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apoorva_srivastva
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Post subject: Tough Modulus + Co-ordinate geomtry DS (700-800 Range) Posted: Wed Jul 01, 2009 4:05 pm |
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Posts: 27
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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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andrew.k.john
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Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range) Posted: Thu Jul 02, 2009 3:01 am |
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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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apoorva_srivastva
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Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range) Posted: Thu Jul 02, 2009 5:06 am |
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Posts: 27
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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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RonPurewal
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Post subject: Re: Tough Modulus + Co-ordinate geomtry DS (700-800 Range) Posted: Mon Jul 13, 2009 7:19 am |
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| ManhattanGMAT Staff |
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Posts: 5031
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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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