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Is x  y > x  z? [#permalink]
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01 Nov 2009, 08:42
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Re: Is x  y > x  z? [#permalink]
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01 Nov 2009, 09:08
C
need to know which is more extreme and what value X is which both statements combined tell you.



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Re: Is x  y > x  z? [#permalink]
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01 Nov 2009, 09:24
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Hussain15 wrote: Is x  y>x  z?
(1) y>z (2) x < 0 Good question. I believe the answer is E. Basically the question asks whether the distance between x and y is greater than the distance between x and z? (1) y>z, means y is farther from 0 than z, which gives us the following possible cases: y0z or yz0 or 0zy or z0y Now depending on the position of x, the distance between x and y may or may not be greater than distance between x and z. (2) x<0. Clearly insufficient as we know nothing about y and z: x0 (1)+(2) x<0 and y>z yx0z in this case x  y>x  z BUT xyz0 in this case x  y<x  z Two different answers, so insufficient. Answer: E.
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Re: Is x  y > x  z? [#permalink]
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02 Nov 2009, 00:52
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Bunuel!! I think you have made the record of fastest century of kudos!! Congrats!! Great explanation! I am not sure about the answer, will post OA in a while.
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Re: Is x  y > x  z? [#permalink]
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23 Aug 2010, 09:34



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Re: Is x  y > x  z? [#permalink]
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25 Aug 2010, 08:14
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In general for these types of questions, if I'm testing values, are the variables different from each other.. eg should I consider the case where x = y, or x = z unless otherwise stated?
Sorry if this is a dumb question.. lol



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Re: Is x  y > x  z? [#permalink]
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25 Aug 2010, 08:28



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Re: Is x  y > x  z? [#permalink]
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02 Sep 2010, 02:02
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I think we need to develop intuition for these kind of questions... from looking at question i was not able to say the answer but after looking at choices, i started to feel it must be E. After less then a minute of thinking i was sure it has to be E.



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Re: Is x  y > x  z? [#permalink]
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02 Sep 2010, 02:17
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It is E. We can make out pretty quickly that these are heavily based on values of X, Y, and Z. So it has to be C or E. You are subtracting Y and Z from a fixed value of X. Having absolute values is not going to help. Nice post but! Thanks.



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Re: Is x  y > x  z? [#permalink]
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06 Sep 2010, 17:25
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Thank you to bunuel again! Super nice explanation!
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Re: Is x  y > x  z? [#permalink]
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23 Apr 2012, 06:49
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New Approach , Bunuel Please correct me if this approach is wrong ... to find if xy>xz ? Squaring both sides (since both sides are +ve) (xy)2(xz)2 >0 (xyx+z)(xy+xz) >0 (zy) (2xyz) >0 From 1 > Red part i negative but cant say anything about green part so insufficient. From 2 > know nothing so insufficient from both also nothing can be concluded so E....
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Re: Is x  y > x  z? [#permalink]
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23 Apr 2012, 07:36
shikhar wrote: New Approach , Bunuel Please correct me if this approach is wrong ...
to find if xy>xz ?
Squaring both sides (since both sides are +ve)
(xy)2(xz)2 >0 (xyx+z)(xy+xz) >0 (zy) (2xyz) >0
From 1 > Red part i negative but cant say anything about green part so insufficient. From 2 > know nothing so insufficient
from both also nothing can be concluded so E.... Yes, you can square both parts of the inequality (since both are nonnegative) and rewrite the question: is (zy)(2xyz)>0? But from y>z you can not say that zy is negative: if y=2 and z=1 then YES but if y=2and z=1 then NO. So, overall, I'd still suggest number line approach.
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Re: Is x  y > x  z? [#permalink]
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23 Apr 2012, 13:28
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Great explanation, I never thought of thinking of the problem in terms of positioning. Great strategy.



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Re: Is x  y > x  z? [#permalink]
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25 Apr 2012, 20:11
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bunuel, would love your feedback on my method
xy > x  z
this applies when both signs are the same, or signs are different.
ie x > y
x > y x > y
a.) (xy) > x  z x + y > x z 2x > z y 2x < z + y
b.) xy > x z y > z y < z
stmt 1:
i did the same thing here: a.) y > z
b.) y < z
or z < y < z
which shows z is negative and not much about. in any case, x is not even mentioned so this is insufficient
stmt 2:
x < 0
again y and z are not mentioned, insufficient
stmt 1 + stmt 2:
attacking (a) from the original problem stem
y > z ?
according to stmt 1 this is false since z < y < z. so (a) from the stem is sufficient.
attacking (b) from the original stem
2x < y + z?
we know z < y < z, so add a z to each
2z < y + z < 0
y+ z < 0
and x < 0
so you have
negative number * 2 < negative number
and since we don't haev any relationship between what 2x would be or y +z would be, it would be insufficient E



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Re: Is x  y > x  z? [#permalink]
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25 Apr 2012, 21:38
pinchharmonic wrote: bunuel, would love your feedback on my method
xy > x  z
this applies when both signs are the same, or signs are different.
ie x > y
x > y x > y
a.) (xy) > x  z x + y > x z 2x > z y 2x < z + y
b.) xy > x z y > z y < z
stmt 1:
i did the same thing here: a.) y > z
b.) y < z
or z < y < z
which shows z is negative and not much about. in any case, x is not even mentioned so this is insufficient
stmt 2:
x < 0
again y and z are not mentioned, insufficient
stmt 1 + stmt 2:
attacking (a) from the original problem stem
y > z ?
according to stmt 1 this is false since z < y < z. so (a) from the stem is sufficient.
attacking (b) from the original stem
2x < y + z?
we know z < y < z, so add a z to each
2z < y + z < 0
y+ z < 0
and x < 0
so you have
negative number * 2 < negative number
and since we don't haev any relationship between what 2x would be or y +z would be, it would be insufficient E There are several flows above. For example: y>z does not mean that z is negative. Consider simple example: y=2 and x=1. y>z simply means that means y is farther from 0 than z: y0z (y<0<z) yz0 (y<z<0) 0zy (y>z>0) z0y (y>0>z)
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Re: Is x  y > x  z? [#permalink]
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26 Apr 2012, 17:06
thanks bunuel, great point i'll think of this type of problem in a more conceptual manner next time



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Re: Is x  y > x  z? [#permalink]
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22 May 2012, 18:06
try assuming values y=3 z=2 ... x can be any no. +ve / ve (take both cases to decide between A & C) but eventualy both did not give answer so E is correct



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Re: Is x  y > x  z? [#permalink]
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22 May 2012, 20:20
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I used the approach below :
1) y>z will give 4 different cases , I assumed only two y=5 , z=2 & y=5 , z= 2 No information about x , so x can be =+ ve or ve , let x = 7 xy=75 =2& xz= 72=5
xy=7(5)=12 & xz = 7(2)=9 contradictory results . Insufficient
2)x<0 does not tell anything about y,z so insufficient.
1&2) Repeating the above values y=5 , z=2 & y=5 , z= 2 x=3 xy= 3+5= 2 , xz = 3+2 = 1 xy= 35= 8 , xz= 32=5
Contradictory results , Insufficient. Hence E.



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Re: Is x  y > x  z? [#permalink]
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25 May 2012, 19:58
picked E purely based on picking numbers approach. took 114 sec (1.54 min) how much time have you guys taken on this sum on avg?



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Re: Is x  y > x  z? [#permalink]
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24 Jun 2012, 10:16
Voted for E
Solved it following way
(A) Y > Z
Case 1 :
[X =4][Y = 3][Z = 2][0][Z = 2][Y=3][X=4]
Case 2 :
[Y = 3][Z = 2][X =1][0][X=1][Z = 2][Y=3]
when x>0
XY = 13 = 2 and XZ = 12 = 1 therefore XY > XZ
XY = 43 = 1 and XZ = 42 = 2 therefore XY < XZ
when x<0
XY = 1+3 = 2 and XZ = 1+2 = 1 therefore XY > XZ
XY = 4+3 = 1 and XZ = 4+2 = 2 therefore XY < XZ
not sufficient
(B) X<0 dont know about Y and Z, therefore nit sufficient
(C) X < 0 and Y > Z
when x<0
XY = 1+3 = 2 and XZ = 1+2 = 1 therefore XY > XZ
XY = 4+3 = 1 and XZ = 4+2 = 2 therefore XY < XZ [not sufficent]
therefore E




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