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# M32-12

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Math Expert
Joined: 02 Sep 2009
Posts: 46329

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17 Jul 2017, 06:28
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Difficulty:

55% (hard)

Question Stats:

42% (00:57) correct 58% (00:51) wrong based on 24 sessions

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

(1) $$|x| \leq 0$$.

(2) $$\sqrt{y^2} \leq 0$$.

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Math Expert
Joined: 02 Sep 2009
Posts: 46329

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17 Jul 2017, 06:28
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1
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.

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Joined: 17 May 2017
Posts: 145
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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
Intern
Joined: 24 Feb 2017
Posts: 38
Schools: CBS '20 (S)
GMAT 1: 760 Q50 V42

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15 Aug 2017, 02:49
1
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.

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).
Intern
Joined: 13 Oct 2017
Posts: 40

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11 Feb 2018, 06:45
Hi Bunuel,

This question confused me a little.

I was able to simplify the question to "xy>0?"

My issue was in translating the statements...

1) [x]<= 0

Converted [x] to x^2<=0
Since x squared cannot be less than 0...x^2 is 0, thus x is 0 and xy=0...meaning the answer to the question is no.

My question for this part is thus...whenever you see [x] or [y] etc... does that always translated into x^2, y^2 etc.?

2. For statement 2 why isn't √y^2 y? I thought this would be y<=0 making 2 insufficient.
Math Expert
Joined: 02 Sep 2009
Posts: 46329

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11 Feb 2018, 08:00
1
ttaiwo wrote:
Hi Bunuel,

This question confused me a little.

I was able to simplify the question to "xy>0?"

My issue was in translating the statements...

1) [x]<= 0

Converted [x] to x^2<=0
Since x squared cannot be less than 0...x^2 is 0, thus x is 0 and xy=0...meaning the answer to the question is no.

My question for this part is thus...whenever you see [x] or [y] etc... does that always translated into x^2, y^2 etc.?

2. For statement 2 why isn't √y^2 y? I thought this would be y<=0 making 2 insufficient.

1. |x| is the absolute value of x. NOT [x]. || is modulus sign.

2. $$\sqrt{x^2}=|x|$$ yes, but $$x^2=|x|$$ IS WRONG.

3. For (2): since $$\sqrt{y^2}=|y|$$, then $$\sqrt{y^2} \leq 0$$ gives: $$|y| \leq 0$$, which is the same info as in (1) but about y.

10. Absolute Value

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Intern
Joined: 13 Oct 2017
Posts: 40

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11 Feb 2018, 09:42
Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

This question confused me a little.

I was able to simplify the question to "xy>0?"

My issue was in translating the statements...

1) [x]<= 0

Converted [x] to x^2<=0
Since x squared cannot be less than 0...x^2 is 0, thus x is 0 and xy=0...meaning the answer to the question is no.

My question for this part is thus...whenever you see [x] or [y] etc... does that always translated into x^2, y^2 etc.?

2. For statement 2 why isn't √y^2 y? I thought this would be y<=0 making 2 insufficient.

1. |x| is the absolute value of x. NOT [x]. || is modulus sign.

2. $$\sqrt{x^2}=|x|$$ yes, but $$x^2=|x|$$ IS WRONG.

3. For (2): since $$\sqrt{y^2}=|y|$$, then $$\sqrt{y^2} \leq 0$$ gives: $$|y| \leq 0$$, which is the same info as in (1) but about y.

10. Absolute Value

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.

Thanks bro.
Re: M32-12   [#permalink] 11 Feb 2018, 09:42
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# M32-12

Moderators: chetan2u, Bunuel

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