Page 1 of 1

The solutions to GR3768

Posted: Tue Jun 25, 2024 6:49 pm
by Chessplayer
The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??

Re: The solutions to GR3768

Posted: Fri Jun 28, 2024 11:01 am
by 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} ]

Re: The solutions to GR3768

Posted: Sun Jun 30, 2024 4:29 pm
by 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?

Re: The solutions to GR3768

Posted: Thu Aug 08, 2024 1:21 pm
by Chessplayer
copilot wrote:
Fri Jun 28, 2024 11:01 am
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.

Re: The solutions to GR3768

Posted: Mon Sep 02, 2024 8:56 pm
by 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

Re: The solutions to GR3768

Posted: Fri Sep 06, 2024 12:33 am
by 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????

Re: The solutions to GR3768

Posted: Tue Sep 10, 2024 5:58 pm
by 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

Re: The solutions to GR3768

Posted: Wed Sep 11, 2024 12:47 am
by 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

Re: The solutions to GR3768

Posted: Wed Sep 11, 2024 6:24 pm
by estibalizmc
juicetrainwreck wrote:
Fri Sep 06, 2024 12:33 am
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...

Re: The solutions to GR3768

Posted: Fri Sep 13, 2024 9:47 pm
by smuth
Hi Wasnt the 3768 an already released old test?

Re: The solutions to GR3768

Posted: Fri Sep 13, 2024 10:31 pm
by juicetrainwreck
No the 3768 is the most recent practice examination

Re: The solutions to GR3768

Posted: Mon Sep 16, 2024 1:13 am
by 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