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

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Current Student
Joined: 04 Feb 2014
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Is x - x^2 > y - y^2?  [#permalink]

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05 Jul 2017, 00:31
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Difficulty:

85% (hard)

Question Stats:

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

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

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

_________________
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Math Expert
Joined: 02 Aug 2009
Posts: 7681
Re: Is x - x^2 > y - y^2?  [#permalink]

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05 Jul 2017, 02:11
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)...
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
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Re: Is x - x^2 > y - y^2?  [#permalink]

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06 Jul 2017, 06:02
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

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Re: Is x - x^2 > y - y^2?  [#permalink]

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06 Jul 2017, 06:13
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

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
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Re: Is x - x^2 > y - y^2?  [#permalink]

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16 Jul 2017, 04:22
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
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Re: Is x - x^2 > y - y^2?  [#permalink]

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11 Dec 2018, 12:07
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...
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Joined: 01 Nov 2015
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Re: Is x - x^2 > y - y^2?  [#permalink]

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25 Apr 2019, 20:04
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)...
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
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Joined: 17 Jul 2018
Posts: 39
Re: Is x - x^2 > y - y^2?  [#permalink]

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25 Apr 2019, 23:02
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)...
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.

Re: Is x - x^2 > y - y^2?   [#permalink] 25 Apr 2019, 23:02
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