Inequality Questions and Answers

Answer: D

Correct answer: Either II or III and I are true

Combine the statements into what they jointly say. \( P = M \) ties two letters together, and \( P < R \le T < S \) runs from there up to S, while \( M \le V \) hangs V off the same point from below: \[ V \ge M = P < R \le T < S. \]

I. \( S > M \): reading right from M gives \( M = P < R \le T < S \), strict twice. True.

IV. \( T < P \): the same run shows \( T > P \), so this is definitely false.

II and III are the pair to think about. All we know of V is \( V \ge P \), from \( M \le V \) and \( P = M \). II claims \( V = P \) and III claims \( P < V \); neither is forced, because V could equal P or stand above it. But those two cases exhaust \( V \ge P \), so exactly one of them holds.

So I is definitely true, IV is definitely false, and either II or III is true - which is the keyed answer.

Answer: C

Correct answer: Only II,III and IV are true

The statements give \( B < P \le S = M \), with \( M \ge F \), \( Q \ge M \) and \( M < O \). Since \( S = M \) the two can be merged, leaving M as the hub that everything else attaches to: B and P climb up to it, F hangs below it, and Q and O stand above it.

ii. \( B < Q \): the run \( B < P \le S = M \le Q \) points one way throughout and opens strict. True.

iii. \( Q \ge M \): this is one of the statements restated, which is always worth checking for - a conclusion that merely repeats a premise is true by definition. True.

iv. \( F < O \): \( F \le M < O \), strict at the second link. True.

i. \( P \le F \): P and F both sit at or below M on different branches, so nothing orders them. With \( M = 10 \), \( P = 9, F = 1 \) and \( P = 1, F = 9 \) are equally allowed. Not determined.

So ii, iii and iv follow and i does not.

Answer: D

Correct answer: II and III are true

The statements give \( F \le B \le A \le D \), with \( C = B \) and \( F < E \).

II. \( D \ge B \): the chain \( B \le A \le D \) reads straight off. True.

III. \( A \ge F \): likewise from \( F \le B \le A \). True.

I. \( C < E \): E enters only through \( F < E \), and C equals B, which stands above F. So C and E both lie above F on separate branches and are not ordered by it. With \( F = 1 \), \( C = 2, E = 9 \) and \( C = 9, E = 2 \) both work. Not determined.

IV. \( E > D \): the same obstacle, further along - D is at the top of the first chain and E is on the branch above F, with nothing joining them. Not determined.

II and III follow. The lesson is that \( F < E \) is a dead end: F is the bottom of the main chain, so anything hanging above F is left free.

Answer: B

Correct answer: Only I is true

The statements give \( P = M \), \( Q \le R \), \( T > Q \) and \( M < R \). R is the high point: M and Q both sit below it, and T is tied only to Q from above.

I. \( M < R \): this is one of the statements repeated verbatim. True.

II. \( R > T \): both R and T stand above Q, so their order is free. With \( Q = 1 \), take \( R = 2, T = 9 \) or \( R = 9, T = 2 \). Not determined.

III. \( P > T \): P equals M, which sits below R; T sits above Q. No chain runs between them. Not determined.

IV. \( P > Q \): P and Q both sit below R on separate branches - the mirror image of the II case. Not determined.

Only I follows. Three of the four conclusions fail for the same structural reason: two elements attached to a common third from the same side are never ordered by that attachment.

Answer: C

Correct answer: Only I and II are true

The statements give \( S = A \), \( Q > A \), \( P \ge Q \), \( R \ge S \) and \( R < T \). Merging S with A gives two branches rising from that point: one through Q to P, the other through R to T.

II. \( Q > S \): \( Q > A = S \). True.

I. \( P > S \): extend it - \( P \ge Q > A = S \), strict in the middle. True.

III. \( T < A \): follow the other branch. \( R \ge S = A \) and \( T > R \), so \( T > R \ge A \), giving \( T > A \). The conclusion is definitely false.

IV. \( A \ge R \): we have \( R \ge A \), which permits equality but does not require it, so the reverse relation cannot be asserted. Not determined - and note this is not a complementary pair with anything else on offer, so no “either…or” answer arises.

Only I and II follow.

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.