If a carton containing a dozen mirrors is dropped, which of the follow

\(12\,\,{\rm{mirrors}}\,\,\,\left\{ \matrix{\\
{\rm{broken}}\,\,{\rm{:}}\,\,b \hfill \cr \\
{\rm{unbroken}}\,\,{\rm{:}}\,\,u \hfill \cr} \right.\)

\(?:\,\,\,b:u\,\,\,\underline {{\rm{impossible}}}\)

The k technique is very useful here!

\(\left( {\rm{A}} \right)\,\,\,\left\{ \matrix{\\
\,b = 2k \hfill \cr \\
u = k \hfill \cr} \right.\,\,\,\left( {k \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,3k = 12\,\,\, \Rightarrow \,\,\,{\rm{viable}}\)

\(\left( {\rm{B}} \right)\,\,\,\left\{ \matrix{\\
\,b = 3k \hfill \cr \\
u = k \hfill \cr} \right.\,\,\,\left( {k \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,4k = 12\,\,\, \Rightarrow \,\,\,{\rm{viable}}\)

\(\left( {\rm{C}} \right)\,\,\,\left\{ \matrix{\\
\,b = 3k \hfill \cr \\
u = 2k \hfill \cr} \right.\,\,\,\left( {k = 3k - 2k \ge 1\,\,\underline {{\mathop{\rm int}} } } \right)\,\,\,\,\,\, \Rightarrow \,\,\,5k = 12\,\,\, \Rightarrow \,\,\,\underline {{\rm{not}}} \,\,{\rm{viable}}\)

The correct answer is therefore (C).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.