A strict inequality satisfies the weak one the conclusion states. True.

Both follow. Note that the same statements, asked about A and K, would give nothing - those two hang below C on separate branches. It is the routing of the question, not the statements alone, that decides.

Answer: C

Correct answer: Either Conclusion I or II is true.

The statements are \( S > O \ge U \ge P \), \( O \le T < R \) and \( T \ge E > A \).

Both conclusions compare T with P, so assemble that run first: \( T \ge O \ge U \ge P \). Every link is weak, so the most that can be said is \[ T \ge P. \]

I claims \( T > P \) and II claims \( T = P \). Neither is forced - taking all four of T, O, U and P equal satisfies the statements and makes II true, while \( T = 10, O = 5, U = 3, P = 1 \) makes I true. But \( T \ge P \) leaves exactly these two possibilities and no third, so one of them must hold.

That is the “either…or” case: the two conclusions form a complementary pair spanning everything the statements allow. The signature to watch for is a chain of purely weak links between the two elements named - one strict link anywhere in it would have settled the matter and made I definitely true.

Answer: A

Correct answer: Only conclusion I is true.

Read the chain from B rightwards: \( B = Q \le P < J \le Y \). A single strict link anywhere in a chain makes the whole relation strict, and there is one at \( P < J \), so \( B < Y \) - conclusion I is true.

Conclusion II claims \( X \ge J \). Here the chain works against us. We know \( J \le Y \) and \( Y > X \), so both J and X sit at or below Y, but the two are approached from the same side and nothing fixes their order relative to each other. Take \( Y = 10 \): then \( J \) could be 9 and \( X \) 2, making \( X < J \); or \( J \) could be 2 and \( X \) 9, making \( X > J \). Both are consistent with every statement, so conclusion II does not follow.

Only the first conclusion holds. The general rule this illustrates: a conclusion follows only when an unbroken chain of relations runs between its two elements all pointing the same way; two elements that both hang below a common third are never ordered by that alone. The statement in this record had been corrupted by a stray < in the stored HTML, which swallowed the middle of the chain; it is restored above.

Answer: E

I. Z ≥ A ≥ B = Q [FALSE]

II. Z ≥ A ≥ B = Q [FALSE]

Answer: D

Correct answer: Both conclusions I and conclusion II are true.

The statements give \( G < R \), \( R = A \), \( A \le S \) and \( R > T \).

Conclusion I claims \( G < S \). Chain it: \( G < R = A \le S \). The strict link at the left end carries through the equality and the weak link, giving \( G < S \). True.

Conclusion II claims \( S > T \). Chain it the other way: \( S \ge A = R > T \). Again a single strict link makes the whole relation strict, so \( S > T \). True.

Both follow, so the answer is that both conclusions are true. Note how R does the work in each case - it is the element every other one is tied to, and finding that pivot is usually the quickest route through a question of this shape. The statements in this record had been destroyed by a stray < in the stored HTML; they are restored above from the relations the surviving working refers to.

Answer: B

Correct answer: Neither conclusion I nor II is true.

The statements give P = U < M > N, with D ≥ P and I ≥ C. Conclusion II asks whether N > U, but the chain reverses at M and both U and N lie below it with no link between them. Conclusion I involves C, which is connected only to I and to nothing else in the problem, so it cannot be reached at all.

Answer: C

Correct answer: Only conclusion II is true, though the statements this depends on are damaged

The statements in this record have been corrupted. What survives reads “P = U < M> N; D ≥ P; I ≥ C”, where a stray < in the stored HTML has swallowed the text between M and N, and conclusion I speaks about an element K that appears nowhere in the statements at all - so at least one further relation, the one introducing K, has been lost with it.

What can still be said is limited. From \( D \ge P \) and \( P = U \) we know D stands at or above both P and U, but nothing connects D to K, so conclusion I cannot be established from what remains - consistent with the key, which treats it as not following. Conclusion II, \( I > P \), must have followed through the missing link: \( I \ge C \) is the only statement mentioning I, so the lost text has to have tied C to P strictly enough to give \( I > P \).

The item needs its original statements restored before the reasoning can be shown in full; unlike the two questions beside it, the damage here removed a relation rather than merely obscuring one, and it cannot be inferred from the surviving text.