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

For Robert to cover minimum distance, Tim needs to cover maximum distance.
But maximum time Tim can run is 6 hours, therefore he can cover maximum distance of = 4*6 = 24 miles.
Hence, the rest is the minimum distance that needs to be covered by Robert = 50-24 = 26 miles.

Maximize Tim's number of miles to minimize Robert's number if miles.

(One part of this question might seem tricky, but that part is irrelevant. The number of miles has to be a whole number. The time each person takes to run those miles does not have to be a whole number.)

T's rate: \((\frac{1mile}{15mins}*\frac{4}{4})=\frac{4mi}{60mins}=\frac{4mi}{1hr}\)

R's rate: \((\frac{1mile}{12mins}*\frac{5}{5})=\frac{5mi}{60mins}=\frac{5mi}{1hr}\)

Both runners must run at least 4 hours and at most 6 hours.

Maximize Tim's # of miles by making him run the maximum number of 6 hours.
Tim can run (6 hours * 4 mph) = 24 miles

Tim is done. Robert must finish the 50-mile total, so Robert runs (50 - 24) = 26 miles

(And while TIM'S time is a whole number, Robert's time is not. Does not matter. Only # of miles run must be a whole number.)

Answer E

Answer

Please make a note of rules very carefully. Atleast 4 hours each has to run , and no once can run for more than 6 hours.

How calculate the speed of Tim and Robert in miles per hour
Tim Speed = 60/15 = 4 miles /hour
robert speed = 60/12 = 5 miles/hour

Now , both have to run for at least 4 hours, , that implies in four hours , distance travelled by both individually will be,
Tim = 16 miles
Robert = 20 miles.

In Question stem it is mentioned , together they traveled 50 miles. 36 miles have already been covered (16+20 ), total of 14 miles left to be covered.

The Q stem asks us to find the minimum miles Robert must run, for that Tim must run for the complete 6 hours.
Therefore Tim distance would be 4miles/hour*6hours = 24 miles

Remaining distance = 50-24 = 26 miles, is the minimum distance traveled by Robert .

Thus E is the correct ans.

Let T and R be the number of minutes Tim and Robert run, respectively. Hence:

DATA:

\(4 \cdot 60\,\,\, \le \,\,\,\,\,T,R\,\,\,\, \le \,\,\,6 \cdot 60\,\,\,\,\,\,\,\left( {\rm{I}} \right)\)

\(\left. \matrix{\\
{\mathop{\rm int}} \,\,\, = \,\,\,T\,\,\min \,\,\,\left( {{{1\,\,{\rm{mile}}} \over {15\,\,\min }}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,T\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,15\,\,\,\,\, \hfill \cr \\
{\mathop{\rm int}} \,\,\,\mathop = \limits^{\left( * \right)} \,\,\,R\,\,\min \,\,\,\left( {{{1\,\,{\rm{mile}}} \over {12\,\,\min }}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,R\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,12 \hfill \cr} \right\}\,\,\,\,\,\,\,\left( {{\rm{II}}} \right)\)

\(\boxed4\,T\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{3 \cdot 5 \cdot \boxed4\,\,\min }}} \right)\,\,\, + \,\,\boxed5\,R\,\,\min \,\,\,\left( {\frac{{1\,\,{\text{mile}}}}{{3 \cdot 4 \cdot \boxed5\,\,\min }}} \right) = \frac{{50 \cdot \boxed{3 \cdot 4 \cdot 5}}}{{\boxed{3 \cdot 4 \cdot 5}}}\,\,{\text{miles}}\)

\(4T + 5R = 50 \cdot 3 \cdot 4 \cdot 5\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,4T = 5\left( {3 \cdot 4 \cdot 5 \cdot 10 - R} \right)\,\,\,\,\mathop \leqslant \limits^{\left( {\text{I}} \right)} \,\,\,\,4 \cdot 6 \cdot 60\)

\(R\,\,\, \geqslant \,\,\,3 \cdot 4 \cdot 5 \cdot 10 - 4 \cdot 6 \cdot 12 = 4 \cdot 6 \cdot \left( {25 - 12} \right) = 4 \cdot 6 \cdot 13\,\,\,\,\,\,\,\left( {{\text{III}}} \right)\)


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.

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.

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

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