BTGmoderatorLU wrote:Source: Princeton Review
If Jim drives k miles in 50 minutes, how many minutes will it take him to drive 10 miles, at the same rate?
A. 500/k
B. k/500
C. 60k
D. 10k
E. 50/k
\[k\,\,{\text{miles}}\,\,\,\, \leftrightarrow \,\,\,\,{\text{50}}\,\,{\text{minutes}}\,\,\,\,\,\,\,{\text{;}}\,\,\,\,\,\,\,\,{\text{10}}\,\,{\text{miles}}\,\,\,\, \leftrightarrow \,\,\,\,{\text{?}}\,\,{\text{minutes}}\]
This is a (direct) proportional problem, therefore a 10-year-child would probably do it (correctly) like that:
\[\frac{?}{{50}} = \frac{{10}}{k}\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \frac{{500}}{k}\]
On the other hand, this is a trivial scenario for UNITS CONTROL, one of the most powerful tools of our method:
\[?\,\,\, = \,\,\,10\,\,{\text{miles}}\,\,\,\left( {\frac{{50\,\,{\text{minutes}}}}{{k\,\,{\text{miles}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{500}}{k}\,\,\,\,\left[ {{\text{minutes}}} \right]\]
Obs.: arrows indicate
licit converter.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
