Examkin

Question

Difficulty: HardQuestion Stem Simplification and Target Rephrasing

If xx and yy are non-zero real numbers, is x3y+xy3<2x2y2x^3 y + x y^3 < 2 x^2 y^2?

(1) ∣x+y∣<∣x−y∣|x + y| < |x - y|
(2) x2y3>0x^2 y^3 > 0

  1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.Answer
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
The correct response is the option stating that Statement (1) alone is sufficient, but Statement (2) alone is not sufficient. Rephrasing the stem x3y+xy3<2x2y2x^3 y + x y^3 < 2 x^2 y^2 by factoring yields xy(x−y)2<0xy(x - y)^2 < 0. Because (x−y)2>0(x - y)^2 > 0 whenever x≠yx \neq y, the inequality holds if and only if xy<0xy < 0. Statement (1) reduces to (x+y)2<(x−y)2(x+y)^2 < (x-y)^2, which simplifies directly to xy<0xy < 0, giving a definitive YES. Statement (2) reduces to y>0y > 0, which leaves the sign of xx (and thus xyxy) unknown.

Step-by-Step Solution

1
Rephrase the target question stem algebraically.
The target inequality x3y+xy3<2x2y2x^3 y + x y^3 < 2 x^2 y^2 can be rearranged as x3y+xy3−2x2y2<0x^3 y + x y^3 - 2 x^2 y^2 < 0. Factoring out xyxy yields xy(x2−2xy+y2)<0xy(x^2 - 2xy + y^2) < 0, which is xy(x−y)2<0xy(x - y)^2 < 0.
Simplifying the stem isolates the core algebraic condition needed to answer the question.
2
Analyze the conditions under which xy(x−y)2<0xy(x - y)^2 < 0 holds.
Since xx and yy are non-zero, if x=yx = y, xy(x−y)2=0xy(x-y)^2 = 0, so the inequality is false (NO). If x≠yx \neq y, (x−y)2>0(x - y)^2 > 0, so the sign of xy(x−y)2xy(x - y)^2 is determined entirely by xyxy. If xy<0xy < 0, then x≠yx \neq y and xy(x−y)2<0xy(x - y)^2 < 0 (YES). If xy>0xy > 0, then xy(x−y)2≥0xy(x - y)^2 \ge 0 (NO). Thus, the target question simplifies to: 'Is xy<0xy < 0?'
A perfect square of a real number is non-negative, so sign evaluation reduces to checking xy<0xy < 0.
3
Evaluate Statement (1): ∣x+y∣<∣x−y∣|x + y| < |x - y|.
Squaring both sides of ∣x+y∣<∣x−y∣|x + y| < |x - y| gives (x+y)2<(x−y)2  ⟹  x2+2xy+y2<x2−2xy+y2  ⟹  4xy<0  ⟹  xy<0(x + y)^2 < (x - y)^2 \implies x^2 + 2xy + y^2 < x^2 - 2xy + y^2 \implies 4xy < 0 \implies xy < 0.
Since Statement (1) directly proves xy<0xy < 0, it gives a definitive YES to the rephrased target question, making Statement (1) sufficient.
4
Evaluate Statement (2): x2y3>0x^2 y^3 > 0.
Since x≠0x \neq 0, x2>0x^2 > 0. Thus x2y3>0x^2 y^3 > 0 simplifies to y3>0y^3 > 0, which means y>0y > 0. However, xx can still be positive (yielding xy>0xy > 0) or negative (yielding xy<0xy < 0).
Knowing only that y>0y > 0 leaves the sign of xyxy undetermined, so Statement (2) is not sufficient.

Key Concept

Data Sufficiency Target Simplification and Algebraic Rephrasing
Rate this question

Topics

Question Stem Simplification and Target Rephrasing

More questions from this topic