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An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 05:48
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An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover? A. \(3\sqrt{2}+3\) B. \(6\sqrt{2}\) C. \(3*3^{\frac 13}\) D. \(3\sqrt{5}\) E. \(9\)
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 08:25
Quote: Hehe is that a good thing or a bad thing? Anyway, this is the first time I encountered such type of question. I can only remember the greatest distance (deluxe Pythagorean theorem). If my memory serves me right we could also use Calculus here. Does the topic multivariable Calculus ring any bell? Hi gmatsaga, The concept of this question is to open up the surfaces and consider two adjacent surfaces as a plane. (Check the diagram below) Thus, using the classical Pythagoras concept, the hypotenuse (or the shortest distance between two points) would be calculated as: \(\sqrt{(3+3)^2+3^2} = 3\sqrt{5}\) well talking about calculus, I would only say out of scope! Regards,
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 06:19
cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) This is a Physics question LOL Anyway, for any dimension of a room that has dimensions a, b and c, the length of the shortest path is: minimum among (1) square root[(a+b)^2 + c^2] (2) square root[(b+c)^2 + a^2] (3) square root](a+c)^2 + b^2\ Since we have dimensions 3, 3 and 3 we can try any of the three: square root [(3+3)^2 + 3^2]= square root [6^2 + 9] = square root [36 +9] = square root [45] = square root [9*5] = 3 square root 5 Now how about some kudos, yes?
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 06:51
gmatsaga wrote: cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) This is a Physics question LOL Anyway, for any dimension of a room that has dimensions a, b and c, the length of the shortest path is: minimum among (1) square root[(a+b)^2 + c^2] (2) square root[(b+c)^2 + a^2] (3) square root](a+c)^2 + b^2\ Since we have dimensions 3, 3 and 3 we can try any of the three: square root [(3+3)^2 + 3^2]= square root [6^2 + 9] = square root [36 +9] = square root [45] = square root [9*5] = 3 square root 5 Now how about some kudos, yes? You are good at googling..! This indeed is a physics problem, but many variants of this problem are asked in various competitive exams. If one can do this, then similar concept can be extended to squares, rectangles or even cylinders. Anyways, it is good problem Regards,



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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 07:08
cyberjadugar wrote: gmatsaga wrote: cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) This is a Physics question LOL Anyway, for any dimension of a room that has dimensions a, b and c, the length of the shortest path is: minimum among (1) square root[(a+b)^2 + c^2] (2) square root[(b+c)^2 + a^2] (3) square root](a+c)^2 + b^2\ Since we have dimensions 3, 3 and 3 we can try any of the three: square root [(3+3)^2 + 3^2]= square root [6^2 + 9] = square root [36 +9] = square root [45] = square root [9*5] = 3 square root 5 Now how about some kudos, yes? You are good at googling..! This indeed is a physics problem, but many variants of this problem are asked in various competitive exams. If one can do this, then similar concept can be extended to squares, rectangles or even cylinders. Anyways, it is good problem Regards, Hehe is that a good thing or a bad thing? Anyway, this is the first time I encountered such type of question. I can only remember the greatest distance (deluxe Pythagorean theorem). If my memory serves me right we could also use Calculus here. Does the topic multivariable Calculus ring any bell?
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Far better is it to dare mighty things, to win glorious triumphs, even though checkered by failure... than to rank with those poor spirits who neither enjoy nor suffer much, because they live in a gray twilight that knows not victory nor defeat.  T. Roosevelt



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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 17:52
cyberjadugar wrote: Quote: Hehe is that a good thing or a bad thing? Anyway, this is the first time I encountered such type of question. I can only remember the greatest distance (deluxe Pythagorean theorem). If my memory serves me right we could also use Calculus here. Does the topic multivariable Calculus ring any bell? Hi gmatsaga, The concept of this question is to open up the surfaces and consider two adjacent surfaces as a plane. (Check the diagram below) Thus, using the classical Pythagoras concept, the hypotenuse (or the shortest distance between two points) would be calculated as: \(\sqrt{(3+3)^2+3^2} = 3\sqrt{5}\) well talking about calculus, I would only say out of scope! Regards, Did I tell you you're good? AMAZING!!!!!
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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13 Jun 2012, 19:15
Thanks for the magical explanation cj ...you deserve some kudos for this!



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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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14 Jun 2012, 00:07
gmatdog wrote: Thanks for the magical explanation cj ...you deserve some kudos for this! gmatsaga wrote: Did I tell you you're good? AMAZING!!!!! Thanks I appreciate the appreciation.



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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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28 Nov 2017, 17:19



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An ant crawls from one corner of a room to the diagonally [#permalink]
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28 Nov 2017, 18:08
cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) Hi Bunuel, chetan2uIf the ant has to reach opposite corner of a room diagonally, shouldn't the answer be \(3\sqrt{3}\). can you please let me know where I am wrong in the below approach. First the ant will crawl along the diagonal of one plane i.e. \(3\sqrt{2}\) and the then along the edge to the opposite corner = 3 So the total diagonal length will be \(d^2 = 3^2 + (3\sqrt{2})^2\) = \(27\)
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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28 Nov 2017, 20:09
rahul16singh28 wrote: cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) Hi Bunuel, chetan2uIf the ant has to reach opposite corner of a room diagonally, shouldn't the answer be \(3\sqrt{3}\). can you please let me know where I am wrong in the below approach. First the ant will crawl along the diagonal of one plane i.e. \(3\sqrt{2}\) and the then along the edge to the opposite corner = 3 So the total diagonal length will be \(d^2 = 3^2 + (3\sqrt{2})^2\) = \(27\) Hi.. I'll attach a figure in sometime, but just try to understand.. Take any two adjacent sides which are perpendicular to each other and sides of each are 3*3...ABCD and BDEF are two sides Now open it ... A.....B......E C.....D......F The ant has to move from C to E.. By opening the two sides you have a rectangle ACFE.. The diagonal CE will be the shortest route.. CE=√(3^2+6^2)=√45=3√5 Hope you could visualise
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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28 Nov 2017, 20:44
chetan2u wrote: rahul16singh28 wrote: cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) Hi Bunuel, chetan2uIf the ant has to reach opposite corner of a room diagonally, shouldn't the answer be \(3\sqrt{3}\). can you please let me know where I am wrong in the below approach. First the ant will crawl along the diagonal of one plane i.e. \(3\sqrt{2}\) and the then along the edge to the opposite corner = 3 So the total diagonal length will be \(d^2 = 3^2 + (3\sqrt{2})^2\) = \(27\) Hi.. I'll attach a figure in sometime, but just try to understand.. Take any two adjacent sides which are perpendicular to each other and sides of each are 3*3...ABCD and BDEF are two sides Now open it ... A.....B......E C.....D......F The ant has to move from C to E.. By opening the two sides you have a rectangle ACFE.. The diagonal CE will be the shortest route.. CE=√(3^2+6^2)=√45=3√5 Hope you could visualise Hi chetan2u.. Thanks for the explanation.. Yes I could visualize it. However, the question never mentions that ant can walk only along the plane. It can take the edge path also. For Ex  it will move from one of Diagonal of ABCD and then the adjoining edge  In this case 3root(2) + 3. Please correct if I am missing something.
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Re: An ant crawls from one corner of a room to the diagonally [#permalink]
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28 Nov 2017, 20:49
rahul16singh28 wrote: chetan2u wrote: cyberjadugar wrote: An ant crawls from one corner of a room to the diagonally opposite corner along the shortest possible path. If the dimensions of the room are 3 x 3 x 3, what distance does the ant cover?
A. \(3\sqrt{2}+3\)
B. \(6\sqrt{2}\)
C. \(3*3^{\frac 13}\)
D. \(3\sqrt{5}\)
E. \(9\) Hi Bunuel, chetan2uIf the ant has to reach opposite corner of a room diagonally, shouldn't the answer be \(3\sqrt{3}\). can you please let me know where I am wrong in the below approach. First the ant will crawl along the diagonal of one plane i.e. \(3\sqrt{2}\) and the then along the edge to the opposite corner = 3 So the total diagonal length will be \(d^2 = 3^2 + (3\sqrt{2})^2\) = \(27\) Hi.. I'll attach a figure in sometime, but just try to understand.. Take any two adjacent sides which are perpendicular to each other and sides of each are 3*3...ABCD and BDEF are two sides Now open it ... A.....B......E C.....D......F The ant has to move from C to E.. By opening the two sides you have a rectangle ACFE.. The diagonal CE will be the shortest route.. CE=√(3^2+6^2)=√45=3√5 Hope you could visualise Hi chetan2u.. Thanks for the explanation.. Yes I could visualize it. However, the question never mentions that ant can walk only along the plane. It can take the edge path also. For Ex  it will move from one of Diagonal of ABCD and then the adjoining edge  In this case 3root(2) + 3. Please correct if I am missing something. What you are missing is word SHORTEST.. Check the two answers, diagonal will always be the Shortest
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Re: An ant crawls from one corner of a room to the diagonally
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