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M32-12

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V
Joined: 02 Sep 2009
Posts: 41908

Kudos [?]: 129394 [0], given: 12197

M32-12 [#permalink]

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New post 17 Jul 2017, 06:28
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A
B
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D
E

Difficulty:

  55% (hard)

Question Stats:

27% (01:28) correct 73% (00:43) wrong based on 11 sessions

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Kudos [?]: 129394 [0], given: 12197

Expert Post
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Kudos [?]: 129394 [1], given: 12197

Re M32-12 [#permalink]

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New post 17 Jul 2017, 06:28
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Official Solution:


Is \((x + y)^2 > x^2 + y^2\)?

First let's rephrase the question: is \(x^2 +2xy+ y^2 > x^2 + y^2\)?

First let's rephrase the question: is \(xy>0\)?

(1) \(|x| \leq 0\). The absolute value of a number cannot be negative, it can be 0 or positive. Thus, from this statement we can deduce that \(x=0\). Therefore, \(xy=0\). We have a NO answer to the question. Sufficient.

(2) \(\sqrt{y^2} \leq 0\). This is the same as \(|y| \leq 0\). So, we have the same as above: \(y=0\). Therefore, \(xy=0\). We have a NO answer to the question. Sufficient.


Answer: D
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Kudos [?]: 129394 [1], given: 12197

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Re: M32-12 [#permalink]

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New post 09 Aug 2017, 02:35
Wow what a question
this made me question my fundamentals for a moment this is really really a high-quality question

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Re: M32-12 [#permalink]

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New post 15 Aug 2017, 02:49
Bunuel wrote:
Official Solution:


Is \((x + y)^2 > x^2 + y^2\)?

First let's rephrase the question: is \(x^2 +2xy+ y^2 > x^2 + y^2\)?

First let's rephrase the question: is \(xy>0\)?

(1) \(|x| \leq 0\). The absolute value of a number cannot be negative, it can be 0 or positive. Thus, from this statement we can deduce that \(x=0\). Therefore, \(xy=0\). We have a NO answer to the question. Sufficient.

(2) \(\sqrt{y^2} \leq 0\). This is the same as \(|y| \leq 0\). So, we have the same as above: \(y=0\). Therefore, \(xy=0\). We have a NO answer to the question. Sufficient.


Answer: D


If I'd answered this question during the start of my prep I'd have got it wrong.

I think the key here is the square root operator \(\sqrt{any number}\) will only refer to the positive root (and not both positive & negative roots).

Had the second statement been \(y^2 \leq 1\), this would give both positive and negative roots (y is between -1 and 1).

Kudos [?]: 9 [0], given: 38

Re: M32-12   [#permalink] 15 Aug 2017, 02:49
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