Inequality Questions and Answers

Answer: B

Correct answer: Only conclusion b follows

The statements are \( A > B \), \( B \le C \), \( D = E \) and \( C > D \). Written as one line they read \[ A > B \le C > D = E, \] which reverses direction at B.

Conclusion II claims \( C > E \). Both lie on the right of the reversal: \( C > D = E \) gives it in one step. Follows.

Conclusion I claims \( A > E \). A sits on the left of the reversal and E on the right, so nothing links them. Concretely, with \( B = 5 \) and \( C = 100 \) we may take \( A = 6 \) and \( D = E = 99 \), which satisfies every statement and makes \( A < E \). Not determined.

Only the second conclusion follows. The trap is that A looks like the largest element because the chain opens with it; but \( A > B \) caps nothing beyond B, and B is the bottom of the chain.

Answer: D

Correct answer: Neither conclusion a nor b follows

The statements are \( A \ge B \), \( C = D \), \( E > F \), \( E \ge D \) and \( B > C \). Assembling them gives \( A \ge B > C = D \), with E standing at or above D and F below E: \[ E \ge D = C, \qquad E > F. \]

Both conclusions compare something in the C–D pair with F, and both fail for the same reason: D and F are attached to E from the same side, D at or below it and F strictly below it, which fixes no order between them.

I. \( D > F \): with \( E = 10 \), take \( D = 9, F = 1 \) and the conclusion holds; take \( D = 1, F = 9 \) and it fails. Both satisfy every statement. Not determined.

II. \( C \ge F \): C equals D, so this is the same comparison in weak form, and the same two cases defeat it. Not determined.

Neither conclusion follows.

Answer: B

Correct answer: Both conclusions a and b follow

The statements are \( A > B \), \( D > E \), \( C = D \), \( E > F \) and \( B \ge C \). They assemble into a single descending chain with no reversal anywhere: \[ A > B \ge C = D > E > F. \]

I. \( B \ge E \): reading from B, \( B \ge C = D > E \) contains a strict link, so in fact \( B > E \) - more than the conclusion asks. Follows.

II. \( B > F \): continue one link further, \( B \ge C = D > E > F \), strict twice over. Follows.

Both follow. When the statements line up into one unbroken chain like this, every comparison along it is decidable, and the only work is checking whether at least one strict link falls between the two elements named.

Answer: D

Correct answer: Neither conclusion a nor b follows

The statements are \( P \le Q \), \( Q > R \), \( T \ge U \), \( S > R \) and \( T = S \). Merging T with S, the picture is R sitting below both Q and S, with P at or below Q and U at or below S: \[ P \le Q > R < S = T \ge U. \] There are two reversals here, at Q and at R.

I. \( R \ge U \): R sits below S and U sits at or below S, so both hang beneath the same element from the same side. With \( S = 10 \), \( R = 9, U = 1 \) and \( R = 1, U = 9 \) both satisfy the statements. Not determined.

II. \( P \le U \): P is capped by Q, U is capped by S, and the two caps are joined only through R, which is below both. Nothing carries across. Not determined.

Neither conclusion follows. Every element here is bounded from above and nothing is bounded from below, which is exactly the arrangement in which almost no comparison can be made.

Answer: C

Correct answer: Either Conclusion a or b follows

The statements give \( C = D \), \( D \le E \le B \) and \( E = F \). Both conclusions compare B with D, so assemble that run: \[ D \le E \le B. \] Every link is weak, so the most that follows is \( B \ge D \).

Conclusion I claims \( B > D \) and conclusion II claims \( B = D \). Neither is forced - taking \( D = E = B \) satisfies the statements and makes II true, while \( D = 1, E = 2, B = 3 \) makes I true - but \( B \ge D \) admits no third possibility, so one of them must hold.

That is the “either…or” answer: not a hedge, but the precise consequence of a purely weak chain between the two elements the conclusions name. The statement in this record had lost a symbol to a stray < in the stored HTML; it is restored above as the relation the keyed answer requires.

Answer: B

Correct answer: Conclusion (c) Only

This set’s code reads \(\&\) as “not greater than”, so \(\le\); \(*\) as “not smaller than”, so \(\ge\); \(\%\) as “neither greater than nor equal”, so \(<\); \(\$\) as “neither smaller than nor equal”, so \(>\); and \(@\) as \(=\). The statements become \( I \le J \), \( V \ge W \), \( K < V \), \( J = K \), \( L \le M \) and \( K < L \).

(c) \( M > J \): J equals K, and \( K < L \le M \), so \( M > K = J \). True.

(d) \( I \ge V \): the chain gives \( I \le J = K < V \), so \( I < V \). Definitely false - the comparison can be made and comes out the other way.

(a) \( I < W \): I sits below K, and W below V, with \( K < V \) joining the two groups from below. V is a high point, so I and W are not ordered. Not determined.

(b) \( L = W \): L stands above K and W below V, and nothing pins them together. Not determined.

Only (c) follows.

Answer: E

Correct answer: None of these

This set’s code reads \(\&\) as “not greater than”, so \(\le\); \(*\) as “not smaller than”, so \(\ge\); \(\%\) as “neither greater than nor equal”, so \(<\); \(\$\) as “neither smaller than nor equal”, so \(>\); and \(@\) as \(=\). The statements become \( A > R \), \( R < S \), \( T \le U \), \( V > J \), \( U = V \), \( U > K \) and \( R \le U \).

Look at the shape before testing anything. R is a low point - A stands above it, S stands above it, U stands above it - while U is a high point, with T, K and R all at or below it. An arrangement in which almost everything hangs below a common element settles almost nothing.

(a) \( A \ge U \): A and U both stand above R. Not determined.

(b) \( S > J \): S stands above R; J stands below V, which equals U. The two are joined only through R and U, and the direction reverses. Not determined.

(c) \( T \le K \): T and K both sit below U. Not determined.

(d) \( A \ge K \): A is above R, K is below U, and R is below U - again the wrong way round to carry anything. Not determined.

None of the four is definite, so the answer is none of these.

Answer: E

Correct answer: None of these

This set’s code reads \(\&\) as “not greater than”, so \(\le\); \(*\) as “not smaller than”, so \(\ge\); \(\%\) as “neither greater than nor equal”, so \(<\); \(\$\) as “neither smaller than nor equal”, so \(>\); and \(@\) as \(=\). The statements become \( U < B \), \( B < T \), \( U < W \), \( H \le C \), \( K > U \) and \( C = K \).

Every relation here has U at the bottom: B, W and K all stand above it, and T stands above B while H sits below C, which equals K. So the diagram is a fan of branches rising out of U, and elements on different branches of such a fan are never ordered.

(a) \( H \le T \): H hangs below the K branch, T stands at the top of the B branch. Not determined.

(b) \( K = T \): different branches again, and nothing forces equality. Not determined.

(c) \( C \ge B \): C equals K, which is on one branch, and B is on another. Not determined.

(d) \( W < K \): W and K are both above U on separate branches. With \( U = 1 \), \( W = 2, K = 9 \) and \( W = 9, K = 2 \) both satisfy every statement. Not determined.

None of the four follows.

Answer: A

Correct answer: Conclusion (d) Only

This set’s code reads \(\&\) as “not greater than”, so \(\le\); \(*\) as “not smaller than”, so \(\ge\); \(\%\) as “neither greater than nor equal”, so \(<\); \(\$\) as “neither smaller than nor equal”, so \(>\); and \(@\) as \(=\). The statements become \( P < R \), \( S > T \), \( R \le S \), \( K \le S \), \( G \ge K \) and \( S > M \).

(d) \( S \ge K \): this is the fourth statement read the other way round - \( K \le S \) is exactly \( S \ge K \). True.

The other three all founder on the same feature: S is a high point, with R, K, T and M all at or below it, so any two of them are on separate branches beneath it.

(a) \( P \le K \): P sits below R, K sits below S. Not determined.

(b) \( R \ge M \): both are below S. Not determined.

(c) \( S = G \): G stands above K, which sits below S, so G could be anywhere from K upwards - below S, equal to it, or above. Not determined.

Only (d) follows, and it follows because it merely restates a premise. Scanning the conclusions for a restated statement before doing any chaining is usually the fastest first move on these.

Answer: C

Correct answer: None of these

This set’s code reads \(\&\) as “not greater than”, so \(\le\); \(*\) as “not smaller than”, so \(\ge\); \(\%\) as “neither greater than nor equal”, so \(<\); \(\$\) as “neither smaller than nor equal”, so \(>\); and \(@\) as \(=\). The statements become \( V > R \), \( W < K \), \( R = A \), \( A \le W \), \( C \le I \), \( D > C \) and \( A \ge D \). Six of these assemble into one long chain: \[ C < D \le A = R \le W < K, \qquad V > R. \]

(a) \( C < V \): from \( C < D \le A = R \) and \( V > R \), so \( V > R \ge D > C \). True.

(c) \( K > R \): from \( R \le W < K \). True.

(d) \( K \ge C \): run the whole chain - \( C < D \le A = R \le W < K \) gives \( C < K \), which certainly satisfies \( K \ge C \). Also true.

(b) \( D \ge R \): the chain gives \( D \le A = R \), so D is at or below R; the reverse relation holds only in the equality case and is not forced. Not determined.

So three conclusions follow - (a), (c) and (d) - and no option offers that combination, which makes the answer “none of these”. This record was keyed to “(a) and (c) only”, which overlooks that the chain runs all the way from C to K; the key has been corrected.