Problem Solving

AAPL wrote:Manhattan Prep
Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8

\[{\text{2}}\,{\text{miles}}\,\,\,\,({\text{go}}\,\,{\text{and}}\,\,{\text{back}})\]
\[{V_{{\text{run}}}} = \frac{{10\,\,{\text{miles}}}}{{1\,\,{\text{h}}}}\,\,\,\,\,\,;\,\,\,\,\,{V_{{\text{swim}}}} = \frac{{6\,\,{\text{miles}}}}{{1\,\,{\text{h}}}}\]
\[?\,\,:\,\,\,\,\frac{{{\text{total}}\,\,{\text{miles}}}}{{{\text{total}}\,\,{\text{minutes}}}} = \frac{{4\,\,{\text{miles}}}}{{?{\,_{{\text{temporary}}\,\,{\text{focus}}}}}}\]

Excellent opportunity to use UNITS CONTROL, one of the most powerful tools of our method!

\[\left. \begin{gathered}
{\text{Run}}:\,\,\,2\,{\text{miles}}\,\,\left( {\frac{{1\,\,{\text{h}}}}{{10\,\,{\text{miles}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\left( {\frac{{60\,\,{\text{minutes}}}}{{1\,\,h}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,12\,\,{\text{minutes}} \hfill \\
{\text{Swim}}:\,\,\,2\,{\text{miles}}\,\,\left( {\frac{{1\,\,{\text{h}}}}{{6\,\,{\text{miles}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\left( {\frac{{60\,\,{\text{minutes}}}}{{1\,\,h}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,20\,\,{\text{minutes}} \hfill \\
\end{gathered} \right\}\,\,\,\, \Rightarrow \,\,\,\,?{\,_{{\text{temporary}}\,\,{\text{focus}}}}\,\, = \,\,\,32\,\,{\text{minutes}}\]
\[{\text{? = }}\frac{{\text{4}}}{{32}} = \frac{1}{8}\]

Obs.: arrows indicate licit converters.

This solution follows the notations and rationale taught in the GMATH method.