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Zorluk: Çok zorAlgebraic Equations and Systems in Data Sufficiency

If uu and vv are real numbers such that u+v≠0u + v \neq 0, what is the value of u−vu+v\frac{u - v}{u + v}?

(1) u2−3uv+2v2=0u^2 - 3uv + 2v^2 = 0
(2) 2u2−5uv+2v2=02u^2 - 5uv + 2v^2 = 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 answer is the option stating that both statements together are sufficient, but neither alone is sufficient. Statement (1) yields two possible ratios (u=vu=v giving 0, or u=2vu=2v giving 1/31/3), so it is not sufficient. Statement (2) also yields two possible ratios (u=v/2u=v/2 giving −1/3-1/3, or u=2vu=2v giving 1/31/3), so it is not sufficient. When combined, u=vu=v and u=v/2u=v/2 would require u=v=0u=v=0, which is forbidden by the condition u+v≠0u+v \neq 0. Therefore, u=2vu=2v is the only valid relation, producing a unique value of 1/31/3.

Adım Adım Çözüm

1
Rephrase the target question stem.
Dividing the numerator and denominator of u−vu+v\frac{u - v}{u + v} by vv (assuming v≠0v \neq 0) shows that knowing the ratio uv\frac{u}{v} uniquely determines the value of the expression.
Simplifying the target expression reduces the problem to determining whether a unique ratio uv\frac{u}{v} exists.
2
Evaluate Statement (1) independently.
Factor u2−3uv+2v2=0u^2 - 3uv + 2v^2 = 0 as (u−v)(u−2v)=0(u - v)(u - 2v) = 0. This implies u=vu = v or u=2vu = 2v. If u=vu = v, then u−vu+v=02v=0\frac{u - v}{u + v} = \frac{0}{2v} = 0. If u=2vu = 2v, then u−vu+v=2v−v2v+v=v3v=13\frac{u - v}{u + v} = \frac{2v - v}{2v + v} = \frac{v}{3v} = \frac{1}{3}.
Since Statement (1) allows two distinct numerical values (00 and 13\frac{1}{3}), Statement (1) alone is NOT sufficient.
3
Evaluate Statement (2) independently.
Factor 2u2−5uv+2v2=02u^2 - 5uv + 2v^2 = 0 as (2u−v)(u−2v)=0(2u - v)(u - 2v) = 0. This implies u=12vu = \frac{1}{2}v or u=2vu = 2v. If u=12vu = \frac{1}{2}v, then u−vu+v=12v−v12v+v=−12v32v=−13\frac{u - v}{u + v} = \frac{\frac{1}{2}v - v}{\frac{1}{2}v + v} = \frac{-\frac{1}{2}v}{\frac{3}{2}v} = -\frac{1}{3}. If u=2vu = 2v, then u−vu+v=13\frac{u - v}{u + v} = \frac{1}{3}.
Since Statement (2) allows two distinct numerical values (−13-\frac{1}{3} and 13\frac{1}{3}), Statement (2) alone is NOT sufficient.
4
Evaluate Statement (1) and Statement (2) combined.
Combine the conditions: Statement (1) requires u=vu = v or u=2vu = 2v; Statement (2) requires u=12vu = \frac{1}{2}v or u=2vu = 2v. Testing u=vu = v in Statement (2) yields 2v2−5v2+2v2=−v2=0  ⟹  v=02v^2 - 5v^2 + 2v^2 = -v^2 = 0 \implies v = 0, which implies u=0u = 0, but u+v≠0u + v \neq 0 rules out u=v=0u = v = 0. Testing u=12vu = \frac{1}{2}v in Statement (1) similarly leads to u=v=0u = v = 0. Thus, the only non-zero solution satisfying both equations is u=2vu = 2v, which gives a unique value of 13\frac{1}{3}.
Combining both statements eliminates the ambiguous cases that violate the stem constraint, establishing a unique value.

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