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An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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Updated on: 30 Apr 2015, 03:44
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An openempty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel? A. \(6\sqrt{2}\) B. \(\sqrt{10}\) C. \(2\sqrt{19}\) D. \(10\) E. \(14\)
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Originally posted by itzmyzone911 on 30 Apr 2015, 01:53.
Last edited by Bunuel on 30 Apr 2015, 03:44, edited 3 times in total.
Edited the question.



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 03:09
If I understood the question correctly I really wonder if thats actually a gmat question. Not only my answer is different from what is proposed (my answer is 9 feet or approximately 9.54 feet) but also the way to solve it required me to use things which are most likely beyond the bounadires of gmat requirements. Basically the way I approached this task was evaluating 2 scenarios: and The first one gave me the lowest answer after conducting analysis on the distance function (sum of lengths of AC and CB)



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 03:52



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 03:53
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Zhenek wrote: If I understood the question correctly I really wonder if thats actually a gmat question. Not only my answer is different from what is proposed (my answer is 9 feet or approximately 9.54 feet) but also the way to solve it required me to use things which are most likely beyond the bounadires of gmat requirements. Basically the way I approached this task was evaluating 2 scenarios: and The first one gave me the lowest answer after conducting analysis on the distance function (sum of lengths of AC and CB) How did you arrive at your ans?? Explain and then we shall see whether this is within or beyond GMAT scope



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 03:58
Bunuel wrote: itzmyzone911 wrote: Attachment: box.PNG An openempty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel? A. \(6\sqrt{2}\) B. \(\sqrt{10}\) C. \(2\sqrt{19}\) D. \(10\) E. \(14\) Similar questions to practice: anantisclingingtoonecornerofaboxintheshapeofa135055.htmls97184708.htmlanantcrawlsfromonecornerofaroomtothediagonally134454.htmlHope it helps. Bunuel you rock ...you seem to have virtually every GMAT problem on earth on your finger tips..



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An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 04:23
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Haha, I'm glad I actually looked at it again, anyway, I made a mistake in my calculations and it is indeed 10 in the end, damn. I'll just explain my way of thinking then: My 2nd picture, the point C is on the line and X is a variable. Length function is \(\sqrt{36+x^2} + \sqrt{36+(2x)^2}\). With the help of derivatives I found that its lowest value is \(2*\sqrt{37}\) when X = 1 (although one could also apply some common sense there since if u try drawing the thing abit differently  as a doublerectangular figure omitting the folding, the shortest distance ends up being a straight line and it happens that since the rectangles have equal lengths, the line will cross the folding in its middle thus X = 1). \(2*\sqrt{37}\) > 12. First picture is similar. Length function is \(F(x) = \sqrt{x^2+4} + \sqrt{36+(6x)^2}\). In order to find minimum value I went with the rule that this \(X_0\) is pretty much a root of an equation \(F'(X_0) = 0\) \(\frac{2*x}{\sqrt{x^2+4}}  \frac{2*(6x)}{\sqrt{36+(6x)^2}} = 0\) \(x* \sqrt{36+(6x)^2} = (6x)* \sqrt{4+x^2}\) solving this you get x = 1,5 as the only fitting answer, y ends up being 10 which is the answer to the question



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 09:14
Zhenek wrote: Haha, I'm glad I actually looked at it again, anyway, I made a mistake in my calculations and it is indeed 10 in the end, damn. I'll just explain my way of thinking then: My 2nd picture, the point C is on the line and X is a variable. Length function is \(\sqrt{36+x^2} + \sqrt{36+(2x)^2}\). With the help of derivatives I found that its lowest value is \(2*\sqrt{37}\) when X = 1 (although one could also apply some common sense there since if u try drawing the thing abit differently  as a doublerectangular figure omitting the folding, the shortest distance ends up being a straight line and it happens that since the rectangles have equal lengths, the line will cross the folding in its middle thus X = 1). \(2*\sqrt{37}\) > 12. First picture is similar. Length function is \(F(x) = \sqrt{x^2+4} + \sqrt{36+(6x)^2}\). In order to find minimum value I went with the rule that this \(X_0\) is pretty much a root of an equation \(F'(X_0) = 0\) \(\frac{2*x}{\sqrt{x^2+4}}  \frac{2*(6x)}{\sqrt{36+(6x)^2}} = 0\) \(x* \sqrt{36+(6x)^2} = (6x)* \sqrt{4+x^2}\) solving this you get x = 1,5 as the only fitting answer, y ends up being 10 which is the answer to the question Well dear, I appreciate your insights...very thorough indeed. However, the question can be solved using basics of Pythagoras theorem...do visit the link suggested by Bunuel at anantcrawlsfromonecornerofaroomtothediagonally134454.html Relevant discussions have taken place along with diagrams to make the point better. Do revert if any doubts..



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An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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30 Apr 2015, 09:28
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Yea I did understand the approach displayed there and I find it very clever. I guess I will elaborate with the changed picture: AB' is the shortest path the ant has to take and its the straight line, thus we got a triangle ADB' where AB' = hypotenuse and is easily found to be equal to \(\sqrt{36+64} = 10\)



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An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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14 Sep 2015, 08:15
itzmyzone911 wrote: Attachment: box.PNG An openempty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel? A. \(6\sqrt{2}\) B. \(\sqrt{10}\) C. \(2\sqrt{19}\) D. \(10\) E. \(14\) Hi I think there is a typo in the question. I believe either width or length needs to be 2, if the answer is indeed the correct answer. Using the data given the question: Basically if we unfold the box, this will give us two rectangles, 1 rectangle with height 2 and base 6, 2nd rectangle with height 2 and base 6. The two rectangles have point A and B as diametrically opposite points. Now to find AB, it's sqrt( (6+6)^2 + 2^2) and that's not the correct answer. Bunuel  I have followed the other posts that you have mentioned, but not sure what I am missing here in this question. Any inputs will be very helpful. Thanks, Chanakya



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Re: An openempty rectangular box with negligible wall thickness is 6 feet [#permalink]
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29 Dec 2017, 16:20
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Re: An openempty rectangular box with negligible wall thickness is 6 feet
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