If xy^2z^3 0? (1) y

Question

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

(1)  y < 0 
(2)  x < 0

abhishekdadarwal2009 · Answer

IMO : A

If

xy^2z^3<0 , is xyz>0?

(1) y<0

(2) x<0

Sol:

from the give info we know that either x or z is negative because y^2 cannot be negative.

so

from 1) y<0 tell that two from x,y,or z is negative that means xyz>0

sufficinet.

2)x<0 we cant say anything about y, if y is <0 then the value become positive and if y>0 then the value becomes negative.

So A is sufficient.

kitipriyanka · Answer

By given equation we know, either x<0 or z <0

By 1, y<0 and either x<0 or z <0, hence xyz>0. Sufficient.
By 2, we have no info about y. Insufficient.

Hence A

Kinshook · Answer

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

Vinit1 · Answer

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

MayankSingh · Answer

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

unraveled · Answer

IMO Answer(A)

Check the snapshot below.

hk2760 · Answer

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

somesh86 · Answer

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

JonShukhrat · Answer

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

freedom128 · Answer

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)

duchessjs · Answer

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

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