Is x + y > xy ?

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by fskilnik@GMATH » Tue Oct 02, 2018 11:23 am

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subh2273 wrote:Is x + y > xy ?

Statement (1):
x > 0 > y
Statement (2):
|y| = x
VERY nice problem, subh2273. Congrats!
$$x + y\,\,\mathop > \limits^? \,\,xy$$
$$\left( 1 \right)\,\,x > 0 > y\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{1 \over 2}, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,x = \left| y \right|\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,x = \left| y \right|\mathop = \limits^{y\, < \,0} - y$$
$$\left. \matrix{
{\rm{FOCU}}{{\rm{S}}_{\,{\rm{LHS}}}}\,\,\,:\,\,\,\,x + y = 0 \hfill \cr
{\rm{FOCU}}{{\rm{S}}_{\,{\rm{RHS}}}}\,\,\,:\,\,\,\,xy = - {y^2}\,\,\mathop < \limits^{y\,\, \ne \,0} \,\,0\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

The correct answer is therefore [spoiler]__(C)______[/spoiler] .


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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Jay@ManhattanReview » Tue Oct 02, 2018 9:11 pm

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subh2273 wrote:Is x + y > xy ?

Statement (1): x > 0 > y
Statement (2): |y| = x
Question: Is x + y > xy?

Let's take each statement one by one.

Statement (1): x > 0 > y

Case 1: Say x = 2 and y = - 1

x + y > xy => 2 - 1 ? 2*-1 => 1 > -2. The answer is Yes.

Case 2: Say x = 1/2 and y = -1

x + y > xy => 1/2 - 1 ? 1/2*-1 => -1/2 = -1/2. The answer is No.

No unique answer. Insufficient.

Statement (2): |y| = x

Case 1: x = 1 and y = -1

x + y > xy => 1 - 1 ? 1*-1 => 0 > -1. The answer is Yes.

Case 2: Say x = 0 and y = 0

x + y > xy => 0 - 0 ? 0*0 => 0 = 0. The answer is No.

No unique answer. Insufficient.

(1) and (2) together

From both the statements, we have y = - x

Thus, x + y > xy => x - x ? x*-x => 0 > -x^2. The answer is Yes. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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