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You gotta register to see it, and I won't re-post it here. [#permalink] New post 17 Nov 2003, 09:41
You gotta register to see it, and I won't re-post it here.
http://www.manhattangmat.com/

I got B.

Anyone get anything different?
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Are the Q on real test as difficult as the ones on Manhattan [#permalink] New post 17 Nov 2003, 09:59
I looked at one question two weeks back and it sounded reasonable, last weeks triangulation was a stretch and this week's probability is out of my skill level. Are any questions on real test that difficult? Is this is a scare tactic from manhattan?
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 [#permalink] New post 17 Nov 2003, 10:23
mantha:
I looked at one question two weeks back and it sounded reasonable, last weeks triangulation was a stretch and this week's probability is out of my skill level. Are any questions on real test that difficult? Is this is a scare tactic from manhattan?

I think this week's problem (If I'm right) was pretty easy. I just figured out the "answer" for a 3x3 square, and plugged 3 into each of the answers-- and only one answer worked. Once you figure out that there are 4 possible formations out of 9 choose 3, it's just arithmetic (assuming I'm correct, maybe not)

As far as whether the real questions are that difficult, well, I haven't seen any on the PowerPrep or retired ETS paper tests, but I think the practice of the freebies is valuable. Further, if you're error-free 20 questions or so into the quant part of the GMAT, the questions tend to test some similar concepts, and approach that level of difficulty, from what I understand.
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same approach.. [#permalink] New post 17 Nov 2003, 10:37
Thank you.

I used similar approach too..

x=3, 4 out of 9 and x=3, 9/16(?) etc.,

But to get to a generic solution without choices needs some advance perm and comb knowledge.

Usually, OG questions can be solved in an alternate time consuming way -similar to the examples for purchase on this site.
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Re: same approach.. [#permalink] New post 18 Nov 2003, 01:56
mantha wrote:
Thank you.

I used similar approach too..

x=3, 4 out of 9 and x=3, 9/16(?) etc.,

But to get to a generic solution without choices needs some advance perm and comb knowledge.

Usually, OG questions can be solved in an alternate time consuming way -similar to the examples for purchase on this site.


All of the Manhattan GMAT questions are relatively "easy" once you see the "insight" involved.

For this week's problem, yes it is easy to solve using substitution. However, it is just as easy to solve generically if you see the simple pattern for the numerator of the probability equation. (if I had created the problem, I would have had two answers that worked for x=3 :twisted: ). I do disagree with one of your statements because IMO you do NOT need to know any "advanced" probability, but simple the combination formula where n = X^2 and K = 4.

To get the resulting expression to match one of the answer choices, you do need to understand how to simplify expressions using factorials. Then it is just algebra to reduce to the proper answer. If you are clever, you should be able to tell, without completely simplifying the expression, which of the answer choices are possible.

As for last weeks problem, this is the perfect type of advanced GMAT question: one that involves no esoteric knowledge, but requires creativity by the solver. About one in 10 of my students solved it quite quickly, 2 after some thought, and the rest couldn't solve it. Since these problems are written specifically to challange those capable of scoring 750 or higher, this is a reasonable result.

All in all, these problems are screened to be quickly solvable by a small percentage of people and are representative of the very hardest questions that someone on track for a 750+ score would encounter. (How do you think they separate the 750s from the 790s?). MGMAT is very careful to publish only problem that test knowledge expected to be known by a GMAT candidate and one that can be solved in less than two minutes by the very best students.

Please remember, these questions are meant to supplement your training, not substitute for them. I think solving these types of problems are good practice because they prepare and condition you to deal with situations that are slightly different than the hundreds of questions you answer over and over again. The questions that "freeze" people on the GMAT are typically not extremely difficult -- just "different" enough to put most people in panic mode.
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 [#permalink] New post 18 Nov 2003, 02:10
stoolfi wrote:
mantha:
I looked at one question two weeks back and it sounded reasonable, last weeks triangulation was a stretch and this week's probability is out of my skill level. Are any questions on real test that difficult? Is this is a scare tactic from manhattan?

I think this week's problem (If I'm right) was pretty easy. I just figured out the "answer" for a 3x3 square, and plugged 3 into each of the answers-- and only one answer worked. Once you figure out that there are 4 possible formations out of 9 choose 3, it's just arithmetic (assuming I'm correct, maybe not)

As far as whether the real questions are that difficult, well, I haven't seen any on the PowerPrep or retired ETS paper tests, but I think the practice of the freebies is valuable. Further, if you're error-free 20 questions or so into the quant part of the GMAT, the questions tend to test some similar concepts, and approach that level of difficulty, from what I understand.


Why do you think this is above your level of difficulty? Seems to me you are psyching yourself out before even trying.

Clearly it was easy for you to solve the x=3 version. Note look at x=4. Imagine sliding a 2x2 box across then down the matrix, counting as you go along. See a pattern? Can you generalize the number of 2x2 boxes for any x? Sure you can.

Now the number of possible choices is simply the number of ways you can pick 4 out of the total number of squares (which is X^2) . You had no problem when x = 3 so you shouldn't have any problem for any arbitrary x.

Now you have number of ways and number of possibles, just divide the first by the second to get the probability, using algebra to simplify.

What was so hard about that that it was over your head?

Don't be intimidated -- simply organize your thoughts and go back to basics! This problem is simple counting, simple probability, and moderate algebraic manipulations.
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 [#permalink] New post 18 Nov 2003, 08:13
AkamaiBrah-- Why do you think this is above your level of difficulty?

I didn' t mean to imply that it was (I think you mistook Mantha's comment for mine, or you're addressing your whole post to mantha). What I meant is that I don't remember seeing any questions (between OG, PowerPrep, and the papertests) that are as difficult as the average of the past three challenge problems.

if I had created the problem, I would have had two answers that worked for x=3

Well, I was very surprised that Manhattan didn't do that, to be honest. But I double checked my answer using 4 as well. Now to take the equation in the right answer and see if I can work it backwards to understand the formula...
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 [#permalink] New post 18 Nov 2003, 16:42
I just went for the general formula instead of picking numbers. I was able to find quickly that the number of 2X2 squares would be (x-1)^2 and of course, the total possible combinations would be C(x^2, 4).

So probability = (x-1)^2/C(x^2, 4)

Simplifying the expression, I got:

((x - 1)*24) / ((x^2)(x^2 - 2)(x^2 - 3)(x + 1))

I hope this answer is correct.

stoolfi wrote:
AkamaiBrah-- Why do you think this is above your
level of difficulty?


I didn' t mean to imply that it was (I think you mistook Mantha's comment for mine, or you're addressing your whole post to mantha). What I meant is that I don't remember seeing any questions (between OG, PowerPrep, and the papertests) that are as difficult as the average of the past three challenge problems.

if I had created the problem, I would have had two answers that worked for x=3

Well, I was very surprised that Manhattan didn't do that, to be honest. But I double checked my answer using 4 as well. Now to take the equation in the right answer and see if I can work it backwards to understand the formula...
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AkamaiBrah..Agreed.. [#permalink] New post 18 Nov 2003, 17:12
Agreed that the questions needed some creative thinking without getting scared in the first place. However, where on the OG do you see such questions? Can you solve these in 3-4 minutes (assuming you spend less time on other questions) in an actual exam? Where does manhattangmat get these questions from and how closely do they resemble the tough questions on gmat?

Anyone who took manhattangmat courses/study material volunteer to offer such insight.

AkamaiBrah, I am assuming that you teach at manhattanmgmt. I would like to converse and see how is your team different from the Kaplan's etc.,

I may not be 750+ candidate but I always thought I should score 50 in Quant with my background and knowledge.

Thanks
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Re: AkamaiBrah..Agreed.. [#permalink] New post 19 Nov 2003, 14:12
mantha wrote:
Agreed that the questions needed some creative thinking without getting scared in the first place. However, where on the OG do you see such questions? Can you solve these in 3-4 minutes (assuming you spend less time on other questions) in an actual exam? Where does manhattangmat get these questions from and how closely do they resemble the tough questions on gmat?

Anyone who took manhattangmat courses/study material volunteer to offer such insight.

AkamaiBrah, I am assuming that you teach at manhattanmgmt. I would like to converse and see how is your team different from the Kaplan's etc.,

I may not be 750+ candidate but I always thought I should score 50 in Quant with my background and knowledge.

Thanks


All of the problems are composed by the instructors. Since all of the instructors have scored at least a 750, all of them have seen the most difficult question you might see on a GMAT so they have a good frame of reference.

ETS will not publish the most difficult questions in an OG edition. Hence, IMO, MGMAT is the only place that offers 750+ level questions. All of the instructors can solve 90% of the challenge problems in 1-4 minutes. All of the problems are carefully screened by the boss for applicability and difficulty (not too hard but not too easy). Most are difficult simply because they apply multiple concepts or require some flash of insight into what exactly you need to calculate to solve the problem. Once my students get over the "fear" of the problem and learn to apply multiple approaches solving the problems, they slowly become adept at solving more and more of them and gain huge confident in Q questions altogether.

Overall, they are more difficult that any question 99% of the people will ever see on a GMAT. However, they are certainly solvable using only basic concepts and, IMO, are a great training aid in building confidence, testing your overall knowledge of concepts, and testing your ability to apply those concepts to a variety of problems.

As for your question of how we differ from Kaplan? All of our instructors are required to have a recent score of 750 or above AND have extensive prior teaching experience either in classroom or corporate settings. Kaplan does not have nearly the same qualification criteria. Most Kaplan instructors would not qualify to teach at MGMAT (if they could, they would because MGMAT pays 3-4 times more per hour than does Kaplan).

Kaplan has the advantage in that they have classes all over the world whereas MGMAT is concentrated in NYC with instructors in LA/OC, SF, Chicago, and Boston. HOwever, MGMAT has much experience in distance teaching and we offer live private instruction via the internet to anyone in the world.
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 [#permalink] New post 19 Nov 2003, 14:21
edealfan wrote:
I just went for the general formula instead of picking numbers. I was able to find quickly that the number of 2X2 squares would be (x-1)^2 and of course, the total possible combinations would be C(x^2, 4).

So probability = (x-1)^2/C(x^2, 4)

Simplifying the expression, I got:

((x - 1)*24) / ((x^2)(x^2 - 2)(x^2 - 3)(x + 1))

I hope this answer is correct.

stoolfi wrote:
AkamaiBrah-- Why do you think this is above your
level of difficulty?


I didn' t mean to imply that it was (I think you mistook Mantha's comment for mine, or you're addressing your whole post to mantha). What I meant is that I don't remember seeing any questions (between OG, PowerPrep, and the papertests) that are as difficult as the average of the past three challenge problems.

if I had created the problem, I would have had two answers that worked for x=3

Well, I was very surprised that Manhattan didn't do that, to be honest. But I double checked my answer using 4 as well. Now to take the equation in the right answer and see if I can work it backwards to understand the formula...


qood job. that is exactly the correct solution and approach I would have used.
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 [#permalink] New post 19 Nov 2003, 15:38
I think there should be more possibilities than (x-1)^2. Let's take the example of x=3. Can't we consider the bulbs at the four corners as one 2 by 2 square!? Similarly if x=4 there could be 3 + 3 + 2 + 3+ 2 + 1 = 14 possibliities!!!


AkamaiBrah wrote:
edealfan wrote:
I just went for the general formula instead of picking numbers. I was able to find quickly that the number of 2X2 squares would be (x-1)^2 and of course, the total possible combinations would be C(x^2, 4).

So probability = (x-1)^2/C(x^2, 4)

Simplifying the expression, I got:

((x - 1)*24) / ((x^2)(x^2 - 2)(x^2 - 3)(x + 1))

I hope this answer is correct.

stoolfi wrote:
AkamaiBrah-- Why do you think this is above your
level of difficulty?


I didn' t mean to imply that it was (I think you mistook Mantha's comment for mine, or you're addressing your whole post to mantha). What I meant is that I don't remember seeing any questions (between OG, PowerPrep, and the papertests) that are as difficult as the average of the past three challenge problems.

if I had created the problem, I would have had two answers that worked for x=3

Well, I was very surprised that Manhattan didn't do that, to be honest. But I double checked my answer using 4 as well. Now to take the equation in the right answer and see if I can work it backwards to understand the formula...


qood job. that is exactly the correct solution and approach I would have used.
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 [#permalink] New post 20 Nov 2003, 05:30
pawargmat wrote:
I think there should be more possibilities than (x-1)^2. Let's take the example of x=3. Can't we consider the bulbs at the four corners as one 2 by 2 square!? Similarly if x=4 there could be 3 + 3 + 2 + 3+ 2 + 1 = 14 possibliities!!!


AkamaiBrah wrote:
edealfan wrote:
I just went for the general formula instead of picking numbers. I was able to find quickly that the number of 2X2 squares would be (x-1)^2 and of course, the total possible combinations would be C(x^2, 4).

So probability = (x-1)^2/C(x^2, 4)

Simplifying the expression, I got:

((x - 1)*24) / ((x^2)(x^2 - 2)(x^2 - 3)(x + 1))

I hope this answer is correct.

stoolfi wrote:
AkamaiBrah-- Why do you think this is above your
level of difficulty?


I didn' t mean to imply that it was (I think you mistook Mantha's comment for mine, or you're addressing your whole post to mantha). What I meant is that I don't remember seeing any questions (between OG, PowerPrep, and the papertests) that are as difficult as the average of the past three challenge problems.

if I had created the problem, I would have had two answers that worked for x=3

Well, I was very surprised that Manhattan didn't do that, to be honest. But I double checked my answer using 4 as well. Now to take the equation in the right answer and see if I can work it backwards to understand the formula...


qood job. that is exactly the correct solution and approach I would have used.


No. You are trying too hard to "outclever" the question.
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 [#permalink] New post 20 Nov 2003, 07:07
AkamaiBrah wrote:
No. You are trying too hard to "outclever" the question.


Akamai, I really don't understand why 2 by 2 bulb square should mean only 2 by 2 bulbs that are together. Please elaborate.



thanks..
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 [#permalink] New post 21 Nov 2003, 05:51
pawargmat wrote:
AkamaiBrah wrote:
No. You are trying too hard to "outclever" the question.


Akamai, I really don't understand why 2 by 2 bulb square should mean only 2 by 2 bulbs that are together. Please elaborate.



thanks..


because 2x2 refers to distance, not number of bulbs.
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 [#permalink] New post 22 Nov 2003, 13:55
pawargmat,

The question asks 2 by 2 squares. If you pick four squares from the corners, it sure is a 2 by 2 square, but is not congruent with the original 2 by 2 square where only four bulbs were present.

So, in short, in a 2 by 2 square, all the bulbs are next to each other and are not separated. Your approach would be correct if the question asked for the number of all 2 by 2 squares.

Hope this helps.

pawargmat wrote:
AkamaiBrah wrote:
No. You are trying too hard to "outclever" the question.


Akamai, I really don't understand why 2 by 2 bulb square should mean only 2 by 2 bulbs that are together. Please elaborate.



thanks..
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 [#permalink] New post 27 Nov 2003, 15:54
Hi AkmaiBrah,

Could you tell me how do you rate the MGMT tests on CD? How do they compare with respect to verbal section?

Thanks
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 [#permalink] New post 27 Nov 2003, 23:07
gmatblast wrote:
Hi AkmaiBrah,

Could you tell me how do you rate the MGMT tests on CD? How do they compare with respect to verbal section?

Thanks


In all honesty, the MGMAT tests on CD are simply licensed versions of the ARCO CD. You would do better to order a used version of the ARCO book from Amazon as it contains more information (some lessons on video) and you get a book.
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Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 05 Dec 2003, 14:44
AkamaiBrah wrote:
gmatblast wrote:
Hi AkmaiBrah,

Could you tell me how do you rate the MGMT tests on CD? How do they compare with respect to verbal section?

Thanks


In all honesty, the MGMAT tests on CD are simply licensed versions of the ARCO CD. You would do better to order a used version of the ARCO book from Amazon as it contains more information (some lessons on video) and you get a book.


Thanks AkmaiBrah,

I really appreciate your feedback. I researched on the web for the ARCO books. It seems, ARCO has more than one book with CD and it is really confusing.

Could you tell me the exact title of the book that I should look for?

I woudl greatly appreciate your help. Thanks
  [#permalink] 05 Dec 2003, 14:44
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