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Difficulty: HardQuestion Stem Simplification and Target Rephrasing

In Data Sufficiency analysis, for any non-zero real numbers xx and yy with ∣x∣≠∣y∣|x| \neq |y|, the target question 'Is x2−y2xy>0\frac{x^2 - y^2}{xy} > 0?' is algebraically equivalent to the rephrased target question 'Do xx and yy have opposite signs if and only if ∣x∣<∣y∣|x| < |y|?'

Answer: Answer

Answer

The statement is TRUE.
The quotient x2−y2xy\frac{x^2 - y^2}{xy} is positive in two distinct cases: when x,yx, y have the same sign and ∣x∣>∣y∣|x| > |y| (in which case both 'opposite signs' and '∣x∣<∣y∣|x| < |y|' are false, making their biconditional true), or when x,yx, y have opposite signs and ∣x∣<∣y∣|x| < |y| (in which case both components are true, making their biconditional true). In cases where the quotient is negative, the biconditional evaluates to false. Thus, the rephrased target question is logically and algebraically identical to the original target question.

Step-by-Step Solution

1
Analyze the sign conditions required for the fraction to be positive.
x2−y2xy>0\frac{x^2 - y^2}{xy} > 0 requires numerator x2−y2x^2 - y^2 and denominator xyxy to have the exact same algebraic sign.
A rational expression AB\frac{A}{B} is strictly positive if and only if A>0,B>0A > 0, B > 0 or A<0,B<0A < 0, B < 0.
2
Evaluate Case 1: Both numerator and denominator are positive (xy>0xy > 0 and x2−y2>0x^2 - y^2 > 0).
xy>0  ⟹  x,yxy > 0 \implies x, y have SAME signs (Opposite signs = False). x2−y2>0  ⟹  ∣x∣>∣y∣x^2 - y^2 > 0 \implies |x| > |y| (∣x∣<∣y∣|x| < |y| = False).
The rephrased question tests 'Opposite signs   ⟺  ∣x∣<∣y∣\iff |x| < |y|'. Here, False   ⟺  \iff False evaluates to TRUE.
3
Evaluate Case 2: Both numerator and denominator are negative (xy<0xy < 0 and x2−y2<0x^2 - y^2 < 0).
xy<0  ⟹  x,yxy < 0 \implies x, y have OPPOSITE signs (Opposite signs = True). x2−y2<0  ⟹  ∣x∣<∣y∣x^2 - y^2 < 0 \implies |x| < |y| (∣x∣<∣y∣|x| < |y| = True).
The rephrased question tests 'Opposite signs   ⟺  ∣x∣<∣y∣\iff |x| < |y|'. Here, True   ⟺  \iff True evaluates to TRUE.
4
Evaluate cases where the fraction is negative to ensure complete equivalence.
If xy>0xy > 0 and x2−y2<0x^2 - y^2 < 0, we get False   ⟺  \iff True (FALSE). If xy<0xy < 0 and x2−y2>0x^2 - y^2 > 0, we get True   ⟺  \iff False (FALSE).
The biconditional statement yields TRUE in exactly the same scenarios where x2−y2xy>0\frac{x^2 - y^2}{xy} > 0, establishing strict algebraic equivalence.

Key Concept

Target Stem Rephrasing via Quotient Sign Decomposition and Logical Biconditional Equivalence
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Topics

Question Stem Simplification and Target Rephrasing

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