It is currently 19 Nov 2017, 00:21

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 15 Apr 2013
Posts: 83

Kudos [?]: 130 [2], given: 61

Location: India
Concentration: Finance, General Management
Schools: ISB '15
WE: Account Management (Other)
Is x^2 + y^2 > 3z [#permalink]

### Show Tags

01 Sep 2013, 11:16
2
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

39% (01:03) correct 61% (00:55) wrong based on 394 sessions

### HideShow timer Statistics

Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0
[Reveal] Spoiler: OA

Kudos [?]: 130 [2], given: 61

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132643 [7], given: 12326

Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

01 Sep 2013, 11:34
7
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 132643 [7], given: 12326

Manager
Joined: 17 Mar 2014
Posts: 70

Kudos [?]: 79 [0], given: 38

Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

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

Kudos [?]: 79 [0], given: 38

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132643 [1], given: 12326

Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

26 Mar 2014, 08:13
1
KUDOS
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.
_________________

Kudos [?]: 132643 [1], given: 12326

Retired Moderator
Joined: 17 Sep 2013
Posts: 387

Kudos [?]: 348 [1], given: 139

Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

31 Oct 2014, 12:42
1
KUDOS
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!
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Kudos [?]: 348 [1], given: 139

Board of Directors
Joined: 17 Jul 2014
Posts: 2672

Kudos [?]: 431 [0], given: 200

Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

08 Mar 2016, 19:24
I knew that somewhere has to be the trick...got fooled and picked A...

Kudos [?]: 431 [0], given: 200

Senior Manager
Joined: 23 Feb 2015
Posts: 474

Kudos [?]: 219 [0], given: 179

Re: Is x^2 + y^2 > 3z [#permalink]

### Show Tags

02 Dec 2016, 03:40
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.
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”

Kudos [?]: 219 [0], given: 179

Re: Is x^2 + y^2 > 3z   [#permalink] 02 Dec 2016, 03:40
Display posts from previous: Sort by