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Source: — Data Sufficiency |
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by shankar.ashwin » Sun Nov 13, 2011 11:56 am
xyz -> +ve (Either all 3 Nos are +ve or exactly 2 are -ve)

Statement 1

y -> -ve -> Either 'x' or 'z' is -ve

y^2 will be +ve and since either 'x' or 'z' is -ve x*z^3 will be < 0 (odd powers) Sufficient.

Statement 2:

You have 2 cases 'y' and 'z' can both be +ve or -ve.

Both +ve -> x*y^2*z^3 > 0
Both -ve -> x*y^2*z^3 < 0 Insuff

A IMO
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by bpdulog » Sun Nov 13, 2011 4:00 pm
I think it's E.

xy raised to 2z will mask the sign of xy since it's positive no matter what.

We also don't know if z is positive or negative.
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by neelgandham » Sun Nov 13, 2011 4:12 pm
bpdulog wrote:I think it's E.

xy raised to 2z will mask the sign of xy since it's positive no matter what.

We also don't know if z is positive or negative.
I am not sure which step are you pointing at but

xy^2z is always > 0.

How ?

a^z is always > 0 if a >0
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by neelgandham » Sun Nov 13, 2011 4:17 pm
ashish1354 wrote:If xyz > 0, is xy∧2z∧3 < 0 ?
(1) y < 0
(2) x > 0
Please solve...
Ashish, Can you please post the question in a way that people wouldn't be confused ?

xy∧2z∧3 < 0 can be

a) x*(y^2)*(z^3) or
b)((xy)^(2*z))^3 or
c)x *((y^(2*z))^3)
d) ((xy)^2)*(z^3)
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by bpdulog » Sun Nov 13, 2011 5:11 pm
neelgandham wrote:
bpdulog wrote:I think it's E.

xy raised to 2z will mask the sign of xy since it's positive no matter what.

We also don't know if z is positive or negative.
I am not sure which step are you pointing at but

xy^2z is always > 0.

How ?

a^z is always > 0 if a >0
If you square a negative or a positive number it will always be positive. So that holds true when you have an even exponent, such as 2z. The negatives will cancel out.

So, if xy were positive, z would be positive.

If xy were negative, z would be negative.
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