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# Is x > y?

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Manager
Joined: 06 Mar 2014
Posts: 239
Location: India
GMAT Date: 04-30-2015

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Updated on: 03 Oct 2014, 14:43
6
00:00

Difficulty:

65% (hard)

Question Stats:

51% (01:34) correct 49% (01:48) wrong based on 308 sessions

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Is x > y?

(1) $$x + 2\sqrt{xy} + y = 0$$

(2) x^2 - y^2 = 0

Originally posted by earnit on 03 Oct 2014, 14:38.
Last edited by Bunuel on 03 Oct 2014, 14:43, edited 1 time in total.
Edited the question.
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03 Oct 2014, 14:48
2
1
Is x > y?

(1) $$x + 2\sqrt{xy} + y = 0$$;

Re-arrange: $$x + y = -2\sqrt{xy}$$;

Square: $$x^2 + 2xy + y^2 = 4xy$$;

Re-arrange: $$x^2 - 2xy + y^2 = 0$$;

$$(x-y)^2=0$$;

$$x=y$$.

x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

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Re: Is x > y?  [#permalink]

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25 Feb 2016, 19:15
1
1
i arrived to the answer choice differently...kind of..

2 - started with this one, since it is easier - clearly insufficient: (x+y)(x-y)=0, so either x=y - which is a definite no, or x=-y. if y=positive, then x<y. if y=negative, then x>y. so 2 outcomes...B and D - out.

now 1.
x+y=-sqrt(4xy)
square everything:
x^2+y^2+2xy = 4xy
x^2+y^2=2xy
x^2+y^2-2xy=0
(x-y)(x-y)=0
we know for sure that x=y. so the answer to the main question is only NO.

A
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25 Aug 2018, 21:51
1
Bunuel wrote:
Is x > y?

(1) $$x + 2\sqrt{xy} + y = 0$$ --> $$(\sqrt{x}+\sqrt{y})^2=0$$ --> $$\sqrt{x}+\sqrt{y}=0$$. The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing $$\sqrt{x}$$ and $$\sqrt{y}$$ seems not quite right.
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Re: Is x > y?  [#permalink]

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26 Aug 2018, 02:32
1
Izzyjolly wrote:
Bunuel wrote:
Is x > y?

(1) $$x + 2\sqrt{xy} + y = 0$$ --> $$(\sqrt{x}+\sqrt{y})^2=0$$ --> $$\sqrt{x}+\sqrt{y}=0$$. The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing $$\sqrt{x}$$ and $$\sqrt{y}$$ seems not quite right.

Good catch. You are absolutely right. Edited the post. Thank you.
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Re: Is x > y?  [#permalink]

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19 Sep 2018, 04:08
Bunuel wrote:
Izzyjolly wrote:
Bunuel wrote:
Is x > y?

(1) $$x + 2\sqrt{xy} + y = 0$$ --> $$(\sqrt{x}+\sqrt{y})^2=0$$ --> $$\sqrt{x}+\sqrt{y}=0$$. The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing $$\sqrt{x}$$ and $$\sqrt{y}$$ seems not quite right.

Good catch. You are absolutely right. Edited the post. Thank you.

Hi Bunuel

I do not understand why we can not infer from the red part that x and y have to be 0? Why do we need to square the whole expression in statement 1 and can not use the square roots of x and y? Could you explain it in more detail?

Thank you!
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Re: Is x > y?  [#permalink]

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19 Sep 2018, 04:10
1
T1101 wrote:

Hi Bunuel

I do not understand why we can not infer from the red part that x and y have to be 0? Why do we need to square the whole expression in statement 1 and can not use the square roots of x and y? Could you explain it in more detail?

Thank you!

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Consider x = y = -1.
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Re: Is x > y?   [#permalink] 19 Sep 2018, 04:10
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