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If zy < xy < 0, is xz + x = z? [#permalink]
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28 Jan 2010, 01:50
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If zy < xy < 0, is xz + x = z? (1) z < x (2) y > 0 For zy < xy < 0 to be true, I am counting two possible scenarios
x y z ve +ve ve 1 +ve ve +ve2
Statement 1 rules out scenario 2 but scenario 1 is possible. Now when i substitute the signs of x and z and take them out from the modulus, i get 
(x + z) + (x) = (z) 2x = 2z x=z    Therefore in the original equation, xz = 0 and x = z
hence sufficient
Statement 2
Again eliminates the possibility of the scenario 2 and hence is sufficient
Am i correct here?
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Re: GMATprep Inequalities [#permalink]
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28 Jan 2010, 02:51
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An easy way to approach this ineq will be to analyze that: xz + x = z means zx = z  x. zx is the distance between z and x on number line. It can only be equal to z  x if both z and x have the same signs. a) z < x  implies that y > 0 because zy < xy. If y > 0 then z < x < 0. Therefore both have same signs. SUFF b) y>0 then z < x < 0. Therefore both have same signs. SUFF D is the answer i think
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Re: GMATprep Inequalities [#permalink]
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28 Jan 2010, 05:14
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This question was discussed before here is my post from there: This is not a good question, as neither of statement is needed to answer the question, stem is enough to do so. If \(zy<xy<0\) is \(xz+x = z\) Look at the inequality \(zy<xy<0\): We can have two cases: A. If \(y<0\) > when reducing we should flip signs and we'll get: \(z>x>0\). In this case: as \(z>x\) > \(xz=x+z\); as \(x>0\) and \(z>0\) > \(x=x\) and \(z=z\). Hence in this case \(xz+x=z\) will expand as follows: \(x+z+x=z\) > \(0=0\), which is true. And: B. If \(y>0\) > when reducing we'll get: \(z<x<0\). In this case: as \(z<x\) > \(xz=xz\); as \(x<0\) and \(z<0\) > \(x=x\) and \(z=z\). Hence in this case \(xz+x=z\) will expand as follows: \(xzx=z\) > \(0=0\), which is true. So knowing that \(zy<xy<0\) is true, we can conclude that \(xz+x = z\) will also be true. Answer should be D even not considering the statements themselves. As for the statements:Statement (1) says that \(z<x\), hence we have case B. Statement (2) says that \(y>0\), again we have case B. Hope it helps.
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Re: GMATprep Inequalities [#permalink]
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31 Jan 2010, 11:40
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wow, I solved this question by myself:)



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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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27 Nov 2013, 07:32
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zaarathelab wrote: If zy < xy < 0, is xz + x = z? (1) z < x (2) y > 0 For zy < xy < 0 to be true, I am counting two possible scenarios
x y z ve +ve ve 1 +ve ve +ve2
Statement 1 rules out scenario 2 but scenario 1 is possible. Now when i substitute the signs of x and z and take them out from the modulus, i get 
(x + z) + (x) = (z) 2x = 2z x=z    Therefore in the original equation, xz = 0 and x = z
hence sufficient
Statement 2
Again eliminates the possibility of the scenario 2 and hence is sufficient
Am i correct here? Rearrange question (x+z)=zx = zx? By property they will be equal when both x and z have the same sign Statement 1 If x>z, then with zy < xy < 0, x and z are both positive. Same sign. Suff Statement 2. If y>0 then same with zy < xy < 0, x and z both positive. Same sign Suff Answer is (D) Hope it helps Cheers! J



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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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17 Dec 2013, 03:17
I solved this question on number line, we can first analyze this eq xz + x = z? This equality is only possible, if x and z are on the same side of the number line. () zx0 or 0xz (+) distance between xandz + distance of x from origin = distance of z from origin given zy<xy<0 which can happen in two case. z x y   + zy negative xy negative < both less than 0 + +  zy negative xy negative < both less than 0 Therefore we don't even need option 1 and 2 to validate this xz + x = z. Answer : D
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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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16 Apr 2014, 21:28
(1) z < x => y > 0 (2) Same as (1)
(1) or (2) x < 0 z < 0
0 < xz abs(xz) + abs(x) = abs(z)? xz + x = z S
D



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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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10 May 2014, 11:04
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using Statement 1 : from the question , zy < xy y(zx)<0  (A) Now statement 1 tells me that (zx)< 0 . This implies Y>0 So, if zy < xy < 0 and Y>0 This implies Z & X < 0 mod (xz) + mod (x) = XZ (since ZX<0) + (X) = Z = Z
Using Statement 2 : From (A), y(zx)<0 Since from Statement 2 we know that y > 0 That implies (zx)< 0 .
Hence D.



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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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14 Mar 2017, 22:37
Bunuel wrote: This question was discussed before here is my post from there:
This is not a good question, as neither of statement is needed to answer the question, stem is enough to do so. Hi Bunuel, your explanation is totally convincing. Can you or someone please confirm that this is an official question? I had thought official questions are always correct.



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Re: If zy < xy < 0, is xz + x = z? [#permalink]
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14 Mar 2017, 22:39




Re: If zy < xy < 0, is xz + x = z?
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