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Quote:
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

Hi Karishma
Can you pls help me with the answer to the above link.
I was able to solve the inequality
My answer after solving inequality is -1<x<0 or x>1
So how can be the answer not E
The point of elimination for option e in the official explanation is as given below:-How can x be 2 when the range is less than 0......
E. −1<x<0. Not necessarily true since x could be 2.


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

Check out this post on must be true questions:
https://anaprep.com/algebra-game-must-b ... questions/
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If you choose 0.9, then B fails.
I just reviewed m09 q22 on the forum. You have changed the answer choices in that thread but not yet on GC CAT.
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Marcab
If you choose 0.9, then B fails.
I just reviewed m09 q22 on the forum. You have changed the answer choices in that thread but not yet on GC CAT.

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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That's what I am saying. Since x cannot take this value then how can B be answer.
How can x>-1 when 0<x<1 is not accepted?
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Hey thanks.
Its crystal clear now.
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Marcab
Hey thanks.
Its crystal clear now.

Yes - same here.

The word in that explanation that helped me the most is "satify". I think the difficulty of this question is good. The learning moment is also exactley what I needed. The language is what confused me on the first attempt. I think it would be understood by more people if the question had the english rephrased to: "... which of the following statements can be satisfied by all possible values of x".

Having said that, I learnt a lot about absolute values on the number plane trying to get my head around this explanation, so maybe it's helping us learn in the best way possible :-D
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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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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.

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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Can you tell me what I am doing wrong here?

There are two cases for x/|x|<x

Negative: x/-x<x -1<x
x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x
x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Thanks!


Bunuel
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one.
Not convinced at all with the OA.

My range is -1<x<0 and x>1.

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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #3 and 6. Thank you.
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WholeLottaLove
Can you tell me what I am doing wrong here?

There are two cases for x/|x|<x

Negative: x/-x<x -1<x
x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x
x is positive when x>0, so:

0<x<1

So why aren't the values considered: -1<x<1?

Thanks!


Bunuel
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one.
Not convinced at all with the OA.

My range is -1<x<0 and x>1.

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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #3 and 6. Thank you.

For \(x>0\) you get \(x>1\) --> \(x>1\) not 0<x<1.
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Thank you for that link. Unfortunately I am still a bit lost.

When we take the negative case (x<0) we have -1<x therefore we join x<0 and -1<x thus: -1<x<0

When we take the positive case (x>0) we have 1<x I would assume we join the two cases thus: x>0, x>1

In your link you made reference to the AND/OR concepts and I am vaguely familiar with them but I am still a bit confused as to why we count the range of the negative cases (which makes sense because we are given that x falls within -1 and 0) but we don't count any # between zero and one but we do count numbers greater than one (is it because only values that fall within both x>0 and X>1 count? And if so, why?)

Thanks!

VeritasPrepKarishma
WholeLottaLove
Hi - I just fixed the question.

There are two cases for x/|x|<x

Negative: x/-x<x -1<x
x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x
x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Check out this link which discusses this question and this issue in detail:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... -and-sets/
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WholeLottaLove
Thank you for that link. Unfortunately I am still a bit lost.

When we take the negative case (x<0) we have -1<x therefore we join x<0 and -1<x thus: -1<x<0

When we take the positive case (x>0) we have 1<x I would assume we join the two cases thus: x>0, x>1

In your link you made reference to the AND/OR concepts and I am vaguely familiar with them but I am still a bit confused as to why we count the range of the negative cases (which makes sense because we are given that x falls within -1 and 0) but we don't count any # between zero and one but we do count numbers greater than one (is it because only values that fall within both x>0 and X>1 count? And if so, why?)

Thanks!

VeritasPrepKarishma
WholeLottaLove
Hi - I just fixed the question.

There are two cases for x/|x|<x

Negative: x/-x<x -1<x
x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x
x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Check out this link which discusses this question and this issue in detail:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... -and-sets/


In simple terms, this is what the concept is all about:

You have 5 numbers in your list: 1, 4, 7, 8, 11

Which of the following is true for all the numbers in your list?
(A) Every number > 0
(B) Every number > 2
(C) Every number > 13

I hope you agree that answer is (A)

Every number is greater than 0. Do you have a problem that 2 is not a part of the numbers you have so how can (A) be the answer? Every number greater than 0 needn't be in your list. The question was what is true for all the numbers in the list.

This is the same concept.

You got the following ranges for x: -1 < x < 0, x > 1

Which of the following must be true?
x > 1
x > -1

Do you see that for all values of x that you got, x must be greater than -1?
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Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

m09 q22

Explanations required for this one.
Not convinced at all with the OA.

My range is -1<x<0 and x>1.


can you please, explain me what does the must be true clause mean in this question ??

0.1 to 0.9999999999 none satisfy this relation and more over they are greater than -1
and when again we have an option called > 1,
why do we choose this to be wrng ?
do we have any value > 1 but still don't satisfy this question ??



please explain, i did understand from -1 to 0 there are values which accept this relation but accpeting this doesn't mean we can omit from 0 to -9 ...


Im confused, im out of nuts . please help me for this :) :)
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Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

m09 q22

Explanations required for this one.
Not convinced at all with the OA.

My range is -1<x<0 and x>1.


can you please, explain me what does the must be true clause mean in this question ??

0.1 to 0.9999999999 none satisfy this relation and more over they are greater than -1
and when again we have an option called > 1,
why do we choose this to be wrng ?
do we have any value > 1 but still don't satisfy this question ??

please explain, i did understand from -1 to 0 there are values which accept this relation but accpeting this doesn't mean we can omit from 0 to -9 ...


Im confused, im out of nuts . please help me for this :) :)

Question: if \(-1<x<0\) or \(x>1\), then which of the following must be true? Notice that \(-1<x<0\) or \(x>1\) is given to be true: x is either from {-1, 0} or from {1, +infinity}

Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\). (Complete solution is here: if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1150594)

The following posts might help:
if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1150609
if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1181573

All Must or Could be True Questions to practice: search.php?search_id=tag&tag_id=193

Hope it helps.
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I did it this way....is it a correct way?
x/|x|<x

(x/|x|)-x<0

x((1/|x|)-1)<0

|x| has to be positive; x can take decimal value (between 0 and 1 zero not included) or whole value.

in that case (1/|x|)-1 can be more than zero and less than zero so accordingly x will be negative or positive as (+)x(-) = less than zero

if 0<|x|<1 then (1/|x|)-1 is positive and to make x((1/|x|)-1)<0, x has to be negative

if x>1 then (1/|x|)-1 is negative and to make x((1/|x|)-1)<0, x has to be positive

Hence, x>-1
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I did it this way....is it a correct way?
x/|x|<x

(x/|x|)-x<0

x((1/|x|)-1)<0


I think it is though it is a little hard to understand. Let me try to write it clearly:

x((1/|x|)-1)<0

Two cases:

Case 1: x < 0 AND (1/|x|-1) > 0
From (1/|x|-1) > 0 we get that |x| < 1 i.e. -1 < x < 1
So combining, we get -1 < x < 0

Case 2: x > 0 AND (1/|x|-1) < 0
From (1/|x|-1) < 0 we get that |x| > 1 i.e. x > 1 or x < -1
So combining, we get x > 1

x could lie between -1 and 0 or it could be more than 1. Hence in any case, it will be more than -1.
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Hi, I tried solving the problem by multiplying both sides by |x| since |x| will always be positive. This makes the equation as x<x|x|.
Case 1: x>0 => x<sq(x)=>x>1
Case 2:x<0=>x<-sq(x)=>x<-1

I believe I am going wrong in multiplying both sides by |x|, however, I fail to understand how since |x| will always be positive and hence I could multiply it.

Please assist.

Thanks



Bunuel
Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

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

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one.
Not convinced at all with the OA.

My range is -1<x<0 and x>1.

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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rules #3 and 6. Thank you.
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