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Thread: GMAT Geometry...etc.

  1. #1
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    GMAT Geometry...etc.

    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.
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  2. #2
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    Frankly I have a great deal of trouble reading your posting.
    But the correct answer to that question is: s=\sqrt{3}~\&~t=1
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  3. #3
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    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
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  4. #4
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    not according to the software

    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.
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  5. #5
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    this is 2 unit circle

    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
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  6. #6
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    thanks

    Makes sense. thanks
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  7. #7
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     \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.

    .
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  8. #8
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    GMAT problem

    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
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  9. #9
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    Never assume such diagrams are drawn to scale. They are done that way to throw you off.
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