Links
About Arcadia
LSAT Shorthand NotationArcadia Prep's GMAT CourseArcadia Prep's LSAT CourseArcadia Prep's SAT CourseArcadia Prep's GRE CourseThe Simply Brilliant GMAT AppAbout the MCATMCAT Physical and Biological SciencesAbout Arcadia Prep's MCAT CourseTerms of SaleTerms of UseAbout Arcadia Apps
LSAT Prep
If xy2z < 0, then which one of the following statements must also be ... ...
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
(B) ...
(C) ...
(D) ...
(E) ...
*This question is included in Introduction to Nova GRE Math
Replies to This Thread: 1 | ----
What if y is an imaginary number?
Reply 1 of 1
Replies to This Thread: 0 | ----
Posted: 02/22/2012 16:07
Replies to This Thread: 0 | ----
Specifically, what if y= (-1)^(1/2)
Replies to This Thread: 0 | ----
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.
Imaginary number is a math concept introduced to make it easy to solve certain problems (e.g., those involving vectors) in physics / engineering.
Replies to This Thread: 1 | ----
If y was raised to a odd number would xyz<0?
Admin
Reply 1 of 1
Replies to This Thread: 0 | ----
Posted: 03/29/2012 17:12
Replies to This Thread: 0 | ----
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.
Replies to This Thread: 0 | ----
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.
AdminReplies to This Thread: 0 | ----
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.
Replies to This Thread: 0 | ----
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.
AdminReplies to This Thread: 0 | ----
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.
Replies to This Thread: 0 | ----
Assuming we were finding which quantity was greater then I don't think B would have been correct.
Replies to This Thread: 0 | ----
Z=-1, x= 1, y= -2
Replies to This Thread: 0 | ----
I need an explanation... I didn't get it
Replies to This Thread: 0 | ----
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.
Replies to This Thread: 0 | ----
What is the reasoning for dividing both sides by y^2?
Replies to This Thread: 0 | ----
Derya,
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.
Niels
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.
Niels
Replies to This Thread: 1 | ----
If xy2z < 0, then which one of the following statements must also be ... ...
Posted: 06/30/2013 19:53
What if y is negative no?
Reply 1 of 1
Replies to This Thread: 0 | ----
Posted: 07/02/2013 11:27
Replies to This Thread: 0 | ----
Ashish, if y < 0, then:
I. xz < 0, is true
III. xyz < 0, is not true.
I. xz < 0, is true
III. xyz < 0, is not true.
Replies to This Thread: 1 | ----
If xy2z < 0, then which one of the following statements must also be ... ...
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 :/
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.
Niels
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.
Niels
Admin
Reply 1 of 1
Replies to This Thread: 0 | ----
Posted: 11/14/2014 10:55
Replies to This Thread: 0 | ----
A mathematician with a respect for history. That's the kind of people we need more.
Replies to This Thread: 0 | ----
If xy2z < 0, then which one of the following statements must also be ... ...
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,
Replies to This Thread: 1 | ----
If xy2z < 0, then which one of the following statements must also be ... ...
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
Admin
Reply 1 of 1
Replies to This Thread: 0 | ----
Posted: 11/14/2014 11:20
Replies to This Thread: 0 | ----
Brittany, since in the explanation, z = 3, which is > 0, statement II is NOT true, and D can be eliminated.