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

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Intern
Joined: 13 May 2009
Posts: 19
If the box shown is a cube, then the difference in length be  [#permalink]

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Updated on: 31 Oct 2013, 01:02
3
6
00:00

Difficulty:

25% (medium)

Question Stats:

75% (00:56) correct 25% (01:19) wrong based on 418 sessions

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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

Attachments

Diagram.doc [23.5 KiB]

Originally posted by lbsgmat on 09 Aug 2009, 11:36.
Last edited by Bunuel on 31 Oct 2013, 01:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Manager
Joined: 18 Jul 2009
Posts: 163
Location: India
Schools: South Asian B-schools
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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09 Aug 2009, 11:47
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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Bhushan S.
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Intern
Joined: 13 May 2009
Posts: 19
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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09 Aug 2009, 12:14
1
Can you please explain in detail?
Retired Moderator
Joined: 05 Jul 2006
Posts: 1701
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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11 Aug 2009, 01:39
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

Manager
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Posts: 74
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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11 Aug 2009, 01:56
Senior Manager
Joined: 29 Jan 2011
Posts: 278
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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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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Posts: 1701
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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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 %?

Senior Manager
Joined: 29 Jan 2011
Posts: 278
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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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
Manager
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Posts: 149
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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

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30 Oct 2013, 19:02
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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Joined: 02 Sep 2009
Posts: 55150
Re: If the box shown is a cube, then the difference in length be  [#permalink]

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31 Oct 2013, 01:03
4
6
lbsgmat wrote:

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

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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09 Jan 2019, 02:30
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Re: If the box shown is a cube, then the difference in length be   [#permalink] 09 Jan 2019, 02:30
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