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# Is \hspace{3} (x^7)(y^2)(z^3) > 0

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Is \hspace{3} (x^7)(y^2)(z^3) > 0 [#permalink]  01 Sep 2008, 17:12
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43% (01:31) correct 57% (00:56) wrong based on 7 sessions
$$Is \hspace{3} (x^7)(y^2)(z^3) > 0$$ ?

1. $$yz < 0$$
2. $$xz > 0$$
[Reveal] Spoiler: OA
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Re: DS trap???? [#permalink]  01 Sep 2008, 17:27
C.

from 1) x UNKNOWN , if x=0 then exprn = 0.
from 2) y unknown. if y=0, then exprn = 0.

together the expression is +ve.
suff.
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Re: DS trap???? [#permalink]  01 Sep 2008, 17:35
IMO C
same explanation as above
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Re: DS trap???? [#permalink]  01 Sep 2008, 20:41
From II
X and Z either both are +ve or -ve
X^7.Z^3 will always be Z +ve
Y^2 will always be Z +ve

So whole expression will always be +ve or Zero if Y=0
But in both cases it can be answered

So B suff
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Re: DS trap???? [#permalink]  01 Sep 2008, 20:55
IMO C.

To be x^7y^2Z^3 > 0 ; x and z both have to have the same sign + y not 0 .

1 ) YZ <0 ; Either (y +ve and z -ve ) or ( Y -ve and Z +ve ) => not sufficient

2 ) xz >0 ; Either ( x and Z +ve ) or ( x and Z -ve) = > i.e x and z same sign but y unknown => not sufficient

Combined, when ( x and z +ve )and y -ve => x^7y^2Z^3 > 0

and when ( x and Z -ve) y +ve = > x^7y^2Z^3 > 0
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Re: DS trap???? [#permalink]  01 Sep 2008, 21:10
arorag wrote:
From II
X and Z either both are +ve or -ve
X^7.Z^3 will always be Z +ve
Y^2 will always be Z +ve

So whole expression will always be +ve or Zero if Y=0
But in both cases it can be answered

So B suff

if y = 0, then the expression is NOT greater than 0
if y not equal to 0, then the expression is greater than 0
how would that make B is sufficient?

If the question asked if $$x^7y^2Z^3 LESS THAN 0$$
then B would be sufficient
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Re: DS trap???? [#permalink]  02 Sep 2008, 00:45
Good trap, keep posting fun questions.
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Re: DS trap???? [#permalink]  02 Sep 2008, 12:05
'B' looks good.

1) not sufficient
2) xz > 0, means both can be negative or positive, making the entire term always positive(y^2 is positive only).
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Re: DS trap???? [#permalink]  02 Sep 2008, 12:08
amitkaneria wrote:
'B' looks good.

1) not sufficient
2) xz > 0, means both can be negative or positive, making the entire term always positive(y^2 is positive only).

Y^2 can be zero too..
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Re: DS trap???? [#permalink]  06 Oct 2011, 03:01
1
KUDOS
arorag wrote:
$$(x^7)(y^2)(z^3) > 0$$ ?
1. $$yz < 0$$
2. $$xz > 0$$

1. $$yz < 0$$
Neither of y and z is 0.

y=1; z=-1;
x=-1; Yes.
x=1; No
Not Sufficient.

2. $$xz > 0$$

$$xy^2z \ge 0$$ {:y can be 0}

So,
the expression may or may not be greater than 0.
Not Sufficient.

Using both;
$$y \ne 0$$
So,
$$xy^2z>0$$
OR
$$x^{(Any \hspace{2} +ve \hspace{2} odd \hspace{2} integer)}y^2z^{(Any \hspace{2} +ve \hspace{2} odd \hspace{2} integer)}>0$$
Sufficient.

Ans: "C"
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Re: DS: Is (x^7)(y^2)(z^3) > 0 [#permalink]  07 Oct 2011, 06:36
I missed the Y=0 trap.... 1+ for C
Re: DS: Is (x^7)(y^2)(z^3) > 0   [#permalink] 07 Oct 2011, 06:36
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# Is \hspace{3} (x^7)(y^2)(z^3) > 0

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