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If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0

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Bunuel
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If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   03 Jan 2021, 15:08
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If \(xyz > 0\), is \(x(y^2)(z^3) > 0\) ?

(1) \(y > 0\)
(2) \(x > 0\)
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A
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If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   03 Jan 2021, 20:12
Given \(xyz > 0\). Means either all x, y, and z are positive or 2 of them are negative.

From 1:

\(y > 0\)
\(y^2\) will always be positive. But value of x and z are unknown. Hence 1 is insufficient.

From 2:

\(x > 0\)
x = +ve. But no info about y and z.
Hence 2 is insufficient.

Combining both
x and y are > 0. Then z has to be +ve.
Hence \(x(y^2)(z^3) > 0\)

C is the answer.
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Re: If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   03 Jan 2021, 20:44
If xyz>0xyz>0, is x(y2)(z3)>0x(y2)(z3)>0 ?

(1) y>0
(2) x>0

1)y>0, no info abt x and z. Insufficient,

2)x>0
y^2 will be +ve, as square of any number is always positive
but no info on z. Insufficient

Combining both
x and y are > 0. Then z has to be +ve.
Hence x(y2)(z3)>0x(y2)(z3)>0

C is the answer.
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If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   04 Jan 2021, 10:05
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Bunuel wrote:
If \(xyz > 0\), is \(x(y^2)(z^3) > 0\) ?

(1) \(y > 0\)
(2) \(x > 0\)


Please enlighten me here
From (1) we get that Y>0
That means X and Z both will be either positive or negative
If both are positive in that case x(y^2)(z^3) > 0 -----True
If both are negative in that case x(y^2)(z^3) > 0 -----True

So in my opinion 1 is sufficient.
Please correct my flaws
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Re: If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   04 Jan 2021, 10:30
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Here option A is correct as ,if Y>0
then either a and z both are positive or both are negative as xyz >0.
In the both the conditions xy^2z^3 will be postiive.
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Re: If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   15 Jan 2021, 22:21
If xyz>0, is x(y^2)(z3)>0?

Since xyz > 0, there are two possibilities:
1. (+)(+)(+) OR
2. (+)(-)(-)

We need to see what happens to these scenarios given the ranges below.

(1) y>0
Scenario 1 checks out if all three are positive so x(y^2)(z3)>0
Scenario 2 checks out if y is positive, but x and z are negative so x(y^2)(z3)>0
Sufficient.

(2) x>0
Scenario 1 checks out if all three are positive so x(y^2)(z3)>0
Scenario 2 does not check out because y^2 ends up positive so you get (+)(+)(-) so x(y^2)(z3) <0.
Insufficient.
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If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   05 Jun 2022, 03:36
xyz>0

Question: x(y^2)(z^3)>0 ??? as we know that x(y^2)(z^3) = xyz(yz^2),
We can rephrase the Question as: xyz(yz^2)>0???

Given that xyz>0 (+) and z^2>0 (+) we only need to know the sign value of y.

Answer A
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Re: If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink] New post   06 Jul 2023, 00:24
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Re: If xyz > 0, is x(y^2)(z^3) > 0 ? (1) y > 0 (2) x > 0 [#permalink]
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