The solutions to GR3768

[Chessplayer]

The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??

[copilot]

Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

[boredmathguy]

There is no solution manual as far as I know. However, if there are specific questions you're having trouble with, maybe you can post them here and people can explain their solutions?

[Chessplayer]

Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

Actually he has all previous GRE math solutions, except for the newly released ones .Also he is not responding to emails..This newly released sample test seems like it is a more realistic review of undergrad math , although I am not sure the actual test I will be taking will resemble anything like this.

[juicetrainwreck]

I'm sure the test will be similar to 3768.

If you study all the previous exams there are some repeat/similar questions. For example there's always a question about dim(u+v) = dim(u) + dim(v) - dim(u intersect v)

there's usually a problem about isomorphic graphs

Etc

[juicetrainwreck]

For number 21, is this valid reasoning?

Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.


p.s. how to get LaTeX to render here????

[brianstonelake]

I'm making video solutions. It'll likely take a couple months, but I'll be posting them relatively regularly on youtube.

I made solutions for the previous test here:
https://youtube.com/playlist?list=PL81I ... X9KLdVwXUx

Hopefully I've learned a little and these will be better. Good luck with the studies!

Here's the playlist for 3768. I'll add videos as I go. Depending on how busy I am with work, it may take a couple months to do the entire exam.
https://www.youtube.com/playlist?list=P ... 9BWnfPxbFu

[juicetrainwreck]

Thanks Ill check it out!

Ive been typing up a solutions manual too

there is another guy on youtube with many of the solutions to gr3768

[estibalizmc]

For number 21, is this valid reasoning?

Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.


p.s. how to get LaTeX to render here????

I'm not sure if that's a valid reason or not. But in case it helps what I did was multiplying by 2 [x+7y \equiv 1 (mod 13)] getting 2x+14y \equiv 2 (mod 13). Since we're in mod 13 the 14y is the same as y, so you get 2x+y \equiv 2 (mod 13). And then sum this to the other equation and you'll get 5x +3y \equiv 7 (mod 13)

About the solutions I wish someone share them too I have my exam at the end of September and I'm more and more anxious So far, the only ones that I've found are the ones from the JoshMathWizz YT chanel but only a few problems are solved...

[smuth]

Hi Wasnt the 3768 an already released old test?

[juicetrainwreck]

No the 3768 is the most recent practice examination

[juicetrainwreck]

I made anki decks for many of the exams, with the solutions provided by IACOLEY and RAMBO. msg if youd like a copy

they are basic cards (front -> back) with timers

[juicetrainwreck]

Here are some of the solutions

awindsor_DOT_info_SLASH_gr3768-solutions_SLASH_

[Chessplayer]

Here are some of the solutions

awindsor_DOT_info_SLASH_gr3768-solutions_SLASH_

Is this a University website?? Can't seem to follow the hyperlink..

[juicetrainwreck]

Replace _DOT_ with .

[jester]

Can anyone explain how number 61 is E? As far as I am aware, the product of the n nth roots of unity is (-1)^(n+1). So for n = 10, the product should be (-1)^11 = -1, giving choice D....