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Difficulty:
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Question Stats:
61% (02:19) correct
39%
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wrong
based on 493
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Is |x − z − y| > x − z + y?
(1) 0 < x < z < y
(2) (x – z – y) is negative
(1) 0 < x < z < y
(2) (x – z – y) is negative
Hi,
I have been doing various GMAT problems, and one major confusion was pertaining to absolute sign problems in data sufficiency (Specifically the one mentioned below)
Using the standard approach for absolute problems, I usually interpreted the question as if x-z-y >0, Then y<0 OR if x-z-y<0, Then x<z.
However my point of confusion starts, when i start looking at the two given statements.
As per the above problem, statement 1 states that x<z - Is this only information sufficient or do we also need the condition that x-z-y<0. I am really confused at this part, in some problems, both the statements are used, and in some statements only one statement is sufficient. I am really not sure how should i go about dealing with the DS statements for these type of absolute problems.
Please clarify my confusion or let me know if i am missing out on something. Thanks you.
I have been doing various GMAT problems, and one major confusion was pertaining to absolute sign problems in data sufficiency (Specifically the one mentioned below)
Using the standard approach for absolute problems, I usually interpreted the question as if x-z-y >0, Then y<0 OR if x-z-y<0, Then x<z.
However my point of confusion starts, when i start looking at the two given statements.
As per the above problem, statement 1 states that x<z - Is this only information sufficient or do we also need the condition that x-z-y<0. I am really confused at this part, in some problems, both the statements are used, and in some statements only one statement is sufficient. I am really not sure how should i go about dealing with the DS statements for these type of absolute problems.
Please clarify my confusion or let me know if i am missing out on something. Thanks you.
03 Sep 2014, 18:23
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Is |x−z−y| > x−z+y?
(1) 0<x<z<y. This implies that x-z-y must be negative, which means that |x-z-y| = -(x-z-y) = y + z - x (recall that when x<0, then |x| = -x). So, the question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? We are told that this is in fact so (x < z). Sufficient.
(2) (x–z–y) is negative --> |x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? But here we cannot answer this question. For example, if x=y=0 and z=1, then the answer would be YES but if x=1, y=2 and z=0, then the answer would be NO. Not sufficient.
Answer: A.
(1) 0<x<z<y. This implies that x-z-y must be negative, which means that |x-z-y| = -(x-z-y) = y + z - x (recall that when x<0, then |x| = -x). So, the question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? We are told that this is in fact so (x < z). Sufficient.
(2) (x–z–y) is negative --> |x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? But here we cannot answer this question. For example, if x=y=0 and z=1, then the answer would be YES but if x=1, y=2 and z=0, then the answer would be NO. Not sufficient.
Answer: A.
12 Jun 2013, 06:17
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Is |x−z−y|>x−z+y?
(1) 0<x<z<y. This implies that \(x-z-y\) is negative (positive number x minus greater number z minus greater number y must be less than 0). Thus \(|x-z-y|=-(x-z-y)=y+z-x\). So, the question becomes: is \(y+z-x>x-z+y\)? --> is \(2z>2x\)? --> is \(z>x\)? The statement gives an YES answer to the question. Sufficient.
(2) (x–z–y) is negative. Almost the same here: \(|x-z-y|=-(x-z-y)=y+z-x\). So, the question becomes: is \(y+z-x>x-z+y\)? --> is \(2z>2x\)? --> is \(z>x\)? We don't know that. Not sufficient.
Answer: A.
Theory on absolute value is HERE.
PS absolute value questions are HERE.
DS absolute value questions are HERE.
700+ absolute value questions HERE.
(1) 0<x<z<y. This implies that \(x-z-y\) is negative (positive number x minus greater number z minus greater number y must be less than 0). Thus \(|x-z-y|=-(x-z-y)=y+z-x\). So, the question becomes: is \(y+z-x>x-z+y\)? --> is \(2z>2x\)? --> is \(z>x\)? The statement gives an YES answer to the question. Sufficient.
(2) (x–z–y) is negative. Almost the same here: \(|x-z-y|=-(x-z-y)=y+z-x\). So, the question becomes: is \(y+z-x>x-z+y\)? --> is \(2z>2x\)? --> is \(z>x\)? We don't know that. Not sufficient.
Answer: A.
Theory on absolute value is HERE.
PS absolute value questions are HERE.
DS absolute value questions are HERE.
700+ absolute value questions HERE.
