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

Given: \(xy^2z^3<0\)
Asked; Is xyz>0

\(xy^2z^3<0\) => x,y,z are non-zero numbers
xz<0 since \(y^2z^2\) is always positive

For xyz>0
Since xz<0 => y<0 for xyz>0

(1) y<0
Since y<0 => xyz>0 since xz<0 and y<0
SUFFICIENT

(2) x<0
If x<0 => z>0 but there is no information about y
NOT SUFFICIENT

IMO A

If \(xy^2z^3<0\), is xyz>0?

\(xy^2z^3<0\)

Lets evaluate possible combinations of x, y, and z which will allow above inequality to be true (negative)
We need to have either 1 -ve or 3 negatives, but 3 negatives are not possible because y^2 can never be negative.. so y may be positive or negative, but only 1 out of x and z can be negative. Both x and z can never be negative together.

x____y____z
-____-____+..... x and y are negative
+____-____-..... y and z are negatrive
-____+____+..... only x is negative
+____+____-.... only z is negative


(1) y<0
Y is negative
So from above combination we can see that when y is negative, it is always combined with 1 other negative
So there will be 2 negatives and 1 positive... their product will always be positive

xyz>0 is true

(1) IS SUFFICIENT

(2) x<0
X is negative
X can be negative alone or along with Y
If X alone is negative, product will be negative, If X and Y are both negative, product will be positive.

(2) IS NOT SUFFICIENT


ANSWER: A - 1 ALONE IS SUFFICIENT

IMO-A

x * y^2 * z^3<0

Check: xyz>0?

x * y^2 * z^3<0 => Either x<0 0r z<0

Statement 1) y<0

Case 1: x<0 & z>0 with y<0
xyz = (-) x (-) x (+) = (+)
Case 2: x>0 & z<0 with y<0
xyz = (+) x (-) x (-) = (+)

Sufficient, xyz>0

Statement 2) x<0
x<0, z>0 & y<>0
xyz = (-) x (-/+) x (+) = (-/+)

Not Sufficient , xyz <>0

IMO Answer(A)

Check the snapshot below.

Analyzing the Qs Stem

Looking at the question we can see that either X or Y needs to be -ve

Why?

- Y is raised to an even power, so it will be positive
- Either X or Y needs to be -ve but not both as ( Negative X Negative = Positive) and the stem says that the product is a negative number.

So to prove that XYZ> 0. all we need to check is if Y is Positive or Negative
If Y is negative or Positive, then we can make a decision on sufficiency

Option A: y< 0 ( Perfect)!
Option B: Well, I mean can't say if X is positive or negative and that has an impact on the final answer, too much ambiguity here

Answer: A

Given,

xy^2z^3 < 0

To find if,

xyz > 0

Let us consider the following cases as per the signs of each of the terms

x y^2 z^3

- - - ----> Not possible as y^2 can't be -ve
+ - - ----> Not possible as y^2 can't be -ve
- + - ----> Not possible as xy^2z^3 is -ve
+ + - ----> Valid possible case
- - + ----> Not possible as y^2 can't be -ve
+ - + ----> Not possible as y^2 can't be -ve
- + + ----> Valid possible case
+ + + ----> Not possible as xy^2z^3 is -ve

Now let us run through the 2 cases through each of our options at hand.

Option 1: y < 0

Case 1:

x = +
y^2 = +
z^3 = -

As per this,

x = +
y = -
z = -

=> xyz > 0

Case 2:

x = -
y^2 = +
z^3 = +

As per this,

x = -
y = -
z = +

=> xyz > 0

Hence in both valid cases for this option, xyz > 0

Option 1 is sufficient.

Option 2: x < 0

Case 1:

x = +
y^2 = +
z^3 = -

This is not a valid case for this option as we are given that x is -ve.

Case 2:

x = -
y^2 = +
z^3 = +

=>
x = -
y = + or -
z = +

=>
xyz < 0 or xyz > 0

Hence,

Option 2 is insufficient.

Answer: A

From the stem we already know that \(xz<0\) and none of \(x, y,\) and \(z\) equal to \(0\). What we are unsure about is the sign of \(y\). To answer whether \(xyz>0\), we need to clarify whether \(y\) is positive or negative.


ST1. \(y<0\)

That’s the much needed piece of information. If both \(x*z\) and \(y\) are negative, then \(y\) will change the sign of the product \(x*y\). Therefore, we can surely conclude that \(xyz>0\).

Sufficient

ST2. \(x<0\)

If \(x\) is negative, then \(z\) must be positive so that \(x*z<0\). However, we still don’t know what is \(y\). If \(y<0\), then \(xyz>0\). If \(y>0\), then \(xyz<0\). We can see that undetermined \(y\) can change the sign of \(xyz\).

Insufficient

Hence A

Let's analyze the known and the question

\(xy^2z^3<0\) --> \(y^2\) is obviously positive, then:
- if \(x\) is \(positive\), then \(z\) must be \(negative\).
- if \(x\) is \(negative\), then \(z\) must be \(positive\).

is \(xyz>0\)?
- if \(x\) is \(positive\) and \(z\) is \(negative\), then \(y\) must be \(negative\).
- if \(x\) is \(negative\) and \(z\) is \(positive\), then \(y\) must also be \(negative\).
Thus, we know that the real question: is y<0?

(1) y<0
This statement EXACTLY addresses the real question.
SUFFICIENT

(2) x<0
We are only certain that if \(x\) is \(negative\), then \(z\) must be \(positive\). We don't know whether \(y\) is \(negative\) or \(positive\)
NOT SUFFICIENT


Answer is (A)

Question: If \(xy^2z^3<0\), is \(xyz>0\)?

\(xy^2z^3<0 \Leftarrow\) either \(x\) or \(y\) can be negative, NOT both.
\(x\) = can be positive or negative
\(y^2\) = will always have a positive value
\(z^3\) = a positive or negative sign of \(z\) will remain when cubed.

(1) \(y<0 \text{ }\) SUFF.
\(-y^2\) = will always have a positive value. Because either \(x\) or \(y\) can be negative, but not both...
\(xyz = (+)(+)(-) = neg\)
OR
\(xyz = (-)(+)(-) = neg\)

(2)\(x<0\) INSUFF.
\(-x =\) will have either a positive or negative sign effect on \(xyz\).
\(xyz = (-)(+)(+) = neg\)
OR
\(xyz = (-)(-)(+) = pos\)

Correct Answer: A

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