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In the case with positive powers, the balance of three variables looks the following way:
  • When \(x>y>z>0\), in the fourth power the same inequality is going to remain, regardless of whether the numbers are integers or decimals below 1.
  • When \(0>x>y>z\), in the fourth power all signs change to the opposite, preserving the order, again regardless of whether the numbers are integers or decimals above -1:
    \(0>-1>-2>-3\), but \(0<(-1)^4<(-2)^4<(-3)^4\)

This means, obviously, that both Option I and Option II are possible, pretty much for any number available.

With Option III the situation is trickier, as it changes the order of ascendance among the variables.
The only case when this is possible is when zero 'cuts into' the sequence of numbers, making some of them positive, and others - negative. For instance:
\(2 > 0 > -1\), but \(16 > 1 > 0\) - so the last two numbers switched places.
However, as can be seen from the example above, zero can only be in the middle - because when zero is in the first place, all the other numbers are negative, and we see the simple case discussed above.
Also, while it's possible to switch places of X and Y, like \( x>y\) with \(1>-2\), but \(x^4<y^4\), as \(1^4<(-2)^4\). However, Z will always travel from the lowest to the highest position, so there's no way to preserve \(z^4\) as the smallest member.
Therefore, Option III is impossible.

The correct answer is therefore C. I and II only.
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If x>y>z, which of the following may be true?

I. x^4>y^4>z^4
II. z^4>y^4>x^4
III. y^4>x^4>z^4


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

Let x = 5, y = 4 and z = 3
x^4 = 625
y^4 = 256
z^4 = 81

Hence, Case 1 i.e, x^4>y^4>z^4 is POSSIBLE

Case II. z^4>y^4>x^4

Let x = 2, y = -3 and z = -4
x^4 = 16
y^4 = 81
z^4 = 256

Hence, Case 2 i.e, z^4>y^4>x^4 is POSSIBLE

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

This case is never possible for the given condition x>y>z
For each value of x and y satisfying the condition y^4 > x^4, y must be less than zero and x must be greater than zero and |y| must be greater than |x|

But with this condition, z^4 will ALWAYS be greater than y^4 as z < y and thus |z| will ALWAYS be greater than |y|
Hence, Case 3 is NEVER POSSIBLE.

Answer : Option C
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If x>y>z , which of the following may be true?


With x=5,y=3, z=1;

x^4>y^4>z^4 (thus (I) may be true)

With x=-1,y=-3, z=-5;

z^4>y^4>x^4 (thus (II) may be true)

No ordered pair (x,y,z) can be found such that

x>y>z
also, y^4>x^4>z^4 (thus (III) will not be true)

(C) will be the right choice
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Deconstructing the Question

We are given \(x>y>z\) and asked which fourth-power orderings may be true.

The key idea is that fourth powers depend only on absolute value:
\(x^4=|x|^4\)

So the order of \(x,y,z\) does not automatically determine the order of \(x^4,y^4,z^4\).

Step-by-step

I can be true.

Example:
\(x=3,\ y=2,\ z=1\)

Then:
\(x^4=81,\ y^4=16,\ z^4=1\)

So:
\(x^4>y^4>z^4\)

II can also be true.

Example:
\(x=-1,\ y=-2,\ z=-3\)

Since \(-1>-2>-3\), the condition \(x>y>z\) is satisfied.

Then:
\(x^4=1,\ y^4=16,\ z^4=81\)

So:
\(z^4>y^4>x^4\)

III cannot be true.

To have \(y^4>x^4>z^4\), we would need \(|y|>|x|>|z|\). But with \(x>y>z\), this ordering cannot be achieved consistently, because making \(|y|>|x|\) forces \(y\) negative in a way that also makes \(z\) too far left, so \(|z|\) will not stay below \(|x|\).

Therefore only I and II may be true.

Answer: C
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