Bunuel wrote:
If x, y and z are real numbers, is y – z < x – z?
(1) y lies between x and z.
(2) z < x < 0
Target question: Is y – z < x – z?This is a good candidate for
rephrasing the target question. Take
y – z < x – z and add z to both sides to get:
y < x.
So,......
REPHRASED target question: Is y < x? Statement 1: y lies between x and z. BE CAREFUL, we cannot conclude that
x < y < z, because it could also be the case that
z < y < xIn both cases, y lies between x and z on the number line. These two possible scenarios yield different answers to the REPHRASED target question.
Case a: if
x < y < z, then we can see that
x < y.
Case b: if
z < y < x, then we can see that
y < x.
Since we cannot answer the
REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: z < x < 0There's no information about y.
So, there's no way to determine whether or not
y < xSince we cannot answer the
REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Statement 1 tells us that EITHER
x < y < z OR
z < y < xStatement 2 tells us that z < x, which RULES OUT the possibility that
x < y < zSo, it MUST be the case that
z < y < x, which means
y < xSo, the answer to the
REPHRASED target question is
"YES. It is definitely the case that y < x"Since we can answer the
REPHRASED target question with certainty, the combined statements are SUFFICIENT
Answer: C
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