Is x^2 + y^2 3z

Question

Is x^2 + y^2 > 3z

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

Bunuel · Accepted Answer

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.

Answer: C.

qlx · Answer

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

Bunuel · Answer

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.

JusTLucK04 · Answer

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!

mvictor · Answer

I knew that somewhere has to be the trick...got fooled and picked A...

TheUltimateWinner · Answer

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.

Basshead · Answer

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

2(x^2+y^2) = 10z 
 x^2 + y^2 = 5z

However, we don't know if  z = 0 . If  z = 0 , then  x^2 + y^2  is equal to 0, not greater. INSUFFICINET.

(2) Clearly insufficient.

(1&2) Combined, we know z = 0. Therefore  x^2 + y^2  is not greater than 3z. SUFFICIENT.

Answer is C.

KeyurJoshi · Answer

x^2 + y^2 > 3z?

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

(x-y)^2 = z
x^2+y^2-2xy = z
x^2+y^2 = z + 2xy

Equating both,
z+ 2xy = 9z - 2xy
8z = 4xy
z = xy/2

Put this value in above eqn x^2+y^2 = 9z - 2xy
x^2+y^2 = 9(xy/2) - 2xy
x^2+y^2 = 5xy/2

So, question becomes, x^2 + y^2 > 3z ?
5xy/2 > 3xy/2
5xy > 3xy ?
5xy-3xy > 0?
2xy > 0?
 xy > 0? 
Not Sufficient bc we dont know x and y.

(2) z = 0
x^2 + y^2 > 3(0) ?
x^2 + y^2 > 0?
Say x=0 y=0 then 0 = 0 NO
Say x=1 y=1 then 2 > 0 YES
Not Sufficient

(1) + (2),
z = 0 
Question was  xy > 0? 
We know that z = xy/2
so, xy/2 = 0
So xy must be 0
Hence xy = 0 and NOT xy > 0.
Sufficient, C.

rvgmat12 · Answer

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).

Please note: the most common answer to this problem is A, largely because those who work on statement 1 first will first expand the quadratics:

x2+2xy+y2=9z

x2−2xy+y2=z
Then combine the two, eliminating the middle term and arriving at:

2x2+2y2=10z

x2+y2=5z
And since 5z looks like it has to be greater than 3z where z is nonnegative, they'll say that the answer has to be "yes." But remember: nonnegative means positive OR zero. Even if you missed that concept upon going through statement 1 ALONE, statement 2 is a chance to save yourself by asking "Why Are You Here?" to the easier statement. It's there to tip you off to the possibility of z = 0.

pudu · Answer

how c?

x^2+ y^2+2xy=9z and x^2+ y^2-2xy=z. 4xy=10z and 2xy= 5z
9z-5z=4z>3z. A

Bunuel · Answer

________________________________________
What if  x=y=z=0  ?

pudu · Answer

thank you. understood...don't know why i am making silly mistakes

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Is x^2 + y^2 > 3z

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