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If the box shown is a cube, then the difference in length be

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If the box shown is a cube, then the difference in length be  [#permalink]

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New post 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: if-the-box-shown-is-a-cube-then-the-difference-in-length-127463.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  [#permalink]

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New post 09 Aug 2009, 10:47
2
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  [#permalink]

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New post 09 Aug 2009, 11:14
1
Can you please explain in detail?
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Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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New post 11 Aug 2009, 00:39
2
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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(1.7-1.4)
= y(.3)
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Re: If the box shown is a cube, then the difference in length be  [#permalink]

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New post 31 Oct 2013, 00:03
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lbsgmat wrote:
Image
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%


AC is the edge (the side) of a cube, suppose it equals to 1;
AB is the diagonal of a face, hence is equals to \(\sqrt{2}\), (either from 45-45-90 triangle properties or form Pythagorean theorem);
BC is the diagonal of the cube itself and is equal to \(\sqrt{1^2+1^2+1^2}=\sqrt{3}\);

Ratio: \(\frac{BC-AB}{AC}=\frac{\sqrt{3}-\sqrt{2}}{1}\approx{1.7-1.4}={0.3}\).

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-box-shown-is-a-cube-then-the-difference-in-length-127463.html
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Re: If the box shown is a cube, then the difference in length be  [#permalink]

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