What fraction of the distance from A to C?

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

What fraction of the distance from A to C?

by BTGmoderatorLU » Wed Jan 31, 2018 6:14 am

If the box shown is a cube, then the difference in lenght 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%

The OA is C.

I'm really confused with this PS question. Experts, any suggestion please? I don't understand how can I solve thi question. Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Wed Jan 31, 2018 6:14 am

DrMaths: Senior | Next Rank: 100 Posts; Posts: 82; Joined: Mon Jan 15, 2018 2:01 am

by DrMaths » Wed Jan 31, 2018 7:43 am

Pythagoras' Theorem works in 3D as well as 2D.
In 2D, hypotenuse^2 = a^2 + b^2
In 3D, hypotenuse^2 = a^2 + b^2 + c^2
Where a, b and c are the orthogonal triangle sides or edges.

With a cube of unit length,
AC = 1
AB = SQRT(1 + 1) = SQRT(2)
BC = SQRT(1 + 1 + 1 = SQRT(3)

So (BC - AB)/AC = (SQRT(3) - SQRT(2))/1
= 1.7 - 1.4 (1 decimal place)
=0.3 (1d.p.) = 30%

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Feb 02, 2018 11:13 am

LUANDATO wrote:

If the box shown is a cube, then the difference in lenght 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%

If we let AC = n, then CB (the diagonal of the cube) = nâˆš3, and AB (the diagonal of a face of the cube) = nâˆš2.

The difference in length between line segment BC and line segment AB is:

nâˆš3 - nâˆš2 = n(âˆš3 - âˆš2) = n(1.7 - 1.4) = 0.3n

Finally we can solve for the difference in length between line segment BC and line segment AB as a fraction of A to C:

0.3n/n = 0.3 = 30%.

Answer: C

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