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# Is x^2 + y^2 > 3z

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Manager
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Is x^2 + y^2 > 3z [#permalink]

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01 Sep 2013, 11:16
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38% (02:16) correct 62% (01:01) wrong based on 222 sessions

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Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0
[Reveal] Spoiler: OA
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Re: Is x^2 + y^2 > 3z [#permalink]

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01 Sep 2013, 11:34
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Expert's post
Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z --> $$x^2+2xy+y^2=9z$$ and $$x^2-2xy+y^2=z$$. Add them up $$2(x^2+y^2)=10z$$ --> $$x^2+y^2=5z$$. If $$x=y=z=0$$, then $$x^2 + y^2=0=3z$$ and the answer is NO but if x, y, and z are different from zero, then the answer is YES. Not sufficient.

(2) z = 0. If $$x=y=z=0$$, then $$x^2 + y^2=0=3z$$ and the answer is NO but if x or y are different from zero, then the answer is YES. Not sufficient.

(1)+(2) From (1) we have that $$x^2+y^2=(non \ negative)+(non \ negative)=5z$$ and since from (2) we have that $$z=0$$, then $$x=y=z=0$$. Thereofre the answer to the question is NO. Sufficient.

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Re: Is x^2 + y^2 > 3z [#permalink]

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26 Mar 2014, 03:40
pavan2185 wrote:
Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0

1) given x^2+y^2+2xy = 9z --> x^2+y^2 = 9z- 2xy

given x^2+y^2-2xy = z --> x^2+y^2 = z+2xy

so 9z- 2xy = z+2xy --> 8z= 4xy --> z = (xy)/2

question now becomes is x^2 + y^2 > (3xy)/2

if x=y = 0 then answer is yes otherwise it is no

2)z= 0
x=y =z = 0 then yes other wise no

1+2

from 1 we know z= (xy)/2 and from 2 we know z = 0

so 0 = (xy)/2 --> xy = 0 so either x or y or both can be 0

if x=0 and y =0 then answer is no
if either one of x and y is not 0 then the answer is yes

I am getting E as the answer, can anyone tell me where is my error if any ?
Thank you
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Re: Is x^2 + y^2 > 3z [#permalink]

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26 Mar 2014, 08:13
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Expert's post
qlx wrote:
pavan2185 wrote:
Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0

1) given x^2+y^2+2xy = 9z --> x^2+y^2 = 9z- 2xy

given x^2+y^2-2xy = z --> x^2+y^2 = z+2xy

so 9z- 2xy = z+2xy --> 8z= 4xy --> z = (xy)/2

question now becomes is x^2 + y^2 > (3xy)/2

if x=y = 0 then answer is yes otherwise it is no

2)z= 0
x=y =z = 0 then yes other wise no

1+2

from 1 we know z= (xy)/2 and from 2 we know z = 0

so 0 = (xy)/2 --> xy = 0 so either x or y or both can be 0

if x=0 and y =0 then answer is no
if either one of x and y is not 0 then the answer is yes

I am getting E as the answer, can anyone tell me where is my error if any ?
Thank you

The scenario in red is not possible: From (1) we have that $$x^2+y^2=5z$$ (is-x-2-y-2-3z-158984.html#p1262747) --> $$x^2+y^2=0$$. Both x and y must be zero in order that to hold true.

Hope it's clear.
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Re: Is x^2 + y^2 > 3z [#permalink]

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31 Oct 2014, 12:42
OE:When possible, start with the easier statement. Let’s consider statement (2). If z = 0, then the question is really whether x2+y2>0. Any number squared is either 0 or positive, so this statement is not sufficient: if either x or y is not zero, the answer is yes, while if both x and y are 0, the answer is no. Statement (1) is trickier, but take statement (2) as a clue. If z = 0, then the two equations say that (x+y)2=0 and (x−y)2 = 0, from which we could conclude that (x+y)=0 and (x-y)=0, meaning that both x and y are 0. If z isn’t 0, however, lots of things could happen; not sufficient. Taken together, we have the scenario described above: x is 0, y is 0, and z is 0, and the answer to the original stimulus is a definitive no: (C). If this question seems hard, it is hard, so don’t fret!
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Re: Is x^2 + y^2 > 3z [#permalink]

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08 Mar 2016, 19:24
I knew that somewhere has to be the trick...got fooled and picked A...
Re: Is x^2 + y^2 > 3z   [#permalink] 08 Mar 2016, 19:24
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