Bunuel wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?
(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.
We consider the image attached.
\(? = x + y\)
\(\left. \begin{gathered}
y + z = 11 \hfill \\
y + w = 5 \hfill \\
\end{gathered} \right\}\,\,\,\,\left( * \right)\)
\(\left( 1 \right)\,\,\underline {B = 16} \,\,\,\left\{ \begin{gathered}
\,\,{\text{Take}}\,\,\left( {x,\underline {y,z,w,k} } \right) = \,\,\left( {0,\underline {4,6,1,5} } \right)\,\,\,{\text{viable}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 0 + 4 = 4 \hfill \\
\,\,{\text{Take}}\,\,\left( {x,\underline {y,z,w,k} } \right) = \,\,\left( {1,\underline {4,6,1,5} } \right)\,\,\,{\text{viable}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 1 + 4 = 5 \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,x + y = 2\,\,\,\, \Rightarrow \,\,\,\,? = 2\,\,\,\,\left( {{\text{immediately}}} \right)\,\,\,\,\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
Attachments
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