Diagonals of a Cube

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edvhou812: Master | Next Rank: 500 Posts; Posts: 143; Joined: Sat Nov 06, 2010 9:06 pmFollowed by: 1 member; Thanked: 10 times

by edvhou812 » Wed May 11, 2011 7:19 pm

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?

A - 10%
B - 20%
C - 30%
D - 40%
E - 50%

Answer: C

This question has been bugging me. The method I used had the following:

AC=x <Side of Cube>
AB=x*sqrt(2) <40-40-90 triangle for the diagonal of a face>
BC=sqrt(3x) <formula for space diagonal of a cube>

However I have gotten the question wrong numerous times with this method because I get the wrong value for BC. Official explanation says I should use the Pythagorean Theorem to determine the value of BC, but I'm perplexed as to why my method did not work. Is anyone able to give some insight?

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Source: Beat The GMAT — Problem Solving | Wed May 11, 2011 7:19 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Wed May 11, 2011 8:47 pm

edvhou812 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?

A - 10%
B - 20%
C - 30%
D - 40%
E - 50%

Answer: C

This question has been bugging me. The method I used had the following:

AC=x <Side of Cube>
AB=x*sqrt(2) <40-40-90 triangle for the diagonal of a face>
BC=sqrt(3x) <formula for space diagonal of a cube>

However I have gotten the question wrong numerous times with this method because I get the wrong value for BC. Official explanation says I should use the Pythagorean Theorem to determine the value of BC, but I'm perplexed as to why my method did not work. Is anyone able to give some insight?

Solution:
Let AC be of length x.
The line segment BC is sqrt(x^2 + x^2 + x^2) = x*sqrt3.
Line segment AB is sqrt(x^2 + x^2) = x*sqrt2.
Difference in length of BC and AB is x*(sqrt3 - sqrt2) = x*(1.732 - 1.414) = x*(0.3)
So, required % is 0.3x/x * 100 = 30%.

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edvhou812: Master | Next Rank: 500 Posts; Posts: 143; Joined: Sat Nov 06, 2010 9:06 pmFollowed by: 1 member; Thanked: 10 times

by edvhou812 » Wed May 11, 2011 9:00 pm

Anurag@Gurome wrote:
edvhou812 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?

A - 10%
B - 20%
C - 30%
D - 40%
E - 50%

Answer: C

This question has been bugging me. The method I used had the following:

AC=x <Side of Cube>
AB=x*sqrt(2) <40-40-90 triangle for the diagonal of a face>
BC=sqrt(3x) <formula for space diagonal of a cube>

However I have gotten the question wrong numerous times with this method because I get the wrong value for BC. Official explanation says I should use the Pythagorean Theorem to determine the value of BC, but I'm perplexed as to why my method did not work. Is anyone able to give some insight?
Solution:
Let AC be of length x.
The line segment BC is sqrt(x^2 + x^2 + x^2) = x*sqrt3.
Line segment AB is sqrt(x^2 + x^2) = x*sqrt2.
Difference in length of BC and AB is x*(sqrt3 - sqrt2) = x*(1.732 - 1.414) = x*(0.3)
So, required % is 0.3x/x * 100 = 30%.

I see. I goofed the space diagonal formula. Thanks for the response.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu May 12, 2011 3:49 am

edvhou812 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?

A - 10%
B - 20%
C - 30%
D - 40%
E - 50%

Answer: C

This question has been bugging me. The method I used had the following:

AC=x <Side of Cube>
AB=x*sqrt(2) <40-40-90 triangle for the diagonal of a face>
BC=sqrt(3x) <formula for space diagonal of a cube>

However I have gotten the question wrong numerous times with this method because I get the wrong value for BC. Official explanation says I should use the Pythagorean Theorem to determine the value of BC, but I'm perplexed as to why my method did not work. Is anyone able to give some insight?

Let s = side of each square face.

It is helpful to have memorized the following:
Diagonal of the cube = âˆš(3s^2).
Diagonal of each square face = sâˆš2.

Let AC = 1.
BC = âˆš(3* 1^2) = âˆš3.
AB = 1*âˆš2 = âˆš2.
BC - AB = âˆš3-âˆš2 â‰ˆ 1.7 - 1.4 â‰ˆ .3.
Difference/AC = .3/1 = 30/100 = 30%.

The correct answer is C.

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djiddish98: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Fri Jan 14, 2011 10:01 am; Thanked: 10 times

by djiddish98 » Thu May 12, 2011 7:48 am

I've seen the BeatTheGMAT flash cards indicate that we should know the actual value of sqrt(3) and sqrt(2), but I'd never seen a question that actually required that knowledge.

So should I know that sqrt(3) = 1.7 and sqrt(2) = 1.4 - or is this an odd question?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu May 12, 2011 8:12 am

djiddish98 wrote:I've seen the BeatTheGMAT flash cards indicate that we should know the actual value of sqrt(3) and sqrt(2), but I'd never seen a question that actually required that knowledge.

So should I know that sqrt(3) = 1.7 and sqrt(2) = 1.4 - or is this an odd question?

Knowing that âˆš3 â‰ˆ 1.7 and that âˆš2 â‰ˆ 1.4 can certainly be helpful.
Consider the following problem:

One approach would be to plug in a value for n.
Let n = 2.
Then 1/(âˆš(n+1) - âˆšn)
= 1/(âˆš3-âˆš2)
â‰ˆ 1/(1.7-1.4)
= 1/.3
= 10/3.

Now we plug n=2 into all the answers to see which yields a value close to 10/3.

Only answer choice E works:
âˆš(n+1) + âˆšn
= âˆš(2+1) + âˆš2
= âˆš3 + âˆš2
â‰ˆ 1.7 + 1.4
â‰ˆ 3.1, which is very close to 10/3.

The correct answer is E.

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djiddish98: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Fri Jan 14, 2011 10:01 am; Thanked: 10 times

by djiddish98 » Thu May 12, 2011 9:36 am

Thanks for the example - it makes sense to come equipped with as many tools in the tool belt as possible, I suppose!

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