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Is -x/(-y)(-z) -ve?? 1) x*y >0 2)xyz>0

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Is -x/(-y)(-z) -ve?? 1) x*y >0 2)xyz>0 [#permalink]

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New post 11 Sep 2006, 10:21
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Is -x/(-y)(-z) -ve??

1) x*y >0
2)xyz>0

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Director
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New post 11 Sep 2006, 10:33
Interesting.

A) Given X and Y are either both pos or neg but the sign of Z is unknown. Stmnt A is NOT suff
B) possible scenarios either X,Y,Z are pos or 2 of them are neg. In either case the stem is neg so B)

B) SUFF

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New post 11 Sep 2006, 10:47
BG wrote:
Interesting.

A) Given X and Y are either both pos or neg but the sign of Z is unknown. Stmnt A is NOT suff
B) possible scenarios either X,Y,Z are pos or 2 of them are neg. In either case the stem is neg so B)

B) SUFF

lets pick numbers x,y and z
x=-1 y=-2 z =3
x-ve y-ve and z+ve
Is -x/(-y)(-z) -ve??

1/(2)*(-3) so it is -ve

if x=1 y=-2 z=-3
-1/(2)*(3) it is -ve
if x=-1 y=2 z=-3
1/-2*3 - ve
Oops it seems like B it is

:oops:
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Re: DS INIQUALITIES [#permalink]

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New post 12 Sep 2006, 08:47
yezz wrote:
Is -x/(-y)(-z) -ve??

1) x*y >0
2)xyz>0

B

signs of xyz or x/yz or xy/z etc will always be same.

so for the sign purpose we can consider -x/(-y)(-z) as (-xyz)

B clearly says that xyz > 0 so
(-xyz) will always be less than 0, so will be -x/(-y)(-z)
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Re: DS INIQUALITIES [#permalink]

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New post 12 Sep 2006, 09:07
ps_dahiya wrote:
yezz wrote:
Is -x/(-y)(-z) -ve??

1) x*y >0
2)xyz>0

B

signs of xyz or x/yz or xy/z etc will always be same.

so for the sign purpose we can consider -x/(-y)(-z) as (-xyz)

B clearly says that xyz > 0 so
(-xyz) will always be less than 0, so will be -x/(-y)(-z)


suppose y,z are both negative, -1, -1 and x is positive 1

-1/(- -1)*(- -1) the entire sum is negative..

IMO it should be C..

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New post 12 Sep 2006, 09:44
nice ralley everyone.

The OA is B

HOPE YOU ENJOYED IT

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New post 14 Sep 2006, 10:22
if we simplified

-x/(-y*-z) to -x/yz = -1(x/yz)

and since x/yz will always be positive then -1(x/yz)
will always be negative

that is the OE

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  [#permalink] 14 Sep 2006, 10:22
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