Quote:
A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20 miles per hour. Five minutes after the passing the cyclist stops, while the hiker continues to walk at the hiker´s rate. How many minutes must the cyclist wait until the hiker catches up?
A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3
\(?\,\,\,:\,\,\,{\text{minutes}}\,\,\,\left( {{\text{last}}\,\,{\text{diagram}}} \right)\)
Let´s use UNITS CONTROL, one of the most powerful tools of our method!
\({\rm{cyclist}}:\,\,\,5\min \,\,\left( {{{20\,\,{\rm{miles}}} \over {60\,\,\min }}\,\matrix{
\nearrow \cr
\nearrow \cr
} } \right)\,\,\, = {5 \over 3}\,\,{\rm{miles}}\)
\({\rm{hiker}}:\,\,\,{5 \over 3}{\rm{miles}}\,\,\left( {{{60\,\,{\rm{min}}} \over {4\,\,{\rm{miles}}}}\,\matrix{
\nearrow \cr
\nearrow \cr
} } \right)\,\,\, = 25\,\,{\rm{minutes}}\)
Obs.: arrows indicate
licit converters.
\(? = 25 - 5\left( * \right) = 20\min\)
\(\left( * \right)\,\,{\rm{used}}\,\,{\rm{while}}\,\,\left( {{\rm{also}}} \right)\,\,{\rm{cyclist}}\,\,{\rm{was}}\,\,{\rm{moving}}!\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
I gt 25 minutes but I amnot able to understand why are we subtracting 5 from it.