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Tim and Robert have entered a race, the rules of which

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Tim and Robert have entered a race, the rules of which

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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

OA E.

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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.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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[quote="fskilnik@GMATH"]
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
Alternate approach: (Risky but many-times-"awarded", as in this case!)

Let (again) T and R be the number of minutes Tim and Robert run, respectively.

To minimize Roberts mileage (our FOCUS), we must minimize R, therefore (by duality) we must maximize T.

The maximum POTENTIAL value of T is 6*60 (minutes), hence let´s check whether this number - and R obtained from it - are viable:
$${\rm{Tim}}\,\,\left( {6\,\,{\rm{h}}} \right)\,\,:\,\,\,\,6 \cdot 60\,\,\min \,\,\,\left( {{{1\,\,\,{\rm{mile}}} \over {15\,\,\,\min }}\,\matrix{
\nearrow \cr
\nearrow \cr

} } \right)\,\,\, = \,\,24\,\,{\rm{miles}}$$
$${?_{{\rm{potencial}}\,\left( {{\rm{Robert}}} \right)}}\,\,\,\mathop \Rightarrow \limits^{\sum {\, = \,50\,\,{\rm{miles}}} } \,\,\,\,26\,\,{\rm{miles}}\,\,\,\left( {{{12\,\,\,\min } \over {1\,\,\,{\rm{mile}}}}\,\matrix{
\nearrow \cr
\nearrow \cr

} } \right)\,\,\, = {2^3} \cdot 3 \cdot 13\,\,\, > \,\,\,\underbrace {{2^3} \cdot 3 \cdot 10}_{4\,\, \cdot \,60}$$

They are! The answer is therefore 26 (miles).

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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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
Since it takes Tim 15 minutes to run a mile, he runs 4 miles per hour. Similarly, it takes Robert 12 minutes to run a mile, he runs 5 miles per hour.

Since each runner must run for at least 4 hours and no runner can run for more than 6 hours, Tim runs at least 4 x 4 = 16 miles and at most 6 x 4 = 24 miles. Similarly, Robert runs at least 4 x 5 = 20 miles and at most 6 x 5 = 30 miles.

Since we want to determine the minimum number of miles Robert runs, we can assume Tim runs the the greatest number of miles he possibly can, which is 24. So Robert runs at least 50 - 24 = 26 miles.

Answer: E

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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