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GMAT Prep - Tough PS Triangle, Coordinate Geometry


 
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albertrahul
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PostPosted: Wed Jul 02, 2008 8:26 pm    Post subject: GMAT Prep - Tough PS Triangle, Coordinate Geometry Reply with quote

Preety good question, though I got it wrong in first attempt.
A is not OA. I'll post OA after discussion.

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simpdimp
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PostPosted: Wed Jul 02, 2008 10:33 pm    Post subject: Reply with quote

Looks like a pretty simple question, let me know if I got it right!

(6+0+x)/3 = 3 and (0+0+y)/3 = 2 --> (3,6)
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asigheartau
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PostPosted: Fri Jul 04, 2008 9:24 am    Post subject: Reply with quote

simpdimp wrote:
Looks like a pretty simple question, let me know if I got it right!

(6+0+x)/3 = 3 and (0+0+y)/3 = 2 --> (3,6)


If the values of x are the result of the average of the x vertices, shouldn't it be (6+0)/2=3 in which case y would be (0+2+Y)/3=??

Can you please detail your solution?

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albertrahul
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PostPosted: Sat Jul 05, 2008 6:50 am    Post subject: Reply with quote

Answer is (3,6).
This is actually a pretty simple question, but verbiage is confusing.

We've been told that values of x and y are average of respective vertices so value for x coordinate will be (6+0+x)/3 = 3.

Hope that helps!

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tolmar
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PostPosted: Sun Jul 13, 2008 12:00 pm    Post subject: Reply with quote

(3,6) to my opinion. This question is just long and cumbersome.
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