swerve wrote:A man walking at a constant rate of 4 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 20 miles per hour. The woman stops to wait for the man 5 minutes after passing him, while the man continues to walk at his constant rate. How many minutes must the woman wait until the man catches up?
A. 16 mins
B. 20 mins
C. 24 mins
D. 25 mins
E. 28 mins
Excellent opportunity for relative velocity (speed) and UNITS CONTROL :
$${{16\,\,{\rm{miles}}} \over {1\,\,{\rm{hour}}}}\,\,\, \cdot \,\,\,\left( {{{1\,\,{\rm{hour}}} \over {\,60\,\,{\rm{minutes}}\,}}} \right)\,\,\, \cdot \,\,\,5\,\,{\rm{minutes}}\,\,\,\,{\rm{ = }}\,\,\,\,{4 \over 3}\,\,{\rm{miles}}\,\,\,\,\,\,\,\left( {{\rm{woman}}\,{\rm{ - }}\,{\rm{man}}\,\,{\rm{distance}}\,{\rm{,}}\,\,{\rm{both}}\,\,{\rm{walking}}\,\,{\rm{during}}\,\,{\rm{5}}\,\,{\rm{minutes}}} \right)$$
$${\rm{only}}\,\,{\rm{man}}\,\,{\rm{walking}}\,\,\,\,:\,\,\,\,\,?\,\,\, = \,\,\,\,{4 \over 3}\,\,{\rm{miles}}\,\,\, \cdot \,\,\,\left( {{{1\,\,{\rm{hour}}} \over {\,4\,\,{\rm{miles}}\,}}} \right)\,\,\,\,\, = \,\,\,\,{1 \over 3}\,\,\,{\rm{hour}}\,\,\, = \,\,20\min \,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{B}} \right)$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
