ZERO

[GmatKiss]

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

1. yz < 0
2. xz > 0

[neelgandham]

Q) (x^7)*(y^2)*(z^3) > 0 ?

1. yz<0 so either

The term (x^7)*(y^2)*(z^3) can be written in the form (x^7) * z * ((yz)^2). So if yz < 0 the sign of the equation depends on the sign of (x^7) * (z). Hence, insufficient


2. xz >0 so either

The term (x^7)*(y^2)*(z^3) can be written in the form (x^4) * (y^2) * ((xz)^3). So if xz < 0 the sign of the equation depends on the sign of (x^4) * (y^2). [spoiler](x^4) * (y^2) >=0 (0, when y =0), Hence insufficient[/spoiler]

Using 1 and 2, we get [spoiler]y != 0[/spoiler], Hence option C

[pemdas]

st(1) is Not Sufficient
st(2) Must be Not Sufficient, the product of x and z can be +ve when x,z<0 or x,z>0

the final will be either +*+*+ >0 or -*+*- >0 Only if y is not 0 and we could answer Yes BUT y is not known

combined st(1&2): yz<0 we know that z can't be 0 and Sufficient
c

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

1. yz < 0
2. xz > 0

[GmatKiss]

OA is C

[vivgoyal]

for b the equation can be rewritten as x6y2z2*xz, so x6 is always +ve, y2 is always +ve, z2 is always +ve and given xz +ve

so the equation is always greater than zero
Hence B

[user123321]

Just B is enough.

user123321

[Brent@GMATPrepNow]

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

1. yz < 0
2. xz > 0

This question is tricky in that it exploits a common belief that the square of a number will always be positive.
Examples: 3^2 = 9, (-5)^2 = 25, 1^2 = 1, etc.
However, 0^2 is not positive. So, the belief falls apart here.

What we can say is: the square of a number will always be greater than or equal to zero.

Now on to the question.

Statement 1: yz < 0
consider 2 cases:
case a) x=1, y=-1, z=1. In this case, (x^7)(y^2)(z^3) is greater than 0
case b) x=-1, y=-1, z=1. In this case, (x^7)(y^2)(z^3) is not greater than 0
INSUFFICIENT

Statement 2: xz > 0
consider 2 cases:
case a) x=1, y=1, z=1. In this case, (x^7)(y^2)(z^3) is greater than 0
case b) x=1, y=0, z=1. In this case, (x^7)(y^2)(z^3) is not greater than 0
INSUFFICIENT

Statements 1& 2
Statement 1 eliminates the possibility that y=0 and z=0
Statement 2 eliminates the possibility that x=0 and z=0
So, we know that no variable equals 0

At this point, we can use two nice rules:
If k does not equal zero, then k^(even #) is positive
If k does not equal zero, then k^(odd #)has the same sign as k

So, if xz > 0 (from statement 1), then (x^7)(z^3) > 0
In other words, (x^7)(z^3) is positive
Also, if y does not equal 0, we know that (y^2) is positive.

So, altogether, we can see that (x^7)(y^2)(z^3) = (positive)(positive) = positive
In other words, (x^7)(y^2)(z^3) > 0

SUFFICIENT

Answer = C

Cheers,
Brent