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# Is xy <0? 1. (x^3 * y^4)/(xy) <0 2. |x| -|y| < |x-y

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Is xy <0? 1. (x^3 * y^4)/(xy) <0 2. |x| -|y| < |x-y [#permalink]

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15 May 2004, 17:30
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Is xy <0?
1. (x^3 * y^4)/(xy) <0
2. |x| -|y| < |x-y|
Senior Manager
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15 May 2004, 18:18
hallelujah1234 wrote:
Is xy <0?
1. (x^3 * y^4)/(xy) <0
2. |x| -|y| < |x-y|

(x^3 * y^4)/(xy) <0

x^2 * y^3<0
y<0, x could go either way! insuff.

|x| -|y| < |x-y|
if, |x| -|y| = |x-y|, both x & y have same sign
for |x| -|y| < |x-y|, x must have greater abs. value than y. but it doesn't say what's their normal signs are. insuff.
it's E.
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15 May 2004, 18:38
I would go with C

From 1) (x^3 * y^4)/(xy) < 0 --> x^2 * y^3 < 0
This means x can be either positive or negative but y HAS to be negative. So possible combination of x,y are:
(x=+) and (y=+) or
(x=-) and (y=+)
Insuff.

From 2) In order for lx-yl to be greater than left side, combinations have to be:
(x=+) and (y=-) or
(x=-) and (y=+)
Insuff.

From both combined, we can see that the only combination is when (x=-) and (y=+).
Therefore, xy < 0 is affirmative
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Paul

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15 May 2004, 18:48
Paul wrote:
I would go with C

From 1) (x^3 * y^4)/(xy) < 0 --> x^2 * y^3 < 0
This means x can be either positive or negative but y HAS to be negative. So possible combination of x,y are:
(x=+) and (y=+) or
(x=-) and (y=+)
Insuff.

From 2) In order for lx-yl to be greater than left side, combinations have to be:
(x=+) and (y=-) or
(x=-) and (y=+)
Insuff.

From both combined, we can see that the only combination is when (x=-) and (y=+).
Therefore, xy < 0 is affirmative

can you pls pick a number & prove it
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15 May 2004, 19:19
Hmmm, I just realized I misinterpreted statement 1 as instances where the left side was > 0. It should have been < 0 but answer is still C

From 1) (x^3 * y^4)/(xy) < 0 --> x^2 * y^3 < 0
This means x can be either positive or negative but y HAS to be negative. So possible combination of x,y are:
(x=+) and (y=-) or
(x=-) and (y=-)
Insuff.

From 2) In order for lx-yl to be greater than left side, combinations have to be:
(x=+) and (y=-) or
(x=-) and (y=+)
Insuff.

From both combined, we can see that the only combination is when (x=+) and (y=-)

Therefore, xy < 0 is still affirmative
Let's pick numbers
x=2
y=-1
From 1) 2^2 * (-1)^3 = -4 which is < 0
From 2) l2l - l-1l < l2 - (-1)l --> 1 < 3
Both combined: 2*(-1) = -2 which is < 0 and question can be answered
y
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Best Regards,

Paul

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16 May 2004, 05:21
b) |x| - |y| < |x-y| only when both signs are same and neither of x and y is zero.
In anycase xy > 0
B is sufficient.
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16 May 2004, 10:42
hallelujah1234 wrote:
Is xy <0?
1. (x^3 * y^4)/(xy) <0
2. |x| -|y| < |x-y|

2 is sufficient to guarantee xy < |xy| => xy < 0.

1 is not.
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16 May 2004, 10:57
Emmanuel wrote:
hallelujah1234 wrote:
Is xy <0?
1. (x^3 * y^4)/(xy) <0
2. |x| -|y| < |x-y|

2 is sufficient to guarantee xy < |xy| => xy < 0.

1 is not.

x= 1; y = -2; xy < 0; |1|-|-2| < |1-(-2)|
x = -1; y = -2; xy > 0; |-1|-|-2| < |-1-(-2)|

Insufficient.

Last edited by hallelujah1234 on 17 May 2004, 09:29, edited 1 time in total.
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16 May 2004, 11:01
hallelujah1234 wrote:
x= 1; y = -2; xy < 0; |1|-|-2| < |1-(-2)|
x = -1; y = -2; xy > 0; |-1|-|-2| < |-1-(-2)|

Insufficient.

Yeah, you're right, I'm wrong. I used wrong equivalence

|x|-|y|<|x-y| <=> (|x|-|y|)^2 < (x-y)^2!!!
Re: DS-107   [#permalink] 16 May 2004, 11:01
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