Notice that we are asked to find which of the options MUST be true, not COULD be true.

Let's see what ranges does \frac{x}{|x|}< x give for x. Two cases:

If x<0 then |x|=-x, hence in this case we would have: \frac{x}{-x}<x --> -1<x. But remember that we consider the range x<0, so -1<x<0;

If x>0 then |x|=x, hence in this case we would have: \frac{x}{x}<x --> 1<x.

So, \frac{x}{|x|}< x means that -1<x<0 or x>1.

Only option which is ALWAYS true is B. ANY x from the range -1<x<0 or x>1 will definitely be more the -1.

Answer: B.

As for other options:

A. x>1. Not necessarily true since x could be -0.5;
C. |x|<1 --> -1<x<1. Not necessarily true since x could be 2;
D. |x|>1 --> x<-1 or x>1. Not necessarily true since x could be -0.5;
E. -1<x<0. Not necessarily true since x could be 2.

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The question above is exactly as it appears in CAT.

Also, notice that x=0.9, does not satisfy x/|x|<x, thus x cannot take this value.
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Consider following:
If x=5, then which of the following must be true about x:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If -1<x<10, then which of the following must be true about x:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY x from -1<x<10 will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If -1<x<0 or x>1, then which of the following must be true about x:
A. x>1
B. x>-1
C. |x|<1
D. |x|>1
E. -1<x<0

As -1<x<0 or x>1 then ANY x from these ranges would satisfy x>-1. So B is always true.

x could be for example -1/2, -3/4, or 10 but no matter what x actually is, it's IN ANY CASE more than -1. So we can say about x that it's more than -1.

Hope it's clear.
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\frac{x}{|x|}<x

Test the values for x.

(1) x = 0 ==> No! Since it is given that x is not 0
(2) x = -1 ==> -1 < -1 ==> No!
(3) x = 1 ==> 1 < 1 ==> No!
(4) x = -2 ==> -1 < -2 ==> No!
(5) x = \frac{-1}{4} ==> -1 < \frac{-1}{4} ===> Yes!
(5) x = 2 ==> 1 < 2 ==> Yes!

Our range: -1 < x < 0 or x > 1

What must always be true in that range? x > -1 always

Answer: B

Appreciate if someone could point out where I am going wrong here.


x / |x| < x

Since x is non zero, dividing by x on both sides


1 / |x| < 1

Taking reciprocal,

|x| > 1

Then I just jumped into Choice D. Didn't even look at the others.

Await your valued views.
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- Stay Hungry, stay Foolish -

But your dividing my the same number x on both sides, whether its positive or negative,
implying both sides will simultaneously be negative together or positive together which doesnt change the sign.

Right or not?
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Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot divide both parts of inequality x / |x| < x by x as you don't know the sign of this unknown: if x>0 you should write 1/|x|<1 BUT if x<0 you should write 1/|x|>1.

Hope it helps.
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A 'must be true' question! They are absolutely straight forward if you get the fundamental but they can drive you crazy if you don't.

"My answer after solving inequality is -1<x<0 or x>1" Perfect. That is the range of x for which the inequality works. So tell me, what values can x take?
-1/2, -1/3, -2/3, 1.4, 2, 500, 123498 etc...
Now the question is "which of the following must be true?"

(A) x>1
Are all these values greater than 1? No.

(B) x>-1
Are all these values greater than -1? Yes. The answer. Note that you dont have to establish that all value greater than -1 should work for the inequality. You only have to establish that all values which work for the inequality must satisfy this condition.

(C) |x|<1
Not true for all values of x.

(D) |x|>1
Not true for all values of x.

(E) -1<x<0
Not true for all values of x.
x can take values 1.4, 2, 500 etc

I wrote a post on this beautiful question sometime back:
http://www.veritasprep.com/blog/2012/07 ... -and-sets/
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\Rightarrow Given, x≠0 & \frac{x}{|x|}< x .................

So, Two cases will be formed here ........ i.e.,

When x < 0 & when x > 0.

Now, First when x<0, in this case we have, \frac{x}{-x}< x
Therefore, x> -1

Now, when x > 0, in this case we have, \frac{x}{x}< x.
Therefore, x> 1.

Hence, from both the conditions above, we can say that x must be greater than -1 . i.e., x> -1

Hence, B.
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We are not saying that x / |x| < x for all x > -1. That defies the whole purpose of breaking down the given inequality into 2 ranges. All we are saying is that for the 2 given ranges, the value of x MUST BE x>-1. You are taking the value of x=0.5, which in the first place is invalid for the given problem.
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