Inequality Questions and Answers

Answer: D

@ - >   # - <   $ - =    % - ≤     & - ≥

A > B = C < D I. A < D  False II. B > D  False

Answer: D

 @ - >   # - <   $ - =    % - ≤     & - ≥ 

 P ≥ Q ≤ R = S  I. Q = S  False   II. P < S   False

Answer: D

@ - >   # - <   $ - =    % - ≤     & - ≥

P ≥ Q > R ≤ S  I. Q ≤ S   False  II. P ≤ S False

Answer: A

@ - >   # - <   $ - =    % - ≤     & - ≥ 

 A > B < C ≤ D  I. B < D    True  II. D > A False

Answer: D

@ - >   # - <   $ - =    % - ≤     & - ≥

A > B ≤ C > D I. D = B   False   II. B > D False

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.

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: S < R

From R > O, O = A and A > S the chain runs R > O = A > S without reversal, so R > S. Everything involving T fails, since S < T points away from the rest of the chain and leaves T unconnected to R, O and A.

Answer: D

Correct answer: <

The relation wanted runs D ≤ G ? F ≤ E, and E > D needs an unbroken chain from D up to E with at least one strict step. Both the outer links are non-strict, so the strictness must come from the middle: G < F gives D ≤ G < F ≤ E and therefore D < E. An equals or a ≤ would leave only E ≥ D, and a > would reverse the chain entirely.

Answer: D

Correct answer: S > E

Following the chain from S: S = G, G ≥ Q, Q ≥ R, R > E, so S ≥ R > E and therefore S > E. The other options all require crossing the reversal at P ≤ V, or comparing two elements that lie on the same side of a ≥ without a strict link between them.