AAPL wrote:Veritas Prep
Tim and Robert have entered a race, the rules of which stipulate that each runner must run for at least 4 hours and no runner can run for more than 6 hours. Together, they must run a total of 50 miles. If it takes Tim 15 minutes to run a mile and Robert 12 minutes to run a mile, what is the minimum number of miles Robert must run if both Tim and Robert must individually run a whole number of miles?
A. 18
B. 20
C. 22
D. 24
E. 26
Let T and R be the number of
minutes Tim and Robert run, respectively. Hence:
DATA:
$$4 \cdot 60\,\,\, \leqslant \,\,\,\,\,T,R\,\,\,\, \leqslant \,\,\,6 \cdot 60\,\,\,\,\,\,\,\left( {\text{I}} \right)$$
\[\left. \begin{gathered}
\operatorname{int} \,\,\, = \,\,\,T\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{15\,\,\min }}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,T\,\,{\text{divisible}}\,\,{\text{by}}\,\,15\,\,\,\,\, \hfill \\
\operatorname{int} \,\,\,\mathop = \limits^{\left( * \right)} \,\,\,R\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{12\,\,\min }}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,R\,\,{\text{divisible}}\,\,{\text{by}}\,\,12 \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\left( {{\text{II}}} \right)\]
\[\boxed4\,T\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{3 \cdot 5 \cdot \boxed4\,\,\min }}} \right)\,\,\, + \,\,\boxed5\,R\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{3 \cdot 4 \cdot \boxed5\,\,\min }}} \right) = \frac{{50 \cdot \boxed{3 \cdot 4 \cdot 5}}}{{\boxed{3 \cdot 4 \cdot 5}}}\,\,{\text{miles}}\]
\[4T + 5R = 50 \cdot 3 \cdot 4 \cdot 5\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,4T = 5\left( {3 \cdot 4 \cdot 5 \cdot 10 - R} \right)\,\,\,\,\mathop \leqslant \limits^{\left( {\text{I}} \right)} \,\,\,\,4 \cdot 6 \cdot 60\]
\[R\,\,\, \geqslant \,\,\,3 \cdot 4 \cdot 5 \cdot 10 - 4 \cdot 6 \cdot 12 = 4 \cdot 6 \cdot \left( {25 - 12} \right) = 4 \cdot 6 \cdot 13\,\,\,\,\,\,\,\left( {{\text{III}}} \right)\]
FOCUS:
\[?\,\,\,:\,\,\,\min \,\,\,\frac{R}{{12}}\,\,\,\,\,\left( * \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\min \,\,\,R\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,? = {{\left( {\frac{R}{{12}}} \right)}_{\,\min }}\,} \right]\,\,\,\,\,\]
DATA-FOCUS CONNECTION:
$$\left. \matrix{
\left( {\rm{I}} \right)\,\,\, \Rightarrow \,\,\,R \ge 4 \cdot 60 = 4 \cdot 3 \cdot 20\,\,\, \hfill \cr
\left( {{\rm{II}}} \right)\,\,\, \Rightarrow \,\,\,R = 3 \cdot 4 \cdot {\mathop{\rm int}} \hfill \cr
\left( {{\rm{III}}} \right)\,\,\, \Rightarrow \,\,\,R \ge 2 \cdot 3 \cdot 4 \cdot 13 \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {\mathop{\rm int}} = 26$$
\[\left( \begin{gathered}
{R_{\,\min }} = \underline {3 \cdot 4 \cdot 26} \,\,\,\, \Rightarrow \,\,\,4T = 5\left( {3 \cdot 4 \cdot 5 \cdot 10 - \underline {3 \cdot 4 \cdot 26} } \right) = \underleftrightarrow {60\left( {50 - 26} \right)} = 60 \cdot 24 \hfill \\
T = 6 \cdot 60\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\left( {\text{I}} \right)\,\,\,{\text{ok}} \hfill \\
\left( {{\text{II}}} \right)\,\,\,{\text{ok}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{R_{\,\min }} = 3 \cdot 4 \cdot 26\,\,\,{\text{viable}}\,\,\,\, \hfill \\
\end{gathered} \right)\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
