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 Post subject: Tough Modulus + Co-ordinate geomtry DS (700-800 Range)
 Post Posted: Wed Jul 01, 2009 4:05 pm 
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Students


Posts: 27
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


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 Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range)
 Post Posted: Thu Jul 02, 2009 3:01 am 
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Course Students


Posts: 13
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.


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 Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range)
 Post Posted: Thu Jul 02, 2009 5:06 am 
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Students


Posts: 27
andrew.k.john wrote:
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


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 Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range)
 Post Posted: Mon Jul 13, 2009 7:19 am 
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ManhattanGMAT Staff


Posts: 5031
andrew.k.john wrote:
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.


yes, nicely done. bravo.


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