New practice book test GR0568 release!

[lime]

This is very helpful from ETS to divulge new practice test GR0568 this autumn. I found this test the same we were attacking last spring. In order to avoid turmoil I think we can start discuss problems from it here.

[lime]

I found that according the statistics in the end of the practice book two problems N43 (26%) and N47 (28%) caused the difficulties for the majority of test takers. Meanwhile, these are only concern of calculation and don't have any tricks inside. That's why we should solve them quickly and accurately.
Problem 43

Problem 47

[benjiemoe]

sweet. this is the test I took in April, 2008. Looking back at it, the questions all seem quite simple. The problem that happens is that on the test one is 1) pressured for time and 2) becomes fatigued towards the end. These are the primary reasons the test can be difficult. When i studied, I spent the majority of my time just looking at individual questions and solving them. On the test of course, I made many more stupid mistakes than in practice. The point is-- to get the max out of studying make sure to take tests all in one 180 minute period rather than spacing them out as this will help you prepare accurately for the real one.

[CoCoA]

I am saving it until 3 weeks before my test, and doing other practice tests and problems until then.

[diogenes]

I found that according the statistics in the end of the practice book two problems N43 (26%) and N47 (28%) caused the difficulties for the majority of test takers. Meanwhile, these are only concern of calculation and don't have any tricks inside. That's why we should solve them quickly and accurately.
Problem 43

Problem 47

Lime your nice solutions are gone.

[mk]

So perhaps someone could explain this to me. The ETS put out a new practice book 0568 this fall. Is that the oen currently on their website, or the one they will send me (since I signed up for the October test) or both? If it is one or both of these, where can I find the old test they used to have on their website?

For those of ou who have taken the test once already, how similar are the new book, old book, and practice tests on the UCSB website compared to recent tests?

[diogenes]

It looks like 9768 is the one ETS has on their website, whereas the one they sent out was 0568.

Sadly there is not a clearing house for old math gre subject tests that I know of (except for lime ).

[sky]

Is this test booklet available on net? Can anybody please scan it and post it somewhere. I hope it doesn't violate any copyrights as ETS has released the booklet.

[lime]

It is available on the official website.

http://www.ets.org/Media/Tests/GRE/pdf/ ... e_book.pdf

[sky]

Thanks lime.
It would be highly appreciated if you can answer my few queries.
I have planned in this way:
1. Dummit and Foote, Abstract Algebra Part I-II
2. Munkres, Topology, Chapters 1-3
3. Royden's Real Analysis Chapter 1-4
4. Linear Algebra by Gilbert Strang

What book should I refer for Calculus. Will Kreyszig do for remaining topics like Probability and Statistics.?
Can you please give me few tips on what to cover and to what extent. Just few guidelines, no need of details.
Should I study transforms. [I am from engineering background and can cover them too if required.]
What is considered to be a good score ? Percentile as well as raw score !!

Good Day !!

[cbreeden]

I found that according the statistics in the end of the practice book two problems N43 (26%) and N47 (28%) caused the difficulties for the majority of test takers. Meanwhile, these are only concern of calculation and don't have any tricks inside. That's why we should solve them quickly and accurately.
Problem 43

Problem 47

Actually, there is a trick for efficiently solving problem 43, which is similar to an exercise from L. Ahlfors complex analysis (I don't have the actually exercise number on hand, but if someone asks, I will look it up and post it):

Since z^5 = 1, we have that 1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9 = 5 + 5z + 5z^2 + 5z^3 + 10z^4 = 5(1 + z + z^2 + z^3 + z^4) + 5z^4.

The trick is that, z(1 + z + z^2 + z^3 + z^4) = z + z^2 + z^3 + z^4 + z^5 = 1 + z + z^2 + z^3 + z^4.
Since z != 1, we conclude that we must have that 1 + z + z^2 + z^3 + z^4 = 0.

So we have 5(1 + z + z^2 + z^3 + z^4) + 5z^4 = 5*0 + 5z^4 = 5e^(8ip/5) = 5e^(3ip/5 + ip) = -5e^(3ip/5). Where p = pie.

I dont know if that's what you had in mind, or if it is common enough to not consider it as a trick, but that's what I had in mind when I first came across the problem. However, by the construction of the problem, it seems apparent that they were looking for something like this.

[owlpride]

In hindsight, it is pretty well-known that 1 + z + z^2 + ... + z^(n-1) = 0 if z^n =1. It follows for example from your argument or from the well-known formula for geometric sums ( =( 1 - z^n)/(1-z)), and it's not too hard to see geometrically on the unit circle either.

I have to admit that I might have not remembered that on the real GRE though. I suspect that I would have tried to estimate the real or imaginary parts of the ugly sum (after reducing powers) and compared them to the answer choices.