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In which quadrant of the coordinate plane does the point [#permalink]
 12 Nov 2009, 12:11
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Re: Quadrant for the point [#permalink]
12 Nov 2009, 14:40
D for me as well
(x,y) lies in first quadrant since both are positive!!
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Re: Quadrant for the point [#permalink]
12 Nov 2009, 16:28
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| i'll take d as well bunuel..post some good inquality ds questions
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Re: Quadrant for the point [#permalink]
12 Nov 2009, 18:58
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| 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 Ans 'D'
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Re: Quadrant for the point [#permalink]
13 Nov 2009, 22:19
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Re: Quadrant for the point [#permalink]
13 Nov 2009, 23:18
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Statement 1:
To determine what quadrant (x,y) is in, we need to see if either value is positive or negative. To test this, it best to just plug in:
|xy| + x|y| + |x|y + xy > 0
(2,3): 6+6+6+6>0 CHECK!
(-2,-3): 6-6-6+6>0 No good
(2,-3): 6+6-6-6>0 No good
(-2,3): 6-6+6-6>0 No good
SUFFICIENT
Statement 2:
-x < -y < |y|
For -y < |y| to remain true, y must be positive.
If y is positive, then x must also be positive for -x < -y to be true.
SUFFICIENT
Answer: D.
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Re: Quadrant for the point [#permalink]
13 Nov 2009, 23:40
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| @bunuel, Can we simplify the stmt 2 as follows x>y>-|y| (mutiply by -1) x > y + |y|.
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Re: Quadrant for the point [#permalink]
13 Nov 2009, 23:57
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Re: Quadrant for the point [#permalink]
15 Nov 2009, 03:20
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?
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Re: Quadrant for the point [#permalink]
16 Nov 2009, 04:33
Bunuel, I am definitely with ctrlaltdel on his request for more problem sets in groups. Thanks!!! 
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Re: Quadrant for the point [#permalink]
15 Mar 2011, 21:09
From (1), it must be Q1, The only value +ve is |xy| and any other quadrant can make the value < 0 depending on size of x or y. From (2), -y < |y| so -y is -ve and |y| is +ve, hence y is +ve, so it can be II or 1st Quadrant. And -x < -y => say -3 < -2 but x > y ( 3 > 2) , as y is +ve so x is +ve, hence it's Q1. So answer is D.
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Re: In which quadrant of the coordinate plane does the point [#permalink]
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.
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Re: In which quadrant of the coordinate plane does the point [#permalink]
18 Jun 2012, 06:40
D - But i took 2 mins! 
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In which quadrant of the coordinate plane does the point [#permalink]
27 Mar 2013, 02:17
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............
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Re: In which quadrant of the coordinate plane does the point [#permalink]
27 Mar 2013, 02:37
In which quadrant of the coordinate plane does the point (x, y) lie? (1) |xy| + x|y| + |x|y + xy > 0Case quadrant I (x,y)=(+,+)|xy| + x|y| + |x|y + xyxy +xy + xy + xy >0The first term is positive, the second the third and the fourt also. The sum of 4 positive integers is >0. so quadrant I is possibleCase 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)yxy -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|>-ycase y>0 y>-yy>0case y<0 -y>-ySo y>0 must be the case here We know that -y >-x and that y>0, we can sum these elements -y+y>-x+00>-xx>0and given that x>0 and that y>0 the point is in the first quadrant SUFFICIENT
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Re: In which quadrant of the coordinate plane does the point [#permalink]
27 Mar 2013, 04:22
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Re: In which quadrant of the coordinate plane does the point [#permalink]
27 Mar 2013, 04:47
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. D.
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Re: In which quadrant of the coordinate plane does the point
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