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Which of the following best approximates the percent by [#permalink]
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10 Oct 2010, 03:58
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Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD? A. 30% B. 43% C. 45% D. 50% E. 70%
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Re: Square Diagonal versus Perimeter [#permalink]
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10 Oct 2010, 04:57
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Re: Square Diagonal versus Perimeter [#permalink]
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10 Oct 2010, 14:44
Great! Thank you!!!



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Re: Square Diagonal versus Perimeter [#permalink]
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27 Nov 2010, 03:19
Could someone please explain to me the wording of the problem? I thought that the phrase: Quote: ... by which the distance from A to C along a diagonal of square ABCD reduces ... means 1.4a/2a = 70% I am a little bit confused here. Thank you.



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Re: Square Diagonal versus Perimeter [#permalink]
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28 Nov 2010, 11:16
Please read the question carefully, the question says  "..the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD?"
So its not asking what percent the diagonal is of the distance around the edge but rather the percent of the difference between the two distances.
Hope this was helpful.



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Re: Square Diagonal versus Perimeter [#permalink]
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29 Nov 2010, 16:39
Using formula for 454590 triangle, diagonal = sqrt(2) of the each side.
Answer: A



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Re: Square Diagonal versus Perimeter [#permalink]
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30 Nov 2010, 13:07
Great discussion, everyone  I just want to point out that (fittingly), gettinit gets it! One of the easiest things for the GMAT to do to make a pretty hard problem very hard is to bait you toward answering the wrong question. I've seen them do this a lot with Geometry problems that involve percents  there's a significant but subtle difference between: Percent OF and Percent GREATER THAN or LESS THAN When you see a percentage problem, make sure you pause to answer the right question because pretty much any percent problem could be asked in either way. For another example that also includes squares and diagonals, you may want to check out: http://www.veritasprep.com/blog/2010/11/gmatchallengequestionthesquaredcircle/
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Re: Square Diagonal versus Perimeter [#permalink]
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30 Nov 2010, 16:46
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You could use pythagorean theorem to solve this. x^2+x^2=y^2 All sides of a square are equal, hence the two x^2. Plug in any number and solve. vivaslluis wrote: Hello,
I've seen the following example that I have doubts to solve:
Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD? a. 30% b. 43% c. 45% d. 50% e. 70%
Thank you



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Re: Square Diagonal versus Perimeter [#permalink]
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30 Nov 2010, 21:08
Hey Trojan, Great call on that  even if you have the xxx*sqrt 2 ratio memorized, I think it's important to know where it comes from. In the a^2 + b^2 = c^2 Pythagorean Theorem, if we know that a = b then it's really 2a^2 = c^2. And deriving that for yourself once or twice means there's very little chance you ever forget it (and you know you can always go back and prove it if you do forget). Thanks for bringing that up  I'm a huge fan of knowledge over memorization!
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Re: Square Diagonal versus Perimeter [#permalink]
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01 Dec 2010, 19:45
Brian great challenge question post it fits this question perfectly!VeritasPrepBrian wrote: Great discussion, everyone  I just want to point out that (fittingly), gettinit gets it! One of the easiest things for the GMAT to do to make a pretty hard problem very hard is to bait you toward answering the wrong question. I've seen them do this a lot with Geometry problems that involve percents  there's a significant but subtle difference between: Percent OF and Percent GREATER THAN or LESS THAN When you see a percentage problem, make sure you pause to answer the right question because pretty much any percent problem could be asked in either way. For another example that also includes squares and diagonals, you may want to check out: http://www.veritasprep.com/blog/2010/11/gmatchallengequestionthesquaredcircle/



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Re: Square Diagonal versus Perimeter [#permalink]
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11 Oct 2013, 06:50
Bunuel wrote: vivaslluis wrote: Hello,
I've seen the following example that I have doubts to solve:
Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD? a. 30% b. 43% c. 45% d. 50% e. 70%
Thank you Le the side of a square be \(a\). Route from A to C along a diagonal AC is \(\sqrt{2}a\approx{1.4a}\); Route from A to C around the edge ABC is \(2a\); Difference is \(2a1.4a=0.6a\) > \(\frac{0.6a}{2a}=0.3=30%\). Answer: A. Hi , i am confused about the denominator in the equation. if the equation is (2a1.4a) then the denominator should be 1.4a ??? how it is 2a??? not geeting...



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Re: Square Diagonal versus Perimeter [#permalink]
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11 Oct 2013, 06:53
sunny3011 wrote: Bunuel wrote: vivaslluis wrote: Hello,
I've seen the following example that I have doubts to solve:
Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD? a. 30% b. 43% c. 45% d. 50% e. 70%
Thank you Le the side of a square be \(a\). Route from A to C along a diagonal AC is \(\sqrt{2}a\approx{1.4a}\); Route from A to C around the edge ABC is \(2a\); Difference is \(2a1.4a=0.6a\) > \(\frac{0.6a}{2a}=0.3=30%\). Answer: A. Hi , i am confused about the denominator in the equation. if the equation is (2a1.4a) then the denominator should be 1.4a ??? how it is 2a??? not geeting... We are comparing to the route from A to C around the edge, which is 2a, so 2a must be in the denominator.
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Re: Which of the following best approximates the percent by [#permalink]
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31 May 2016, 21:04
vivaslluis wrote: Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD?
A. 30% B. 43% C. 45% D. 50% E. 70% let the sides be 2 units. original distance=2+2=4units changed distance=2sq.root2 %change=change dist.original dist./original dist. =(2sq.root24)/4==.2955 reduced by ~30% Ans A




Re: Which of the following best approximates the percent by
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