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

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

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New post 23 Jul 2019, 08:00
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A
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E

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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:20
If xy2z3<0xy2z3<0, is xyz>0xyz>0?

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


E has to be the answer. Because we have to know the value of both X and Z
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:25
Stmt 1 :)
is y is squared power does value does not matter as it will always be positive.
not sufficient

stmt 2)
x is -ve y always positive
z?
not sufficient

together both also not sufficient as z can be positive or negative.
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:26
2
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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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post Updated on: 23 Jul 2019, 09:33
1
Quote:
If xyˆzˆ3<0, is xyz>0?

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


xyˆzˆ3<0: this means that x and z are different signs and all are ≠ 0.
[1] x>0 and z<0: +(-)=(-)… (-)y>0, y must be negative;
[2] x<0 and z>0: (-)+=(-)… (-)y>0, y must be negative;

(1) y<0; sufficient.
(2) x<0; we need y, insufficient.

Answer (A).

Originally posted by exc4libur on 23 Jul 2019, 08:29.
Last edited by exc4libur on 23 Jul 2019, 09:33, edited 1 time in total.
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:31
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IMO answer is A:

given xy^2z^3 <0
as y^2>0, xz^3<0, this implies x and z have opposite signs. xz<0
we only need sign of y to know the result xyz > 0 or xyz <0

from1:directly get y<0 suff
from2: not suff, all we know is sign of z from this
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:35
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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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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:36
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If\(xy^2z^3<0\), is xyz>0?

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

given \(xy^2z^3<0\)
thus \(xz^3<0\) as y^2 will always be positive.
to determine \(xyz>0\), we can check if \(y<0\) then (-)(-) will make positive
as xz is also zero
(1) y<0
thus this will make entire expression positive hence sufficient
(2) x<0
No help as we know either x or z sign make no use to us. We need y sign
Hence not sufficient
Thus answer is A
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:39
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If xy2z3<0, is xyz>0?

This is an easy question.
Before we see the options, let's understand the question.
We are given that xy2z3<0

Which means that Y^2 is positive so for xy2z3<0 to be negative either of X or Z is negative.

Now for us to prove xyz>0, We have two options either all are positive or two are negative and 1 is positive. The first option is out as we know that either of X or Z is already negative. So we are now looking if Y is negative or not.

Let's get to POE.

(1) y<0
This is exactly what we are looking for this proves us that XYZ > 0.
Hence this is sufficient.

(2) x<0
This is not sufficient alone as we already knew that X/ Z is going to be negative, the result still depends on Y.

Hence the answer is A.
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:40
1
A.

We are basically given xz<0, so for xyz>0 , we need y<0

St 1: y <0 sufficient
St 2: x<0 doesn't help. Not sufficient

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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:43
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The answer is A.

This is a question about positives and negatives.

If xy2z3<0, is xyz>0?

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

Since y2 is always positive, for xy2z3 to be negative, then either x or z has to be negative (either one, not both). Also, to answer the question of is xyz positive, since we know either x or z are negative (there is already a negative there), we need to know whether y is positive or negative to answer the question.

From (1) we know that y is negative. Since y is negative, then we can answer that xyz is positive "Yes" based on the conditions above. We don't need to know which variable X or Z is negative, we just need to know that y is negative to answer, so 1 is sufficient.

From (2) alone, we don't know anything about whether y is positive or negative so we can't answer the question. 2 is insufficient.

So answer is A.
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post Updated on: 23 Jul 2019, 22:38
1
\(x*y^2*z^3\)= negative
Thus, x and z will always have opposite signs. -> (a)
Also none of x, y and z equals to 0.

We need to find whether the product of xyz is positive.

(1) y < 0
When x is +ve and z is –ve:
xyz > 0

When x is –ve and z is +ve:
xyz > 0

Sufficient

(2) x<0
Given x is –ve, so from (a) z is +ve

When y < 0
xyz > 0

When y > 0
xyz < 0

Not Sufficient

Answer A

Originally posted by Sayon on 23 Jul 2019, 08:43.
Last edited by Sayon on 23 Jul 2019, 22:38, edited 1 time in total.
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post Updated on: 24 Jul 2019, 11:01
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If xy2z3<0, is xyz>0?

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

from given info
we can say that either x or z is -ve and y may be -ve which makes xy2z3<0
#1
y<0 ; either of x or z is -ve so sufficient that xyz>0 is possible
#2
x<0 dont know about y and z so insufficient

IMO A

Originally posted by Archit3110 on 23 Jul 2019, 08:44.
Last edited by Archit3110 on 24 Jul 2019, 11:01, edited 1 time in total.
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post Updated on: 24 Jul 2019, 03:50
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If \(xy^{2}z^{3}<0\), is xyz>0

Since \(y^{2}\) is always > 0 => \(xz^{3}<0\).
\(z^{3}\) will have the same sign as z, therefore xz<0
Therefore, to find out about xyz, we need to know the sign of y.

(1) y<0, y sign is given, therefore sufficient
(2) x<0, y sign is not given, therefore insufficient

Answer is A

Originally posted by berdibekov on 23 Jul 2019, 08:46.
Last edited by berdibekov on 24 Jul 2019, 03:50, edited 1 time in total.
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:52
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x.y^2.z^3 <0 ==> x.z^3<0 ==> x.z<0

To find whether xyz>0
we already have x.z<0

Ask is to only find whether y is positive or negative.

Statement 1 is sufficient

Hence, A is the answer.
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post Updated on: 23 Jul 2019, 08:57
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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
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Originally posted by Kinshook on 23 Jul 2019, 08:56.
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:56
1
If x*(y^2)*z^3<0, is xyz>0?

From the given inequality we can conclude that since y has power 2, it is always positive.
This means that x*z<0, either x positive and z negative or x negative and z positive.
We are asked whether xyz is positive.
Since we already know that xy<0 all we need is the value of y.

(1) y<0

Sufficient, as stated above.

(2) x<0

Not sufficient. We don't need to know the value of x.


IMO A
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 08:59
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\(y^{2}\) must be positive. Since x\(y^{2}\)\(z^{3}\) is negative, then either x or z is negative, however, both cannot be negative or the total product would be positive. We don't need to know if x or z is negative or positive since we know one or the other is. If we know y is negative then we know the overall total must be positive (two negatives * one positive = positive total).
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 09:04
1
Upon analyzing the question stem, we can infer that X & Z are of different signs, since the product of y^2 (+ve), X & Z is negative.

Statement 1

Y is negative. then the product of XZ (-ve) and Y (-ve) will be positive.

Statement 2

X is negative. Then, we know that Z is +ve but we cannot further conclude whether product of XYZ will be negative.
Case 1. XZ (-ve) & Y (-ve) = +ve

Case2. XZ (-ve) & Y (+ve) = -ve
(2) x<0x<0
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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New post 23 Jul 2019, 09:11
Our answer is going to be affected by 3 cases to prove if xyz >0 :
x y z
+ + - = result will be xyz <0
- + + = result will be xyz < 0
+ + + = result will be xyz >0
Have taken y positive as y^2 : will always be positive
So we need all 3 x,y,z > 0 to prove question stem

Statement 1 : y <0 : It actually doesn't have any effect because its going to be positive in any case: So in sufficient
Statement 2 : As discussed, we also need to know what is z which is not given : x < 0 gives 2 results , so insufficient

Combining 1 and 2 : Still we don't have any information for Z , so Option E
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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0   [#permalink] 23 Jul 2019, 09:11

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