Inequality Questions and Answers

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.

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: C

Correct answer: None of these

Translating the code: \(@\) is “greater than”, \(\#\) is “greater than or equal to”, \(\%\) is “less than or equal to”, \(*\) is “equal to” and \(\$\) is “less than”. The statements read \( R < S \), \( S > T \) and \( T \ge U \). Chaining the last two gives \( S > T \ge U \), so \( S > U \). Conclusion III is \( S @ U \), that is \( S > U \) - true. But conclusion I is \( U \% S \), that is \( U \le S \), and \( S > U \) certainly makes \( U \le S \) hold - so I is true as well. Conclusion II is \( T < R \); here \( R < S \) and \( T < S \) both point at S from below, which fixes no order between R and T, so II does not follow. Conclusions I and III are true and II is not, and no option offers that combination - hence “None of these”. The trap is reading \(\le\) as requiring equality to be possible; a strict inequality satisfies it.

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: 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: 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: C

Correct answer: < , =, = , <

The symbols give F < O = U = N < D. From F < O = N comes F < N, so F > N is definitely false; and from U = N < D comes U < D, so U > D is definitely false. Both statements are required to be false, which needs the opposite relation established in each case rather than merely left open.

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: E

Correct answer: P, N, A, L

Filling the blanks gives P < N < A > L, from which A > N > P and therefore A > P. The expression asked for is one making A < P definitely false, and establishing the opposite relation is what does that - leaving the two unrelated would make it undetermined rather than false.