It is currently 20 Sep 2017, 04:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 13 May 2009
Posts: 18

Kudos [?]: 45 [2], given: 1

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

### Show Tags

09 Aug 2009, 11:36
2
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

70% (01:01) correct 30% (01:13) wrong based on 356 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
[Reveal] Spoiler: OA

Attachments

Diagram.doc [23.5 KiB]

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.

Kudos [?]: 45 [2], given: 1

Manager
Joined: 18 Jul 2009
Posts: 168

Kudos [?]: 113 [2], given: 37

Location: India
Schools: South Asian B-schools
Re: If the box pictured to the right is a cube, then the differe [#permalink]

### Show Tags

09 Aug 2009, 11:47
2
KUDOS
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

Kudos [?]: 113 [2], given: 37

Intern
Joined: 13 May 2009
Posts: 18

Kudos [?]: 45 [1], given: 1

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

### Show Tags

09 Aug 2009, 12:14
1
KUDOS
Can you please explain in detail?

Kudos [?]: 45 [1], given: 1

SVP
Joined: 05 Jul 2006
Posts: 1741

Kudos [?]: 418 [2], given: 49

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

### Show Tags

11 Aug 2009, 01:39
2
KUDOS
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%

Kudos [?]: 418 [2], given: 49

Manager
Joined: 14 Jun 2009
Posts: 96

Kudos [?]: 32 [0], given: 9

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

### Show Tags

11 Aug 2009, 01:56
agree answer is 30%

Kudos [?]: 32 [0], given: 9

Senior Manager
Joined: 29 Jan 2011
Posts: 349

Kudos [?]: 240 [0], given: 87

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

### Show Tags

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 %?

Kudos [?]: 240 [0], given: 87

SVP
Joined: 05 Jul 2006
Posts: 1741

Kudos [?]: 418 [0], given: 49

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

### Show Tags

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 %?

From the available answer choices

Kudos [?]: 418 [0], given: 49

Senior Manager
Joined: 29 Jan 2011
Posts: 349

Kudos [?]: 240 [0], given: 87

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

### Show Tags

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 %?

From the available answer choices

Yes, sure you can if you make a guess..but i wanted to understand so i asked

Kudos [?]: 240 [0], given: 87

Manager
Joined: 16 Mar 2011
Posts: 170

Kudos [?]: 46 [0], given: 13

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

### Show Tags

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

Kudos [?]: 46 [0], given: 13

Senior Manager
Joined: 10 Mar 2013
Posts: 273

Kudos [?]: 114 [0], given: 2405

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]

### Show Tags

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)

Kudos [?]: 114 [0], given: 2405

Math Expert
Joined: 02 Sep 2009
Posts: 41631

Kudos [?]: 124106 [4], given: 12071

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

### Show Tags

31 Oct 2013, 01:03
4
KUDOS
Expert's post
4
This post was
BOOKMARKED
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
_________________

Kudos [?]: 124106 [4], given: 12071

Re: If the box shown is a cube, then the difference in length be   [#permalink] 31 Oct 2013, 01:03
Similar topics Replies Last post
Similar
Topics:
4 If a cube of side length 24 units is cut into 512 smaller cubes of equ 7 28 May 2017, 21:00
7 There are 5 boxes of different weights... 4 08 Jun 2016, 09:48
5 If the length of an edge of cube X is twice the length of an 7 04 Jul 2017, 06:23
2 If the length of an edge of cube X is twice the length of an 3 29 Jul 2014, 09:21
12 If the box shown is a cube, then the difference in length 9 06 Feb 2017, 14:11
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.