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# Is x > y ?

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Is x > y ? [#permalink]

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14 May 2012, 10:34
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Is x > y ?

(1) x = 4y^2
(2) x > 1
[Reveal] Spoiler: OA

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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]

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16 May 2012, 00:57
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Is x > y ?

(1) x = 4y^2. If x=y=0 then the answer is NO but if x=4 and y=1 then the answer is YES. Not sufficient.

(2) x > 1. Not sufficient.

(1)+(2) $$x = 4y^2$$ --> $$y=\frac{\sqrt{x}}{2}$$ (or $$y=-\frac{\sqrt{x}}{2}=negative$$). Now, since $$x>1$$ then $$x>\sqrt{x}$$, so $$x>(\frac{\sqrt{x}}{2}=y)$$. Sufficient.

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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]

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16 May 2012, 02:26
Thanks Bunuel, could you point to other similar inequalities problems.
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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]

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16 May 2012, 02:30
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AbhiJ wrote:
Thanks Bunuel, could you point to other similar inequalities problems.

DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=64

Hope it helps.
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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]

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06 Mar 2014, 15:22
Hi Bunuel,

i got 2 approaches yielding different results. Can you please take a look?

1. Approach

x = 4y^2 ; 1 < x

=> 1 < 4y^2 => 1/4 < y^2

2. Approach

x = 4y^2 ; 1 < x

=> x/x = (4y^2)/x
1 = (4y^2)/x

but if 1 < x then (4y^2)/x < x => 4y^2 < 1 => y^2 < 1/4

Where is the flaw ?

Bunuel wrote:
Is x > y ?

(1) x = 4y^2. If x=y=0 then the answer is NO but if x=4 and y=1 then the answer is YES. Not sufficient.

(2) x > 1. Not sufficient.

(1)+(2) $$x = 4y^2$$ --> $$y=\frac{\sqrt{x}}{2}$$ (or $$y=-\frac{\sqrt{x}}{2}=negative$$). Now, since $$x>1$$ then $$x>\sqrt{x}$$, so $$x>(\frac{\sqrt{x}}{2}=y)$$. Sufficient.

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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]

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07 Mar 2014, 00:50
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mariofelix wrote:
Hi Bunuel,

i got 2 approaches yielding different results. Can you please take a look?

1. Approach

x = 4y^2 ; 1 < x

=> 1 < 4y^2 => 1/4 < y^2

2. Approach

x = 4y^2 ; 1 < x

=> x/x = (4y^2)/x
1 = (4y^2)/x

but if 1 < x then (4y^2)/x < x => 4y^2 < 1 => y^2 < 1/4

Where is the flaw ?

Bunuel wrote:
Is x > y ?

(1) x = 4y^2. If x=y=0 then the answer is NO but if x=4 and y=1 then the answer is YES. Not sufficient.

(2) x > 1. Not sufficient.

(1)+(2) $$x = 4y^2$$ --> $$y=\frac{\sqrt{x}}{2}$$ (or $$y=-\frac{\sqrt{x}}{2}=negative$$). Now, since $$x>1$$ then $$x>\sqrt{x}$$, so $$x>(\frac{\sqrt{x}}{2}=y)$$. Sufficient.

$$\frac{(4y^2)}{x} < x$$ --> $$4y^2 < x^2$$, not $$4y^2 < 1$$.
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Re: Is x > y ? [#permalink]

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08 Mar 2014, 04:36

Stat 1: x = 4y^2
Stat 2: x>1

1+2: since x > 1( x^2 is always greater that x) and y^2 is also positive. It follows that lxl = 4ly^2l
hence, x> y

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Re: Is x > y ? [#permalink]

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20 Aug 2015, 10:04
To solve these types of inequality questions, it's useful to see the ranges where x>y, x<y, and where x=y.

1) To start off, we know that $$x = 4y^2$$. If we wanna see where x=y, set both variables equal to each other.

x = 4x^2
1 = 4x (dividing by y means we have to consider the case when x = 0. Notice in this case that x = y.
x = 1/4.

So our critical values here are y = 0 and x = 1/4.

<-------(0)--------(1/4)-------->

If x<0 (for instance, -1, then y = 4), x<y
if 0<x<1/4 (for instance, 1/8, then y =1/16) x>y
If x>1/4 (for instance, x = 1, y= 4) x<y

So, for x to be greater than y, x has be greater than 0 and less than 1/4. This is not enough information on its own. Insufficient.

2) is obviously not enough information on its own. Insufficient.

However, if we know that x>1, we know that it's outside this range and therefore that x<y.

We need both pieces of information. Therefore answer: C

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Is x > y ? [#permalink]

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17 Feb 2016, 12:08
Ans C

Please check the solution in the picture attached
Attachments

Capture.PNG [ 932.48 KiB | Viewed 3885 times ]

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Re: Is x > y ? [#permalink]

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13 Mar 2016, 08:02
For any inequality we must the check the bound (-1,0) and (0,1)
THEY BEHAVE FUNNY WHEN SQUARED....
WRITE THAT DOWN...
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Re: Is x > y ? [#permalink]

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10 Aug 2016, 05:37
Top Contributor
AbhiJ wrote:
Is x > y ?

(1) x = 4y^2
(2) x > 1

(1)
when y=$$\frac{1}{2}$$,then x=4$$(\frac{1}{2})^2$$=1,so x>Y
When y=$$\frac{1}{4}$$,then x=4$$(\frac{1}{4})^2$$=$$\frac{1}{4}$$,x=y,Not Sufficient

(2) No information about y ,Clearly Not sufficient

(1)+(2) when x>1 ,x is always greater than y in terms of the equation from Statement (1),Sufficient

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Re: Is x > y ? [#permalink]

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11 Aug 2016, 04:38
Is x > y ?

(1) x = 4y^2 ---> It represents the equation of parabola symmetric towards Ist and 4th quadrant , x can be +ve or zero
if any +ve value , y> x , for x=0 , x=y=0; ( INSUFF)

(2) x > 1 --- standalone not sufficient

by combining we know that x is value greater than > 1 , and it is always smaller than y

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Re: Is x > y ? [#permalink]

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23 Aug 2016, 03:22
Is x > y ?

(1) x = 4y^2
(2) x > 1

(1) x = 4y^2

if y =0 => x =0 No
if y =1 & x= 4 yes
if y= 1/2 x = 1 yes

(2) x > 1

both combined yes

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Re: Is x > y ? [#permalink]

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Re: Is x > y ?   [#permalink] 02 Sep 2017, 01:48
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