The solutions to GR3768

 Posts: 16
 Joined: Tue Jun 25, 2024 6:42 pm
The solutions to GR3768
The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??
Re: The solutions to GR3768
Try here. https://www.mathsub.com/solutions/
I also found this answer on the internet, not sure if its relevant:
We have realvalued 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 realvalued 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} ]
I also found this answer on the internet, not sure if its relevant:
We have realvalued 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 realvalued 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} ]

 Posts: 8
 Joined: Thu Mar 23, 2023 1:48 pm
Re: The solutions to GR3768
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?

 Posts: 16
 Joined: Tue Jun 25, 2024 6:42 pm
Re: The solutions to GR3768
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.copilot wrote: ↑Fri Jun 28, 2024 11:01 amTry here. https://www.mathsub.com/solutions/
I also found this answer on the internet, not sure if its relevant:
We have realvalued 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 realvalued 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} ]

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
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
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

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
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????
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????

 Posts: 2
 Joined: Sat Jun 04, 2016 8:30 pm
Re: The solutions to GR3768
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
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

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
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
Ive been typing up a solutions manual too
there is another guy on youtube with many of the solutions to gr3768

 Posts: 1
 Joined: Sun Sep 08, 2024 11:31 am
Re: The solutions to GR3768
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)juicetrainwreck wrote: ↑Fri Sep 06, 2024 12:33 amFor 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????
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
Hi Wasnt the 3768 an already released old test?

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
No the 3768 is the most recent practice examination

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
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
they are basic cards (front > back) with timers

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
Here are some of the solutions
awindsor_DOT_info_SLASH_gr3768solutions_SLASH_
awindsor_DOT_info_SLASH_gr3768solutions_SLASH_

 Posts: 16
 Joined: Tue Jun 25, 2024 6:42 pm
Re: The solutions to GR3768
Is this a University website?? Can't seem to follow the hyperlink..juicetrainwreck wrote: ↑Fri Sep 20, 2024 6:09 pmHere are some of the solutions
awindsor_DOT_info_SLASH_gr3768solutions_SLASH_

 Posts: 37
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
Replace _DOT_ with .
Re: The solutions to GR3768
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....