DSGB wrote:
How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
\(400\,{\rm{km}}\,\,\,\,\,\left\{ \matrix{\\
\,\,\left( {{\rm{real}}\,\,{\rm{speed}}\,,\,\,{\rm{real}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_R}\,\,,\,\,{T_R}} \right) \hfill \cr \\
\,\,\left( {{\rm{hypothetical}}\,\,{\rm{speed}}\,,\,\,{\rm{hypothetical}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_H}\,\,,\,\,{T_H}} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\,\,\left[ {\,{{{\rm{km}}} \over {\rm{h}}}\,} \right]\,\,\,,\,\,\,\,\left[ {\,{\rm{h}}\,} \right]\,\,\,} \right)\)
\(? = {T_R}\,\,\,\left[ {\rm{h}} \right]\)
\(\left( 1 \right)\,\,\left\{ \matrix{\\
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,0.5\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3\, \hfill \cr \\
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,1\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3.5\,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,{V_H} = {V_R} + 20\,\,\,\left( * \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{T_R} - {T_H} = 1\,\,\,\,\,\,\left[ {\rm{h}} \right]\)
Now it´s time for
UNITS CONTROL, one of the most powerful tools of our method:
\({{{\rm{km}}} \over {\,\,\,{{{\rm{km}}} \over {\rm{h}}}\,\,}} = {\rm{h}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{\\
{T_R} = {{400} \over {{V_R}}} \hfill \cr \\
{T_H}\,\mathop = \limits^{\left( * \right)} \,\,{{400} \over {{V_R} + 20}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{{400} \over {{V_R}}} - {{400} \over {{V_R} + 20}} = 1\)
\({{400\left( {{V_R} + 20} \right)} \over {{V_R}\,\,\left( {{V_R} + 20} \right)}} - {{400\,\,{V_R}} \over {\left( {{V_R} + 20} \right)\,\,{V_R}}} = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,400 \cdot 20 = {V_R}\,\,\left( {{V_R} + 20} \right)\)
\({V_R}^2 + 20{V_R} - 400 \cdot 20 = 0\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{of}}\,\,{\rm{roots}}}^{{\rm{product}}} \,\,\,\,\,\,\,\left( {{c \over a} = } \right)\,\,\,{{ - 400 \cdot 20} \over 1} < 0\,\,\,\, \Rightarrow \,\,\,\,\,{V_R}\, > 0\,\,\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)
\(\left( {\,\,\,\left. \matrix{\\
{V_R}\,\,{\rm{unique}}\,\, \hfill \cr \\
{\rm{400}}\,{\rm{km}} \hfill \cr} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,{T_R}\,\,\,{\rm{unique}}\,\,\,} \right)\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)