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Absolute Coordinates

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Source: — Data Sufficiency |
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by vineeshp » Sun Apr 24, 2011 7:11 am
Is the OA A?

If yes, soln follows.
Last edited by vineeshp on Sun Apr 24, 2011 7:18 am, edited 1 time in total.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by vineeshp » Sun Apr 24, 2011 7:18 am
THis is what I did. My answer is A now.

Stmt 1:
|xy| + x|y| + |x|y + xy > 0

Choose x = 1, y=1
Stmt holds good.

Choose x = -1, y=1
Stmt does not hold good.

Choose x = 1, y= -1
Stmt does not hold good.

Choose x = -1, y= -1
Stmt does not hold good.

It can be safely concluded that unless x and y have both got positive sign, this inequality does not hold good. (I tried a few other values too)

Stmt 2.
-y < |y|

Implies y is positive.
But what about x.
-y > -x ==> x > y

Let x = 4, y=3
Stmt holds good.

x=-2, y=-3
Stmt still holds good.
Hence Not sufficient.
Vineesh,
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by akshatgupta87 » Sun Apr 24, 2011 7:27 am
The answer is D
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by manpsingh87 » Sun Apr 24, 2011 7:40 am
akshatgupta87 wrote:Q.)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

Someone explain...
1) |xy| + x|y| + |x|y + xy > 0;

case 1) consider x>0; y>0; we have
xy+xy+xy+xy>0 4xy>0; which is true as both x and y are positive;
case 2) consider x>0; y<0; we have
-xy-xy+xy+xy>0;
0>0 which is false;
case 3) x<0; y>0;
-xy+xy-xy+xy>0;
0>0; which is false;
case 4) x<0; y<0;
xy-xy-xy+xy>0;
0>0; which is false;

hence out of all possible case only 1 case holds true, i.e. x>0; and y>0; hence x and y lies in the first quadrant. therefore 1 is sufficient.

2) -x<-y<|y|;
-y<|y|;
if y>0;
-y<y; which is true;
if y<0
-y<-y which is false; hence y is positive;
now consider -x<-y; if x becomes negative then -(-x) becomes positive, therefore our inequality -x<-y won't hold true, because positive quantity is always greater than negative;
therefore x should be positive;
as x and y should be positive therefore it must lie in the first quadrant, hence 2 alone is also sufficient to answer the question..!!!

hence D
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by vineeshp » Sun Apr 24, 2011 9:32 am
Ya I made a mistake. I resolved that y is positive and then assumed a negative value for y.

My initial answer was fine.
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by Ian Stewart » Sun Apr 24, 2011 12:37 pm
akshatgupta87 wrote:Q.)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|

Someone explain...
Since |ab| = |a|*|b|, we can rewrite the inequality in Statement 1:

|x|*|y| + x|y| + |x|y + xy > 0
|y|(|x| + x) + y(|x| + x) > 0
(|y| + y)(|x| + x) > 0

If y is negative (or zero), the first factor above would be zero, and if x is negative (or zero), the second factor above would be zero. So if either y or x is negative or zero, the inequality will not be true, so they must both be positive.

From Statement 2, if -y < |y|, then y is positive, since -y would equal |y| if y was negative or zero. If -x < -y, then x > y, so if y is positive, x is as well.

The answer is D.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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