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If the box shown is a cube, then the difference in length be
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Updated on: 31 Oct 2013, 00:02
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If the box shown is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C? A. 10% B. 20% C. 30% D. 40% E. 50% OPEN DISCUSSION OF THIS QUESTION IS HERE: iftheboxshownisacubethenthedifferenceinlength127463.html
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Originally posted by lbsgmat on 09 Aug 2009, 10:36.
Last edited by Bunuel on 31 Oct 2013, 00:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If the box pictured to the right is a cube, then the differe
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09 Aug 2009, 10:47
Ans is 22%AB = (sqrt3) * lenght of cube CD = (sqrt2) * lenght of cube hence AB/CD 1 = 0.22 ~ 22%
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Re: If the box pictured to the right is a cube, then the differe
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09 Aug 2009, 11:14
Can you please explain in detail?



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Re: If the box pictured to the right is a cube, then the differe
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11 Aug 2009, 00:39
lbsgmat wrote: If the box pictured to the right is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C? 10% 20% 30% 40% 50% main diagonal of a cube = bc = sqrt(3x^2) = xsqrt3 diagonal of a square = xsqrt2 difference = (sqrt 3 sqrt 2)x = (1.7  1.4)x approx = 0.3x. distance A to C = x thus answer is 30%



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Re: If the box pictured to the right is a cube, then the differe
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11 Aug 2009, 00:56
agree answer is 30%



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Re: If the box pictured to the right is a cube, then the differe
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19 Jun 2011, 15:28
How do you get 30% i am not sure from the fraction?
AB^2 = x^2 + x^2
Therefore, AB = root(2) x 1
BC^2 = [root(2) x] ^2 + x^2
Therefore, BC = root(3) x 2
2  1
[root(3)  root(2)]x 3
We know , AC = x 4
3/4 =>
0.3
Now how is this 30%???? We are asked for fraction so how did this get converted to %?



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Re: If the box pictured to the right is a cube, then the differe
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20 Jun 2011, 22:39
siddhans wrote: How do you get 30% i am not sure from the fraction?
AB^2 = x^2 + x^2
Therefore, AB = root(2) x 1
BC^2 = [root(2) x] ^2 + x^2
Therefore, BC = root(3) x 2
2  1
[root(3)  root(2)]x 3
We know , AC = x 4
3/4 =>
0.3
Now how is this 30%???? We are asked for fraction so how did this get converted to %? From the available answer choices



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Re: If the box pictured to the right is a cube, then the differe
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20 Jun 2011, 22:44
yezz wrote: siddhans wrote: How do you get 30% i am not sure from the fraction?
AB^2 = x^2 + x^2
Therefore, AB = root(2) x 1
BC^2 = [root(2) x] ^2 + x^2
Therefore, BC = root(3) x 2
2  1
[root(3)  root(2)]x 3
We know , AC = x 4
3/4 =>
0.3
Now how is this 30%???? We are asked for fraction so how did this get converted to %? From the available answer choices Yes, sure you can if you make a guess..but i wanted to understand so i asked



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Re: If the box pictured to the right is a cube, then the differe
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20 Jun 2011, 23:00
First pythagorem to find CD = x^2 + x^2 = xsqrt2
Using CD as the leg of another triangle, we want to find BC: xsqrt2^2 + x^2 = sqrt(3x^2) = xsqrt3
Now they ask for the difference in length BC  AB and they want you to calculate its proportion to AC. AB = CD, so we can use what we just calculated:
xsqrt3  xsqrt2 (BC  AB) / x (AC) = sqrt3  sqrt 2 = approx. 0,3



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Re: If the box pictured to the right is a cube, then the differe
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30 Oct 2013, 18:02
I also agree that it is 30%
y = AC d = BC k = AB
k = \sqrt{2*y^2}
d = \sqrt{k^2 + y^2}
BC = y*\sqrt{3} AB = y*\sqrt{2} BC  AB = y(\sqrt{3}\sqrt{2}) = y(1.71.4) = y(.3)



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Re: If the box shown is a cube, then the difference in length be
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31 Oct 2013, 00:03



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Re: If the box shown is a cube, then the difference in length be
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09 Jan 2019, 01:30
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