Data Sufficiency

If n is a prime number greater than 2, is 1/x > 1?

(1) x^n<x<x^(1/n)

(2)x^(n−1)>x^(2n−2)

Source: Manhattan

Mo2men wrote:If n is a prime number greater than 2, is 1/x > 1?

(1) x^n<x<x^(1/n)

(2)x^(n−1)>x^(2n−2)

1/x > 1 only if x is a POSITIVE FRACTION BETWEEN 0 AND 1.
Question stem, rephrased:
Is x a positive fraction between 0 and 1?

A prime number greater than 2 must be ODD.
Any odd exponent greater than 1 will have the same effect on the two statements.
Thus, the two statements can be rephrased with n=3, as follows:
x³ < x < x^(1/3)
x² > x�.

Statement 1: x³ < x < x^(1/3)
Case 1: x=-8
Here, x³=-512 and x^(1/3) = -2, with the result that the inequality in red becomes:
-512 < -8 < -2.
Case 2: x=1/8
Here, x³=1/512 and x^(1/3) = 2, with the result that the inequality in red becomes:
1/512 < 1/8 <1/2.
Since the answer to the question stem is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: x² > x�
Since the inequality in blue implies that x is NONZERO, x² > 0.
Thus, we can safely divide each side by x²:
x²/x² > x�/x²
1 > x²
x² < 1.
Here, it's possible that x is a positive fraction between 0 and 1 (in which case the answer to the question stem is YES) or a negative fraction between -1 and 0 (in which case the answer to the question stem is NO).
INSUFFICIENT.

Statements combined:
Of the two ranges for x that satisfy Statement 2 -- 0<x<1 and -1<x<0 -- only the first will satisfy Statement 1.
Thus, for both statements to be satisfied, x must be a positive fraction between 0 and 1, implying that the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.

Piggybacking a bit, the second statement can be written this way:

x� � ¹ > (x� � ¹)²

We know that n is an odd prime, or n = 2m + 1, where m is some positive integer whose value we don't care about. Substituting, we have

x²��¹ � ¹ > (x²��¹ � ¹)²

or

x²� > (x²�)²

or

x²� > x��

Since x ≠ 0 (or else the inequality above is impossible), we can safely divide both sides by x²�, which must be positive (it's a nonzero square), to get 1 > x²�, or 1 > x > -1 with x ≠ 0.