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In each of the cases X > -1, therefore X > -1 is true for all possible X . Note that -10 and -0.5 can serve as counter-examples for other options. The correct answer is B.

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Isn't the correct answer -1<x<0 and x>1? Using formula B with x=0.9 gives you 1<0.9 which doesn't work.

The question is expecting the value of X and not necessarily for all values of X.

The answers are x > 1 and -1 < x < 0.

Hence, for any value of x within these ranges, x > -1.

Actually the question stem states: which of the following MUST be true about X..

MUST means for any values, right? There is no limitations on the Value of X, so if one can find one case that doesn't satisfy our inequality, then the answer is wrong. And X>-1 includes values 0<x<1 For example if the stem had a note: X is an integer that would make the question clear.

IMHO the answer choices or the question stem should be reviewed.

I've come across this question and the stimulus still contains "about integer". This is the question I saw

If \frac{X}{|X|} < X , which of the following must be true about integer X ? ( X \ne 0 )

(C) 2008 GMAT Club - m09#22

* X > 2 * X > -1 * |X| < 1 * |X| = 1 * |X|^2 > 1

According to my understanding B cannot be the correct answer choice. As other members pointed out the solution is

The answers are x > 1 and -1 < x < 0. Since X is an integer x cannot take any values from -1 < x < 0 so the inequality only makes sense when x > 1 --> x >=2

IMO if you change option A for X >=2, I think it should be the correct answer choice. What do you think?

I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X. I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move .

pleonasm wrote:

dzyubam wrote:

I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.

IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.

I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X. I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move .

pleonasm wrote:

dzyubam wrote:

I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.

IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.

Shouldn't the answer option B be: X \in (-1,0) \cup (1,\infty)

Even the solution describes the same. If X falls in the range (0,1) the statement is not true.

I think the trick here is wording. Think about all the possible values of X.

They are all greater than -1. So, B doesn't define the set of possible values of X, but all the possible values of X are greater than -1. So it is ture.

You're right, guys. The question still needs revision. Here's what I've come up with. Is there any problem if we change the question stem and options to the following? I've removed the "integer" from the stem:

If \frac{X}{|X|} < X , which of the following must be true about X ? ( X \ne 0 )

(C) 2008 GMAT Club - m09#22

* X > 2 * X > -1 * |X| < 1 * |X| = 1 * |X|^2 > 1

The explanation stays the same. -0.5 can serve as a counter example for A, D and E. If you plug -0.5 into the inequality from the stem, you will see that it holds true. -10 could be probably removed from the OE. In order to answer the question, you have to solve the inequality from the question stem. The OE solves the modulus inequality. Is there anything wrong with the way the OE solves the inequality that I don't see?

What do you all think about the changes?
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I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.
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I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.

IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X. I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move .

pleonasm wrote:

dzyubam wrote:

I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.

IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.

Yeah ok that's what I thought. Thanks for the clarification. It's the modulus tag that makes the x look a bit different.
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In the land of the night, the chariot of the sun is drawn by the grateful dead