I agree with you on zero, but you need to consider different fractions.

Here's how I'd break this one down:

- This is a Yes/No question, sufficiency is attained if a statement allows us to answer 100% YES or 100% NO. It doesn't matter HOW we answer it, but a statement should REMOVE the other answer as a possibility.

- Givens: x is NOT equal to 1. What can x be? Negative integers, Fractions (both + and -), Positive integers, and Zero.

- The QUESTION is whether x^2 / (x-1) > x?

Let's consider some real world numbers.

If x = 2, then x^2 / (x-1) = 4, since 4 > 2, the answer is YES.
If x = 4, then x^2 / (x-1) = 16/3, since 16/3 > 4, the answer is YES.

Obviously, if x is a positive integer, the answer will always be YES.

Let's look at some different types of numbers:

If x = 1/2, then x^2 / (x-1) = 1/4 / (-1/2) = 1/4 x -2 = -2/4 = -1/2; since -1/2 < 1/2, the answer is NO.

If x = 3/2, then x^2 / (x-1) = 9/4 / 1/2 = 9/4 x 2 = 18/4; since 18/4 > 3/2, the answer is YES.

Now that we've seen some patterns with real-world numbers, we can address the statements.

1) x is not an integer. As we saw, if x = 1/2, the answer is NO. If x = 3/2, the answer is YES. Not sufficient.

2) x is positive. Our same two examples make this one insufficient as well.

If we combine, it still will not be sufficient since BOTH 1/2 and 3/2 are NOT integers and ARE positive, and yet yield two different answers: one YES, and one NO.

The answer is (E).

vipulgoyal wrote:If x is not equal to 1, is x²/(x-1) greater than x?
(1) x is not an integer.
(2) x is positive.

CRITICAL POINT approach:

CRITICAL POINTS occur when an inequality is UNDEFINED or when the left side is EQUAL to the right side.
x²/(x-1) is undefined when x=1.
x²/(x-1) = x when x=0.
Thus, the critical points are x=0 and x=1.

To determine the ranges where x²/(x-1) > x, test one value to the left and right of each critical point.

x<0:
If we plug x=-1 into x²/(x-1) > x, we get:
(-1)²/(-1-1) > -1
-1/2 > -1.
This works.
Thus, x<0 is a valid range.

0<x<1:
If we plug x=1/2 into x²/(x-1) > x, we get:
(1/2)²/(1/2 - 1 ) > 1/2
-1/2 > 1/2.
Doesn't work.
Thus, 0<x<1 is NOT a valid range.

x>1:
If we plug x=2 into x²/(x-1) > x, we get:
(2)²/(2-1) > 2
4 > 2.
This works.
Thus, x>0 is a valid range.

Result:
x²/(x-1) > x if x<0 or x>1.

Question stem, rephrased:
Is x<0 or x>1?

Statements combined:
It's possible that x=1/2, in which case the answer to the question stem is NO.
It's possible that x=1.5, in which case the answer to the question stem is YES.
Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.