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Is x > y ? [#permalink]
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14 May 2012, 11:34
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Is x > y ? (1) x = 4y^2 (2) x > 1
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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]
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16 May 2012, 01:57
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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]
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16 May 2012, 03:26
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, 03:30



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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]
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06 Mar 2014, 16:22
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 ? Thanks in advance 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.
Answer: C.



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Re: Is x > y ? (1) x = 4y^2 (2) x > 1 [#permalink]
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07 Mar 2014, 01:50
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 ? Thanks in advance 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.
Answer: C. \(\frac{(4y^2)}{x} < x\) > \(4y^2 < x^2\), not \(4y^2 < 1\).
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Re: Is x > y ? [#permalink]
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08 Mar 2014, 05:36
easy to eliminate ADB
answer is C
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, 11:04
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, 13:08
Ans C Please check the solution in the picture attached
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Re: Is x > y ? [#permalink]
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Re: Is x > y ? [#permalink]
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10 Aug 2016, 06:37
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), SufficientCorrect Answer C
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Re: Is x > y ? [#permalink]
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11 Aug 2016, 05:38
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
(C) is the answer



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Re: Is x > y ? [#permalink]
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23 Aug 2016, 04:22
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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