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{\Large{
\hfil Math 120 A, B Winter 2013 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil January 31, 2013 \hfil \pp
\hfil Answers \hfil \pp
}}

There were two versions of the exam.  

Version A - In problem 1, Shin walks 2 km/hr.

\begin{enumerate}

\item (a) $\dsps 3+\sqrt{75}$ hours after starting.
(b) $\dsps \frac{34}{3}$ hours after starting.

\item (a) 4.736113 km (b) $\dsps \frac{168}{65}$ km west of the center.

\item \[
\mbox{area} = \left\{ \begin{array}{cc}
\frac{1}{2}x(18-x) & \mbox{if $0 \le x \le 4$} \\
28 + \frac{1}{2}(x-4)(8+\frac{1}{2}x) & \mbox{ if $4 \le x \le 10$} 
\end{array}
\right.
\]

\item (a) $12x+4$ (b) There are no solutions.

\end{enumerate}

Version B - In problem 1, Shin walks 2.5 km/hr.

\begin{enumerate}
\item (a) $\dsps 1+\frac{\sqrt{48}}{3}$ hours after starting.
(b) $\dsps \frac{25}{9}$ hours after starting.

\item (a) 6.6564023 km (b) $\dsps \frac{40}{13}$ km west of the center.

\item \[
\mbox{area} = \left\{ \begin{array}{cc}
\frac{1}{2}x(18-x) & \mbox{if $0 \le x \le 4$} \\
28 + \frac{1}{2}(x-4)(10+\frac{1}{3}(x-4)) & \mbox{ if $4 \le x \le 10$} 
\end{array}
\right.
\]

\item (a) $8x-2$ (b) $\frac{9}{4}$ is the only solution.


\end{enumerate}

\end{document}


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