lalania1 wrote:

A length of rope is cut into three different lengths. What is the length of the shortest rope?

(1) The combined length of the longest two pieces is 6 feet.

(2) The combined length of the shortest two pieces is 3 feet.

All lengths are presented in feet.

\({\rm{rope}}\left( s \right):\,\,s < i < l\,\,\,\,\left\{ \matrix{

\,? = s\,\,\left( {{\rm{shortest}}} \right) \hfill \cr

\,i\,\,\left( {{\rm{intermediate}}} \right) \hfill \cr

\,l\,\,\left( {{\rm{longest}}} \right) \hfill \cr} \right.\)

Let go straight to (1+2). A

BIFURCATION proves the correct answer is (E):

\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{

\,l + i = 6 \hfill \cr

s + i = 3 \hfill \cr} \right.\,\,\,\,\,\left[ {{\rm{feet}}} \right]\)

\(\left\{ \matrix{

\,{\rm{Take}}\,\,\left( {l,i,s} \right){\rm{ = }}\left( {4,2,1} \right)\,\,\,\, \Rightarrow \,\,? = 1\,\,{\rm{viable}} \hfill \cr

\,{\rm{Take}}\,\,\left( {l,i,s} \right){\rm{ = }}\left( {3.5,2.5,0.5} \right)\,\,\,\, \Rightarrow \,\,? = 0.5\,\,{\rm{viable}} \hfill \cr} \right.\)

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

Regards,

Fabio.

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