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
GMATPrep Geometry Bugger
This topic has expert replies
-
- Junior | Next Rank: 30 Posts
- Posts: 15
- Joined: Mon Oct 29, 2007 12:43 pm
- Thanked: 2 times
-
- Senior | Next Rank: 100 Posts
- Posts: 42
- Joined: Wed Jun 20, 2007 8:45 am
- Thanked: 1 times
-
- Junior | Next Rank: 30 Posts
- Posts: 18
- Joined: Wed Jan 30, 2008 8:50 pm
Hey can you tell me how you arrived on the answer D?
What is the explanation to the OA?
What is the explanation to the OA?
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Problem solving diagrams ARE drawn to scale unless they specifically state otherwise.tutonaranjo wrote:unless it specifically says "figure not drawn to scale" you can use the diagram for reference
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.
![Image](https://i.imgur.com/YCxbQ7s.png)
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
- Senior | Next Rank: 100 Posts
- Posts: 42
- Joined: Wed Jun 20, 2007 8:45 am
- Thanked: 1 times
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
![Smile :)](./images/smilies/smile.png)
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
![Smile :)](./images/smilies/smile.png)
-
- Junior | Next Rank: 30 Posts
- Posts: 15
- Joined: Mon Oct 29, 2007 12:43 pm
- Thanked: 2 times