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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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3
6 00:00

Difficulty:   25% (medium)

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

### HideShow timer Statistics 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, 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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2
Ans is 22%

AB = (sqrt3) * lenght of cube
CD = (sqrt2) * lenght of cube
hence AB/CD -1 = 0.22 ~ 22%
_________________
Bhushan S.
If you like my post....Consider it for Kudos 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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1
Can you please explain in detail?
Retired Moderator B
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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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%
Manager  Joined: 14 Jun 2009
Posts: 74
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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agree answer is 30%
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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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 %?
Retired Moderator B
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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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 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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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 Manager  Joined: 16 Mar 2011
Posts: 149
Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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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
Manager  Joined: 10 Mar 2013
Posts: 189
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Re: If the box pictured to the right is a cube, then the differe  [#permalink]

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

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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Posts: 10952
Re: If the box shown is a cube, then the difference in length be  [#permalink]

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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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