Help with one more GMAT problem solving question

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This is the other problem i have no idea how to go about solving, any help with explanation would be greatly appreciated.

The problem is taken from KAPLAN diagnostic exam, the first one

What is the range of straight-line distances between any two vertices of a cube with a volume of 64?

A) 2(square root 2 - 1)
B) 2(sqaure root 3 - 1)
C) 4(square root 3 - 1)
D) 4
E) 8(square root 3 - 1)

the square root is just for the one number only

the answer is E 8 (square root 3 - 1)

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relaxin99 wrote:This is the other problem i have no idea how to go about solving, any help with explanation would be greatly appreciated.

The problem is taken from KAPLAN diagnostic exam, the first one

What is the range of straight-line distances between any two vertices of a cube with a volume of 64?

A) 2(square root 2 - 1)
B) 2(sqaure root 3 - 1)
C) 4(square root 3 - 1)
D) 4
E) 8(square root 3 - 1)

the square root is just for the one number only

the answer is E 8 (square root 3 - 1)
I think the answer should be C unless i am missing something. 4 (square root 3 - 1). Are you sure about the OA?

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by mals24 » Wed Sep 17, 2008 2:44 am
agree with parallel_chase

even I'm getting C as the answer

The maximum distance will be the diagonal = 4 sq root 3
The minimum distance will be the length of a side = 4

Range = max-min = 4 sq root 3 - 4 = 4(sq root 3-1)

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by gdrea3 » Wed Sep 17, 2008 5:19 am
I got C as an answer too--Volume of a cube is s^3 since all sides are equal. Therefore, all the sides=4. The longest length in a quadrilateral is the DIAGONAL, and a diagonal in a cube splits a sides into 2 isoceles triangles.

Draw a diagram of a cube with the diagonal splitting the front into a triangle. Using the pythagoreom theorem you will see a pattern in the length of the sides and the diagonal--1:1:square root of 2. Since every side is equal to 4, multiply the length of the sides by 4 so the length of the diagonal is 4(square root of 2)

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by relaxin99 » Mon Sep 22, 2008 10:39 pm
Yes you guys are both right, maybe thats why i had such a hard time....THANK YOU skeeee :o

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by lunarpower » Tue Sep 23, 2008 2:17 am
yeah, this one is definitely (c).

you should MEMORIZE the following, and PUT IT ON FLASH CARD(S):

if SIDE of a cube is "s", then:
FACE DIAGONAL of a cube = s√2
MAIN DIAGONAL of the cube = s√3


the face diagonal shouldn't be news - it's just an application of the 45°-45°-90° triangle template - but the main diagonal can be very time-consuming to figure out if you don't memorize it.

flash cards are your friends.
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by kaf » Mon Mar 16, 2009 2:57 pm
Hi Lunarpower
if SIDE of a cube is "s", then:
FACE DIAGONAL of a cube = s√2
MAIN DIAGONAL of the cube = s√3
Can you please explain what this means

thanks

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by lunarpower » Tue Mar 24, 2009 4:37 pm
kaf wrote:Hi Lunarpower
if SIDE of a cube is "s", then:
FACE DIAGONAL of a cube = s√2
MAIN DIAGONAL of the cube = s√3
Can you please explain what this means

thanks
take a look at a cube. ideally, you should look at a real, physical model of a cube; if you don't have one available, google around for a VERY good drawing of one.

* the longest diagonal in a cube is called the main diagonal. this goes through the interior of the cube, between opposite vertices. it does NOT go along the sides of the cube.
if you imagine spinning the cube on one of its vertices, as you'd do with a top, then this diagonal will go from the bottom vertex (the one on which the cube is spinning) to the top vertex.
this diagonal has length equal to √3 times the side length of the cube.

* if you just look at ONE of the square faces of the cube, and ignore everything else, then the diagonal of that square face is called a face diagonal.
the length of this diagonal, as can be deduced from a 45-45-90 triangle template, is √2 times the side length of the cube.
Ron has been teaching various standardized tests for 20 years.

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The Cube

by jaimito01 » Tue Mar 24, 2009 8:43 pm
Here as mentioned before:

AC = 4
BZ = 4sqrt(2)
AZ = 4sqrt(3)
Attachments
cUBE.jpg

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by logitech » Tue Mar 24, 2009 9:09 pm
We are very luck to have lunarpower on this forum.
LGTCH
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