Inequality Questions and Answers

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: K ≥ H

From H ≤ G and G = K it follows that H ≤ K, which is the same statement as K ≥ H. Nothing links F to H or to G except through K, and the inequality there runs the wrong way - K > F tells us F is below K but says nothing about how F compares with H, which is also below K.

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

Correct answer: Either I or II and III is true

Translating: M < R, R ≤ T, T = N, so M < R ≤ N. Since R ≤ N, either N = R or N > R, which makes conclusions I and II a complementary pair - exactly one holds, so either follows. And N ≥ R > M gives N > M, so III is true outright.

Answer: D

Correct answer: All are true

In this set the code reads: \(@\) is “not smaller than”, so \(\ge\); \(\#\) is “not greater than”, so \(\le\); \(\%\) is “neither greater nor equal”, so \(<\); \(*\) is “neither smaller nor greater”, so \(=\); and \(\$\) is “neither smaller nor equal”, so \(>\). Note that this is not the same code as the neighbouring sets use - check the direction before translating. The statements become \( J \le N \), \( K \ge N \) and \( T > K \), which join into the single chain \[ J \le N \le K < T. \]

I. \( J \% T \) asks whether \( J < T \). The chain runs from J all the way to T, and it contains the strict link \( K < T \), so the whole relation is strict. True.

II. \( T \$ N \) asks whether \( T > N \). Reading the chain backwards from T gives \( T > K \ge N \), again strict. True.

III. \( N @ J \) asks whether \( N \ge J \). That is the first statement read the other way round. True.

All three follow, so the answer is that all are true.