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# If x, y and z are real numbers, is y – z < x – z?

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Re: If x, y and z are real numbers, is y – z < x – z? [#permalink]
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GMATPrepNow wrote:
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 < x
In 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 < 0
So, there's no way to determine whether or not y < x
Since 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 < x
Statement 2 tells us that z < x, which RULES OUT the possibility that x < y < z
So, it MUST be the case that x < y < z, which means x < y

So, the answer to the REPHRASED target question is "NO. It is definitely NOT the case that y < x"
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

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Hi Brent,

Nice solution! I do think you've made a mistake in the highlighted part though. I believe you meant to say:

Statement 2 tells us that z < x, which RULES OUT the possibility that x < y < z
So, it MUST be the case that z < y < x, which means y < x

So, the answer to the REPHRASED target question is "YES. It is DEFINITELY the case that y < x"
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