Ravloo

No problemo :) Thanks for the percentile numbers!

Hi Shweta, here is the breakdown of my verbal prep: The 6-7 verbal sections I did were from the GRE Big Book (go google!). After doing each section, as I reviewed the answers I would note down every word I did not know. Then I put the 250 vocabulary words into a Quizlet.com word list (great resource!) and used this to study. Another similar resource you may want to try is the word list of Dictionary, Encyclopedia and Thesaurus - The Free Dictionary. If I had more time I would have definitely used these tools to increase my vocab (it beats the hell out of old school flash cards!) Although I suggest trying to figure out relationships on your own, I posted on this forum on a couple of the analogies which gave me trouble as well. What else.. I made a list of gre prep websites I found. To be honest though I did not really use them due to time, I basically dabbled when I could. It was nice to have a change of scenery for studying as memorizing words isn't my idea of party time. Here is a compiled list: GRE prep - Antonyms - majortests.com Test Prep: GMAT, GRE, LSAT, MCAT, SAT, ACT, and More gre Flashcards Flash Matching Quiz - GRE Vocabulary Words Online games - SAT and GRE Games and Info Other then that the GRE PowerPrep tests are a must. The above link to princetonreview.com is great as it has a full length practice test. Also there is a Barron's Pass key textbook which is nice and lightweight for carrying around to study on the road. For Quant, there are TONS of great problems that people have posted on this forum. So I'd suggest identify your weaknesses and search/post away. Also if you find people who have well-written questions, look at more posts by them. Good luck!

Maybe you should try practicing in timed environments? Use a timer where you can see the countdown to emulate that pressure the GRE gives. Otherwise you can try to improve concentration? Note that I'm sure grad schools take into account careless errors and don't disregard non-perfect scores, as we aren't robots! So if you are making one or two mistakes your score is still above average.

Hi there. I just finished my second GRE attempt this morning and scored a 1450 (q800/v650). Goldust I appreciate your advice. It helped me make the decision to take the GREs again. I feel I can offer some general advice now. If you read my last post you can see the mistakes I made, and now it's confirmed. This time I scheduled the exam for Saturday morning and got ample rest the night before. Woke up with a clear mind and positive attitude. I strongly believe that this accounted for most of my improvement in score. Next strategy I used which played a significant role was timing myself. This time I finished the quantitative section with about three minutes to spare. I wanted to spend another minute here and there to make sure I wasn't tricked by the question, but instead I kept my reasonably confident answer choice and moved on. This is crucial, unless you really have that speedy Math in you and can make up time. Note that I don't regret not spending extra time even though I had the extra three minutes, as I think if I had I would have been very pressured towards the end, and it would have also been a gamble. Finally, practice-wise I didn't really do anymore Math problems since the last test, and I reviewed about 6-7 Verbal sections. I also memorized about 250 vocab words which I found mostly from the Verbal sections I did. I hope that helps anyone who reads this and thanks again to all the forum members. Now I have started applying once again and have a question about how to figure out my percentiles since waiting for the test result may result in missed deadlines. If anyone can help with the percentiles for the following scored I'd greatly appreciate it: Verbal: 650 Quant: 800 Analytical: 4 Cumulative: 1450

I just came across a similar antonym question. Vacillation: a) perpetual activity b) rapid growth c) motionless balance d) accurate focus e) minimal response I put d, which is similar to a in hoper's questions. The correct answer however is c, similar to c in hoper's question. So in both cases they are looking for the physical antonym of vacillation. I think our guesses are incorrect because...aim and focus have to do with unwavering attention, rather than unwavering decision making. As in.. you can vacillate between answer choices c & d, but it doesn't make sense to vacillate between extreme focus and lack of attentiveness.

Thanks! Very clever!

obey - you should always post the solution if you find the answer to your own question :) For the second question... I'd start with some algebra: 72.42 = 24k + kn/100 7242 = 2400k + kn 7242 = k(2400 + n) Then since k&n are positive integers, at most k could be 3 because 4*2400 would be more than 7242. If k is 1, then "2400 + n =7242" but this can't be since n Then k+n=3+14=17

Here is the question: Jim and Mary have to invite 5 of their friends to their party. Jim has 7 friends, Mary has 6 friends, and they have no friends in common. At least one friend of each has to be invited. How many possible combinations of friends can they invite? First can someone verify this method (not sure about the answer): (Total ways to choose 5 friends) - (ways to have only Jim's friends) - (ways to have only Mary's friends) Which would be 13C5 - 7C5 - 6C5 = 1260 Also how would you "build up" the solution instead of using the (total - exceptions) trick?

10. EMOLLIENT: SUPPLENESS:: (D) desiccant: dryness x contributes to y 11. DRAW: DOODLE:: (B) travel: ramble y is to x without a definite direction 12. CONSPICUOUS: SEE: (E) transparent: understand x is easy to y 13. IMMATURE: DEVELOPED:: (E) incipient: realized with time, something x can be y 14. PERSPICACITY: ACUTE:: © caprice: whimsical being y leads to x

Since the rule is 68-95-99.7, wouldn't the answer to problem one be 95% rather than 96%?

Aasho, Answer should be D, it was answered in this thread: http://www.www.urch.com/forums/gre-math/107029-statistics-problem.html

Thanks!

Dwarrior, Cool clipmarks thing.. Can you elaborate on your solution? Based on your picture it looks like 30 > k > 15, so the answer would be B. And if you flip it, then still 15 Or did you mean the graph is right.. but you just filled in the wrong answer?

We have to factor in that the number "can be same also." For example, you can pick 5, and then another 5. So using rain's method we would get: 1 - 5/10 * 5/10 = 1 - 25/100 = 3/4 Here is another way to think about it. We have four cases. OO, EE, OE, EO First see that they are equally likely. Let's take the OO case. There are 5 evens and 5 odds. So selecting one odd has a prob. of 1/2. After our first selection, since the numbers can also be the same, there are still 5 evens and 5 odds. So selecting the second odd has a prob. of 1/2. Similarly we see that all of the four events are equally likely. Then since the probability of what we want: EE, OE, EO, makes up 3 of 4 equally likely cases, our probability is 3/4. mmmt, since you like counting I will dedicate this last solution to success on your exam tomorrow :) First remember that our number can be the same. So there are 10*10 = 100 ways to choose any two numbers. Now we have 5 evens, so there are 5*5 = 25 ways to choose two evens. How to choose one even and one odd? The first number is either even OR odd, so there are 10 possibilities. The second choice must NOT be what the first choice was, so we only have 5 possibilities for the second number. Then we have 10*5 = 50 ways of getting an even and odd number. Then (possibilities to get at least one even)/(total possibilities) = (25+50)/100 = 3/4

Regarding #8: Remember median is equal to the middle number, or the average of the middle two numbers if the total amount of numbers is even. Ex: Median of {2, 4, 6, 10, 12} will be 6. Mean of {2, 4, 6, 10, 12} = 2+4+6+10+12 = 34/5 = 6.8 I think #8 does not have enough information to solve. Case 1: Class 1: 22 students received a score of 52 1 student received a score of 53 22 students received a score of 54 Then the median is 53. Class 2: 16 students scored 39 2 students scored 40 16 students scored 41 Then median is mean average of 40 and 40, which is 40. Combined: 45+34 = 79 total students 40th score should be median Then median would be 52. Case 2 Same as Case 1, except in Class 1 22 students receive score of 20 instead of 52. Our scores in increasing order would then look like: 20, 20, ...., 39, 39, ...., 40, 40, 41, 41, .... etc. If you count it, the 40th number is 40. Then median would be 40.