lego2401

Erin, I think testmagic has developed alot of brand equity in the past couple of years. It actually means something and has to do with what it does/did for many of us. It gives a perception of the portal where people come together to make their studying experience easier, and where their problems can be solved. Also, many of us old timers have become less frequent users (because we're busy with school now thanks to this site ;) ), but whenever someone tells me they're studying for the GMAT, I tell them to use testmagic. So you have to consider the viral marketing aspect as well. Finally, in terms of urch being easier to remember, I would disagree. Although its only 4 letters, urch doesn't mean much to most people. Its an abreviation, and even if it were pronouced as one word, people will make mistakes with it. Test Magic on the other hand, is easy and catchy. The way memory works, we tend to remember by association. So someone studying for a GMAT, for example, will find it easy to relate and associate. They wont forget the name, and it actually means something to them. I hope this helps. Just my thoughts as a Marketer.

Erin, can we put this thread up as a sticky/anouncement?

August First any time in the 9-11 PM EST sounds great.

Cocoray, your more than welcome, and congrats once again. Now go celebrate your amazing score :) :)

How much longer till we get there ? :)

cocoray, congratulations on the extremely amazing score :) :) As for retaking the test, I believe thats your greed talking my friend, we all have some of that. I got a 760, should I retake the test because there iare 40 more points that I can earn? Because others, even here on the forum have done it? Most importantly, because on my practice test scores, I did a 770 and because I just as you, just as everyone else, may have made a silly mistake somewhere or gotten bored by the end of the exam ? The answer to both of us, is a big NO :) There is absolutely no difference between a 750, a 760 or even a, 800. At that point it boils down to a couple of question here and there, are we to be ones who take scores, IQs etc seriously? So someone with a 160 (based on a highly one dimensional test that took a couple of hours) really is much smarter than a guy with a 140? I always wondered how people with a 160 IQ who actually believe that are actually smart. Im not saying any of us are like that, but retaking the test would make us like that. 750 is an astonishing score that you should be extremely proud of, and whos to say you wont actually score lower next time. Could be higher, could be lower, as I said at this level its a matter of a couple of questions, 15 mins of extra sleep, etc. Accept what you have! (PS you must have made more than one silly mistake on the Q section, a 50 is atleast 4 mistakes later in the test or one in the first ten and 3 later). Finally, I notice that so many people here, especially those from outside the US, put way too much emphasis on the GMAT. Its just ONE factor, not all of them. If the top ten schools were to take 28 maths and 30 verbal questions so seriously as an indicator of who to accept, and marginalize all other factors, then I'd totally question the intellectual abilities of their admissions officers and thus their credebility as academic institutions. You can have a 600 GMAT but an amazing way of doing things and a great personality (emotional intelligence). You could also be a techie with a Q 50 working against an art history major with a Q of 47. Which of these scores is harder to accomplish? Finally, what can you do with a 760 or 790 that you can't do with a 750? If I were an admission officer at a business school, and I saw that an applicant who got a 750 retook the test, I am sure to send that application straight to the shredder. Why? Because that person would not be fit as a manager. Manager think of opportunity costs and know where to say enough. That applicant on the other hand, wasted valuable time reprearing for the test when they could have spent that time doing something more useful, be it at work, at home, or in society. I hope you think likewise.

Agreed with Y=3 We have 2X + Y = 7 or 17 or even 27 => 2X can be 0,2,4,6,8 or 18. Now in the second case, we have 2X + 7 = 11 (if 2X is 2,4 or 6) or 2x + 7 + 1 = 11 (if 2X = 8) 2x + 7 + 2 = 11 (if 2X = 18) Starting with the last case, if 2X = 18, the equation doesnt add up to 11. if 2X = 8, the equation doesnt add up either. which leaves us with 2X + 7 = 11 => 2X = 4 => X=2 so Y = 3 Ans. C

Great work! Congrats!

Official Guide to the GMAT. You can buy it on the ETS website.

TEST NAME | SCORE | V % | Q % | Kaplan Diagnostic 680 Princeton Review 720 KAPLAN CAT 1 (Q Only) 50 KAPLAN CAT 2 (Q Only) 50 PP1 (After OG) 760 Q 50 V 42 PP2 (After OG) 770 Q 51 V 44 ACTUAL GMAT 760 Q 50 V 44

Congratulations !!!!!

If I told you we had two people, and one of them needs to be chosen to be the goal keeper of a team, then how many options do we have? 2 Since the question says only 2 people are fit to be in goal, that gives us 2 possible people. Now lets look at the other groups. For every one of the 2 goal keepers, there will be a group of 2 playing defense. The two defenders can be any of 8 people (10 - the ones who only play goal). So how many possible groups of 2 defenders are there? The product rule says 8*7 (8 cause we have 8, and once we've picked 1 of them, there are only 7 left to chose from for the next defender). But there is one problem with the 8*7 = 56 count. This count assumes that Jim and Joe is different from a group with Joe then Jim. That would be true if order mattered, but in this case, it doesn't (in a password on the other hand, XY would be different from YX). So how do we resolve this order problem? Well, once we chose 2 people, how many way are there to arrange them? 2! ways. So we divide 8*7= 56 by 2! and that gets rid of all repetitions (XY and YX will be counted once instead of twice). From here, mathematicians (god bless them) developed the concept of the Combination, or 8C2. 8C2 = 8*7/2! Back to the big picture, we said we have two ways to chose the goalies. So far we have 2*(56/2!) Now we also need 2 midfielders. Once the defense has been picked, how many players left to chose from? 6 So its 6*5/2! (same logic as above). So far we have 2*(56/2!)*(30/2!) Finally, we need once forward. How many people are still not on the team? 4 So there are 4 players to chose one forward from, ie 4 options. (and once the player is picked 3 players sit on the bench and don't get to party after). So what do we have now? 2 way to choose a goal keeper for every goal keeper, 56/2! ways to choose defense. Then 30/2! ways to choose midfield. Then 4 ways to choose forward. total is 2*(56/2!)*(30/2!)*4 = 3360.

This doesnt work because you are not accounting for order within the groups. In this question, order matters when u look at the groups, ie XY in Defense is different from XY in offense. But within the groups, XY and YX is the same group. With your method, you dont account for such a situation. Therefore you have to divide by 2! for each group of 2 people. dividng a permuation such as 8P2 by 2! is really 8C2. so we say 2C1 * 8C2 * 6C2 * 4C1 = 3360, and this numbe doesn't double count XY and YX as same group.

On another note, I will always be around the forum from time to time. If you have any questions you'd like help with, I'd be more than happy to do my best to help. Feel free to PM me and I will try to reply to all messages as time permits.

Thank you so much to everyone for the comments. I am so, so sorry I havent posted anything about my experience or preparation yet, I had a crazy week. I promise a long, detailed posts tonight :)

I came across this spreadsheet the other day (unfortunately only after finishing my GMAT :) ). It has a list of all the questions in the [tooltip=Official Guide]OG[/tooltip] in all sections, and rates their level of difficulty by ETS standards. Hope you guys find it useful. Lego. OG Question Info.zip

These questions have been going around on the forum for a while, so I decided to create a small turorial on them. We have 3 groups of 2. Lets split it up In the first group, we have 6 people to chose from for the first person. That leaves us with 5 more to chose from, so total of 30 groups. For each of the 30 groups, in the second group, we have 4 to chose * 3 for the second person, which is 12. So we have 30*12. In the final group, 2 people left, then 1 for the second person. So In total, we have 30*12*2 groups = 720. But, we have 4 overcounts here. firstly, in the 3 groups, these can be ordered in 3! ways. We have AB CD EF is the same as CD AB EF. So we divide 720 by 3! to get 120. Now we have a second problem, which is that within each group, order is not important, ie AB is the same as BA. So we divide each group by 2! ways to order it. So since we have 3 groups, thats 120/ (2!*2!*2!) = 120/8 = 15. So the Ans is 15. I used this long and tedious approach to explain the logic behind the solution. In reality, this is simply 6C2 * 4C2 * 2C2 / 3! = 15 Overal, we have 6 dudes. A,B,C,D,E,F. We have AB,AC,AD,AE,AF,BC,BD,BE,BF,CD,CE,CF,DE,DF,EF. Can any one think of any more groups? Can also be seen as, take A as first person. A group can be made with A in 5 ways. Then a group can be made with B in 4 ways. Then a group with C in 3 ways, and so on, upto F, who's already been counted in all groups so we dont count him again. We get 5 + 4 + 3 + 2 + 1 = 15. A final way to approach this problem. The classic round robin tournament method, how many games can the teams play. We've done this one before on the forum. Draw a 6*6 matrix. Look at it. We have a total of 36 boxes. The middle diagonal has the teams playing against themselves, so we get rid of it. It contains 6 items. So Out of a total of 36, remove 6. Leaves us with 30. Now look at the matrix again. Notice that the items to the left of the diagonal are the mirror images of the ones to the right. i.e., We have A playing E, then E playing A. So out of those 30 games, 1/2 are repetitions. So we end up with 15 distinct games. Apply the same concept here. In fact, the formula becomes, for X iterms, total # of ways = (X*X - X)/2 = X(X-1)/2. In this case 6(5)/2 = 15. Can we have the OA pls.......

Congratulations Dude!!!!!!!!!!!!!!!!!! That is sooo amazing, wow. Seriously dude, hats off, you rule![clap] [clap] :notworthy :notworthy I knew u'd burn it :)

well said dude...........

Not quite. If 2/8 is the probability of picking Jane, then we cant pick another freshman. That mean that now we have 7 left, but the 7th person is also a freshman, so we cant choose them. In this case, probability of a senior = 2/6, since we the rules stipulate that the lottery will pull one of the 6 NONFRESHMAN candidates. 2 of those are seniors, so the new probability is 2/6. So Prob(Freshman & Senior) = 2/8 * 2/6 = 1/12 + 2/6*2/8 = 1/6 Why twice? Because as stated before, we have two possible scenarios that are favorable. the first is where the first pick is a freshman and the second is a senior. The second is where the first is a senior and the second a freshman. Nice try though.........

Andy, welcome to the forum. Post these questions in the Verbal Forum and people will answer them. This forum is only for people who have finished their GMAT to post scores, experiences, etc.

1- 11*10*9*8/11*10*9 = 8 2- 4/7 Yes, from my experience, this is pretty much the standard of prob/comb/perm Qs on the GMAT. No matter which range your in, it wont get more difficult than this. Don't work too much on these questions, you'll waste valuable time that can be used to enhance other skills for the GMAT. This is pretty much as hard as it gets.

Congratulations man!!!! I'm sure you can improve your verbal if you want to, which part of verbal are you weak in ?

WOHOOO !!!!!!!! Way to go dude, I knew u'd kick ass! Now its time to take on the world dude! :tup::tup:[clap][clap]

Dude, sorry about the last response, in fact your 100% correct. Just looked at it again this morning, nothing ambigious about it. Shouldn't have done any solving yesterday, gmat is enough for one day...........