anupamadw wrote:

Bunuel wrote:

Is |x| < 1 ? Is |x| < 1 --> is -1 < x < 1.

(1) x/|x| < x.

Two cases:

A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). Since we consider the case when \(x<0\), then we'd have \(-1<x<0\). Answer YES.

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\). Answer NO.

Not sufficient.

(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.

(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.

Answer: C.

Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source.

Hi Bunuel

I fail to understand above solution.

We have to find whether -1< x<1

From 1 ) we know that x/|x| < x holds true for

-1<x<0 and x>1 so we know that x is not in range 0<x<1

is this not sufficient to answer NO ....?

Again from 2) We know -infinity < x<0 then answer is NO

Sorry, could you please tell me what am I missing...how are you getting YES and NO for each?

Let me try to explain.

From 1,

x/|x| < x ---> 2 cases:

--- when x \(\geq\)0 --> |x| = x ---> x/x < x --> x > 1 . Thus a "no" for is |x| < 1

--- when x < 0 --> |x| = -x ---> x/-x < x --> x > -1 . For x = 0.5, then a "yes" for |x|<1 but if x = 5, then a "no" for is |x|<1. Thus NOT sufficient.

From 2,

x/|x| < 1 , again 2 cases,

--- when x \(\geq\)0 --> |x| = x ---> x/x < 1 --> 1<1 . Thus x \(\geq\)0 is not a possible scenario.

--- when x <0 --> |x| = -x ---> x/-x < 1 --> 1>-1 . This is true for ALL x and thus x <0 satisfies this. For x = -0.5, you get a "yes" for is |x|<1 but for x = -4, you get a "no" for is |x| <1 . Thus NOT sufficient.

Alternately, if you want you can also test cases to come up with ranges for x. But this might be a bit more time consuming.

For your analysis of statement 1, the text in red is not correct. Refer to the explanation mentioned above.

For statement 2, you are correct that x<0 but you need to check whether -1<x<0 is satisfied or not. You get 2 different answers when you take x = -0.5 or x = -4.

Hope this helps.