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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| [#permalink] New post 27 Oct 2008, 08:14
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A
B
C
D
E

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Is xy < 0?

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

Pls explain
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Thanks
rampuria

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Re: DS [#permalink] New post 27 Oct 2008, 09:02
1). (X^3 * Y^5)/ (X*Y^2) <0

This can happen when both X and Y are -ve or when only Y is -ve.


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

This can happen only when each are of different signs.

Thus B.

What is OA
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Re: DS [#permalink] New post 27 Oct 2008, 15:36
i think its E

for 2 .. its also true for both x and y as -ve

whts the OA ?
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Re: DS [#permalink] New post 27 Oct 2008, 16:14
rampuria wrote:
Is xy < 0?

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

Pls explain


for xy<0, only one of the two has to be negative.

(1) gives \(\frac{(X^3 * Y^5)}{(X*Y^2)}\) ===>\(x^2*y^3\)

so \(x^2*y^3\) < 0. This proves y is negative. But x can be +ve or -ve, we do not know so xy may or may not be <0

(2) |X| - |y| < |X-Y|
insuff ..

Together : not suff

So E.
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Re: DS [#permalink] New post 30 Oct 2008, 07:30
E. If correct, explanation follows...

For condition (X^3 * Y^5)/ (X*Y^2) <0

Consider x=-1, y=-2, then above equation evaluates to value <0, product xy is positive.
Consider x=1, y=-1, then above equation evaluates to value <0, product xy is negative.

So 1 is not enough.


For condition |X| - |y| < |X-Y|
Consider X=-1, y=-2, to satisfy above equation. Product xy is positive.
Consider X=5, y=-7, to satisfy above equation. Product xy is negetive.

So 2 is not enough.

Since 1 and 2 are union. They are not enough together as well.
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Re: DS [#permalink] New post 30 Oct 2008, 07:46
i get E as well

|x|-|y|<|x-y|

lets say x=-1 y=-2

-1<1 xy positive

x=1 y=-2

1-2=-1< 3 but xy is negative
Re: DS   [#permalink] 30 Oct 2008, 07:46
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