GMAT Geometry...etc.

rick81 · #1 $Old$ November 6th, 2009, 05:07 PM

I have been studying for the GMAT and GRE. I have taken a few GMAT CAT practice tests from the GMATprep software I downloaded from the GMAT website. I have managed to test to the level where the CAT software is giving me what it considers difficult questions. I have been stumped on a few and even after going back to review them I either don't see how to arrive at the correct answer or I haven't figured out the fastest way to solve the question. On average I should on spend 2 minutes per question. I will post a few of the problems below. The answer with the blue square is the correct answer. The filled in answer is what I choose. Thanks in advance for those that can help me out.

Plato · #2 $Old$ November 6th, 2009, 05:27 PM

Frankly I have a great deal of trouble reading your posting.
But the correct answer to that question is: $s=\sqrt{3}~\&~t=1$

PatrickFoster · #3 $Old$ November 6th, 2009, 05:39 PM

I agree with Plato, the answer should be $s = \sqrt{3}$ . Perhaps there is an error in the software's answers.

Incidentally, and unless I'm missing something, this problem shouldn't take more than 5 seconds to solve.

Patrick

rick81 · #4 $Old$ November 6th, 2009, 05:44 PM

I also answered s=square root of 3 but according to the software the answer is s=1. I think we both assumed the larger triangle was being bisected evenly.

I worked the problem using pythagoreon's thrm and the sides of the right triangle in the negative quadrant to find the length of the radius =2. Then the larger equilateral triangle has sides 2,2, 2 times square root 2 applying the knowledge that the ratio of a right equilateral triangle is 1:1:square root 2. Then S = (2 times square root 2) - (square root 3).

Thats where I am stuck. How do I get from subtracting these two square roots to the answer 1? Is there a simpler approach.

bigwave · #5 $Old$ November 6th, 2009, 05:58 PM

the graph shown is somewhat decieving, it is a 2 unit circle
(not a unit circle which is usually shown)

P is $\frac{5\pi}{6}$ it appears to be more like $\frac{3\pi}{4}$

therefore on a unit circle P would be $(\frac{\sqrt{-3}}{2},\frac{1}{2})$
and Q $-\frac{\pi}{2}$ from P would be $(,\frac{1}{2}\frac{\sqrt{-3}}{2})$

So on a 2 unit circle P $(\sqrt{-3},\ 1)$ and Q $(1, \sqrt{3})$

therefore s = 1

rick81 · #6 $Old$ November 6th, 2009, 11:51 PM

Makes sense. thanks

aidan · #7 $Old$ November 7th, 2009, 08:38 AM

$\angle POQ$ is 90 degrees
$\angle (-\sqrt{3},1),(0,0),(s,t)=$ 90 degress

Similar Triangles:
$\triangle (0,0),(-\sqrt{3},1),(-\sqrt(3),0)$
is equal to
$\triangle (s,t),(0,0),(s,0)$

Thus the y-coord of P is equal to the x-coord of Q.

.

bjhopper · #8 $Old$ November 7th, 2009, 07:16 PM

Posted by Rick.81
The diagram you showed is depicted wrongly . the two points should be at different elevations If thats the way it was depicted I think it unfair.Putting in the perpendicular mark between the radii is a clue which leads to a correct solution and it is all in knowing the properties of 30-60-90 triangles.

bjhopper

Debsta · #9 $Old$ November 7th, 2009, 07:24 PM

Never assume such diagrams are drawn to scale. They are done that way to throw you off.