GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2018, 02:22

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

Is x^2 + y^2 > 3z

Author Message
TAGS:

Hide Tags

Manager
Joined: 15 Apr 2013
Posts: 72
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
4
9
00:00

Difficulty:

95% (hard)

Question Stats:

47% (01:50) correct 53% (01:59) wrong based on 447 sessions

HideShow timer Statistics

Is x^2 + y^2 > 3z

(1) (x + y)^2 = 9z and (x - y)^2 = z
(2) z = 0
Math Expert
Joined: 02 Sep 2009
Posts: 49986
Re: Is x^2 + y^2 > 3z  [#permalink]

Show Tags

01 Sep 2013, 11:34
7
3
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.

_________________
General Discussion
Manager
Joined: 17 Mar 2014
Posts: 68
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
Math Expert
Joined: 02 Sep 2009
Posts: 49986
Re: Is x^2 + y^2 > 3z  [#permalink]

Show Tags

26 Mar 2014, 08:13
1
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.
_________________
Retired Moderator
Joined: 17 Sep 2013
Posts: 349
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
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..

Board of Directors
Joined: 17 Jul 2014
Posts: 2655
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...
Senior Manager
Joined: 23 Feb 2015
Posts: 426
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.”

Non-Human User
Joined: 09 Sep 2013
Posts: 8446
Re: Is x^2 + y^2 > 3z  [#permalink]

Show Tags

02 Apr 2018, 04:57
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Is x^2 + y^2 > 3z &nbs [#permalink] 02 Apr 2018, 04:57
Display posts from previous: Sort by