finding the relation between 2 numbers

[mkhanna]

If x>y^2 (ie. to the power of 2) >z^4, which of the following statements could be true?
I) x>y>z
II) z>y>x
III) x>z>y

Answer choices:
1) I only
2) I and II only
3) I and III only
4) II and III only
5) I, II, and III

[Stuart@KaplanGMAT]

If x>y^2 (ie. to the power of 2) >z^4, which of the following statements could be true?
I) x>y>z
II) z>y>x
III) x>z>y

Answer choices:
1) I only
2) I and II only
3) I and III only
4) II and III only
5) I, II, and III

When we see a relationship question involving exponents, we should immediately think about negatives and fractions, which do wild and wonderful things when we raise them to powers.

Let's begin by understanding the question: we want to know which statements "could be true". Almost certainly the best approach is going to be to try to pick numbers to make the statements true - as soon as we find 1 example, we know that statement is part of the solution.

For roman numeral questions, we generally start by looking at the answer choice patterns. Here, I shows up most often, so let's start by evaluating statement I.

I) x > y > z

If we pick x=1000000, y=500 and z=1 we certainly satisfy our original rule (since 1000000 > 250000 > 1). Therefore, I "could be true".

Eliminate choice (4).

The remaining choices equally include II and III, so let's evaluate whichever one seems simplier. I'd go with II.

II) z > y > x

We still have to hold to our original restriction (x > y^2 > z^4). So, let's think of numbers that get smaller when we raise them to big exponents: positive fractions.

Accordingly, we can pick x=1/4, y=1/3 and z=1/2 (since 1/4>1/9>1/16) to show that it's possible for z>y>x.

Eliminate choices (1) and (3).

Sadly, we still have two choices remaining, so we must also check III.

III) x > z > y

With this mixed-up arrangement we're going to have to be more creative.

Making x the biggest is simple enough - pick x=1000000 again.

The trick is making z > y. We know that y^2 > z^4. Let's ask ourselves, "when is a number bigger but a power smaller?" Let's answer ourselves just as we did for statement II - when we use positive fractions.

So, for example, we could once again let z = 1/2 and y = 1/3.

1/2 is bigger than 1/3, however 1/9 > 1/16.

Our final numbers: x=1000000, z=1/2, y=1/3. These numbers are legal because:

x > y^2 > z^4

(1000000 > 1/9 > 1/16).

Therefore, III also "could be true"... choose (5) I, II and III.