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In which quadrant of the coordinate plane does the point (x,

Expert replies
by RBBmba@2014 » Fri May 08, 2015 2:18 am
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|


OA : D

P.S: I got this but it took around 3.5 mns to solve, especially the first statement. I considered each quadrant individually for the first statement. So, looking for any alternative smarter and faster approach, if any.Thanks!
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Source: — Data Sufficiency |

by GMATGuruNY » Fri May 08, 2015 4:36 am
RBBmba@2014 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|
Statement 1: |xy| + x|y| + |x|y + xy > 0
Here, it is clear that xy≠0.
Test EASY CASES in |xy| + x|y| + |x|y + xy.

Case 1: x=1, y=1 --> |1*1| + 1|1| + |1|1 + 1*1 = 4.
Case 2: x=-1, y=1 --> |(-1)(1)| + -1|1| + |-1|1 + -1*1 = 0.
Case 3: x=1, y=-1 --> |1*-1| + 1|-1| + |1|-1 + (1)(-1) = 0.
Case 4: x=-1, y=-1 --> |(-1)(-1)| + (-1)|-1| + |-1|(-1) + (-1)(-1) = 0.

Only Case 1 satisfies the constraint that |xy| + x|y| + |x|y + xy > 0.
Thus, (x,y) must be (+, +), with the result that it lies in Quadrant I.
SUFFICIENT.

Statement 2: -x < -y < |y|
Here, it is clear that y≠0, implying that |y| > 0.
Thus:
-x < -y < 0 < |y|

Multiplying by -1 and flipping the inequality symbols, we get;
x > y > 0 > -|y|.
Since x > y > 0, (x,y) must lie in Quadrant I.
SUFFICIENT.

The correct answer is D.
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by DavidG@VeritasPrep » Fri May 08, 2015 4:41 am
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|
Interesting question. Because the first statement looks a bit convoluted, I'll start with the second statement. Perhaps that will give me clues about how to proceed with the first.

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

If we multiply by (-1) we'll get x > y > -|y|
The only way y > -|y| will be if y is positive. If x > y, x is positive too.
If x and y are both positive, we know we're in quadrant I. Sufficient


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

Now I'll go into statement 1 simply wondering if it's possible to get a result other than x is positive and y is positive.

Say x = -1 and y = 1. We'll get |-1| + -1*|1| + |-1|*1 + (-1)(1) or 1 - 1 -1 -1, which will not be greater than 0. If we say x = 1 and y = -1, we'll get the same result. So we can't have one negative. (Intuitively, this makes sense. If one of the two variables is negative, x|y|, |x|y|, and xy will all be negative.)

Let's try to make them both negative. x = -1 and y = -1. Now we get 1 +(-1) + (-1) + 1, which = 0. Not greater than 0. (If both terms are negative, now x|y| and |x|y will be negative, and |xy| and xy will be positive. Consequently, the terms will always sum to 0.)

So I know that both x and y must be positive. Again, I'll know that I'm in quadrant I. Sufficient.

Answer is D
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