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Sguha0305
Quote:
If you're not convinced, you might TEST SOME VALUES.
Case a: x = 1, y = 2 and z = 4. In this case, the length of the smallest piece is 1
Case b: x = 2, y = 1 and z = 5. In this case, the length of the smallest piece is 2
So, the combined statements are NOT SUFFICIENT


So in a way you are saying that the smallest length is 1. Also in your assumption X is the smallest then the second case (Case b) is not even valid.
I am sure, I am missing something.
Will be great if you could explain the fallacy here.
TIA

My bad!
I forgot that I said x is the length of the shortest piece.
I've edited my response accordingly.

Cheers and thanks,
Brent
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Sguha0305
Quote:
If you're not convinced, you might TEST SOME VALUES.
Case a: x = 1, y = 2 and z = 4. In this case, the length of the smallest piece is 1
Case b: x = 2, y = 1 and z = 5. In this case, the length of the smallest piece is 2
So, the combined statements are NOT SUFFICIENT


So in a way you are saying that the smallest length is 1. Also in your assumption X is the smallest then the second case (Case b) is not even valid.
I am sure, I am missing something.
Will be great if you could explain the fallacy here.
TIA
I believe the catch in this is the decimal values.
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lalania1
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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