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# If the box pictured to the right is a cube, then the differe

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If the box pictured to the right is a cube, then the differe [#permalink]  09 Aug 2009, 11:36
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44% (01:47) correct 55% (00:30) wrong based on 3 sessions
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%
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Re: If the box pictured to the right is a cube, then the differe [#permalink]  09 Aug 2009, 11:47
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Ans is 22%

AB = (sqrt3) * lenght of cube
CD = (sqrt2) * lenght of cube
hence AB/CD -1 = 0.22 ~ 22%
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Bhushan S.
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Re: If the box pictured to the right is a cube, then the differe [#permalink]  09 Aug 2009, 12:14
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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]  11 Aug 2009, 01:39
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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

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Re: If the box pictured to the right is a cube, then the differe [#permalink]  11 Aug 2009, 01:56
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Re: If the box pictured to the right is a cube, then the differe [#permalink]  19 Jun 2011, 16:28
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]  20 Jun 2011, 23:39
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 %?

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Re: If the box pictured to the right is a cube, then the differe [#permalink]  20 Jun 2011, 23:44
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 %?

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]  21 Jun 2011, 00:00
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
Re: If the box pictured to the right is a cube, then the differe   [#permalink] 21 Jun 2011, 00:00
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