Inequality Questions and Answers

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.

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: E

Correct answer: C, A, D, B, E

Filling the blanks gives C < A ≤ D = B > E. The last two terms then read D = B and B > E, so D > E follows directly. What matters is that D is on the left of the equals and E on the right of the greater-than, so the two are joined without any reversal between them.

Answer: E

Correct answer: Both B and C

With the chain A > B = C > D ? E ≥ F = G, both required relations need D to stand at or above E. A ≥ gives D ≥ E ≥ F = G and B = C > D ≥ E; an equals gives the same with D = E. Either therefore establishes E < B and D ≥ G together, while a ≤ or a < at that point breaks both.

Answer: C

Correct answer: Only I and II are true

Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ A \ge P > E < F \le S. \] E is a low point: the chain falls to it from the left and rises from it to the right, so nothing carries across E.

I. \( S > E \): both lie on the right of the pivot, joined by \( E < F \le S \), which is strict. True.

II. \( A > E \): both lie on the left, joined by \( A \ge P > E \), again strict. True.

III. \( F > P \): P sits left of the pivot and F sits right of it, so no chain connects them. With \( E = 1 \), \( P = 2, F = 9 \) works and so does \( P = 9, F = 2 \). Not determined.

I and II follow, III does not.

Answer: D

Correct answer: All I, II and III are true

Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ P < W = Q > S \ge A. \] Q is a high point here, but every conclusion compares an element with Q or W rather than reaching across the peak, so each has an unbroken run to use.

I. \( A < Q \): reading down the right branch, \( Q > S \ge A \), strict at the first link. True.

II. \( Q > P \): the left branch, \( P < W = Q \). True.

III. \( W > A \): W equals Q, so this is conclusion I again with the peak named by its other letter: \( W = Q > S \ge A \). True.

All three follow. The comparison the chain would not settle is P against A - the two ends, one on each side of the peak - and the question does not ask for it.

Answer: A

Correct answer: Only either I or II and III are true

Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ L > N \le T = D < A. \] N is a low point, so L on its left and A on its right are not connected by any chain.

III. \( A > N \): both sit on the right of the pivot, joined by \( N \le T = D < A \), which is strict at the last link. True.

Now take I and II together, since that is what the answer turns on. I claims \( L > A \) and II claims \( L \le A \). Neither can be established, because L and A stand on opposite sides of the pivot: with \( N = 1 \) we may have \( L = 9, A = 2 \) or \( L = 2, A = 9 \). But the two claims are exhaustive - L is either greater than A or it is not, and “not greater” is exactly \( \le \) - so one of them must hold. That is the “either…or” case, and it arises here from a complementary pair rather than from the statements settling anything.

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

Answer: E

Correct answer: All I, II and III are true

Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \( M \le Q \), \( Q = K \), \( A > K \) and \( V \ge A \). Since Q and K are equal they can be treated as one element, which lines the rest up as \[ V \ge A > K = Q \ge M. \]

That single descending chain settles all three conclusions.

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

II. \( A > Q \): from \( A > K = Q \). True.

III. \( A > M \): the run \( A > K = Q \ge M \) points one way and contains a strict link. True.

All three follow. Spotting the equality early is what makes this quick: an \(=\) link lets two letters be merged, and a chain with nothing but one reversal-free direction settles every comparison along it.

Answer: A

Correct answer: None is true

Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ E = C < A \ge R \le S. \] A is a high point: the chain climbs to it from the left and descends from it to the right. Two elements that both sit below a common third are in no fixed order, and every conclusion here asks for exactly such a comparison.

I. \( S > A \): S is reached from A by descending to R and climbing again, so S may be anywhere at or above R. With \( A = 10, R = 1 \), S could be 2 or 20. Not determined.

II. \( R < C \): R and C both stand below A on separate branches. Not determined.

III. \( R \le E \): E equals C, so this is conclusion II again in weak form, and it fails for the same reason. Not determined.

None of the three follows.

Answer: C

Correct answer: either conclusion I or II is true

Translating: H = W, W ≤ R, so R ≥ H. That leaves exactly two possibilities, R = H or R > H, which are conclusions I and II. Neither can be established alone, but between them they exhaust the cases - which is what an either-or conclusion means.