If x + y 0, is xy

Question

If  x + y > 0 , is  xy < 0 ?

(1)  x^{2y} < 1

(2)  x + 2y < 0

DavidTutorexamPAL · Answer

From x+y>0, we know that x >- y. This means that either 1) x and y are both positive, 2) x is positive and y is negative but x>|y|, 3) x is negative and y is positive and |x|>y. Which one is it? 1) tells us that -1

militarytocorp · Answer

Hi david From x+y>0, we know that x > y . 1) tells us that -10. But if 00 - sufficient! Answer B. _______________ Can u plz explain statement 1 again ... x+y greater than 0 equals x greater than y..?

ahmadreza.rbn · Answer

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

fskilnik · Answer

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.

DavidTutorexamPAL · Answer

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

Mo2men · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions