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TheGmatTutor
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Bunuel
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TheGmatTutor
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Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct? So the whole statement

(x^7)(y^2)(z^3)

must be greater than 0, because the product is either
- + -

or

+++

So I thought S2 was sufficient.
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I think this question would be more interesting if they specified that x,y, and z are non-zero integers.
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TheGmatTutor
I think this question would be more interesting if they specified that x,y, and z are non-zero integers.

Haha it wouldn't be more interesting; it would be easier. :P

They got me too
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Is (x^7)(y^2)(z^3)>0?

(1) yz<0
(2) xz>0

From GMAT Club Test - m25 - Q34

The OA is C but I think it's E
Statement (1) : clearly insufficient since we don't have the sign of Z (since it has an odd exponent)
Statement (2) : clearly insufficient since Y can be 0
Both (1) and (2) : well yes Y can't be 0 but we still can't tell the sign of Z ! Consider this example : X=4 , Y= -1, Z= -1 ; we will have both statements right but the original expression will be negative

Am I wrong ?
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Barkatis
From GMAT Club Test - m25 - Q34

Is \((x^7)(y^2)(z^3) \gt 0\) ?

1. \(yz \lt 0\)
2. \(xz \gt 0\)




The OA is C but I think it's E
Statement (1) : clearly insufficient since we don't have the sign of Z (since it has an odd exponent)
Statement (2) : clearly insufficient since Y can be 0
Both (1) and (2) : well yes Y can't be 0 but we still can't tell the sign of Z ! Consider this example : X=4 , Y= -1, Z= -1 ; we will have both statements right but the original expression will be negative

Am I wrong ?

Inequality \(x^7*y^2*z^3>0\) to be true:
I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

(1) \(yz<0\) --> \(y\neq{0}\) (II is satisfied). Don't know about \(x\) and \(z\). Not sufficient.

(2) \(xz>0\) --> \(x\) and \(z\) are either both positive or both negative (I is satisfied). Don't know about \(y\). Not sufficient.

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

Answer: C.

As for your doubt: we are not interested in the sign of \(z\), we need \(x\) and \(z\) to be be either both positive or both negative. Next, your example is not valid: x=4, y=-1, z=-1 --> yz=4>0 and xz=-4<0 and we are given that \(yz<0\) and \(xz>0\).

Hope it helps.
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Bunuel
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Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct?

No, not correct. \(y^2\) is not always positive, it's never negative: \(y^2\geq{0}\).

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (note that both positive or both negative excludes the possibility of either of them to be zero) AND \(y\) must not be zero. Because if \(y=0\), then \(x^7*y^2*z^3=0\).


Thanks a ton Bunuel , this example is the perfect for learning that while considering signs we should consider +ve , -ve and zero as well . Superb collection .
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x^7*y^2*z^3>0
y is positive, we need to know regarding x and z
1) no information about x
insufficient

2) xz>0

either of them is negative hence

the answer to our question is no

sufficient

hence B
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kraizada84
x^7*y^2*z^3>0
y is positive, we need to know regarding x and z
1) no information about x
insufficient

2) xz>0

either of them is negative hence

the answer to our question is no

sufficient

hence B

OA for this question is C, not B.

(2) \(xz>0\) means that\(x\) and \(z\) are either both positive or both negative. So, \(x^7*z^3 > 0\) but y can still be zero and in this case \(x^7*y^2*z^3=0\), hence this statement is not sufficient.

Refer to the complete solution above.

Hope it helps.
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TheGmatTutor
Is (x^7)(y^2)(z^3) > 0 ?

(1) yz < 0

(2) xz > 0

S1 is insufficient because we need the sign of x.

S2 is SUFFICIENT, because x^7*z^3 will always be positive, and y^2 will be positive, so the whole product (x^7)(y^2)(z^3) will be positive. Answer: B.

(m25#34)


Asked: Is (x^7)(y^2)(z^3) > 0 ?
Q. xz>0 & y<>0?

(1) yz < 0
\(y, z \neq 0\)
NOT SUFFICIENT

(2) xz > 0
Since y may or may not be 0
NOT SUFFICIENT

(1) + (2)
(1) yz < 0
\(y, z \neq 0\)
(2) xz > 0
Since xz>0 and \(y \neq 0\)
SUFFICIENT

IMO C
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