If x 0, is x

Question

If  x ≠ 0 , is  x < |x|  ?

(1)  |x|< 1

(2)  \frac{x}{|x|} < x

TarunKumar1234 · Answer

If x ≠ 0 , is x < |x| ? we need to get if x <0. Stat1: |x|< 1 -1 < x < 1; Not sufficient Stat2: \frac{x}{|x|} < x x*(\frac{1}{|x|} - 1) <0; if x >0 then, (\frac{1}{|x|} - 1) < 0, so x >1. Another way, if x <0 then, (\frac{1}{|x|} - 1) > 0. so, -1< x < 0. Not sufficient Combining both, -1< x < 0. Sufficient So, I think C.

gmatophobia · Answer

Pre-thinking on this question before jumping into the options is immensely helpful.

We know that \(x \neq{0}\) and we have to find out if x < |x|

|x| is always +ve, so the only time when x < |x| is when x is -ve.

So the complex question now boils down to - Is x < 0

Statement 1

|x| < 1

This means x lies between -1 and 1.

Hence x can be negative or it can be positive. The statement alone is not sufficient.

Statement 2

\(\frac{x}{|x|} < x\)

As |x| is always non negative and we know that \(x \neq{0}\), lets multiply both sides by |x|

x < |x| * x

x (1 - |x|) < 0

Case 1

x < 0 & (1-|x|) > 0

|x| < 1

Combining

-1 < x < 0

Case 2

x > 0 & (1-|x|) > 0

|x| > 1

Combining

x > 1

Similar to statement 1, we have two ranges for x , one positive and one negative and this statement alone is not sufficient.

Combining

The common range is -1 < x < 0.

Is x negative - yes !

The combined statement is sufficient to answer the question.

Option C

Bunuel · Answer

Official Solution: If x ≠ 0 , is x < |x| ? For x < |x| to be true, x must be less than 0 because if x > 0 , then x = |x| . Hence, the question essentially asks whether x < 0 . (1) |x|< 1 The above implies that -1 < x < 1 . So, x may or may not be negative. Not sufficient. (2) \frac{x}{|x|} < x If x < 0 , then |x| = -x , and the above can be rewritten as \frac{x}{-x} < x , which yields -1 < x . Since we consider the case when x < 0 , we get the final range for this case as -1 < x < 0 . If x > 0 , then |x| = x , and the above can be rewritten as \frac{x}{x} < x , which yields 1 < x . Since we consider the case when x > 0 , we get the final range for this case as x > 1 . Hence, we deduce that \frac{x}{|x|} < x is true when -1 < x < 0 or x > 1 . Thus, x may or may not be negative. Not sufficient. (1)+(2) The intersection of the ranges from (1) and (2) is -1 < x < 0 . Therefore, the answer to the question of whether x is negative is YES. Sufficient. Answer: C

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If x 0, is x < |x| ? (1) |x|< 1 (2) x/|x| < x

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