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

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

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New post 26 Feb 2017, 08:16
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A
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C
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E

Difficulty:

  25% (medium)

Question Stats:

70% (00:57) correct 30% (00:57) wrong based on 83 sessions

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Kudos [?]: 135410 [0], given: 12692

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

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New post 26 Feb 2017, 09:45
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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
There's no information about y.
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 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"
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Last edited by GMATPrepNow on 19 Oct 2017, 15:03, edited 1 time in total.

Kudos [?]: 2740 [0], given: 364

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Re: If x, y and z are real numbers, is y – z < x – z? [#permalink]

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New post 26 Feb 2017, 20:53
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


Question: Is y – z < x – z?
Question: Is y < x ?

Statement 1: y lies between x and z.
But y may be greater than or smaller than x
NOT SUFFICIENT

Statement 2: z < x < 0
No mention of y hence
NOT SUFFICIENT

Combining the two statements
z < x < 0 and
y lies between x and z
i.e. y must be smaller than x hence
SUFFICIENT

Answer: option C
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Kudos [?]: 2400 [0], given: 51

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Kudos [?]: 1 [1], given: 51

Re: If x, y and z are real numbers, is y – z < x – z? [#permalink]

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New post 19 Oct 2017, 14:06
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KUDOS
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
There's no information about y.
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

Answer: C

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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"

Kudos [?]: 1 [1], given: 51

Expert Post
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Joined: 11 Sep 2015
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Kudos [?]: 2740 [0], given: 364

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Re: If x, y and z are real numbers, is y – z < x – z? [#permalink]

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New post 19 Oct 2017, 15:03
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Thanks BalysLTU!!

I have edited my answer accordingly.

Cheers,
Brent
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Kudos [?]: 2740 [0], given: 364

Re: If x, y and z are real numbers, is y – z < x – z?   [#permalink] 19 Oct 2017, 15:03
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