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If x/|x|<x which of the following must be true about x?

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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 29 Mar 2019, 23:32
In this x = 0 doesn't satisfy the condition. Then how is B the correct answer? Are we supposed to ignore failing values in range?
I am aware that there are values <1 that satisfy the condition, but one among them i.e., 0 fails to satisfy it, in this case are we not supposed to ignore this answer choice?
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 29 Mar 2019, 23:44
2
Hi

yes we need to ignore 0 as correctly pointed out by you.

The main point in this question is to understand what the question is asking.
which of the following must be true about x?

It means the question is asking about a Set of Values, which contains all the value of x which satisfy the given inequality.
But the Big Idea is that the set may contain other values also which does not satisfy the inequality.

It simply means the set must contains all the value of x satisfying the inequality but vice versa is not required.

only after solving the inequality as explained above,
-1<x<0 , when x is negative , or
x>1 when x is positive.

So option B x>-1 is the only the set of values which contains all the above required values of x.

Another way to understand the logic behind the question.
Which of the following must be true for multiple of 4?
a) it is an even no

yes it is right as all multiple of 4 must be even number.
But we do have many even numbers such as 6, 10 etc which are not multiple of 4.

There is a difference between
1)" which of the following must be true about x"- it contains extra non solution points also , and
2) "which is the solution set of x" - it contains only the solution set points



ajmekal wrote:
In this x = 0 doesn't satisfy the condition. Then how is B the correct answer? Are we supposed to ignore failing values in range?
I am aware that there are values <1 that satisfy the condition, but one among them i.e., 0 fails to satisfy it, in this case are we not supposed to ignore this answer choice?

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If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 05 May 2019, 07:19
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.


Hi Bunuel

I have understood all the explanations except the below part

x<0 --> |x|=−x--> x/−x<x --> −1<x--> −1<x<0

in case of x<0, the equation should turn into (-x)/(-x)<-x , as we are assuming X to be negative so through out the equation we have to replace X with -X.
I am bit confused here as only when we assume X as negative lxl = -x and then we should put -x throughout the equation

but here we are only assuming denominator as -x whereas in numerator and on the RHS of inequality we are keeping positive X.

This is a fundamental doubt if you can help explain it will be helpful while coming across other modulus questions like this .

Thanks in advance..
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 05 May 2019, 07:25
cruiseav wrote:
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.


Hi Bunuel

I have understood all the explanations except the below part

x<0 --> |x|=−x--> x/−x<x --> −1<x--> −1<x<0

in case of x<0, the equation should turn into (-x)/(-x)<-x , as we are assuming X to be negative so through out the equation we have to replace X with -X.
I am bit confused here as only when we assume X as negative lxl = -x and then we should put -x throughout the equation

but here we are only assuming denominator as -x whereas in numerator and on the RHS of inequality we are keeping positive X.

This is a fundamental doubt if you can help explain it will be helpful while coming across other modulus questions like this .

Thanks in advance..


Please read the whole discussion: https://gmatclub.com/forum/if-x-x-x-whi ... l#p2231315
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 05 May 2019, 11:17
chetan2u wrote:
Mudit27021988 wrote:
chetan2u

I have a totally different doubt and usually I face it whenever I solve modulus questions.

we say |X|=-X, when X<0
|X|=X , when X>0

Why don't we replace these values in actual X i.e. why do we open only |X| if X<0. Like in this question why don't we substitute
X/|X| , if X<0 would lead to -X/|X| which will give -1. In this question the result is same but I have seen questions where these substitution leads to different result

Thanks

Posted from my mobile device



Hi..
When you say x is negative, the negative sign is already there in the variable.
So if x<0, then -x>0..
Let us take an example..
Say x is -2..
So X/|X| will be -2/|-2|=-2/2=-1..
But if you change the sign that is x/|X|=-x/|X|=-(-1)/|-1|=1/1=1



Hi chetan2u

Two doubts are bugging me...Sorry i might sound repetitive, but i need to clear this fundamental doubt

Part- 1 X/|X| if we take the case where X<0 and hence |X| opens us as -x

so now the equation should become as (-x)/-x =1, as now we are working with condition of X<0, so |X| in denominator should be replaced with -X....But why still it written as |X| in the denominator

Part-2- Following the above, in case of x<0, the equation should turn into (-x)/(-x)<-x , as we are assuming X to be negative so through out the equation we have to replace X with -X.
I am bit confused here as when we assume X as negative (condition when lxl = -x ), then we should put -x throughout the equation

but here we are only assuming denominator as -x whereas in numerator and on the RHS of inequality we are keeping positive X.

This is a fundamental doubt if you can help explain it will be helpful while coming across other modulus questions like this .

Thanks in advance..
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 05 May 2019, 20:23
cruiseav wrote:
chetan2u wrote:
Mudit27021988 wrote:
chetan2u

I have a totally different doubt and usually I face it whenever I solve modulus questions.

we say |X|=-X, when X<0
|X|=X , when X>0

Why don't we replace these values in actual X i.e. why do we open only |X| if X<0. Like in this question why don't we substitute
X/|X| , if X<0 would lead to -X/|X| which will give -1. In this question the result is same but I have seen questions where these substitution leads to different result

Thanks

Posted from my mobile device



Hi..
When you say x is negative, the negative sign is already there in the variable.
So if x<0, then -x>0..
Let us take an example..
Say x is -2..
So X/|X| will be -2/|-2|=-2/2=-1..
But if you change the sign that is x/|X|=-x/|X|=-(-1)/|-1|=1/1=1



Hi chetan2u

Two doubts are bugging me...Sorry i might sound repetitive, but i need to clear this fundamental doubt

Part- 1 X/|X| if we take the case where X<0 and hence |X| opens us as -x

so now the equation should become as (-x)/-x =1, as now we are working with condition of X<0, so |X| in denominator should be replaced with -X....But why still it written as |X| in the denominator

Part-2- Following the above, in case of x<0, the equation should turn into (-x)/(-x)<-x , as we are assuming X to be negative so through out the equation we have to replace X with -X.
I am bit confused here as when we assume X as negative (condition when lxl = -x ), then we should put -x throughout the equation

but here we are only assuming denominator as -x whereas in numerator and on the RHS of inequality we are keeping positive X.

This is a fundamental doubt if you can help explain it will be helpful while coming across other modulus questions like this .

Thanks in advance..



Hi,

x<0 means x is negative, but when you add a negative sign ahead of x, it becomes positive..
so -x >0..

For example let x = -5, so when you add a negative sign ahead, -x=-(-5)=5.

Thus dont add a negative sign when there is no MOD.

But if it is |x| then it has to be -x as x is negative and MOD cannot be negative..

SAme example..
x=-5..|-5|=5...
But |x|=x here means |-5|=-5...5=-5..NO
So, |x|=-x, that is |-5|=-(-5)...5=5
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If x/|x|<x which of the following must be true about x?  [#permalink]

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New post 12 Jun 2019, 05:10
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.


Alternative Solution



Since |x| is always +, we can cross-multiply it.

0 < x (|x| - 1)

If x = +
0 < x(x-1) => x>1 [Using Wavy line trick]

If x = -
0 < x(-x -1) => 0 > x(x+1) => -1 < x < 0 [Using Wavy line trick]

Only B covers both the ranges above, hence ANSWER: B
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Re: If x/|x| < x, which of the following must be true about  [#permalink]

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