Is xy < 0? 1). (X^3 * Y^5)/ (X*Y^2) <0 2). /X/ - /y/

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Is xy < 0? 1). (X^3 * Y^5)/ (X*Y^2) <0 2). /X/ - /y/ [#permalink]

24 Apr 2006, 12:44

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Is xy < 0?

1). (X^3 * Y^5)/ (X*Y^2) <0

2). /X/ - /y/ < /X-Y/

Please explain your approach. I am NOT just interested in knowing the answer. Thank You

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Re: DS - Inequality [#permalink]

24 Apr 2006, 12:52

Ethan wrote:

Is xy < 0?

1). (X^3 * Y^5)/ (X*Y^2) <0

2). /X/ - /y/ < /X-Y/

Please explain your approach. I am NOT just interested in knowing the answer. Thank You

from i) x^2*y^3 < 0
so x could be +ve or -ve so insuff.

from ii) it looks like triangle inequality reversed, but it should be enough

I would go with B

because either x or y must be negative for ii inequality to hold

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Re: DS - Inequality [#permalink]

24 Apr 2006, 13:09

conocieur wrote:

because either x or y must be negative for ii inequality to hold

Sorry, conocieur, but that's not true. Just consider x = 1, y = 2.
|1| - |2| = -1 < |1-2| = +1

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[#permalink]

24 Apr 2006, 13:50

Guys, Thanks for your inputs. I dont have any OA for this one. I to am going for 'E'

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[#permalink]

24 Apr 2006, 22:28

St1:

We can reduce the equation to: (x^2)(y^3) < 0

We know x^2 is always positive, so y^3 must be negative and so y must be negative. But x can be positive or negative, so xy can be positive or negative.

Insufficient.

St2:
Consider:
x = 2, y = -1, then |x|-|y| = 1 and |x-y| = 3 then |x|-|y| < |x-y| and xy = negative.
x = 1, y = 3, then |x|-|y| = -2 and |x-y| = 2 then |x|-|y| < |x-y| and xy = positive.

So this inequality can be satisfied but xy can be positive or negative. Insufficient.

Using both St1 and St2,

y must be negative. x can be negative or positive.

Consider
x = 2, y = -1, then |x|-|y| = 1 and |x-y| = 3 then |x|-|y| < |x-y| and xy = negative.
x = -1, y = -2, then |x|-|y| = -1 and |x-y| = 1 then |x|-|y| < |x-y| and xy = positive

Insufficient.

Ans should be E

[#permalink] 24 Apr 2006, 22:28

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Is xy < 0? 1). (X^3 * Y^5)/ (X*Y^2) <0 2). /X/ - /y/

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Is xy < 0? 1). (X^3 * Y^5)/ (X*Y^2) <0 2). /X/ - /y/

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