If x + y > 0, is xy < 0? (1) x^2y < 1 (2) x + 2y < 0

If x+y>0, is xy<0?
We should see whether x & y have different signs or not

(1) x^2y<1

From this statement we can understand that -1<x<1 since the even power resulted in a number less than 1
Till now x can be both positive and negative.
if x is for example -0.5, y can only a positive integer (if y=1 -> x^2y =1/4<1 , if y=-1 -> x^2y=4 NOT SATISFYING THE PREMISE)
Take +0.5 for x. y should be a positive integer (if y=1 -> x^2y= 1/4 , if y=-1 -> x^2y=4 NOT SATISFYING THE PREMISE)
so we have 2 options for xy, Negative and Positive NOT SUFFICIENT
(2) x+2y<0
Try to draw lines x+2y=0 and x+y=0 in xy-coordinate plane . When applying the > < signs to the drawing, the only mutual area appears in quadrant II and IV where always xy<0 SUFFICIENT

\(x + y > 0\,\,\,\left( * \right)\)

\(xy\,\,\mathop < \limits^? \,\,0\)

\(\left( 1 \right)\,\,{x^{2y}} < \,\,\,1\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 2};{1 \over 2}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( { - {1 \over 3};{1 \over 2}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,x + 2y = \left( {x + y} \right) + y < 0\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,y < 0\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x > 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


The correct answer is therefore (B).


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

Regards,
Fabio.

Thanks for the question! I made a mistake, now fixed.
Take a look at the explanation now!

I'm little bit confused here.

If x = 2 & y =-1, does not it satisfy the conditions?

2+ (-1) > 0 & 2^-2 = (1/2)^2 = 1/4 < 1..........Answer to question is yes.............So x does not need to be between 1 & -1

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