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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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13 Aug 2016, 13:34
I am not really sure what the question is, because the question stem answers the question and Statements 1 and 2 are useless to answer the question.
The question is: Is XZ + X = Z?
In other words: Is XZ = Z  X?
Since XZ must be positive (not zero because ZY<XY<0), the question becomes:Is Z>X?
In this sense, the actual question is: Is: Z<X<0 ? or is 0<X<Z ?
However, the statements are useless, because from the question stem we already know that since ZY<XY<0 If Y is +ve, Z<X<0 If Y is ve, 0<X<Z In other words Z>X
The statements merely confirm what we already know..



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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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09 Oct 2016, 22:16
I like to maximize my usage of number plugging. Here is my method: ZY<XY<0 => ZY  XY < 0 => Y*(ZX) < 0 (i) Y<0 => ZX>0 => Z>X. Also, ZY<0 => Z>0 & X>0 => Z>X>0 & (ii) Y>0 => ZX<0 => Z<X. Also, XY<0 => X<0 & Z<0 => Z<X<0
Also, XZ+X=Z => XZ = Z  X
STATEMENT1: Z<X => Z<X<0 => 2<1<0 => 1+2 = 1 & 2  1 = 21 = 1 => 1 = 1. Sufficient
STATEMENT 2: Y<0 => Z>X>0 => 2>1>0 => 12 = 1 = 1 & 2  1 = 21 = 1 => 1 = 1. Sufficient. D



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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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31 Oct 2016, 13:38
VeritasPrepKarishma wrote: rgtiwari wrote: If zy < xy < 0, is  x  z  + x = z?
(1) z < x (2) y > 0 You can solve such questions easily by restating '< 0' as 'negative' and '> 0' as 'positive'. zy < xy < 0 implies both 'zy' and 'xy' are negative and zy is more negative i.e. has greater absolute value as compared to xy. Since y will be equal in both, z will have a greater absolute value as compared to x. When will zy and xy both be negative? In 2 cases: Case 1: When y is positive and z and x are both negative. Case 2: When y is negative and z and x are both positive. Question: Is  x  z  + x = z ? Is  x  z  = z  x ? Is  z  x  = z  x ? (Since  x  z  =  z  x ) We rewrite the question only for better understanding. Now think, when will  z  x  = z  x ? It happens when x and z have the same sign. In case both have the same sign, they get subtracted on both sides so you get the same answer. In case they have opposite signs, they get added on LHS and subtracted on RHS and hence the equality doesn't hold. So if we can figure whether both x and z have the same sign, we can answer the question. As we saw above, in both case 1 and case 2, x and z must have the same sign. This implies that the equality must hold and you don't actually need the statements to answer the question. You can answer it without the statements (this shouldn't happen in actual GMAT). Hence answer must be (D). but for the equality to hold true done we need to know for sure that /z/ > /x/ and that they have the same sign??



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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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01 Nov 2016, 02:27
yezz wrote: VeritasPrepKarishma wrote: rgtiwari wrote: If zy < xy < 0, is  x  z  + x = z?
(1) z < x (2) y > 0 You can solve such questions easily by restating '< 0' as 'negative' and '> 0' as 'positive'. zy < xy < 0 implies both 'zy' and 'xy' are negative and zy is more negative i.e. has greater absolute value as compared to xy. Since y will be equal in both, z will have a greater absolute value as compared to x.When will zy and xy both be negative? In 2 cases: Case 1: When y is positive and z and x are both negative. Case 2: When y is negative and z and x are both positive. Question: Is  x  z  + x = z ? Is  x  z  = z  x ? Is  z  x  = z  x ? (Since  x  z  =  z  x ) We rewrite the question only for better understanding. Now think, when will  z  x  = z  x ? It happens when x and z have the same sign. In case both have the same sign, they get subtracted on both sides so you get the same answer. In case they have opposite signs, they get added on LHS and subtracted on RHS and hence the equality doesn't hold. So if we can figure whether both x and z have the same sign, we can answer the question. As we saw above, in both case 1 and case 2, x and z must have the same sign. This implies that the equality must hold and you don't actually need the statements to answer the question. You can answer it without the statements (this shouldn't happen in actual GMAT). Hence answer must be (D). but for the equality to hold true done we need to know for sure that /z/ > /x/ and that they have the same sign?? Both points have been considered. The first sentence of the solution states: "zy < xy < 0 implies both 'zy' and 'xy' are negative and zy is more negative i.e. has greater absolute value as compared to xy. Since y will be equal in both, z will have a greater absolute value as compared to x." Since zy > xy, zy > xy Divide by y on both sides z > x Also, note the other highlighted sentence: "Now think, when will  z  x  = z  x ? It happens when x and z have the same sign."
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If zy < xy < 0, is  x  z  + x = z? [#permalink]
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25 Feb 2017, 13:11
Bunuel wrote: This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.
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.
Answer: D.
Hope it helps. why in case A, have you expanded xz as x+z, but x and y as x and y?? shouldnt x and y be expanded as x and y as well?



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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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26 Feb 2017, 03:22
OreoShake wrote: Bunuel wrote: This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.
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.
Answer: D.
Hope it helps. why in case A, have you expanded xz as x+z, but x and y as x and y?? shouldnt x and y be expanded as x and y as well? A. If \(y<0\) > when reducing we should flip signs and we'll get: \(z>x>0\). Since \(z>x\) > x  z < 0, thus \(xz=(xz) =x+z\); Since \(x>0\) and \(z>0\) > \(x=x\) and \(z=z\).
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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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13 Dec 2017, 20:58
Bunuel wrote: This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.
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.
Answer: D.
Hope it helps. Hi Buneul, Does the below explanation make sense? since xz + x=z We can break this down to 1) (z)=(xz)+(x) 2) (z)=(xz)(x) Hence, 1) z=2xz > z=x 2) x=0 In both cases, xz + x=z holds true and hence sufficient.



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Re: If zy < xy < 0, is  x  z  + x = z? [#permalink]
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23 Dec 2017, 11:49
rgtiwari wrote: If zy < xy < 0, is  x  z  + x = z?
(1) z < x (2) y > 0 To think alternatively this is a very easy question hardly a minute, Given IF zy<xy<0 this means that both zy and xy have opposite signs There can be two cases Either y is positive then both z and x are negative Here z<xZX<0 Cae 2 y is negative . Here both x and y are positive and for inequality to be true now x<zZX>0 Rewriting IS  x  z  + x = z,,,= IS((( zx  = z  x))) Which basically means that we have to find whether case 2 hold or CASE 1 St 1 Z<X  Case 1 holds here and therefore  zx  is equal to z  x ((((CASE 1 ZX<0))) you can substitute some numbers to check SUFFICIENT(((( Z=6 , X=2 ,,,, LHS =4 AND RHS = 6  2 =4 ST 2 Y>0 CASE 1 HOLDS AGAIN HENCE SUFFICIENT




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