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by heshamelaziry » Tue Nov 24, 2009 9:57 pm
2010gmat wrote:is x^7*y^2*z^3 > 0?

1. yz <0
2. xz >0

IMO C.

1) y can be + and z -, or y- and z +. also, no info about the sign of x. Insuff

2) both z and x are positves or negatives. No info about the sign of Y. Insuff

Combo:

from 2, if x and z are negatives will satisfy statement 2 and and if y is + this will satisfy statement 1. So x is -, y is +, z is - will be sufficient to answer the question "YES"
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by papgust » Tue Nov 24, 2009 10:28 pm
IMO it's B. This a little tricky one.

1) As Hesham said, its not sufficient.

2) But here xz > 0
Either x and z should be positive or x and z should be negative.

In either case it is positive. Eventhough no information is given about y, in the questions stem it is given as y^2 --> Which means that Regardless of what sign y has, y^2 is going to be positive in all cases.

So x^7*y^2*z^3 > 0 (YES).

Hence, B
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by papgust » Tue Nov 24, 2009 10:32 pm
My mistake.. Apologies.

I didn't consider y as 0 for 2nd stmt.

So, C should be the answer as hesham has shown.
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by 2010gmat » Wed Nov 25, 2009 12:23 am
gr8...i too missed y =0

seemed so simple...but taught me that i shud never take a single question lightly...
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by beatthegmat2910 » Wed Nov 25, 2009 11:34 am
I solved this question a little differently.

x^7.y^2.z^3

Now, irrespective of what the sign of x,y or z is , the square will always be positive.

So, (x^2.y^2.z^2) .x^4.x.z
The underlined part will always be positive.
so, that leaves the product of xz that we don't know the sign of.
Option 2 gives xz > 0, hence 2 alone is sufficient to solve this question. Hence IMO (B).

Can you please tell me whats the OA.
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by bigmikefr » Wed Nov 25, 2009 8:40 pm
The answer is B and here is why: In order for x^7*y^2*z^3 to be greater than 0, x and y need to have the same sign, that is they should both be negative or positive for the number to be positve cuz y^2 willl always be positive. Rephrased another way, "Do x & y have the same sign?"

(1) yz <0 this means, either z or y could be negative. Insufficient

(2) xz >0 Bingo!!!! This is just what we want. They are both negative or positive
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by 2010gmat » Wed Nov 25, 2009 9:53 pm
oa is C,...and rightly so...

2 says xz > 0 ...but wat if y = 0??

the expression wud become = 0 and not > 0....
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by Stuart@KaplanGMAT » Wed Nov 25, 2009 11:59 pm
beatthegmat2910 wrote: Now, irrespective of what the sign of x,y or z is , the square will always be positive.
Close - but no cigar.

The square of a number will always be non-negative, which isn't the same as positive.

When you're dealing with exponents, you always want to keep in mind the properties of -1, 0, 1 and positive fractions, which can behave differently from other numbers. In this particular case we need to remember that 0^2=0, so if y=0 the answer will be "no".

As others have noted, (1) tells us that y can't be 0, so combined with (2) we know that the answer to the question is a definite "yes": choose C.
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