I'm very curious as to what document class could produce a document like this. I'd like to use it myself but I don't have access to the TeX source. Has anyone seen something similar (or the same)? And is there some central gallery for TeX document classes?
'Inverse' questions like this are very tricky :) It could have been produced using any number of documentclass However, if I had to bet, my money would go on the exam documentclass which provides many useful features for typesetting exams- the only one I have used in the MWE is the questions environment, but there is a lot more you can do with it (see the documentation for details).
I've used the mdframed package for the framing, but it might have been done using an \fbox
\documentclass[11pt]{exam}
\usepackage{mdframed}
\setlength{\parindent}{0mm}
\begin{document}
\begin{mdframed}
{\bfseries Math 114E}
{\itshape Practice Midterm 2 Solutions \hfill March 18, 2012}
\end{mdframed}
\begin{questions}
\question The equation of the sphere is $(x-3)^2+y^2+(z+4)^2=25$. To
find a place, we need a normal vector and a point on the plane. To find the
normal, we take the gradient of our surface to get
\[
\vec{u}=(2(x-3),2y,2(z+4))=(8,6,0)
\]
So (dividing by $2$) our normal vector is $\vec{u}=(4,3,0)$, and our point
is $(7,3,-4)$, so we are all set.
\end{questions}
\end{document}
The same effect could easily have been created using the article documentclass as well. This does not have a question environment built-in though, so I created one based on the enumerate environment, using the enumitem to do the heavy lifting for me.
\documentclass[11pt]{article}
\usepackage{mdframed}
\usepackage{enumitem}
\newlist{questions}{enumerate}{5}
\setlist[questions]{label*=\arabic*.}
\setlength{\parindent}{0mm}
\begin{document}
\begin{mdframed}
{\bfseries Math 114E}
{\itshape Practice Midterm 2 Solutions \hfill March 18, 2012}
\end{mdframed}
\begin{questions}
\item The equation of the sphere is $(x-3)^2+y^2+(z+4)^2=25$. To
find a place, we need a normal vector and a point on the plane. To find the
normal, we take the gradient of our surface to get
\[
\vec{u}=(2(x-3),2y,2(z+4))=(8,6,0)
\]
So (dividing by $2$) our normal vector is $\vec{u}=(4,3,0)$, and our point
is $(7,3,-4)$, so we are all set.
\end{questions}
\end{document}
In both cases, tweaks to the page geometry could be achieved using the geometry package.
cvmodern. – doncherry Mar 19 '12 at 09:42