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Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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Is x = y  z ? (1) x + y = z (2) x < 0
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Last edited by Bunuel on 11 Feb 2012, 12:04, edited 2 times in total.
Edited the question and added the OA



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Re: Abs equation from GMATPrep [#permalink]
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From 1, we get x= zy => x= yz
Thus, x = yz
Statement 2 does not give us anything more.
So, A.



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Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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Re: Abs equation from GMATPrep [#permalink]
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Bunuel wrote: yz=x? > yz must be >=0... Brilliant, thank you! :^)



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Re: Abs equation from GMATPrep [#permalink]
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pm4553 wrote: eresh wrote: From 1, we get x= zy => x= yz
Thus, x = yz
Statement 2 does not give us anything more.
So, A. For Abs Q's, you'll always have 2 solutions; A is insuff. Oh...My bad...got that now :D



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Re: Inequality Problem [#permalink]
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is x=yz?
given: (1) x+y=z (2) x<0
solving (1) first:
y=zx x=(zx)z x=x
take x=1, z=2, y=1 1=12 (no) take x=1, z=2, y=3 x=yz? 1=32=1 YES
so what solving for x=x meant was that x MUST be negative for the equation to be true, if it is positive then it is not true (since in that case, x would not equal x).
hence the answer is C.



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Is IxI = y –z? [#permalink]
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Re: Is IxI = y –z? [#permalink]
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(1) x = z  y So x = x = (z  y) = y z only if x is negative Here we don't know that. Insufficient (2) Insufficient, no information about y and z (1) + (2) x is negative, Sufficient. Answer  C
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Re: Is IxI = y –z? [#permalink]
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05 Jun 2011, 23:31
another way of looking at this numerical can be, x = positive meaning is y>z being asked here. a x= zy means x can be <0 ,= 0 or >0. Hence not sufficient. b gives no idea of y>z or y<z. a+b clearly indicated y<z. Hence sufficient. C it is.
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Re: Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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Hey Bunuel I am not very sure of what the question is asking ... Can you please explain the question....
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Re: Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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Re: Is X= Y Z? [#permalink]
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sayak636 wrote: Is X= Y Z?
1. X+Y= Z 2. X< 0 (1) Can be rewritten as X = Y + Z, so X = Y + Z, which would be equal to Y  Z, if and only if \(Y+Z\leq0\). Obviously, we don't know that, so (1) insufficient. (2) Cannot be sufficient, it doesn't say anything about Y and Z. (1) and (2) together: X = Y + Z < 0, therefore X = Y  Z, sufficient. Answer: C
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Re: Abs equation from GMATPrep [#permalink]
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05 Oct 2012, 04:44
Bunuel wrote: Is \(x=yz\)?
Note that \(yz\) must be \(\geq{0}\), because absolute value (in our case \(x\)) can not be negative.
Generally question asks whether \(yz\geq{0}\) and whether the difference between them equals to \(x\).
(1) \(x=yz\) if \(x>0\) > \(yz\) is negative > no good for us; if \(x\leq{0}\) > \(yz\) is positive > good. Two possible answers not sufficient;
(2) \(x<0\) Not sufficient (we need to know value of yz is equal or not to x)
(1)+(2) Sufficient.
Answer: C. Hi bunuel, I am not able to understand the solution for this problem. Can you kindly explain the highlighted areas. Note that yz must be \geq{0}, because absolute value (in our case x) can not be negative. Generally question asks whether yz\geq{0} and whether the difference between them equals to x. (1) x=yz if x>0 > yz is negative > no good for us; if x\leq{0} > yz is positive > good. Waiting for reply.
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Re: Abs equation from GMATPrep [#permalink]
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05 Oct 2012, 05:12
fameatop wrote: Bunuel wrote: Is \(x=yz\)?
Note that \(yz\) must be \(\geq{0}\), because absolute value (in our case \(x\)) can not be negative.
Generally question asks whether \(yz\geq{0}\) and whether the difference between them equals to \(x\).
(1) \(x=yz\) if \(x>0\) > \(yz\) is negative > no good for us; if \(x\leq{0}\) > \(yz\) is positive > good. Two possible answers not sufficient;
(2) \(x<0\) Not sufficient (we need to know value of yz is equal or not to x)
(1)+(2) Sufficient.
Answer: C. Hi bunuel, I am not able to understand the solution for this problem. Can you kindly explain the highlighted areas. Note that yz must be \geq{0}, because absolute value (in our case x) can not be negative. Generally question asks whether yz\geq{0} and whether the difference between them equals to x. (1) x=yz if x>0 > yz is negative > no good for us; if x\leq{0} > yz is positive > good. Waiting for reply. Look at \(x=yz\): the left hand side is absolute value (x), which cannot be negative, hence the right hand side (yz) also cannot be negative. Therefore must be true that \(yz\geq{0}\). Next, for (1) given that \(x=yz\). Now, if \(x>0\), or if \(x\) is positive, then we'll have that \(positive =yz\) > \(negative=yz\). But as we concluded above \(yz\) cannot be negative, hence this scenario is not good. Hope it's clear.
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Re: Abs equation from GMATPrep [#permalink]
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The question poses as x being the centerpiece variable but Bunuel turns it on its face and makes yz the main subject. Which makes all the difference with data pt 1 when u look at it as yz=x. You immediately see that the right side has to be ve for the LEft side to be +ve. Brilliant approach. Bunuel wrote: Is \(x=yz\)?
Note that \(yz\) must be \(\geq{0}\), because absolute value (in our case \(x\)) can not be negative.
Generally question asks whether \(yz\geq{0}\) and whether the difference between them equals to \(x\).
(1) \(x=yz\) if \(x>0\) > \(yz\) is negative > no good for us; if \(x\leq{0}\) > \(yz\) is positive > good. Two possible answers not sufficient;
(2) \(x<0\) Not sufficient (we need to know value of yz is equal or not to x)
(1)+(2) Sufficient.
Answer: C.



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Re: Abs equation from GMATPrep [#permalink]
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Bunuel wrote: Is \(x=yz\)?
Note that \(yz\) must be \(\geq{0}\), because absolute value (in our case \(x\)) can not be negative.
Generally question asks whether \(yz\geq{0}\) and whether the difference between them equals to \(x\).
(1) \(x=yz\) if \(x>0\) > \(yz\) is negative > no good for us; if \(x\leq{0}\) > \(yz\) is positive > good. Two possible answers not sufficient;
(2) \(x<0\) Not sufficient (we need to know value of yz is equal or not to x)
(1)+(2) Sufficient.
Answer: C. Bunuel, I was wondering if we can square the sides and then evaluate: Is \(x=yz\) Is \(x^2= (yz)^2\) Statement 1: \(x+y = z\) \(x = zy\) squaring both sides... \(x^2 = (zy)^2 = (yz)^2\) Statement 1 alone seems to satisfy. Can you please point out my mistake?



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Re: Abs equation from GMATPrep [#permalink]
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02 Oct 2013, 01:28
emailmkarthik wrote: Bunuel wrote: Is \(x=yz\)?
Note that \(yz\) must be \(\geq{0}\), because absolute value (in our case \(x\)) can not be negative.
Generally question asks whether \(yz\geq{0}\) and whether the difference between them equals to \(x\).
(1) \(x=yz\) if \(x>0\) > \(yz\) is negative > no good for us; if \(x\leq{0}\) > \(yz\) is positive > good. Two possible answers not sufficient;
(2) \(x<0\) Not sufficient (we need to know value of yz is equal or not to x)
(1)+(2) Sufficient.
Answer: C. Bunuel, I was wondering if we can square the sides and then evaluate: Is \(x=yz\) Is \(x^2= (yz)^2\) Statement 1: \(x+y = z\) \(x = zy\) squaring both sides... \(x^2 = (zy)^2 = (yz)^2\) Statement 1 alone seems to satisfy. Can you please point out my mistake? The question asks whether x=yz. This cannot be translated to is x^2=(yz)^2. Consider this \(2\neq{13}\) but 2^2=(13)^2.
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Re: Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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12 Aug 2015, 10:04
DenisSh wrote: Is x = y  z ?
(1) x + y = z (2) x < 0 Question : is x = yz. Rephrasing it : is x^2 = (yz)^2. Because root(x^2) = x. Option 1 : x+y = z. i.e x = (zy). x^2 = (zy)^2 = (yz)^2 . Hence isn't 1 sufficient ?



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Is x = y  z ? (1) x + y = z (2) x < 0 [#permalink]
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12 Aug 2015, 10:32
RPgolucky wrote: DenisSh wrote: Is x = y  z ?
(1) x + y = z (2) x < 0 Question : is x = yz. Rephrasing it : is x^2 = (yz)^2. Because root(x^2) = x. Option 1 : x+y = z. i.e x = (zy). x^2 = (zy)^2 = (yz)^2 . Hence isn't 1 sufficient ? Be careful with squaring variables in PS or DS. \(x^2 = (yz)^2\) , yes but this does not mean that x=yz . Example, x = 5, y = 2, z = 7, in this case \(x^2 = (yz)^2\) > x = yz but if x =  5, y = 2, z = 7, in this case \(x^2 = (yz)^2\) > x = zy You are correct in saying that \(\sqrt{x^2}\)= x , thus \(\sqrt{(yz)^2}\) = yz In other words, x = yz and this will have the following cases that will give you either a "yes" or a "no". x = yz > \(\pm\) x = \(\pm\) (yz) and you will have the following cases: x = (yz) x= (yz) x = (yz) x = (yz) Thus this statement is not sufficient.




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