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Is x − z − y > x − z + y?
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Updated on: 15 Feb 2018, 09:33
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60% (02:20) correct 40% (02:27) wrong based on 277 sessions
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Is x − z − y > x − z + y? (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 xzy >0, Then y<0 OR if xzy<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 xzy<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.
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Originally posted by gmat66 on 12 Jun 2013, 06:01.
Last edited by Bunuel on 15 Feb 2018, 09:33, edited 3 times in total.
Renamed the topic, edited the question and added the OA.




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Re: Is x − z − y > x − z + y?
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03 Sep 2014, 18:23
Is x−z−y > x−z+y?(1) 0<x<z<y. This implies that xzy must be negative, which means that xzy = (xzy) = 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 > xzy = (xzy) = 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.
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Re: Is x − z − y > x − z + y?
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12 Jun 2013, 06:17
Is x−z−y>x−z+y?(1) 0<x<z<y. This implies that \(xzy\) is negative (positive number x minus greater number z minus greater number y must be less than 0). Thus \(xzy=(xzy)=y+zx\). So, the question becomes: is \(y+zx>xz+y\)? > 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: \(xzy=(xzy)=y+zx\). So, the question becomes: is \(y+zx>xz+y\)? > 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.
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Re: Is x − z − y > x − z + y?
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27 Jun 2013, 17:06
Is x−z−y>x−z+y?
(1) 0<x<z<y x−z−y>x−z+y Because x is less than z is less than y, xzy is negative, so: (xzy)>xz+y x+z+y>xz+y z+y>2xz+y 2z>2x z>x z>x which is given to us in the statement. SUFFICIENT
(2) (x–z–y) is negative x−z−y>x−z+y (xzy)>xz+y x+z+y>xz+y 2z>2x z>x However, we are not given any more information. INSUFFICIENT
(A)



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Re: Is x − z − y > x − z + y?
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08 Jul 2016, 23:12
Bunuel I have one doubt here x−z−y>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative x−z−y>x−z+y (1) and x+z+y>x−z+y (2) solving 1 we y<0 solving second we get x<z now if we see option A it gives us hint that basically xzy = (xzy) = y + z  x. The question becomes: is y + z  x > x  z + y? so I thought Its sufficient Now in case of B we already know that mode is negative ... so why cant it be sufficient Please help me understand what M i missing here



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Re: Is x − z − y > x − z + y?
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08 Sep 2016, 08:23
mbaprep2016 wrote: Bunuel I have one doubt here x−z−y>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative x−z−y>x−z+y (1) and x+z+y>x−z+y (2) solving 1 we y<0 solving second we get x<z now if we see option A it gives us hint that basically xzy = (xzy) = y + z  x. The question becomes: is y + z  x > x  z + y? so I thought Its sufficient Now in case of B we already know that mode is negative ... so why cant it be sufficient Please help me understand what M i missing here I think the key here is to note that statement one DOES NOT satisfy the condition of y < 0 (the statement says that 0<x<y<z) and thus, we need to only take care of the other case and can ignore the y < 0 case which makes statement 1 sufficient. In other words, if we solve the xzy, we can have two cases  one positive and one negative as you specified above. Now, if we talk about statement 2, it says that (x–z–y) is negative (less than zero). So, this basically means that our expression in the absolute expression (in the question stem is negative) or in other words, x−z−y =  (xzy). BUT, this does not answer the main question we have in the question stem. Knowing that the absolute value of an expression is negative does NOT answer the main question at hand which is whether x−z−y > x−z+y? Hope the information helps. All the best to you !!
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Re: Is x − z − y > x − z + y?
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08 Mar 2017, 22:08
AmoyV wrote: Is \(x−z−y > x−z+y\)?
(1) \(0<x<z<y\) (2) \((x–z–y)\) is negative Official solution from Veritas Prep. Answer: A Explanation: For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If \(x  z  y\) is positive, then the question is really “Is \(x  z  y > x  z + y\)?" which can be simplified by adding/subtracting terms to “Is \(y < 0\)” In other words, if you can prove that xzy is positive you then only need to determine the sign of y to answer the question. However if \(x  z  y\) is negative then you need to answer this question: “Is \( (x  z  y) > x  z + y\)?” Simplifying that question by adding/subtracting terms you see that the question is really “Is \(x < z\)?” In statement 1, you learn cleverly that indeed \((x – z – y)\) is negative because if you subtract progressively bigger positive numbers from an original positive number, the result will always be negative. Given that \((x – z – y)\) is negative then you need to answer the question “Is \(x < z?\)” and this is expressly given in the statement. Statement (1) is sufficient. Statement (2) is tempting you with information that you already know from statement (1). By itself, it is not sufficient because you do not know if \(x < z\). Answer is (A).
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Re: Is x − z − y > x − z + y?
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09 Jul 2018, 21:26
Let's pick a number: 0<x<z<y X=1; Z=2; y=10 Is x−z−y>x−z+y? 1210>12+10 or,11>11 NO Again, Let's pick a number: 0<x<z<y X=1; Z=2; y=3 Is x−z−y>x−z+y? 123>12+3 or,4>2 Yes Thus, Statement 1 is insufficient, I think.



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Re: Is x − z − y > x − z + y?
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10 Jul 2018, 09:32
akash7gupta11 wrote: Let's pick a number: 0<x<z<y X=1; Z=2; y=10 Is x−z−y>x−z+y? 1210>12+10 or,11>11 NO Again, Let's pick a number: 0<x<z<y X=1; Z=2; y=3 Is x−z−y>x−z+y? 123>12+3 or,4>2 Yes Thus, Statement 1 is insufficient, I think.



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Re: Is x − z − y > x − z + y?
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10 Jul 2018, 21:00
akash7gupta11 wrote: Let's pick a number: 0<x<z<y X=1; Z=2; y=10 Is x−z−y>x−z+y? 1210>12+10 or,11>11 NO Again, Let's pick a number: 0<x<z<y X=1; Z=2; y=3 Is x−z−y>x−z+y? 123>12+3 or,4>2 Yes Thus, Statement 1 is insufficient, I think. Hello Please check the highlighted part. 12+10 will be 9, not 11.



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Re: Is x − z − y > x − z + y?
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25 Sep 2018, 13:59
For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If x  z  y is positive, then the question is really “Is x  z  y > x  z + y?" which can be simplified to “Is y < 0?” by adding/subtracting terms. In other words, if you can prove that xzy is positive you then only need to determine the sign of y to answer the question. However if x  z  y is negative then you need to answer this question: “Is  (x  z  y) > x  z + y?” Simplifying that question by adding/subtracting terms, you see that the question is really “Is x < z?” Quote: In statement 1, you learn cleverly that indeed (x – z – y) is negative because if you subtract progressively bigger positive numbers from an original positive number, the result will always be negative. Given that (x – z – y) is negative then you need to answer the question “Is x < z?” and this is expressly given in the statement. Statement (1) is sufficient. Quote: Statement (2) is tempting you with information that you already know from statement (1). By itself, it is not sufficient because you do not know if x < z. Answer is (A).
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Re: Is x − z − y > x − z + y?
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27 May 2019, 23:11
When you simplify x − z − y > x − z + y? You get two possibilities i.e. “Is y<0” or “Is x<z?” St1 is in confirmation with Y is not less than zero so we have a definite "No" hence st 1 is sufficient. St 2 leads us nowhere hence insufficient Option A is the correct answer
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Re: Is x − z − y > x − z + y?
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