Is |x| < 1 ? (1) x/|x| < x (2) x/|x| < 1

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.
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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.
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I have tried to use the line method to solve this problem.
1. x/!x!<xSince !x! is always positive, I can multiply it each side and get x<x.!x! and after substracting x-x.!x!<0
Now taking x as common factor, x(1-!x!)<0
x will have three factors:0,1,-1
Now refer to the attached figure to get the region where the given condition will be true.
Since the inequality is of "less than 0" type, we shold consider "negative portion"
we will get two possibilities which are -1<x<0 and x>1. Hence not suffecient.

2. x/!x!<1. Repeating the same logic above we get x-!x!<0. This will be true only if x<0. Still not suffeceint.

Combining both statements, we can eliminate the portion x>1, which we got in statement 1.
Leaving us only with -1<x<0, which tells us that !x! will always be a fraction less than 1. Hence suffeceint.

Hi,

There is similar question out there
https://gmatclub.com/forum/if-x-is-not- ... 86140.html

Now here Bunuel has explained two approaches for solving the question
https://gmatclub.com/forum/if-x-is-not- ... 86140.html

Hope this helps

Probus
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Is |x| < 1 ?

(1) x/|x| < x
\(\frac{x}{|x|}- x <0\)
\(\frac{x - x|x|}{|x|} < 0\)
Since |x| > 0
x - x|x| < 0
x (1- |x|) < 0
x (|x| -1) > 0
If x>0; |x| > 1; x>1
If x<0; |x| < 1; -1<x<0
Either x>1 or -1<x<0
NOT SUFFICIENT


(2) x/|x| < 1
x/|x| - 1 < 0
x - |x| < 0
If x>0; x-x<0;
If x<0; x + x < 0; x<0
|x| may or may not be <1
NOT SUFFICIENT

(1) + (2)
(1) x/|x| < x
If x>0; |x| > 1; x>1
If x<0; |x| < 1; -1<x<0
(2) x/|x| < 1
x<0
-1<x<0
|x|<1
SUFFICIENT

IMO C
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do not multiply or cancel anything until and unless we know the exact sign of the x (or any other unknown stated in the question )

Is |x| < 1 ?
(1) x/|x| < x
(2) x/|x| < 1

Solution:
(1) x/|x| < x --> insuff
=> x < x|x|
=> x|x|-x > 0
=> x(|x|-1) > 0
` ++>0 => x >0 & |x|>1 --> no
or -->0 => x<0 &|x|<1 --> yes

(2) x/|x| < 1 --> insuff
=> x<|x|
so, x is always -ve
=> x can be -1<x<0 --> yes
or x=<-1 --> no
combing (1) & (2)=>
x< 0 from (2), so |x| <1 from (1) --> suff
Answer: C

The statement asks whether the value of x is between -1 and 1.

A:
x can fall into four domain,
x<-1
-1<x<0
0<x<1
1<x

for first and third the statement holds true, and not for second and fourth.

B:
It simply states that x<1.

Both the statement is insufficient. But by clubbing them still, we cannot say
because x<-1 satisfies and -1<x<0 too.

A bit confused why C is correct answer!
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Hello, Romil042. I just happened upon this question today and scrolled down to find your query. I think I can help clarify your confusion. I will respond inline below.


So far, so good.


This statement at the end is inaccurate. Did you test values or make assumptions?

1) x < -1; test x = -2

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

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

\(\frac{-2}{2} < -2\)

\(-1 < -2\) X

Since -1 is NOT less than -2, we know that x CANNOT be less than -1.

2) -1 < x < 0; test x = -0.5

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

\(\frac{(-0.5)}{|(-0.5)|} < (-0.5)\)

\(\frac{-0.5}{0.5} < -0.5\)

\(-1 < -0.5\) √

We have found that x CAN be between -1 and 0.

3) 0 < x < 1; test x = 0.5

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

\(\frac{(0.5)}{|(0.5)|} < (0.5)\)

\(\frac{0.5}{0.5} < 0.5\)

\(1 < 0.5\) X

Since 1 is NOT less than 0.5, x CANNOT be between 0 and 1.

4) 1 < x; test x = 2

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

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

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

\(1 < 2\) √

We have found that x CAN be greater than 1. In short, we have two valid types of inputs based on Statement (1): x is either between -1 and 0 or it is greater than 1. Of course, we CANNOT answer the original question conclusively with this information.

a) If -1 < x < 0, |x| < 1

b) If x > 1, |x| > 1

Thus, Statement (1) on its own is NOT sufficient.


Of course, I hope you can see now that by combining the two statements, we can eliminate b) above, so it must be true that |x| < 1, and the answer is (C).

Good luck with your studies.

- Andrew
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