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

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Director
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Is xy < 0? 1). (X^3 * Y^5)/ (X*Y^2) <0 2). |X| - |y| [#permalink] New post 20 May 2007, 04:54
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A
B
C
D
E

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(N/A)

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Is xy < 0?
1). (X^3 * Y^5)/ (X*Y^2) <0
2). |X| - |y| < |X-Y|
SVP
SVP
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 [#permalink] New post 20 May 2007, 05:11
(B) for me :)

Is xy < 0?
<=> Sign(x) != Sign(y) ?

From 1
(X^3 * Y^5)/ (X*Y^2) <0 (as the inquation exists, x and y are different from 0)
<=> X^2 * Y^3 < 0
=> Y < 0 but X can be postive or negative.

Thus,
o Sign(x)=Sign(y)
or
o Sign(x)!=Sign(y)

INSUFF

From 2
|X| - |y| < |X-Y|

implies that:
o x > 0 > y
or
o y > 0 > x

So: Sign(x)!=Sign(y)

SUFF.
Manager
Manager
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Joined: 23 Mar 2007
Posts: 177
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 [#permalink] New post 20 May 2007, 12:28
Fig wrote:
(B) for me :)

Is xy < 0?
<=> Sign(x) != Sign(y) ?

From 1
(X^3 * Y^5)/ (X*Y^2) <0 (as the inquation exists, x and y are different from 0)
<=> X^2 * Y^3 <0> Y < 0 but X can be postive or negative.

Thus,
o Sign(x)=Sign(y)
or
o Sign(x)!=Sign(y)

INSUFF

From 2
|X| - |y| <X> 0 > y
or
o y > 0 > x

So: Sign(x)!=Sign(y)

SUFF.


statement 2 is also insufficient. lets take examples.

x = -3 , y = -4
so lx-yl > lxl - lyl

x=3 , y=4
again lx-yl > lxl - lyl

Answer has to be E
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1798
Followers: 10

Kudos [?]: 108 [0], given: 0

 [#permalink] New post 20 May 2007, 12:37
abhinava wrote:
Fig wrote:
(B) for me :)

Is xy < 0?
<=> Sign(x) != Sign(y) ?

From 1
(X^3 * Y^5)/ (X*Y^2) <0 (as the inquation exists, x and y are different from 0)
<=> X^2 * Y^3 <0> Y < 0 but X can be postive or negative.

Thus,
o Sign(x)=Sign(y)
or
o Sign(x)!=Sign(y)

INSUFF

From 2
|X| - |y| <X> 0 > y
or
o y > 0 > x

So: Sign(x)!=Sign(y)

SUFF.


statement 2 is also insufficient. lets take examples.

x = -3 , y = -4
so lx-yl > lxl - lyl

x=3 , y=4
again lx-yl > lxl - lyl

Answer has to be E


Yes... U are right :)... Miscalculating :).... Thanks to have corrected me :)
Director
Director
User avatar
Joined: 14 Jan 2007
Posts: 779
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Kudos [?]: 83 [0], given: 0

 [#permalink] New post 20 May 2007, 16:12
Thanks guys OA is 'E'.
Manager
Manager
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Joined: 23 Mar 2007
Posts: 177
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Kudos [?]: 1 [0], given: 0

 [#permalink] New post 21 May 2007, 09:22
Fig wrote:
abhinava wrote:
Fig wrote:
(B) for me :)

Is xy < 0?
<=> Sign(x) != Sign(y) ?

From 1
(X^3 * Y^5)/ (X*Y^2) <0 (as the inquation exists, x and y are different from 0)
<=> X^2 * Y^3 <0> Y < 0 but X can be postive or negative.

Thus,
o Sign(x)=Sign(y)
or
o Sign(x)!=Sign(y)

INSUFF

From 2
|X| - |y| <X> 0 > y
or
o y > 0 > x

So: Sign(x)!=Sign(y)

SUFF.


statement 2 is also insufficient. lets take examples.

x = -3 , y = -4
so lx-yl > lxl - lyl

x=3 , y=4
again lx-yl > lxl - lyl

Answer has to be E


Yes... U are right :)... Miscalculating :).... Thanks to have corrected me :)



I myself miscalculate a lot of times :) But it is good that we are helping out each other.
  [#permalink] 21 May 2007, 09:22
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