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\begin{document}

\noindent
\vfil \noindent
\large
\hfil Math 120 A Spring 2011 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil April 21, 2011 \hfil \pp
\hfil Answers \hfil \pp
\normalsize

There were two versions of the exam.

\medskip

Version A - In problem 1, Vera starts from a point 120 feet due north of the water main.

\begin{enumerate}

\item (a) $x=10t, y=120-2t$ (b) Solve the equation $(12t)^2 = (10t)^2+(120-2t)^2$ for t.  The one positive solution is 13.899748 seconds.
\item (a) Let $M(t)$ be the amount of money Helga makes if she works $t$ hours.  Then
\[
M(t)=
\begin{cases}
10t & \mbox{if $0 \le t \le 8$,} \\
80+15(t-8) & \mbox{if $8 \le t \le 12$,} \\
140+20(t-12) & \mbox{if $12\le t$.}
\end{cases}
\]
(b) 11.4285 hours.

\item 8.19288 km

\item The maximum possible area is 625 cm$^2$.

\end{enumerate}

Version B - In problem 1, Vera starts from a point 100 feet due south of the water main.

\begin{enumerate}
\item (a) $x=15t, y=-100+2t$ (b) Solve the equation $(14t)^2=(15t)^2+(-100+2t)^2$ for $t$.  Since this equation
has no real solutions, Vera's feet never get wet.

\item (a) Ket $M(t)$ be the amount of money Helga makes if she works $t$ hours.
Then
\[
M(t)=
\begin{cases}
9t & \mbox{if $0\le t\le 8$,}\\
72+13.5(t-8) & \mbox{if $8 \le t \le 12$,}\\
126+18(t-12) & \mbox{if $t>12$.}
\end{cases}
\]

(b) 10.2857 hours

\item 8.94427 km

\item The maximum possible area is 6400 cm$^2$.

\end{enumerate}


\end{document}


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