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655-705 (Hard)|   Algebra|   Inequalities|      
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Bunuel
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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 11 Dec 2018, 02:05
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55% (hard)

Question Stats:

54% (01:35) correct 46% (01:43) wrong based on 145 sessions

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

(1) xz = 3x
(2) xz = -3x
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B
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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 11 Dec 2018, 05:25
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Bunuel
Given z > 0, is x*z > 0?

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

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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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 17 Dec 2018, 00:13
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Archit3110
Bunuel
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
Show more

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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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 17 Dec 2018, 00:23
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KB04
Archit3110
Bunuel
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!
Show more

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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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 17 Dec 2018, 00:33
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Bunuel
KB04
Bunuel
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.
Show more

Thank you, Bunuel!

I see where I was going wrong.
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Given z > 0, is x*z > 0? (1) xz = 3x (2) xz = -3x 12 Sep 2020, 21:47
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