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...
Absolute Coordinates
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Wed Sep 01, 2010 11:11 am
- Thanked: 1 times
- Followed by:2 members
- vineeshp
- Legendary Member
- Posts: 965
- Joined: Thu Jan 28, 2010 12:52 am
- Thanked: 156 times
- Followed by:34 members
- GMAT Score:720
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.
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,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
-
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Wed Sep 01, 2010 11:11 am
- Thanked: 1 times
- Followed by:2 members
- manpsingh87
- Master | Next Rank: 500 Posts
- Posts: 436
- Joined: Tue Feb 08, 2011 3:07 am
- Thanked: 72 times
- Followed by:6 members
1) |xy| + x|y| + |x|y + xy > 0;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...
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
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
- vineeshp
- Legendary Member
- Posts: 965
- Joined: Thu Jan 28, 2010 12:52 am
- Thanked: 156 times
- Followed by:34 members
- GMAT Score:720
Ya I made a mistake. I resolved that y is positive and then assumed a negative value for y.
My initial answer was fine.
My initial answer was fine.
Vineesh,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Since |ab| = |a|*|b|, we can rewrite the inequality in Statement 1: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...
|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
ianstewartgmat.com
ianstewartgmat.com