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Zorluk: Çok zorStatement Independence Evaluation and Statement Combination

If aa and bb are non-zero real numbers, is a>ba > b?

(1) a2b<ab2a^2 b < a b^2
(2) ab+ba<0\frac{a}{b} + \frac{b}{a} < 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. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Cevap
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Cevap

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
The correct option states that both statements together are sufficient, but neither statement alone is sufficient. Statement (1) simplifies to ab(a−b)<0ab(a - b) < 0, which is insufficient by itself because the sign of (a−b)(a - b) depends on whether abab is positive or negative. Statement (2) simplifies to ab<0ab < 0 (since a2+b2>0a^2 + b^2 > 0), which alone does not reveal whether aa or bb is the larger number. When evaluated together, knowing ab<0ab < 0 allows us to divide ab(a−b)<0ab(a - b) < 0 by the negative value abab and flip the inequality sign, yielding a−b>0a - b > 0 (i.e., a>ba > b) definitively.

Adım Adım Çözüm

1
Analyze the target question and simplify Statement (1) algebraically.
Statement (1) can be rewritten as a2b−ab2<0a^2 b - a b^2 < 0, which factors into ab(a−b)<0ab(a - b) < 0.
Factoring isolates the relationship between abab and (a−b)(a - b).
2
Test Statement (1) independently.
If a=1a = 1 and b=2b = 2, then ab(a−b)=2(−1)=−2<0ab(a - b) = 2(-1) = -2 < 0, giving a<ba < b (No). If a=2a = 2 and b=−1b = -1, then ab(a−b)=−2(3)=−6<0ab(a - b) = -2(3) = -6 < 0, giving a>ba > b (Yes).
Since Statement (1) permits both a 'Yes' and a 'No' response depending on the signs of aa and bb, it is NOT sufficient alone.
3
Evaluate Statement (2) strictly in isolation without using information from Statement (1).
Combining terms gives a2+b2ab<0\frac{a^2 + b^2}{ab} < 0. Because a2+b2>0a^2 + b^2 > 0 for non-zero real numbers, the denominator must be negative, so ab<0ab < 0.
Knowing ab<0ab < 0 implies aa and bb have opposite signs, but it does not specify which variable is positive and which is negative (e.g., a=2,b=−1  ⟹  a>ba = 2, b = -1 \implies a > b, whereas a=−1,b=2  ⟹  a<ba = -1, b = 2 \implies a < b). Thus, Statement (2) alone is NOT sufficient.
4
Combine Statement (1) and Statement (2).
From Statement (2), ab<0ab < 0. From Statement (1), ab(a−b)<0ab(a - b) < 0. Dividing ab(a−b)<0ab(a - b) < 0 by the negative quantity abab requires flipping the inequality sign, yielding a−b>0a - b > 0, or a>ba > b.
This establishes a definitive 'Yes' answer to the question, making the statements together sufficient.

Anahtar Kavram

Statement Independence Evaluation and Statement Combination in Data Sufficiency
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