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# M13-33

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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15 Sep 2014, 23:50
2
3
00:00

Difficulty:

65% (hard)

Question Stats:

48% (01:35) correct 52% (00:34) wrong based on 94 sessions

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Is the length of the diagonal of the rectangle bigger than $$\sqrt{6}$$?

(1) The shorter side of the rectangle is 2.

(2) The longer side of the rectangle is 3.

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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15 Sep 2014, 23:50
Official Solution:

Statement (1) by itself is sufficient. The longer side of the rectangle is bigger than 2, thus the diagonal is bigger than $$\sqrt{2^2 + 2^2} = \sqrt{8} \gt \sqrt{6}$$.

Statement (2) by itself is sufficient. The diagonal is bigger than $$3 = \sqrt{9} \gt \sqrt{6}$$.

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Intern
Joined: 01 Nov 2016
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01 Nov 2016, 10:24
Each statement is in itself sufficient :

considering statement 1. shortest side is 2,hence longer side is at least greater than 2 ie diagonal greater than at least square root of 8 hence sufficient.

considering statement 2. longest side is 3,hence shorter can be anything but the diagonal is greater than at least square root of 9 ,hence sufficient.
Intern
Joined: 09 Jul 2016
Posts: 17
GMAT 1: 730 Q50 V39

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29 May 2017, 09:39
I don't think this question is any higher than 600 level
Intern
Joined: 17 Nov 2017
Posts: 1

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16 Apr 2018, 02:53
I think this is a high-quality question and I don't agree with the explanation. Statement 1 : The shorter side is 2, hence the diagonal is longer than √(4+4) = √8. Since, √8 > √6, Sufficient.

Statement 2 : The longer side is 3, hence the diagonal is shorter than √(9+9) = 3√2. Since, 3√2 > √6, Insufficient.
Math Expert
Joined: 02 Sep 2009
Posts: 52294

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16 Apr 2018, 03:57
Shalvi194 wrote:
I think this is a high-quality question and I don't agree with the explanation. Statement 1 : The shorter side is 2, hence the diagonal is longer than √(4+4) = √8. Since, √8 > √6, Sufficient.

Statement 2 : The longer side is 3, hence the diagonal is shorter than √(9+9) = 3√2. Since, 3√2 > √6, Insufficient.

No, that's not right.

For (2) if the longer side is 3, then the diagonal must be greater than $$\sqrt{3^2+ 0^2}=\sqrt{9}$$.
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Re: M13-33 &nbs [#permalink] 16 Apr 2018, 03:57
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# M13-33

Moderators: chetan2u, Bunuel

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