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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x

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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x  [#permalink]

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New post 11 Dec 2018, 01:05
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

46% (01:03) correct 54% (01:39) wrong based on 68 sessions

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Re: Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x  [#permalink]

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New post 11 Dec 2018, 04:25
1
Bunuel wrote:
Given z > 0, is x*z > 0?

(1) xz = 3x
(2) xz = -3x


x*z>0

z>0

from 1:
xz=3x
or
z=3
xz=3x
to be valid x can be any value of integer +/- ( 1,1/9, 1/3 ...) so xz can be either > or < 0 not sufficient

2.
xz=-3x
z=-3

given z>0 so x would be -ve value or say -1

so xz <0

IMO B
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Re: Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x  [#permalink]

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New post 16 Dec 2018, 23:13
Archit3110 wrote:
Bunuel wrote:
Given z > 0, is x*z > 0?

(1) xz = 3x
(2) xz = -3x


x*z>0

z>0

2.
xz=-3x
z=-3

given z>0 so x would be -ve value or say -1

so xz <0

IMO B


I have hard time understanding S2.

xz = -3x
xz+3x=0
x(z+3)=0
x=0; z=-3 (this cannot be true since we are given that z>0.)

The above case was true when x = 0.

But if x were negative - say as suggested by archit - (-1), then

xz = -3x
-1 * z = -3 * -1
z = -3

z is still negative. That means x cannot be -1 or any negative value for that matter.

If x were positive - say 1.
1 * z = -3 * 1
z = -3

Is there any other value I can try?

Bunuel, could you please help me understand what is the gap in my understanding!


Thank you!
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Re: Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x  [#permalink]

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New post 16 Dec 2018, 23:23
1
KB04 wrote:
Archit3110 wrote:
Bunuel wrote:
Given z > 0, is x*z > 0?

(1) xz = 3x
(2) xz = -3x


x*z>0

z>0

2.
xz=-3x
z=-3

given z>0 so x would be -ve value or say -1

so xz <0

IMO B


I have hard time understanding S2.

xz = -3x
xz+3x=0
x(z+3)=0
x=0; z=-3 (this cannot be true since we are given that z>0.)

The above case was true when x = 0.

But if x were negative - say as suggested by archit - (-1), then

xz = -3x
-1 * z = -3 * -1
z = -3

z is still negative. That means x cannot be -1 or any negative value for that matter.

If x were positive - say 1.
1 * z = -3 * 1
z = -3

Is there any other value I can try?

Bunuel, could you please help me understand what is the gap in my understanding!


Thank you!


From (2) we got that x = 0. So, why are you considering other values for x?


Given z > 0, is x*z > 0?

Since z > 0, then the question basically asks whether x > 0.


(1) xz = 3x
x(z - 3) = 0.
x = 0 or z = 3. Notice that if z = 3, then x can be any number, negative, positive or 0 (if z = 3, then xz = 3x holds true irrespective of the value of x).
Not sufficient.

(2) xz = -3x.
x(z + 3) = 0.
x = 0 or z = -3. Since we know that z > 0, then z = -3 is not possible and thus x = 0, which gives a NO answer to the question whether x > 0. Sufficient.

Answer: B.
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Re: Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x  [#permalink]

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New post 16 Dec 2018, 23:33
Bunuel wrote:
KB04 wrote:
Bunuel wrote:
Given z > 0, is x*z > 0?

(1) xz = 3x
(2) xz = -3x


x*z>0

z>0


(2) xz = -3x.
x(z + 3) = 0.
x = 0 or z = -3. Since we know that z > 0, then z = -3 is not possible and thus x = 0, which gives a NO answer to the question whether x > 0. Sufficient.

Answer: B.


Thank you, Bunuel!

I see where I was going wrong.
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Re: Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x &nbs [#permalink] 16 Dec 2018, 23:33
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