If xy2z < 0, then which one of the following statements must also be true?

  I. xz < 0

 II. z < 0

III. xyz < 0

(A)   None

*This question is included in Introduction to Nova GRE Math, question #8

The solution is

Posted: 02/22/2012 15:55
What if y is an imaginary number?
Posted: 02/22/2012 16:07
Specifically, what if y= (-1)^(1/2)
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Posted: 02/24/2012 18:13
Wayne, thanks for coming up with a creative scenario. GRE math questions only test algebra and geometry, and do not test advanced concepts like imaginary numbers. You can safely assume you are working with real numbers, or integers if it is so specified.

Imaginary number is a math concept introduced to make it easy to solve certain problems (e.g., those involving vectors) in physics / engineering.
Posted: 03/29/2012 15:31
If y was raised to a odd number would xyz<0?

Posted: 03/29/2012 17:12
If y is raised to an odd number, say 3, and x•y^3•z < 0, then we don't have enough information to say xyz < 0.
Posted: 04/20/2012 02:37
If we consider x= 1 y = 2 and z=-3 then xz<0, z<0, xyz<0 so all the three answers will be true.
Posted: 04/20/2012 10:57
Nitin, the question asks which choices will be true for all scenarios, not just for one particular scenario. So your answer would be incorrect in a different scenario.
Posted: 04/24/2012 05:16
in this question, i understood that y must be a positive number. therefore in order for it to be less than than 0, i thought x or y has to be a negative number, which cannot be determined. as a result, it would make xyz<0 as well as xz.
Posted: 04/24/2012 16:37
Jacob, y is not always a positive number. y^2 is. To make x•y^2•z a negative number, therefore, x•z must be negative. That's the only thing we can conclude from the three statements.
Posted: 06/14/2012 20:31
Assuming we were finding which quantity was greater then I don't think B would have been correct.
Posted: 06/28/2012 22:40
Z=-1, x= 1, y= -2
Posted: 10/05/2012 07:36
I need an explanation... I didn't get it
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Posted: 10/05/2012 14:40
Nik, there is explanation that you can access from the app. Tap on Show Correct Answer after you make your choice, before you go to the next question.
Posted: 11/24/2012 12:28
What is the reasoning for dividing both sides by y^2?
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Posted: 11/25/2012 01:54
You refer to the in-app explanation:
After the conclusion that all even powers including ^2, regardless of the base number have a positive product, the decision is made to rule y^2 out of the equation which yields xz<0, thus only xz will determine the final positive product, thus both x and z must be either both positive or both negative numbers.

Posted: 06/30/2013 19:53
What if y is negative no?
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Posted: 07/02/2013 11:27
Ashish, if y < 0, then:
I. xz < 0, is true
III. xyz < 0, is not true.
Posted: 07/15/2013 18:18
These are the type of questions that confuse me. Even after I asked to see the solution I'm still lost :/
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Posted: 07/16/2013 02:46
Ivorye, this question depends on basic math knowledge like number properties. You could get the book "Foundation Math" by Anthony Croft and Robert Davison, and make sure this is part of your foundation.
I prefer to read books about math history as well it 'humanizes' the stack of rules they throw at you and you stop feeling like an ape doing tricks and you learn to know some of histories greatest men like Euclid.

Posted: 11/14/2014 10:55
A mathematician with a respect for history. That's the kind of people we need more.
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Posted: 07/19/2013 21:18
This question is so so tricky, especially when using the term (must be true) which should apply only for one specific choice, yes Z can be less than zero but not always, because but it can be positive and the X is the negative, and vise versa, so for sure XZ less than zero, because one of the them is negative, the other is positive, (-) x (+) = - < 0,
Posted: 11/14/2014 09:55
I am confused on why D was eliminated. Since -12 is less than 0, wouldn't that make the equation xy^2z
Posted: 11/14/2014 11:20
Brittany, since in the explanation, z = 3, which is > 0, statement II is NOT true, and D can be eliminated.

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