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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