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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]
 20 May 2007, 05:54
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Is xy < 0?
1). (X^3 * Y^5)/ (X*Y^2) <0
2). |X| - |y| < |X-Y|
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SVP
Joined: 01 May 2006
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(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.
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Manager
Joined: 23 Mar 2007
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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
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SVP
Joined: 01 May 2006
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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
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Director
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Thanks guys OA is 'E'.
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Manager
Joined: 23 Mar 2007
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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.
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