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Difficulty: Very hardStatement Independence Evaluation and Statement Combination

If xx and yy are non-zero real numbers, is xy>1\frac{x}{y} > 1?

(1) x2y>xy2x^2 y > x y^2
(2) x−yx>0\frac{x - y}{x} > 0

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  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. Statements (1) and (2) TOGETHER are NOT sufficient.Answer

Answer

Statements (1) and (2) TOGETHER are NOT sufficient to determine whether xy>1\frac{x}{y} > 1.
The choice stating that statements (1) and (2) together are not sufficient is correct because combining both statements permits two valid cases: case 1 (x=3,y=2x=3, y=2) yields a 'Yes' answer to the stem, while case 2 (x=−1,y=2x=-1, y=2) satisfies both statements but yields a 'No' answer to the stem.

Step-by-Step Solution

1
Rephrase the target question
The condition xy>1\frac{x}{y} > 1 is equivalent to x−yy>0\frac{x - y}{y} > 0, meaning x−yx - y and yy must have the same sign.
Simplifying the question target allows direct testing of signs.
2
Evaluate Statement (1) independently
Statement (1) states x2y>xy2  ⟺  xy(x−y)>0x^2 y > x y^2 \iff x y (x - y) > 0. If x=3,y=2x = 3, y = 2, then xy(x−y)=6>0xy(x-y) = 6 > 0 and xy=1.5>1\frac{x}{y} = 1.5 > 1 (Yes). If x=−2,y=−3x = -2, y = -3, then xy(x−y)=6>0xy(x-y) = 6 > 0 and xy=23<1\frac{x}{y} = \frac{2}{3} < 1 (No). Insufficient.
Testing positive and negative pairs shows Statement (1) alone yields both Yes and No.
3
Evaluate Statement (2) independently without carrying over Statement (1) information
Statement (2) states x−yx>0\frac{x - y}{x} > 0, meaning xx and x−yx - y have the same sign. If x=3,y=2x = 3, y = 2, 13>0\frac{1}{3} > 0 and xy=1.5>1\frac{x}{y} = 1.5 > 1 (Yes). If x=2,y=−3x = 2, y = -3, 52>0\frac{5}{2} > 0 and xy=−23<1\frac{x}{y} = -\frac{2}{3} < 1 (No). Insufficient.
Statement (2) evaluated strictly in isolation also yields both Yes and No.
4
Evaluate Statements (1) and (2) combined
From Statement (1), xy(x−y)>0xy(x-y) > 0. From Statement (2), x−yx>0  ⟹  x(x−y)>0\frac{x-y}{x} > 0 \implies x(x-y) > 0. Combining these implies y>0y > 0. Since y>0y > 0, if x>0x > 0, then x−y>0  ⟹  x>y>0  ⟹  xy>1x - y > 0 \implies x > y > 0 \implies \frac{x}{y} > 1 (Yes). However, if x<0x < 0, since y>0y > 0, x−y<0x - y < 0, satisfying x(x−y)>0x(x-y) > 0 and xy(x−y)>0xy(x-y) > 0. For example, x=−1,y=2x = -1, y = 2 satisfies both statements but gives xy=−0.5<1\frac{x}{y} = -0.5 < 1 (No). Insufficient.
Even combined, two distinct cases exist (x>y>0x > y > 0 giving Yes, and x<0<yx < 0 < y giving No).

Key Concept

Statement Independence Evaluation and Statement Combination with Real Number Sign Traps
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Topics

Statement Independence Evaluation and Statement Combination

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