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|

Please Explain

What does "CRACK THE CODE" MEAN IN MANHATTAN LANGUAGE!!

Regards,
Apoorva Srivastva

Hi Apoorva. I don't know exactly what you mean by your second question about "crack the code" but I can help explain the CAT question for you.

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

First we must recognize that the absolute value of each of these 4 terms is equal. Then we can test cases.

If either x or y is negative (but not both), then the last term is negative, and one of the middle terms are negative, thus making the entire expression |xy| + x|y| + |x|y + xy = 0. We can eliminate quadrants 2,4.

If both x and y are negative, then the middle two terms are negative, making the expression = 0 once again, so we can eliminate quadrant 3.

This leaves quadrant 1. x,y must be greater than 0, so fact (1) is SUFFICIENT.
(note that neither x or y can be 0, which is very important in proving sufficiency of determining a quadrant. This will also be the case for fact 2)

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

Let's break this one up into -y < |y| and -x < -y, and examine each one seperately.

If -y < |y|, we can conclude that y > 0, since a y<=0 would imply -y = |y|, which is not the case.

If -x < -y, we can say that x > y by multiplying both sides by -1.

By combining x > y and y > 0, we can conclude that both x,y > 0, so we are once again in the first quadrant and fact (2) is SUFFICIENT.

great explanation...thanks mate

yes, nicely done. bravo.

For all those who would love to solve this question algebraically :


We can say that |xy| = |x| * |y|

So as per stat(1)

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

This can only be possible when x and y are positive values. Since of x or y or both were negative the LHS will become 0. Hence Stat(1) is sufficient.

(2) –x < -y < |y|

Add y to both the sides

-x + y < 0 < y + |y|

Since y + |y| is > 0 it means y is positive, since had it been a negative value then the LHS would become 0.

Since y is + value and –x + y is <0 it means x is also + value since if x were –ve value then the LHS will be >0.

"Crack the code" in ManhattanGMAT world means to look at a mathematical equation and determine what it means more conceptually.

For example, if a Data Sufficiency question gives me this statement:
1) x > x^2

I am "cracking the code" if I look at this and realize that x must be a fraction between 0 and 1, because those are the only numbers that become smaller when squared.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Hi...I had a question with regard to the explanation given with regard to the evaluation of (1) |xy| + x|y| + |x|y + xy > 0

Manhattan GMAT's Explanation :
(1) SUFFICIENT: The key to evaluating this statement is to see which values of x and y actually satisfy it (“crack the code”). To do so, consider all possibilities for the signs of x and y.
• x > 0, y > 0: The left side becomes xy + xy + xy + xy = 4xy, which is a positive number; the statement is satisfied.
• x < 0, y > 0: The left side becomes xy – xy + xy – xy = 0, so the statement is not satisfied.
• x > 0, y < 0: The left side becomes xy + xy – xy – xy = 0, so the statement is not satisfied.
• x < 0, y < 0: The left side becomes xy – xy – xy + xy = 0, so the statement is not satisfied.
• Either x or y (or both) is 0: The left side becomes 0 + 0 + 0 + 0 = 0, so the statement is not satisfied.
Therefore, statement (1) can be rephrased simply as “Both x and y are positive.” The point (x, y) is thus in the first quadrant.

According to me ,

• x > 0, y < 0: The left side becomes |xy| - xy - xy – xy = |xy| -3xy , as |y| should be -y as y<0.
• x < 0, y > 0: The left side becomes |xy| – xy - xy – xy = |xy| -3xy , as |x| should be -x as x <0.

I am not sure what does |xy| evaluates to ?

I am confused with regard to modulus

I am referring the below link
http://www.manhattangmat.com/strategy-s ... -value.cfm

Please explain

You're not plugging in -y for |y| correctly. Check your work. I'm happy to help you with your question about modulus if you can explain in more detail what your question is..

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Tim Sanders
Manhattan GMAT Instructor

Hi Tim ,

Please ignore my earlier post.

Thanks

done :)

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Tim Sanders
Manhattan GMAT Instructor