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m21 q30 [#permalink]

02 Jul 2010, 05:38

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Is x^3y^2z>0?
1. yz>0
2. xz<0

a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d. EACH statement ALONE is sufficient
e. Statements (1) and (2) TOGETHER are NOT sufficient

[Reveal] Spoiler:

Statement (1) by itself is insufficient. It allows x=0 .
Statement (2) by itself is insufficient. It allows y>0 and y<0 .
Statements (1) and (2) combined are sufficient. S1 and S2 give that is not 0 and thus x^3y^2z>0 .
The correct answer is C.

Can someone please simplify this for me?

I am completely stumped even after reading the explanation

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Re: m21 q30 [#permalink]

02 Jul 2010, 06:55

yeahwill wrote:

[Reveal] Spoiler:

Can someone please simplify this for me?

I am completely stumped even after reading the explanation

Because x has an odd exponent, x^3 can be negative or positive or 0
Because y has an even exponent y^2 will be positive or 0
z can be positive or negative or 0

1. yz > 0 means both are positive or both negative but we know nothing about x, which can be positive, negative, or 0

2. xz< 0 means one is negative and one is positive which would make you think you have sufficient information because y^2 will also be positive, but it can also be zero in which case you don't have enough information

Together we know nothing is zero.

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Re: m21 q30 [#permalink]

02 Jul 2010, 07:19

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yeahwill wrote:

Is x^3y^2z>0?
1. yz>0
2. xz<0

[Reveal] Spoiler:

Can someone please simplify this for me?

I am completely stumped even after reading the explanation

You copied question m21q30 incorrectly. Original question m21q30 is below:

Is x^2*y^3*z>0?

Inequality x^2*y^3*z>0 to be true: 1. y and z must be either both positive or both negative, so they must have the same sign (in this case y^3*z will be positive) AND 2. x must not be zero (in this case x^2 will be positive).

(1) yz>0 --> y and z are either both positive or both negative (first condition). But we don't know about x (second condition). Not sufficient.

(2) xz<0 --> x\neq{0} (second condition). Don't know about the signs of y and z (first condition). Not sufficient.

(1)+(2) Both conditions satisfied. Sufficient.

Answer: C.

Check similar question: data-sufficiency-exponents-question-95626.html#p736291.

For the question you've posted:

Is x^3y^2z>0?

(1) yz>0. If all three are positive than obviously x^3y^2z>0, but if y and z are positive and x is negative then x^3y^2z<0. Two different answers. Not sufficient.

(2) xz<0 --> x and z have opposite signs --> x^3z<0. Now y^2 can be either positive or zero so x^3y^2z can never be positive, it can be either negative if y>0 or zero if y=0. Hence the answer to the question "is x^3y^2z>0" is NO. Sufficient.

Answer: B.

Hope it helps.
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Re: m21 q30 [#permalink]

02 Jul 2010, 23:59

Bunuel, thanks a lot for catching my mistake.

the equation that I pasted in the query was exactly what I was solving for
Wow! this does not bode well for my exam :oops:

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Re: m21 q30 [#permalink]

13 Jun 2012, 12:09

Hi,

x^3y^2z>0?

Using (1),
yz>0,
expression reduces to x^3y(yz), (removing the positive values)
or x^3y, we don't know the sign of y. Insufficient.

Using (2) ,
xz<0
expression reduces to x^2(xz)y^2, (removing the positive values)
or xz which is less than zero. But still y can be 0.
even in that case the given expression will be less than or equal to 0. Sufficient.

Thus, Answer is (B),

_____________________________________________________
x^2y^3z>0?

Using (1),
yz>0,
expression reduces to x^2y^2(yz), (removing the positive values)
or yz, which is greater than 0, but x can be equal to zero. Insufficient.

Using (2) ,
xz<0
expression reduces to x(xz)y^3,
or xy^3*(-ve), no idea about signs of x & y. Insufficient.

Using (1) & (2),
we have, x^3y^2z>0

Thus, Answer is (C),

Regards,
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Re: m21 q30 [#permalink] 13 Jun 2012, 12:09

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m21 q30

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