If x y z, which of the following could be true?

Question

If x > y > z, which of the following could be true?

I.  x^2 > y^4 > z^2

II.  z > y^4 > x^2

III.  x^2 > z^2 > y^4

A. I only 
B. II only
C. I and II only
D. I and III only 
E. I, II and III

AndrewN · Answer

We have to be careful here not to make assumptions, since the given inequality does not tell us anything about the nature of the unknowns—they could be any combination of negatives, 0, positives, fractions, or integers. It may help to pull apart each compound inequality to check for a true or false statement. ---------- I. Looking at the first inequality, it should be clear that if x > z, it could be true that x^2 > z^2 Thus, we can ignore that separate inequality. But what about the others? x^2 > y^4 > z^2 becomes x^2 > y^4 AND y^4 > z^2 Since y is the linchpin, why not think of it as something simple, such as 1? Now, the first inequality seems simple, even in compound form: x^2 > (1)^4 > z^2 x^2 > 1 > z^2 As long as z is a fraction or 0, we can see how the first inequality could be true , without doing any additional math. Get rid of answer choice (B). ---------- II. To assess the second inequality, we need to look at the ends: z > ... x^2 It should be clear that z cannot be negative, 0, or any value 1 or greater. Why? Because as soon as z is greater than or equal to 1, the other unknowns have to be larger, per the original inequality, and squaring them or taking them to the fourth power will only increase the space between them and z . So, we have to consider whether z could be some fraction among other fractions. What about something simple, such as x=\frac{3}{10} y=\frac{2}{10} z=\frac{1}{10} z > y^4 > x^2 (\frac{1}{10}) > (\frac{2}{10})^4 > (\frac{3}{10})^2 \frac{1}{10} > \frac{2^4}{10^4} > \frac{9}{100} \frac{1}{10} > \frac{16}{10000} > \frac{9}{100} This is not true because of the latter inequality. But we could also set z ahead of time to be slightly greater than y to the fourth power. We could make it 17/10000, for example. We would just need to ensure that x , when squared, would be less than 16/10000. To do so, we could just take the square root: x^2 < \frac{16}{10000} \sqrt{x^2} < \sqrt{\frac{16}{10000}} x < \frac{4}{100} We know that this inequality will be invalid, because y itself was set at 2/10, or 20/100, and, again, x must be greater than y . Considering everything above, then, the second inequality cannot be true , so strike answer choices (C) and (E) off the list. ---------- III. Finally, with the third inequality, we can think of x as some large integer just to ease the burden of juggling a compound inequality—we already know that x is the greatest value of the three unknowns. Thus, x^2 > z^2 > y^4 becomes z^2 > y^4 As long as the above inequality could be true , we know the third inequality could work. With an understanding of how fractions behave in exponential functions, we do not need to test numbers, although we can. Consider: Let z = 1/3 and y = 1/2 (\frac{1}{3})^2 > (\frac{1}{2})^4 \frac{1}{9} > \frac{1}{16} Since this is true, and again, with no restrictions on what types of numbers the unknowns can be, we can say that the third inequality could be true . Since I and III could be true, the answer must be (D) . I had fun with this one. It forces us to be careful not to make assumptions about just what types of numbers we are dealing with. Good luck with your studies. - Andrew

Kinshook · Answer

Asked: If x > y > z, which of the following could be true?

I. \(x^2 > y^4 > z^2\)
If x=4,y=3,z=2; COULD BE TRUE
II. \(z > y^4 > x^2\)
z>xˆ2; z>0; If z>1; the inequality is NOT possible. 0<z<1; x, y & z are fractions.
Tried a few specific cases.
NOT POSSIBLE
III. \(x^2 > z^2 > y^4\)
If x=2; y = .5; z=.4; COULD BE TRUE

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II and III

IMO D

AndrewN · Answer

Kinshook , I have read enough of your Quant posts to know that you know your stuff, but the  highlighted  part above is problematic. I wonder if you went through the problem or posted a solution too fast.

In my view, inequality II is the tricky one, the one that probably pushes the question into 700-level territory. Understand, I am not drawing attention to an error in your post to be mean—I make mistakes, too. It is just that I feel as if an Expert ought to guide the community to an accurate answer through sound reasoning, and with that aim in mind, I made a decision.

Thank you for your contributions to GMAT Club.

- Andrew

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If x > y > z, which of the following could be true?

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If x > y > z, which of the following could be true?

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