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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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23 Jul 2019, 07:00
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62% (01:11) correct 38% (01:27) wrong based on 271 sessions

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

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

 This question was provided by Math Revolution for the Game of Timers Competition

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

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23 Jul 2019, 07:26
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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Re: If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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23 Jul 2019, 07:35
2
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  [#permalink]

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Updated on: 23 Jul 2019, 07:57
1
Given: $$xy^2z^3<0$$

$$xy^2z^3<0$$ => 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, 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  [#permalink]

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Updated on: 23 Jul 2019, 20:27
1
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  [#permalink]

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

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23 Jul 2019, 09:35
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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23 Jul 2019, 10:19
1
1
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

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

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23 Jul 2019, 11:38
1
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.

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

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23 Jul 2019, 14:53
1
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
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If xy^2z^3 < 0, is xyz>0? (1) y < 0 (2) x < 0  [#permalink]

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23 Jul 2019, 16:45
1
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

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

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24 Jul 2019, 03:27
1
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
$$xyz = (-)(-)(+) = pos$$

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