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Here's what I did:

What I need to find: Is x/|x| < x?
1> -3/x < 0:
=> 3/x > 0 => Doesn't help us understand the value of x/|x| => Insufficient

2> x/|x| > 0
=> x/(-x) > 0 or x/x > 0
=> -1 < 0 or 1 > 0
=> Both statements are correct but still don't give us a value for x/|x| => Insufficient.

Let me know if this method makes sense.
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Bunuel, could you please explain this.
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Bunuel, could you please explain this.

If x is a non-zero number, Is \(\frac{x}{|x|}<x?\)

(1)\(\frac{−3}{x}<0\). This implies that x is positive. In this case, |x| = x. So, the question becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(2) \(\frac{x}{|x|}>0\). This also implies that x is positive (If x were negative, then we'd get x/(-x) = -1, which is less than 0, not more, as given here). x > 0, implies that |x| = x. So, the question, as above, becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(1)+(2) Both statements give the same info: x > 0. If 0 < x <= 1, then the answer to the question will be NO but if x > 1, the answer to the question will be YES. Not sufficient.


Answer: E.
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mzaid
Bunuel, could you please explain this.

If x is a non-zero number, Is \(\frac{x}{|x|}<x?\)

(1)\(\frac{−3}{x}<0\). This implies that x is positive. In this case, |x| = x. So, the question becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(2) \(\frac{x}{|x|}>0\). This also implies that x is positive (If x were negative, then we'd get x/(-x) = -1, which is less than 0, not more, as given here). x > 0, implies that |x| = x. So, the question, as above, becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(1)+(2) Both statements give the same info: x > 0. If 0 < x <= 1, then the answer to the question will be NO but if x > 1, the answer to the question will be YES. Not sufficient.

Answer: E.


Bunuel, I am not getting the second option. When you say X is negative then wont x/|x|[/fraction] = -x/-x =1. Am I missing anything here.
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Bunuel
mzaid
Bunuel, could you please explain this.

If x is a non-zero number, Is \(\frac{x}{|x|}<x?\)

(1)\(\frac{−3}{x}<0\). This implies that x is positive. In this case, |x| = x. So, the question becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(2) \(\frac{x}{|x|}>0\). This also implies that x is positive (If x were negative, then we'd get x/(-x) = -1, which is less than 0, not more, as given here). x > 0, implies that |x| = x. So, the question, as above, becomes: is \(\frac{x}{x}<x?\)? Or, which is the same: is 1 < x. We just know that x is positive: it can be more or less than 1. Not sufficient.

(1)+(2) Both statements give the same info: x > 0. If 0 < x <= 1, then the answer to the question will be NO but if x > 1, the answer to the question will be YES. Not sufficient.

Answer: E.


Bunuel, I am not getting the second option. When you say X is negative then wont x/|x|[/fraction] = -x/-x =1. Am I missing anything here.

No. If x is negative, then |x| = -x, so x/|x| becomes x/(-x).
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Bunuel, Sorry but I dont understand why we are not considering the numerator x as negative. As in first statement the negative x doesn't fulfill the condition, so we considered it positive. What is lacking in my concept here ?
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Bunuel, Sorry but I dont understand why we are not considering the numerator x as negative. As in first statement the negative x doesn't fulfill the condition, so we considered it positive. What is lacking in my concept here ?

I guess, you asking why we are not changing x to -x, knowing that x is negative. x (in numerator) itself denotes a negative number, so no need to replace it. For example, if given that x + y = 10, and x is negative, are you rewriting this to -x + y = 10?
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