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GMATPrep Geometry Bugger

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by stellategang » Mon Feb 04, 2008 10:33 am
Nothing indicates in the question that ACO is a triangle. If ABCO could be a quadrilateral then this question is impossible to answer. Does anyone feel me on this?

I put E because I thought ABCO could be a quadrilateral . . . the OA is D, which I assume presumes that ACO is a triangle although nothing is stated.

Thanks for the help
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Source: — Data Sufficiency |
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tip

by tutonaranjo » Mon Feb 04, 2008 5:23 pm
unless it specifically says "figure not drawn to scale" you can use the diagram for reference
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OA explanation?

by meghamehta15 » Mon Feb 04, 2008 6:37 pm
Hey can you tell me how you arrived on the answer D?

What is the explanation to the OA?
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Re: tip

by Stuart@KaplanGMAT » Mon Feb 04, 2008 6:52 pm
tutonaranjo wrote:unless it specifically says "figure not drawn to scale" you can use the diagram for reference
Problem solving diagrams ARE drawn to scale unless they specifically state otherwise.

On the other hand, data sufficiency diagrams are NOT necessarily drawn to scale. It is NOT safe to rely on the way things look, we can only rely on explicitly stated information.

However, there are a few exceptions to this general rule, one of which is that if a line looks like it's straight, we can rely on it being straight.

Here's the scoop taken directly from the test directions (emphasis mine):
Figures

A figure accompanying a data sufficiency question will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).

Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.

You may assume that the positions of points, angles, regions, etc. exist in the order shown and that angle measures are greater than zero.

All figures lie in a plane unless otherwise indicated.
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solution

by tutonaranjo » Tue Feb 05, 2008 2:54 pm
Name angles like this>
angle BAO = x
angle BOA = x (triangle is isoceles because BO is a radius and it is = CO = AB)
angle OBC = y
angle BCO = y (triangle is isoceles, two sides are radii)
angle BOC = 180 - 2y

Solve this.
angle CAO+ACO+AOC = 180 ... translated this means
x + y + x + 180-2y = 180 solves to>
2x=y
so actually, angles OBC and BCO are 2x

Now...
name angle BDC = z (this was previously named 180-2y)

statement 1 says angle COD = 60
60+z+x = 180 (the are all supplementary right?) another equation>
4x+z = 180 (from triangle BOC angles add 180)

x+z = 120 (equation 1)
-4x-z = -180 (equation 2 multiplied by -1)
-3x=-60
x=20
SUFFICIENT

Statement 2 gives you the value of 2x, so you can figure X

:)
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by stellategang » Tue Feb 05, 2008 6:12 pm
tutonaranjo and Stuart,

Thanks for the insight, will be using that approach from now on :)
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