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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 120 A, B - Winter 2011 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil January 27, 2011 \hfil \pp
\hfil Answers \hfil \pp

\normalsize

\vfil

There were two versions of the exam in use.

\vfil

Version A - In problem 1, Josephine starts 7 km west of the buoy.

\begin{enumerate}
\item 2.5482 hours
\item (a) $29.1666$
\\
(b) \[
\mbox{area}(x)=
 \left \{ \begin{array}{cc}
 50m & \mbox{if $0 \le m \le 0.5$,} \\
50-\frac{25}{2m}
& \mbox{if $0.5 \le m \le 1$.} 
\end{array}
\right.
\]

\item (a) $\dsps x=3-\frac{6}{\sqrt{90}} t, y = 1+\frac{18}{\sqrt{90}} t$
\\
(b) 
\[
\mbox{distance}(t) = \sqrt{\left( 5- \left(3-\frac{6}{\sqrt{90}} t\right)\right)^2 +
\left(8-\left(1+\frac{18}{\sqrt{90}} t\right)\right)^2}\]

\item (a) $-6x-3h+9$ (b) $x=-5$ is the only solution.

\end{enumerate}

\vfil

Version B - In problem 1, Josephine starts 3 km east of the buoy.

\begin{enumerate}
\item 2.40036 hours
\item (a) 46.2857 
\\
(b) 
\[
\mbox{area}(x)=
 \left \{ \begin{array}{cc}
 72m & \mbox{if $0 \le m \le 0.5$,} \\
72-\frac{18}{m}
& \mbox{if $0.5 \le m \le 1$.} 
\end{array}
\right.
\]
\item (a) $\dsps x=-4+\frac{12}{\sqrt{80}}t,
y = 2+\frac{24}{\sqrt{80}}t$
\\
(b)
\[
\mbox{distance}(t) = \sqrt{\left(1-\left(-4+\frac{12}{\sqrt{80}}t\right)\right)^2
+ \left(1-\left(2+\frac{24}{\sqrt{80}}t\right)\right)^2}
\]

\item (a) $-8x-8h+20$ (b) $x=-1$ is the only solution.


\end{enumerate}


\end{document}


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