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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
Thanks for the detailed explanation.

Very nicely done!
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Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

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

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

OR: \(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.



Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
Thanks Bunuel, It's terrific explanation...too detail..concept is now crystal clear!!
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
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Nice explaination. I had used numbers and came to wrong conclusion. I am inclined to say that for all mod questions I should consider positive and negative like Bunuel suggested
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gmcbsj wrote:
Hi,

I had answered this question incorrectly:

Quote:
If x is not equal to 0, is |x| less than 1?

(1) x / |x| < x
(2) |x| > x


It seemed to me that (1) was enough. Since x is not 0, |x| must be greater than 0, and it should be safe to reorder the inequality like this by multiplying both sides by |x|:

x < |x| * x

And my thinking went: if |x| * x is greater than x, then x must be positive (if it weren't, then |x| * x would be a more negative number than x). And also x must be >= 1, for if it were between 0 and 1 then |x| * x would be a smaller number than x.

So, (1) should be sufficient -- (A). But according to the answer explanation, the answer is (C). Have I made a mistake in my logic here?


Hi, and welcome to Gmat Club.

You are right inequality \(x<|x|*x\) holds true when \(x>1\), but it's not the only range, it also holds true when \(-1<x<0\). So from (1) we can not be sure whether \(|x|<1\).

You should have spotted that your range for (1) is not correct when dealing with statement (2): \(|x| > x\), which basically implies that \(x\) is negative, \(x<0\). So you would have from (1) \(x>1\) and from (2) \(x<0\): statements clearly contradict each other, but on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

For complete solution refer to the posts above.

Hope it helps.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
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gmat1220 wrote:
If x is not equal to 0, is |x| less than 1?

(1) (x/|x|) < x


(2) |x| > x


I would like to do this graphically.
To prove -1<x<1 (x not equal to zero)

Stmt1: x/|x| < x , x(|x|-1)>0 As can be seen from PINK side of graph of x(|x|-1)>0, x can be less than 1 or greater than 1. Not sufficient.

Stmt2:
|x|>x . As can be seen from BLACK side of graph that x is less than 1 but it is also less than -1. Remember we have to prove that -1<x<1

Combining, overlapping area has the equation -1<x<1.

OA. C
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
Hi Bunuel - Firstly thanks for the wonderful collection and the explanation you provided in the word format.

1. In this kind of inequalities involving ‘mod’ values is it ever advisable to first square both side and then proceed ? I saw somewhere this approach was quite fast to solve the problem.
2. In this specific example as it involves negative & decimal , squaring and multiplying makes things erroneous .

Please suggest if there has any general rule/comment on this?

Thanks,
VCG
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Rephrase of stem: Is x a positive or negative fraction.

Following AD/BCE strategy, 2 looks easy so lets start with that.

2. This means that x is negative. But, x could be integer. insuff
1. This one is a bit complicated. If x = positive integer, this ineq holds and it does not for a positive fraction. So, for positive numbers, x=integer. For x = negative integer, this ineq does not hold. It does hold for negative fractions. Insuff.

Combining 1 and 2, x = negative and x = integer. Satisfies the stem. C.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?


Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.



Hi Bunuel...I still didnot got it..red area says -1<x<0 but we need that -1<x<1
Please explain
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verycoolguy33 wrote:
Hi Bunuel - Firstly thanks for the wonderful collection and the explanation you provided in the word format.

1. In this kind of inequalities involving ‘mod’ values is it ever advisable to first square both side and then proceed ? I saw somewhere this approach was quite fast to solve the problem.
2. In this specific example as it involves negative & decimal , squaring and multiplying makes things erroneous .

Please suggest if there has any general rule/comment on this?

Thanks,
VCG


A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.
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Eshika wrote:
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?


Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.



Hi Bunuel...I still didnot got it..red area says -1<x<0 but we need that -1<x<1
Please explain


The question asks "is \(-1<x<1\)?" We got that \(x\) is in the range \(-1<x<0\) (red area). Now, as ANY \(x\) from this range (from red area) is indeed in \(-1<x<1\), then we can answer YES to our original question.

Hope it's clear.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
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bunuel,

i had a slightly slightly different method for statement 1. Do you mind letting me know if the steps I took are OK?

also, what level question is this? My test is coming up soon and I've done all the OG and quant review 2nd edition problems. I wonder if going through these two samurai guides PS/DS would be the best use of my time in the final month and half.


(1).
x/|x| < x
before even going to x<0 and x>0, i knew that |x| is positive, so I multiplied it across
x < x*|x|

case x < 0

divide both sides by x, inequality must flip because x is negative
1 > |x| <-- this says YES to what we asked for


case x > 0

again, divide both sides by x, inequality won't flip
1 < |x| <-- this says NO

insufficient

(2). clearly says x has to be negative, therefore using (1) and (2) it is sufficient.
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pinchharmonic wrote:
bunuel,

i had a slightly slightly different method for statement 1. Do you mind letting me know if the steps I took are OK?

also, what level question is this? My test is coming up soon and I've done all the OG and quant review 2nd edition problems. I wonder if going through these two samurai guides PS/DS would be the best use of my time in the final month and half.


(1).
x/|x| < x
before even going to x<0 and x>0, i knew that |x| is positive, so I multiplied it across
x < x*|x|

case x < 0

divide both sides by x, inequality must flip because x is negative
1 > |x| <-- this says YES to what we asked for


case x > 0

again, divide both sides by x, inequality won't flip
1 < |x| <-- this says NO

insufficient

(2). clearly says x has to be negative, therefore using (1) and (2) it is sufficient.


Though this didn't affect the final answer but the complete solution for (1) would be:
If x<0 and 1 > |x| then 1>-x --> -1<x and since x<0 then -1<x<0.
If x>0 and 1 < |x| the 1<x.

As for the question I'd say it's >650 level.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
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Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

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

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:
1 for \(x>0\) --> so we would have: \(1<x\);
Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.


Hi Bunuel,

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1.
But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.
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arjuntomar wrote:
Hi Bunuel,

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1.
But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.


From \(x>\frac{x}{-x}\) we don't dividing the inequality by some negative number, all we do is just reduce fraction. Since \(\frac{x}{-x}=-1\) (the same way as \(\frac{2}{-2}=-1\)) then from \(x>\frac{x}{-x}\) we have that \(x>-1\).

Hope it's clear.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]
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Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

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

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:
1 for \(x>0\) --> so we would have: \(1<x\);
Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.



Hey Bunuel,

Love your answers. But I want to suggest another method that is giving me some trouble when combining statesment (1) & (2)

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

Since \(0<|x|\) (x cannot equal 0), then we can rewrite statement 2, \(x<|x|\) as \(\frac{x}{|x|}< 1\).

We then subtract statment (1) and (2) as

1) \(\frac{x}{|x|}-\frac{x}{|x|}< x-1\) to get \(0< x-1\) or \(1<x\) showing that x is outside the boundary of \(-1<x<1\) and making the combined statements suffient. But \(1<x\), the derived statement of (1) and (2), contradicts statement (2), which claims that \(0<x\). What am I doing wrong?

Thank you!
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alphabeta1234 wrote:
Hey Bunuel,

Love your answers. But I want to suggest another method that is giving me some trouble when combining statesment (1) & (2)

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

Since \(0<|x|\) (x cannot equal 0), then we can rewrite statement 2, \(x<|x|\) as \(\frac{x}{|x|}< 1\).

We then subtract statment (1) and (2) as

1) \(\frac{x}{|x|}-\frac{x}{|x|}< x-1\) to get \(0< x-1\) or \(1<x\) showing that x is outside the boundary of \(-1<x<1\) and making the combined statements suffient. But \(1<x\), the derived statement of (1) and (2), contradicts statement (2), which claims that \(0<x\). What am I doing wrong?

Thank you!


You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, you cannot subtract \(\frac{x}{|x|}< 1\) from \(\frac{x}{|x|}< x\) since their signs are NOT in the opposite direction.

Hope it's clear.
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