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\noindent
\vfil \noindent
\large
\hfil Math 120 A, B - Winter 2010 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil January 28, 2010 \hfil \pp
\hfil Answers \hfil \pp

\normalsize

\vfil

There were two versions of the exam. 

\vfil

In version A, in problem 1, Aiman will reach the point (5,2) after walking for 2 hours.

\begin{enumerate}

\item (a) $x=2.5t, y=t$ (b) $x=-3+3.6t, y=-4+2t$ (c) 1.38717 hours
\item 2.5298 km
\item \[
f(x) = \left\{ \begin{array}{ll}
4 + \frac{5}{6}x & \mbox{if $0 \le x \le 16$} \\
9+\sqrt{25-(x-21)^2} & \mbox{if $16 \le x \le 21$} \\
14-(x-21) & \mbox{if $21 \le x \le 29$} 
\end{array}
\right.
\]

\item (a) The largest value is 14.08333, and the smallest value is -10.
(b) The expression simplifies to $-12x-10$.

\end{enumerate}

\vfil

In version B, in problem 1, Aiman will reach the point (4,6) after walking for 4 hours.

\begin{enumerate}
\item (a) $x=t, y=1.5t$ (b) $x=-2+2t, y=5-0.8t$ (c) 0.55262 hours
\item 4.64238 km
\item \[
f(x) = \left\{ \begin{array}{ll}
\frac{3}{8}x+2 & \mbox{ if $0 \le x \le 8$} \\
5+\sqrt{9-(x-11)^2} & \mbox{ if $8 \le x \le 11$} \\
8-(x-11) & \mbox{ if $11 \le x \le 16$} 
\end{array}
\right.
\]

\item (a) The largest value is 15.5 and the smallest value is -69. 
(b) The expression simplifies to $-8x-12$.
\end{enumerate}


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