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In which quadrant of the coordinate plane does the point (x,y) lie?

(1) |xy| + x|y| + |x|y + xy > 0 (2) -x < -y < |y|

1. Given condition is true only if both X and Y are positive, so (X,Y) is in I quadrant. SUFFICIENT 2. Given condition is true only if both X and Y are positive, so (X,Y) is in I quadrant. SUFFICIENT

If this is 600-700 level question, i cant imagine 700-800level qs. Bunuel, We want more problems on co-ordinate DS and inEquality DS. Is it possible to post it in sets and then you set the time for each set...say 10qs 15mins...something like that?

Re: In which quadrant of the coordinate plane does the point [#permalink]

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18 Jun 2012, 05:58

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Bunuel wrote:

In which quadrant of the coordinate plane does the point (x,y) lie?

(1) |xy| + x|y| + |x|y + xy > 0 (2) -x < -y < |y|

A remark regarding statement (1):

Since |xy|=|x||y|, the given expression can be written as (|x|+x)(|y|+y)>0. If either x or y is non-positive, the given expression equals 0. Otherwise, it is positive. So, necessarily, both x and y must be positive, and Statement (1) is sufficient. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In which quadrant of the coordinate plane does the point [#permalink]

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27 Mar 2013, 02:37

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In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0 Case quadrant I (x,y)=(+,+) \(|xy| + x|y| + |x|y + xy\) \(xy +xy + xy + xy >0\) The first term is positive, the second the third and the fourt also. The sum of 4 positive integers is >0. so quadrant I is possible Case quadrant II (x,y)=(-,+) The first term is positive(as always will be), the second is negative, the third is positive, the fourth is negative \(|(-x)y| + (-x)|y| + |(-x)|y + (-x)y\) \(xy -xy + xy - xy =0\) and not >0 so quadrant II is not possible Case quadrant III (x,y)=(+,-) The first term is positive(as always will be), the second is positive, the third is negative, the fourth is negative \(|x(-y)| + x|(-y)| + |x|(-y) + x(-y)\) \(xy + xy - xy - xy =0\) not >0 so quadrant III is not possible Case quadrant IV (x,y)=(-,-) The first term is positive(as always will be), the second is negative, the third is negative, the fourth is positive \(|(-x)(-y)| + (-x)|(-y)| + |(-x)|(-y) + (-x)(-y)\) \(xy -xy - xy + xy =0\) and not >0 so quadrant IV is not possible

SUFFICIENT

(2) -x < -y < |y|

\(|y|>-y\) case y>0 \(y>-y\) \(y>0\) case y<0 \(-y>-y\) So \(y>0\) must be the case here We know that -y >-x and that y>0, we can sum these elements \(-y+y>-x+0\) \(0>-x\) \(x>0\) and given that x>0 and that y>0 the point is in the first quadrant

SUFFICIENT _________________

It is beyond a doubt that all our knowledge that begins with experience.

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27 Mar 2013, 04:47

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mun23 wrote:

In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0 (2) -x < -y < |y|

Need help............

From F.S 1, for x,y>0, we can see that the sum will always be positive. For cases where x and y have opposite signs, the term |x|y and |y|x will cancel out each other and similarly, the terms |xy| and xy. Thus it will always be 0 hence not greater than 0. For the case where both x,y<0; the terms |x|y and |y|x will add upto give -2xy and the other two will give 2xy , thus again a 0. Thus Only in the first quadrant, is the given condition possible. Sufficient.

From F.S 2, we know that -y<|y|. Thus we can conclude that y>0. This leads to only the first or the second quadrant. Also, we have -x<-y or x>y. As, in the second quadrant, x<0, thus this is possible only in the first quadrant.Sufficient.

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29 Jul 2014, 17:17

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