Is |x – 1| < 1?
(1) (x – 1)² > 1
(2) x < 0

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.


(D) If we take the square root of both sides of the inequality in Statement (1), we get:
|x – 1| > 1.

This tells us that 1 is definitely not greater than |x – 1|, so Statement (1) is sufficient.

Remember: When you take the square root of a variable expression that has already been squared, you must take the absolute value of the squared expression in the process. A simple example is a² = 9. Let’s solve this:
a² = 9
|a| = 3
a = 3 or a = -3.

The same rule holds when dealing with inequalities, as we saw above.

Let's evaluate Statement (2) by itself. x < 0 implies that x – 1 < 0 , so |x – 1| = 1 – x .
Is |x – 1| < 1 in this case? Plug in |x – 1| = 1 – x.
Is 1 – x < 1?
Is -x < 0?
Is x > 0? NO. Because statement (2) defines x < 0 . We have the definite answer, so Statement (2) is also sufficient.

Since both statements are sufficient individually, the correct answer is choice (D).
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Why are the questions "Is |x – 1| < 1 ?" and "Is x > 0?" the same?

They are the same under the condition that x – 1 < 0, so that we simplify the absolute value: |x – 1| = 1 – x .

In Statement (2) we show that x < 0 implies x – 1 < 0, which turns "Is |x – 1| < 1?" into "Is x > 0?". The latter contradicts Statement (2), x < 0.


If to speak about an equivalent inequality for |x – 1| < 1 with no additional conditions, then it is
-1 < x – 1 < 1
0< x < 2

In other words its solution is that x must be between 0 and 2.

We refer to the principal square root, which results in a positive value only. We can use it if the both parts of an inequality are positive. This operation does not change the sign of the inequality.

(1) (x – 1)² > 1
becomes
√((x – 1)²) > √1
|x – 1| > 1