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In which quadrant of the coordinate plane does the point
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Updated on: 27 Jun 2013, 21:17
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41% (02:30) correct 59% (02:24) wrong based on 929 sessions
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Originally posted by Bunuel on 12 Nov 2009, 11:11.
Last edited by Bunuel on 27 Jun 2013, 21:17, edited 1 time in total.
Added the OA




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Re: Quadrant for the point
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13 Nov 2009, 22:57




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Re: Quadrant for the point
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13 Nov 2009, 22:18
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 + xy + xy + xy > 0 (2,3): 6+6+6+6>0 CHECK! (2,3): 666+6>0 No good (2,3): 6+666>0 No good (2,3): 66+66>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
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12 Nov 2009, 13:40
D for me as well
(x,y) lies in first quadrant since both are positive!!



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Re: Quadrant for the point
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12 Nov 2009, 15:28
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + 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
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12 Nov 2009, 17:58
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + 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
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13 Nov 2009, 21:19



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Re: Quadrant for the point
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13 Nov 2009, 22:40
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + 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
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15 Nov 2009, 02:20
If this is 600700 level question, i cant imagine 700800level qs. Bunuel, We want more problems on coordinate 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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16 Nov 2009, 03:33
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
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15 Mar 2011, 20:09
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
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18 Jun 2012, 04:58
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + xy > 0 (2) x < y < y A remark regarding statement (1): Since xy=xy, the given expression can be written as (x+x)(y+y)>0. If either x or y is nonpositive, 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
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27 Mar 2013, 01:37
In which quadrant of the coordinate plane does the point (x, y) lie? (1) xy + xy + xy + xy > 0Case quadrant I (x,y)=(+,+)\(xy + xy + xy + xy\) \(xy +xy + xy + xy >0\) 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 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 \((x)y + (x)y + (x)y + (x)y\) \(xy xy + xy  xy =0\) 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
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Re: In which quadrant of the coordinate plane does the point
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27 Mar 2013, 03:47
mun23 wrote: In which quadrant of the coordinate plane does the point (x, y) lie?
(1) xy + xy + xy + 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 xy and yx 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 xy and yx 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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31 Jul 2014, 04:12
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + xy > 0 (2) x < y < y Hello Bunuel Please Help. I dont know what i am missing in 2nd statement.  x  y  yFrom 2: for sure y will always be positive. now by number plunging let y= 5 then trying for x = 2 , 2 ,10 , 10 when x =2 then 2 > 5 i.e. x > y it does not satisfy the Equation above x < y < y whereas when x=10 then 10 < 5 i.e. x < y which satisfies the equation. so how can x be +ve Always. What am i missing here. Please help Thankyou



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Re: In which quadrant of the coordinate plane does the point
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31 Jul 2014, 06:25
niyantg wrote: Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + xy > 0 (2) x < y < y Hello Bunuel Please Help. I dont know what i am missing in 2nd statement.  x  y  yFrom 2: for sure y will always be positive. now by number plunging let y= 5 then trying for x = 2 , 2 ,10 , 10 when x =2 then 2 > 5 i.e. x > y it does not satisfy the Equation above x < y < y whereas when x=10 then 10 < 5 i.e. x < y which satisfies the equation. so how can x be +ve Always. What am i missing here. Please help Thankyou From (2) we get that y must be positive. So, we have that x < y > y < x > x is greater than y, which we know is positive so x also must be positive.
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Re: In which quadrant of the coordinate plane does the point
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01 Aug 2014, 04:05
Thankyou Bunuel Got my mistake !! Bunuel wrote: niyantg wrote: Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + xy > 0 (2) x < y < y Hello Bunuel Please Help. I dont know what i am missing in 2nd statement.  x  y  yFrom 2: for sure y will always be positive. now by number plunging let y= 5 then trying for x = 2 , 2 ,10 , 10 when x =2 then 2 > 5 i.e. x > y it does not satisfy the Equation above x < y < y whereas when x=10 then 10 < 5 i.e. x < y which satisfies the equation. so how can x be +ve Always. What am i missing here. Please help Thankyou From (2) we get that y must be positive. So, we have that x < y > y < x > x is greater than y, which we know is positive so x also must be positive.



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Re: In which quadrant of the coordinate plane does the point
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18 Aug 2014, 00:38
Bunuel wrote: In which quadrant of the coordinate plane does the point (x,y) lie?
(1) xy + xy + xy + xy > 0 (2) x < y < y Statement I is sufficient You can use a binary table. x.........y... xy + xy + xy + xy +2......+2.........16 +2.......2.........0 2.......+2.........0 2........2.........0 The statement is only valid when x and y are +ve hence Ist quadrant. Statement II is sufficient: If y> y then y has to be positive. If y is positive and x< y OR x>y then x is also positive Hence answer is D
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Re: In which quadrant of the coordinate plane does the point
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19 Aug 2014, 21:28
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 + xy + xy + xy > 0 (2,3): 6+6+6+6>0 CHECK! (2,3): 666+6>0 No good (2,3): 6+666>0 No good (2,3): 66+66>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: In which quadrant of the coordinate plane does the point
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28 Aug 2014, 01:56
1. xy + xy + xy + xy > 0xy+ xy + xy + xy > 0 ; x(y+y)+x(y+y)>0 ; (y+y)(x+x)>0 if y <= 0 then y+y=y+y=0 =>(y+y)(x+x)=0 then y is absolutely positive. same for x. => sufficient2. x < y < yy<y ; if y < 0 then y=y which is not the case then y > 0. y>0 then y<0, and x < y then x < 0 and x > 0. => sufficientAnswer: D
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