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ZY<XY<0 is XZ + x = Z [#permalink]
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25 Apr 2007, 04:18
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ZY<XY<0 is XZ + x = Z (1) Z<X (2) Y<0 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifzyxy0isxzxz101210.html
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(D) for me
ZY<XY<0
=>ZY  XY < 0
<=> Y*(ZX) < 0
Implies that :
o Y < 0 and ZX > 0 <=> Z > X
or
o Y > 0 and ZX < 0 <=> Z < X
As, ZY<XY<0, the 2 cases above become:
o Y < 0 and Z > X > 0 (case A)
or
o Y > 0 and Z < X < 0 (case B)
XZ + X = Z ?
From 1
Z < X => We are forced to be in the case B.
That implies:
o Z < 0
o X < 0
o XZ > 0
So,
XZ + X
= (XZ) + (X)
= Z
= Z as Z < 0
SUFF.
From 2
Y < 0 => We are forced to be in the case A.
That implies:
o Z > 0
o X > 0
o ZX > 0 <=> XZ < 0
So,
XZ + X
= ((XZ)) + (X)
= Z
= Z as Z > 0
SUFF.
Last edited by Fig on 25 Apr 2007, 07:22, edited 1 time in total.



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my answer [#permalink]
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25 Apr 2007, 06:16
ZY<XY<0 is XZ + x = Z
1.) Z<X
2.) Y<0
What is a good approach here?
Thanks!
The question is asking is XZ + X = Z
XZ + X = Z
XZ = Z  X
This is true only when X and Z are the same sign, or essentially when they are both either greater than zero or less than zero. Since the assumption is that they are both less than zero when multiplied by Y, then if Y is positive, they are both negative, and if Y is negative, they are both positive.
1. Z < X
All this tells us is that Y < 0
2. Y < 0
This also tells us that Y < 0
IF Y < 0 then Z and X must be positive. Thus they are the same sign, thus the answer is D. However, I would say that if Y was positive, then XZ + X = Z would still be true. Y can't be zero because then XY and ZY can't be less than zero.
My question is, will there be questions like this on the Real Gmat, where the data given is not required to solve the question?



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The question is from GMAT PREP so presumably it's representative of the real GMAT.
I hope not to see it on the real thing though  I hate this type of Q and have a limit on how many scenarios I'm willing to write out before moving onto the next question!
I'm guessing it might be a 50/51 threshold question, I didn't get it but got a 50 on the practice test.



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Forgot to say Thanks a million guys for your help!



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doc14 wrote: Forgot to say Thanks a million guys for your help!
You are welcome !
If u want to practice more, try these ones :
http://www.gmatclub.com/phpbb/viewtopic.php?t=39533



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Re: my answer [#permalink]
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25 Apr 2007, 08:23
jayt696969 wrote: ZY<XY<0 is XZ + x = Z 1.) Z<X 2.) Y<0 What is a good approach here?
Thanks!
The question is asking is XZ + X = Z
XZ + X = Z
XZ = Z  X
This is true only when X and Z are the same sign, or essentially when they are both either greater than zero or less than zero. Since the assumption is that they are both less than zero when multiplied by Y, then if Y is positive, they are both negative, and if Y is negative, they are both positive.
1. Z < X All this tells us is that Y < 0
2. Y < 0 This also tells us that Y < 0
IF Y < 0 then Z and X must be positive. Thus they are the same sign, thus the answer is D. However, I would say that if Y was positive, then XZ + X = Z would still be true. Y can't be zero because then XY and ZY can't be less than zero.
My question is, will there be questions like this on the Real Gmat, where the data given is not required to solve the question?
This is a necessary, but not sufficient condition for the statement in the question to be true. It must also be true that z x is positive



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Re: GMATPREP Modulus Q [#permalink]
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15 Jun 2009, 12:53
Im a little lost here. If Z<X and Y<0 then these would be possible numbers:
Y = 1 Z = 2 X = 10
Then ZY would be 2 and XY would be 10 making XY<ZY (the question states that ZY<XY). What is wrong with this conclusion?



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Re: GMATPREP Modulus Q [#permalink]
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25 Nov 2009, 08:30
Fig, I understand your explanation but shouldn't the statements themselves be always true? If so, the two scenarios Y>0 and Y<0 are contradictory? Am I missing something here?



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Fig wrote: (D) for me ZY<XY<0 =>ZY  XY < 0 <=> Y*(ZX) < 0 Implies that : o Y < 0 and ZX > 0 <=> Z > X or o Y > 0 and ZX < 0 <=> Z < X As, ZY<XY<0, the 2 cases above become: o Y < 0 and Z > X > 0 (case A) or o Y > 0 and Z < X < 0 (case B) XZ + X = Z ? From 1Z < X => We are forced to be in the case B. That implies: o Z < 0 o X < 0 o XZ > 0 So, XZ + X = (XZ) + (X) = Z = Z as Z < 0 SUFF. From 2Y < 0 => We are forced to be in the case A. That implies: o Z > 0 o X > 0 o ZX > 0 <=> XZ < 0 So, XZ + X = ((XZ)) + (X) = Z = Z as Z > 0 SUFF. Hi How do you conclude this Y > 0 and Z < X < 0 (case B) if Y>0, then ZX<0 , the case can be z=2, X=3 and ZX<0 which dont agree with z<x<0 please advise



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ISBtarget wrote: Hi
How do you conclude this
Y > 0 and Z < X < 0 (case B)
if Y>0, then ZX<0 , the case can be z=2, X=3 and ZX<0 which dont agree with z<x<0
please advise This is not a good question, for two reasons: 1. Neither of statement is needed to answer the question, stem is enough to do so. I don't know if such kind of situation can occur in real GMAT. Maybe someone can clarify this? 2. Statements contradict. This will never occur in real GMAT. 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. Next:Statement (1) says that \(z<x\), hence we have case B, which implies that \(y>0\). BUT Statement (2) says that \(y<0\). So statements are contradictory, which will never occur in GMAT. Hope it helps.
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Re: GMATPREP Modulus Q [#permalink]
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05 May 2010, 08:08
the given condition itself is sufficient without considering 1 and 2.
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Re: GMATPREP Modulus Q [#permalink]
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06 May 2010, 13:17
ZY<XY<0 is XZ + x = Z 1.) Z<X 2.) Y<0 What is a good approach here?  A simpler approach :: make a few cases subcases x y z case 1 +  + 2  +  case1:: if x,y +ve and y negative thn z>x to satisfy zy<xy frm question part 1) given z<x exact opposite of wht we just did i.e x,z negative and y positive take numbers say x=3 , z=4 , y =+1 so 4=4 substitute in the equation ::same goes for any othr number .. hence part 1 satisfies similary do for part 2..m too tired to write [:)]
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Re: my answer [#permalink]
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30 Jan 2014, 14:21
jayt696969 wrote: ZY<XY<0 is XZ + x = Z 1.) Z<X 2.) Y<0 What is a good approach here?
Thanks!
The question is asking is XZ + X = Z
XZ + X = Z
XZ = Z  X
This is true only when X and Z are the same sign, or essentially when they are both either greater than zero or less than zero. Since the assumption is that they are both less than zero when multiplied by Y, then if Y is positive, they are both negative, and if Y is negative, they are both positive.
1. Z < X All this tells us is that Y < 0
2. Y < 0 This also tells us that Y < 0
IF Y < 0 then Z and X must be positive. Thus they are the same sign, thus the answer is D. However, I would say that if Y was positive, then XZ + X = Z would still be true. Y can't be zero because then XY and ZY can't be less than zero.
My question is, will there be questions like this on the Real Gmat, where the data given is not required to solve the question? I think you almost got it but you are missing the case when both are zero For XZ = Z  X both x and z must have the same sign OR be both equal to zero. Funny thing is that as you mention we already know that y is different from zero since we are told that 0>xy>zy, so it will work in either case Anyone else thinks that this question is flawed? Cheers J



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Re: ZY<XY<0 is XZ + x = Z [#permalink]
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30 Jan 2014, 22:26




Re: ZY<XY<0 is XZ + x = Z
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30 Jan 2014, 22:26







