fskilnik wrote:
GMATH practice exercise (Quant Class 14)
\(A = {{\left( {x + y + 1} \right)} \over {x + y}} \cdot {{\left( {x + z + 2} \right)} \over {x + z}} \cdot {{\left( {x + w + 3} \right)} \over {x + w}} = \left( {1 + {1 \over {x + y}}} \right)\left( {1 + {2 \over {x + z}}} \right)\left( {1 + {3 \over {x + w}}} \right)\,\,\,\mathop > \limits^? \,\,\,8\)
\(\left( 1 \right)\,\,\,x,y,z,w\,\, \ge 1\,\,\,\,{\rm{ints}}\,\,\,\, \Rightarrow \,\,\,\,\,2 \le x + y\,\,,x + z\,\,,\,\,x + w\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\,{1 \over {x + y}} \le {1 \over 2} \hfill \cr
\,\,{1 \over {x + z}} \le {1 \over 2}\,\,\,\,\, \Rightarrow \,\,{2 \over {x + z}}\,\, \le \,\,2 \cdot {1 \over 2}\,\,\,\, \hfill \cr
\,\,{1 \over {x + w}} \le {1 \over 2}\,\,\,\,\, \Rightarrow \,\,{3 \over {x + w}}\,\, \le \,\,3 \cdot {1 \over 2} \hfill \cr} \right.\,\,\,\,\,\,\left( * \right)\)
\(A\,\,\,\mathop < \limits^{\left( * \right)} \,\,\,\left( {1 + {1 \over 2}} \right)\left( {1 + 2 \cdot {1 \over 2}} \right)\left( {1 + 3 \cdot {1 \over 2}} \right) = {3 \over 2} \cdot 2 \cdot {5 \over 2} = {{15} \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)
\(\left( 2 \right)\,\,\,x < y < z < w\,\,\,\,::\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y;z;w} \right) = \left( {1;2;3;4} \right)\,\,\,\,\mathop \Rightarrow \limits^{{\rm{by}}\,\,\left( 1 \right)} \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y;z;w} \right) = \left( {{1 \over 4};{1 \over 3};{1 \over 2};1} \right)\,\,\,\, \Rightarrow \,\,\,A = \left( {1 + {{12} \over 7}} \right)\left( {1 + {8 \over 3}} \right)\left( {1 + {{12} \over 5}} \right) > 2 \cdot 3 \cdot 3\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\)
The correct answer is (A).
We follow the notations and rationale taught in the
GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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