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

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

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12 Feb 2012, 21:50
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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%
[Reveal] Spoiler: OA

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12 Feb 2012, 22:38
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enigma123 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%

Any idea how to solve this guys?

AC is the edge of the cube. Let's say its length is 'a'.
AB is just the diagonal of a face of the cube i.e. the diagonal of the square whose each side is of length 'a'. Using pythagorean theorem, we know that AB =$$\sqrt{2}a$$

Now think of the two dimensional triangle ABC (it is right angled at A)
AC = a and AB = $$\sqrt{2}a$$
Again using pythagorean theorem, $$BC^2 = a^2 + (\sqrt{2}a)^2$$
$$BC = \sqrt{3}a$$

So, $$(BC - AB)/AC * 100 = (\sqrt{3} - \sqrt{2}) * 100 = (1.732 - 1.414) * 100 = apprx 30%$$

By the way, you would probably be given the value of root 3.
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Re: If the box shown is a cube, then the difference in length [#permalink]

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13 Feb 2012, 04:52
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%
Attachment:

Cube.PNG [ 2.46 KiB | Viewed 7363 times ]

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

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

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23 Feb 2012, 07:09
looking at the figure CB is the longest diagonal of cube and we know longest diagonal of cube is
CB = X*rt3 ( assumed side of cube asX)
for AB we will use pythogoras theorem we will get Xrt2 ( X is the side of cube)
and AC is X
we have to find X*rt3 - Xrt2 divded by X will give us rt3 - rt2 that equal to close to 0.30 or 30%

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

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20 Jun 2013, 06:39
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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21 Jul 2014, 07:08
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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19 Jan 2015, 01:43
Since it is a cube, we know that the diagonal of the cube(CB) is s(sqrt(3)) and the diagonal of the square (AB) is s(sqrt(3))
with s being the side (which is the same in both cases because it is a cube.

What we are essentially looking for is (s(sqrt(3))-s(sqrt(2)))/s=x/100
we can divide out the s from the left side to get sqrt(3)-sqrt(2)=x/100
sqrt(3)=1.73 sqrt(2)=1.41 (I recommend memorizing the square roots of 1-10 as they come up on the test and are very difficult to figure out without a calculator)
thus .32=x/100 ------> x=32 and the closest option is C (30%)
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If the box shown is a cube, then the difference in length [#permalink]

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20 Jan 2015, 21:59
Let AC = 1, then

AB $$= \sqrt{1+1} = \sqrt{2}$$

BC $$= \sqrt{1+1+1} = \sqrt{3}$$

Fraction $$= \frac{\sqrt{3} - \sqrt{2}}{1} * 100 = (1.732 - 1.414) * 100 = 32%$$

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

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17 Jul 2016, 00:45
enigma123 wrote:
Attachment:
Cube.PNG
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%

enigma123 wrote:
Attachment:
Cube.PNG
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=side of the cube = x
AB= hypotenuse of the base (which is a square with side x)= $$x\sqrt{2}$$
BC=deluxe hypotenuse or diagonal of the cube =$$x\sqrt{3}$$
BC-AB= $$x\sqrt{3}$$ - $$x\sqrt{2}$$ = x(1.73-1.41)=0.32x
$$\frac{(BC-AB)}{AC}=\frac{0.30x}{x} ==>0.32==>32$$%

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

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06 Feb 2017, 14:11
If we simplify the question it is about to find value of x in below equation:
AC*x = (BC - AB)

suppose the side of cube is 1.

we know that value of diagonal of cube is √3 of any side
and diagonal of nay square part of cube is √2 of any side
then =>

1*x = 1(√3 - √2)
x ≈ 0.30
x ≈ 30% Ans

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Re: If the box shown is a cube, then the difference in length   [#permalink] 06 Feb 2017, 14:11
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