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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 07:00
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If \(xy^2z^3 < 0\), is \(xyz>0\)? (1) \(y < 0\) (2) \(x < 0\)
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 07:26
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.



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 07:35
Quote: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0 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



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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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Updated on: 23 Jul 2019, 07:57
Given: \(xy^2z^3<0\) Asked; Is xyz>0
\(xy^2z^3<0\) => x,y,z are nonzero 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
Originally posted by Kinshook on 23 Jul 2019, 07:56.
Last edited by Kinshook on 23 Jul 2019, 07:57, edited 1 time in total.



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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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Updated on: 23 Jul 2019, 20:27
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
Originally posted by Vinit1 on 23 Jul 2019, 08:18.
Last edited by Vinit1 on 23 Jul 2019, 20:27, edited 2 times in total.



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 08:43
IMOA
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



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 09:35
IMO Answer(A) Check the snapshot below.
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 10:19
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



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 11:38
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



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 14:53
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 \(xyz>0\). SufficientST2. \(x<0\) 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 \(xyz\). InsufficientHence A
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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23 Jul 2019, 16:45
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)



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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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24 Jul 2019, 03:27
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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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0
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24 Jul 2019, 03:27






