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: D

Correct answer: Only III is true

This set’s code reads: \(\$\) is “not smaller than”, so \(\ge\); \(@\) is “neither smaller nor equal”, so \(>\); \(\#\) is “neither greater nor equal”, so \(<\); \(\&\) is \(=\); and \(*\) is “not greater than”, so \(\le\). The statements become \( A \ge M \), \( P > L \), \( K < P \) and \( A \ge L \).

III. \( L \le A \) is the fourth statement read the other way round: \( A \ge L \) is precisely \( L \le A \). True.

I. \( K < L \): K and L both sit below P, so nothing orders them. Not determined.

II. \( A > P \): A and P are connected only through L, which both stand above - \( A \ge L \) and \( P > L \). Not determined.

IV. \( M < P \): M hangs below A, and A is itself unordered against P, so M is further adrift still. Not determined.

Only III follows - and it follows for the simplest possible reason, being a statement restated. Checking each conclusion against the statements for an exact restatement before reasoning about chains often saves the work.

Answer: B

Correct answer: Only I is true

This set’s code reads \(\$\) as \(\ge\), \(@\) as \(>\), \(\#\) as \(<\), \(\&\) as \(=\) and \(*\) as \(\le\). The statements become \( W \le N \), \( K \le V \), \( Y > V \) and \( W > K \).

I. \( Y > K \): the run \( Y > V \ge K \) points one way and opens strict. True.

II. \( W \ge N \): the first statement says \( W \le N \), which allows equality but does not compel it, so the reverse relation cannot be asserted. Not determined.

III. \( W > Y \): W stands above K and Y stands above V, which is itself above K - so both are above K, on separate branches, and unordered. Not determined.

IV. \( W > V \): the same structure, one link shorter. Not determined.

Only I follows. Conclusion II is the one to watch: reversing a weak inequality is never valid, because \( W \le N \) leaves open that W is strictly smaller.

Answer: A

Correct answer: None is true

This set’s code reads \(\$\) as \(\ge\), \(@\) as \(>\), \(\#\) as \(<\), \(\&\) as \(=\) and \(*\) as \(\le\). The statements become \( T \le Y \), \( S < M \), \( Y \ge S \) and \( M > K \). Drawn out, this is two loose fragments joined only at S: T and Y stand above S, while M stands above S as well and K below M. Nothing runs from one fragment through to the other in a single direction.

I. \( K < S \): K and S both sit below M, so their order is free. Not determined.

II. \( Y > M \): Y and M both sit above S. Not determined.

III. \( T < M \): T is only bounded above, by Y, and Y itself is unordered against M. Not determined.

IV. \( Y > K \): Y stands above S and K below M, and the two are three links apart with a reversal between them. Not determined.

None of the four follows. Every failure here is the same one - two elements attached to a common third from the same side.

Answer: B

Correct answer: I, II are true

This set’s code reads \(\$\) as \(\ge\), \(@\) as \(>\), \(\#\) as \(<\), \(\&\) as \(=\) and \(*\) as \(\le\). The statements become \[ S > L < M = B \le Q. \] L is a low point, so S on its left is cut off from everything on its right.

I. \( Q \ge M \): from \( M = B \le Q \). True.

II. \( B > L \): from \( L < M = B \), strict. True.

III. \( S > Q \): S and Q sit on opposite sides of the pivot at L, so neither bounds the other. Not determined.

IV. \( L > Q \): the chain gives the opposite - \( L < M = B \le Q \), so \( L < Q \). Definitely false.

I and II follow. Note the contrast between III and IV: III fails because the comparison cannot be made at all, IV because it can be made and comes out the other way.

Answer: A

Correct answer: None is true

This set’s code reads \(\$\) as \(\ge\), \(@\) as \(>\), \(\#\) as \(<\), \(\&\) as \(=\) and \(*\) as \(\le\). The statements become \[ C < P \le L > E \ge M. \] L is a high point, and every conclusion asks about a pair that straddles it or hangs below it on separate branches.

I. \( M < P \): M sits at or below E, and P at or below L, with the peak between them. Not determined.

II. \( M > L \): the chain gives \( M \le E < L \), so \( M < L \). Definitely false.

III. \( P < E \): both stand below L on separate branches. Not determined.

IV. \( L < C \): the chain gives \( C < P \le L \), so \( L > C \). Definitely false.

None of the four follows - two because the comparison is impossible, two because it comes out the opposite way.