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transylvaner

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Everything posted by transylvaner

  1. Hi all! I took my GRE last week and scored 780Q/460V.... disappointig.... I am applying at 7, top 10 and 4, top 20 econ PhD programs in the US. So, I do not know if I sould retake or not the exam. Since it was a totally disappointing experience, I have now a huge aversion against this test. (i.e. I prepared 3 months for the test, learnt Barron's word list, and scored, on average, Q800/V550 on the practice test). Actually, I am afraid that at the second attempt, I would not be able to get a higher score or I would get even lower than 780... that would look even worse. Do you know guys if there is any cutoff value for GRE below which applications are not considered? I am affraid that my application will not even be considered because the 780 Q GRE. I have good GPAs, a lot of math classes, research and teaching experience, published papers, best thesis awards for BSc, MA and MSc thesises, good recommendation letters....Basically, GRE is the main weakness of my application.....Thanks in advance!
  2. Hi! There is no rule! Just try it by plugging numbers into the formula in order to check the intuition! Bests!
  3. Hi guys! Could you tell me about the strategy you apply for RC ? How can I increase my speed and precision? :mad: My problem is that I cannot really focus under time pressure and, hence, I cannot read carefully the passage! Thanks a lot!
  4. There are in total 58 balls in the jar. A few of them have red color or both red&green (lets note this by R). Others are green or both red&green (G). However, both R and G contain balls which have both colors (lets note this by B). Thus we have the following three groups: (1) balls which are only red: R-B (2) balls which are only green G-B (3) those balls which are both green and red: B Consequently, the total number of balls in the jar 58 = (R-B)+(G-B)+B=R+G-B (a1) We know that 2/7 of red are also green => B=(2/7)*R (a2), respectively 3/7 of green are also red => B=(3/7)*G (a3). From (a2) R=(7/2)*B, from (a3) G=(7/3)*B. By plugging these into (a1) you get: 58=(7/2)*B+7/3)*B-B. By solving the equation you get B=12. Thus the probability to draw a ball which have both colors is 12/58=6/29 :) Good luck!
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