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If x, y and z are real numbers, is y – z < x – z?
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Updated on: 19 Oct 2017, 15:03
Top Contributor
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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Originally posted by GMATPrepNow on 26 Feb 2017, 09:45.
Last edited by GMATPrepNow on 19 Oct 2017, 15:03, edited 1 time in total.
Re: If x, y and z are real numbers, is y – z < x – z?
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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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Re: If x, y and z are real numbers, is y – z < x – z?
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19 Oct 2017, 14:06
1
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"
close
70% (01:32) correct 
