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: C

The correct statement is \(a \lt b + c\).

Decode the given statement. '×' (written X in the list) means 'greater than' and 'Δ' means 'less than', so \(a \times b\,\Delta c\) says \(a \gt b\) and \(b \lt c\).

In option C, '\(\lt\)' means 'not less than' and '+' means 'not greater than', so it reads \(a \ge b\) and \(b \le c\). Both follow from \(a \gt b\) and \(b \lt c\).

Option A says \(a = c\) and \(c \lt b\), which contradicts \(b \lt c\). Option B says \(b \ge a\), which contradicts \(a \gt b\). Option D says \(c \le b\), which again contradicts \(b \lt c\).

Answer: D

The statement that is wrong is B > D.

Rearrange the two conditions:

\[ \begin{aligned} A + B = C + D &\Rightarrow A - C = D - B \\ A + D \gt B + C &\Rightarrow A - C \gt B - D \end{aligned} \]

Substituting the first into the second gives \(D - B \gt B - D\), so \(D \gt B\). Therefore B > D must be wrong. It also follows that \(A - C = D - B \gt 0\), so A > C is true.

A > B and C > D are not fixed by the data either way, so they cannot be called wrong; only B > D contradicts the conditions.

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.