
It is currently Tue Feb 25, 2020 11:47 am

View unanswered posts  View active topics

Page 1 of 1

[ 2 posts ] 

Author 
Message 
questioner

Post subject: GMAT Algebra Posted: Thu Feb 03, 2011 9:19 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

If x³ > y² > z, which of the statements could be true? I. x < y < z II. x < z < y III. y < x < z
A. I only B. III only C. I and II only D. II and III only E. I, II and III
(E) Choice I is possible: x = 3, y = 4, z = 5 Choice II is possible: x = 3, y = 5, z = 4 Choice III is possible: x = 4, y = 3, z = 6 The correct answer is E.  Is there any other method to solve this question? How can I choose the figure to plug in?


Top 


Gennadiy

Post subject: Re: math (test 5, question 36): inequalities Posted: Fri Feb 04, 2011 2:30 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

This is a tough question indeed and requires laconic reasoning. When we deal with exponents and inequalities the usual things to keep in mind are the properties of powers:
x³ > y² > z 1. negative/zero/positive Any square is a nonnegative number, while any cubed number has the same sign as the number itself.
Since x³ > y² and y² is a nonnegative number, then x³ > 0 and so x > 0. On the other hand, y can be positive or negative while y² will still have the same value.
2. greater than 1 / 1 / greater than 1 If 0< x < 1 then x³ < x² < x If x = 1 then x³ = x² = x = 1. If x > 1 then x³ > x² > x
Now, keep in mind those properties and analyze the options onebyone. We try to fix such values that fit or find a contradiction along the way. Let's start with the easiest one.
x³ > y² > z III. y < x < z Since x is positive so is z. Remember, that y can be positive or negative, while y² remains the same. If y is negative, then y < x holds. Now, consider the second part: x < z. On the other hand the original statement defines x³ > z. So x < z < x³. So we must pick a number for x and then pick z within the range. If x = 2, then 2 < z < 8. Let z be 3. So 8 > y² > 3. Let y be 2. Therefore the option III can be true.
x³ > y² > z I. x < y <z Let's write the both lines in one: x < y < z < y² < x³ We can see, that if we fix the values for x and y, then we can easily plug some value for z in between y and y². x < y < y² < x³ Let's fix the value for x first and then pick y in the range. Let x be 3. 3 < y < y² < 27 So y can be 4 and z can be any number in between 4 and 16. Therefore the option I can be true.
x³ > y² > z II. x < z < y This is the toughest one, but we use the same logic. x < z < y² < x³ and z < y Let's fix the value of x first. Let it be 3. 3 < z < y² < 27 and z < y Let y be 5. 3 < z < 25 < 27 and 3 < z < 5. z can be 4. Therefore the option II can be true.
Though it might seem to be quite long in writing, it's not that long in reasoning, especially when you have skills in dealing with inequalities and keep in mind the properties described in the beginning.


Top 



Page 1 of 1

[ 2 posts ] 

Who is online 
Users browsing this forum: No registered users and 5 guests 

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum

