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

 It is currently 21 May 2019, 03:49 ### 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
Your Progress

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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # Is x - x^2 > y - y^2?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Current Student G
Joined: 04 Feb 2014
Posts: 223
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3
WE: Project Management (Manufacturing)
Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

1 00:00

Difficulty:   85% (hard)

Question Stats: 53% (02:29) correct 47% (02:10) wrong based on 92 sessions

### HideShow timer Statistics

Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

_________________
Kudos if you like my post
Math Expert V
Joined: 02 Aug 2009
Posts: 7681
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

anurag16 wrote:
Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

$$x-x^2>y-y^2....... x-y+y^2-x^2>0......$$..
$$(x-y)-(x-y)(x+y)>0...... (x-y)(1-(x+y))>0$$..
So two case..
A) both POSITIVE..
x-y>0 or x>y AND
1>x+y..
B) both NEGATIVE
X<y and x+y>1..

Let's see if any statement fulfills any of the above cases..
1) x>y..
see A, NOTHING about if 1>x+y
Insufficient
2) $$x^2>y^2$$..
It just tells us that NUMERIC value of (x> y)...
Nothing about values with sign..
Insuff

Combined..
We can surely say by adding two statements that..
x+x^2>y+y^2 but nothing about given equation..
You can find various combinations in FRACTION or INTEGERS, in + or - that will give us answer as YES and NO
Insufficient
_________________
Current Student G
Joined: 04 Feb 2014
Posts: 223
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3
WE: Project Management (Manufacturing)
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

Hi chetan2u,

Thanks for the reply. Can you please explain why statement B is insufficient by taking some values?
x^2>y^2 implies mod x> mod y

So in x - x^2 > y - y^2 will always give a no

x=2 y=1 NO
x=-2 y=-1 NO
x=-2 y=1 NO
x=2 y=-1 NO

Thanks in advance.
_________________
Kudos if you like my post
Math Expert V
Joined: 02 Aug 2009
Posts: 7681
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

1
anurag16 wrote:
Hi chetan2u,

Thanks for the reply. Can you please explain why statement B is insufficient by taking some values?
x^2>y^2 implies mod x> mod y

So in x - x^2 > y - y^2 will always give a no

x=2 y=1 NO
x=-2 y=-1 NO
x=-2 y=1 NO
x=2 y=-1 NO

Thanks in advance.

Hi,
What are you missing on is that they can be fractions too..
X=1/2, Y=1/4 YES
Whenever variables are given, check for -,+,0, fractions. We know this but tend to miss while doing a Q
_________________
Intern  S
Joined: 16 Apr 2017
Posts: 45
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

anurag16 wrote:
Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

It can be easily derived:

1)x > y :
x= -2 and y = -3 ==> answer Yes
x = 3 and y = 2 ===> answer no
insufficient
2) x^2>y^2==> x^2-y^2 >0
therefore from original condition x-y> x^2-y^2>0==>x-y>0===> x>y...same as statement 1st which proved insufficient

Since statement 1 and 2 are same and insufficient answer will always be E
_________________
KUDOS please, if you like the post or if it helps Manager  B
Joined: 12 Mar 2018
Posts: 124
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

Supermaverick wrote:
anurag16 wrote:
Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

It can be easily derived:

1)x > y :
x= -2 and y = -3 ==> answer Yes
x = 3 and y = 2 ===> answer no
insufficient
2) x^2>y^2==> x^2-y^2 >0
therefore from original condition x-y> x^2-y^2>0==>x-y>0===> x>y...same as statement 1st which proved insufficient

Since statement 1 and 2 are same and insufficient answer will always be E

statement 2 is not equal to stat1, consider x and y as negative integers...
Intern  B
Joined: 01 Nov 2015
Posts: 1
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

chetan2u wrote:
anurag16 wrote:
Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

$$x-x^2>y-y^2....... x-y+y^2-x^2>0......$$..
$$(x-y)-(x-y)(x+y)>0...... (x-y)(1-(x+y))>0$$..
So two case..
A) both POSITIVE..
x-y>0 or x>y AND
1>x+y..
B) both NEGATIVE
X<y and x+y>1..

Let's see if any statement fulfills any of the above cases..
1) x>y..
see A, NOTHING about if 1>x+y
Insufficient
2) $$x^2>y^2$$..
It just tells us that NUMERIC value of (x> y)...
Nothing about values with sign..
Insuff

Combined..
We can surely say by adding two statements that..
x+x^2>y+y^2 but nothing about given equation..
You can find various combinations in FRACTION or INTEGERS, in + or - that will give us answer as YES and NO
Insufficient

Hi,

But if we add both the inequalities we get
x+ x^2 > y + y^2
Now after rearranging
-1*(x-y)> (x-y)(x+y)
As from statement 1 we know that x>y so x-y is +ve, thus
-1>x+y
now putting this in the main equation i.e.
(x-y)(1-(x+y))>0
we get that it will be always greater than 0. So, answer should be C instead.

Please correct me if I am wrong.

Regards,
Avi
Intern  B
Joined: 17 Jul 2018
Posts: 39
Concentration: Finance, Leadership
Re: Is x - x^2 > y - y^2?  [#permalink]

### Show Tags

aviaggarwal.ism wrote:
chetan2u wrote:
anurag16 wrote:
Is x - x^2 > y - y^2?

(1) x > y
(2) x^2 > y^2

$$x-x^2>y-y^2....... x-y+y^2-x^2>0......$$..
$$(x-y)-(x-y)(x+y)>0...... (x-y)(1-(x+y))>0$$..
So two case..
A) both POSITIVE..
x-y>0 or x>y AND
1>x+y..
B) both NEGATIVE
X<y and x+y>1..

Let's see if any statement fulfills any of the above cases..
1) x>y..
see A, NOTHING about if 1>x+y
Insufficient
2) $$x^2>y^2$$..
It just tells us that NUMERIC value of (x> y)...
Nothing about values with sign..
Insuff

Combined..
We can surely say by adding two statements that..
x+x^2>y+y^2 but nothing about given equation..
You can find various combinations in FRACTION or INTEGERS, in + or - that will give us answer as YES and NO
Insufficient

Hi,

But if we add both the inequalities we get
x+ x^2 > y + y^2
Now after rearranging
-1*(x-y)> (x-y)(x+y)
As from statement 1 we know that x>y so x-y is +ve, thus
-1>x+y
now putting this in the main equation i.e.
(x-y)(1-(x+y))>0
we get that it will be always greater than 0. So, answer should be C instead.

Please correct me if I am wrong.

Regards,
Avi

Hi Avi, you made a mistake here
x+ x^2 > y + y^2
Now after rearranging
-1*(x-y)> (x-y)(x+y)
As from statement 1 we know that x>y so x-y is +ve, thus
-1>x+y

x+x^2>y+Y^2
=> x-y>-(x^2-Y^2)
=> (x-y) > -(x+y)(x-y)
As x>y, this is reduced to 1>-(x+y) or (x+y)>-1

Now putting this in expression (x-y)(1-(x+y)) we cant say if the second term (1-(x+y) is positive or negative. For x+y between -1 and <1, it will be positive, For x+y>1, it will be negative. Therefore Insufficient.

Please hit +1 Kudos if the answer helps.  Re: Is x - x^2 > y - y^2?   [#permalink] 25 Apr 2019, 23:02
Display posts from previous: Sort by

# Is x - x^2 > y - y^2?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

#### MBA Resources  