BTGmoderatorLU wrote:Manhattan Prep
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?
A. R-4
B. R/(R+4)
C. R/(R-8)
D. 8/(R-8)
E. 2R - 4
Let´s REALLY solve the problem, using the WINNING TRIAD and UNITS CONTROL, one of the most powerful tools of our method!
DATA:
$${\rm{Alex}}\,\,\left( A \right)\,\,:\,\,\,{{4\,\,{\rm{miles}}} \over {1\,\,\,{\rm{h}}}}\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,T\,\,\,\,{\rm{h}}$$
$${\rm{Brenda}}\,\,\left( B \right)\,\,:\,\,\,{{R\,\,{\rm{miles}}} \over {1\,\,\,{\rm{h}}}}\,\,\,\,\,\,\,\left( {R > 8} \right)\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\left( {T - 1} \right)\,\,{\rm{h}}$$
FOCUS:
$${\rm{?}}\,\,\,{\rm{:}}\,\,\,\,T = f\left( R \right)$$
DATA-FOCUS CONNECTION:
$$\underbrace {\left( {T - 1} \right)\,\,{\rm{h}}\,\,\,\left( {{{R\,\,\,{\rm{miles}}} \over {1\,\,\,{\rm{h}}}}\matrix{
\nearrow \cr
\nearrow \cr
} } \right)}_{{\rm{distance}}\,\,{\rm{travelled}}\,\,{\rm{by}}\,\,B}\,\,\,\,\,\mathop = \limits^{{\rm{question}}\,\,{\rm{stem}}} \,\,\,\,\,2\,\, \cdot \,\,\,\underbrace {T\,\,{\rm{h}}\,\,\,\left( {{{4\,\,\,{\rm{miles}}} \over {1\,\,\,{\rm{h}}}}\matrix{
\nearrow \cr
\nearrow \cr
} } \right)}_{{\rm{distance}}\,\,{\rm{travelled}}\,\,{\rm{by}}\,\,A}\,\,$$
Obs.: arrows indicate
licit converters.
$$\left( {T - 1} \right)R = 8T\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,:\,\,\,T\left( {R - 8} \right) = R\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{R\, \ne \,8} \,\,\,\,\,\,\,\,?\,\,\,:\,\,T = {R \over {R - 8}}$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.