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Discussion

Statement:

V $ R, W % K, R @ A, A & W, C & I, D $ C, A * D

Conclusion:

(a) C % V (b) D * R (c) K $ R (d) K * C

  • A.Conclusion (a), (b) and (c) Only
  • B.Conclusion (b), (d) Only
  • C.Conclusion (a) and (c) Only
  • D.Conclusion (a) and (d) Only
  • E.None of 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.

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