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Manager  Joined: 15 Apr 2013
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Location: India
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Is x^2 + y^2 > 3z  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 46% (01:50) correct 54% (01:53) wrong based on 292 sessions

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Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: Is x^2 + y^2 > 3z  [#permalink]

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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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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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Joined: 02 Sep 2009
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Re: Is x^2 + y^2 > 3z  [#permalink]

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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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1
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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I knew that somewhere has to be the trick...got fooled and picked A...
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Joined: 23 Feb 2015
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Re: Is x^2 + y^2 > 3z  [#permalink]

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Bunuel wrote:
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.

Hi Bunuel, is my explanation ok? Thank you brother...
statement 1:
x^2+y^2+2xy=9z
x^2+y^2-2xy=z
if we add them then we will get the following
2x^2+2y^2=10z
---> x^2+y^2=5z
Now the question is :
IS x^2+y^2>3z?
So, the rephrase question is-----
IS 5z>3z?
if z=1, then 5*1>3*1?
5>3?
yes, 5 is always greater than 3,

again,
5z>3z?
if we put the value of z=-1, then we get...
5*(-1)>3*(-1)?
-5>-3?
No, -5 is always smaller than -3
So, statement 1 is not sufficient...

Statement 2:

if you put the value of Z=0, then the question stem will be.....

x^2+y^2>3*0?
---> x^2+y^2>0?

if we put positive value for x and y, it always give positive result and also if we put negative value for x and y, it also give positive value as a nature of POSITIVE power. But, if we let x=0, and y=0, then the equation will be----
0^2+0^2>0?
0+0>0?
0 >0?
No,
---> not sufficient.

(1)+(2)
5z>3z?
5*0>3*0?
0>0?
No,
---->sufficient.
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Re: Is x^2 + y^2 > 3z  [#permalink]

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_________________ Re: Is x^2 + y^2 > 3z   [#permalink] 22 Jun 2019, 07:28
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