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If x is different from 0, is |x| < 1 ?

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If x is different from 0, is |x| < 1 ?  [#permalink]

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If x is different from 0, is |x| < 1 ?

(1) x*|x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks

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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 29 Jan 2012, 08:42
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carcass wrote:
If x is different from 0, is |x| < 1 ?

(1) x |x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks


If x is different from 0, is |x| < 1 ?

(1) x*|x|<x --> if \(x>0\) then we can divide both parts by it and we'l get \(|x|<1\) (so we would have an YES answer. Complete range for this case would be: \(0<x<1\)) BUT if \(x<0\) then we can also divide both parts by it though this time we should switch the sign of the inequality as we are dividing by negative value: \(|x|>1\) (so we would have a NO answer. Complete range for this case would be: \(x<-1\)). Not sufficient. (Basically we should know whether \(x\) is positive or negative).

(2) |x| > x --> well this basically tells that \(x\) is negative, as if \(x\) were positive (or zero) then \(|x|\) would be equal to \(x\). Still insufficient to say whether \(|x|<1\).

(1)+(2) From (2) \(x<0\) thus from (1) \(|x|>1\) and the answer to the question is NO (solution based on both statements is \(x<-1\)). Sufficient.

Answer: C.

Similar question: if-x-is-not-equal-to-0-is-x-less-than-1-1-x-x-x-86140.html

Hope it helps.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 29 Jan 2012, 09:32
I suppose that you have put in red my logic on stat 1 because not so clear.

Stat (1) : x > 1 or -x > 1.........x < -1

Stat (2) : x < x or -x > x.........x < -x

This is the logic that conduct me to say that the 2 stats are insufficient. I don't understand completely why in the 2 one |x| > x ........ x is < 0 ??'


I also read the topic that you suggested; your third reply is very helpful but the doubt remain on stat ( 2)

Thanks
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 29 Jan 2012, 09:58
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carcass wrote:
I suppose that you have put in red my logic on stat 1 because not so clear.

Stat (1) : x > 1 or -x > 1.........x < -1

Stat (2) : x < x or -x > x.........x < -x

This is the logic that conduct me to say that the 2 stats are insufficient. I don't understand completely why in the 2 one |x| > x ........ x is < 0 ??'


I also read the topic that you suggested; your third reply is very helpful but the doubt remain on stat ( 2)

Thanks

OK. First of all: I marked "|x|<1" in red for (1), because you don't get that |x| < 1 from this statement. If it were so then then you'd have an YES answer right away (as the question asks exactly about the same thing "is |x|<1")

Next, "is \(|x|<1\)?" means "is \(-1<x<1\)?"

(1) x*|x|<x holds true in two cases for \(x<-1\) and \(0<x<1\):
-----(-1)----(0)----(1)----
So as you can see this one is not sufficient to answer the question.

(2) |x| > x. Absolute value property: when \(x\geq{0}\) then \(|x|=x\) and when \(x<{0}\) (so when \(x\) is negative) then \(|x|=-x\). For the second case \(|x|=positive>x=negative\). That' why \(|x|>x\) means that \(x<0\) (or as I wrote if \(x\) were positive we would have \(|x|=x\) not \(|x|>x\)). You can also consider a simple example: \(|-3|>-3\) --> \(-3<0\). Thus the range for (2) is:
-------------(0)--------
Also insufficient to answer the question.

(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)---- --> \(x<-1\). Which makes the answer to the question "is \(-1<x<1\)?" NO.

Check Absolute Value chapter of Math Book for more on this topic: math-absolute-value-modulus-86462.html

Hope it's clear.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 29 Jan 2012, 11:18
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 27 Jul 2013, 09:50
Notice that they are asking absvalue(x) < 1 --> right away you should be thinking about the 5 cases: X<-1; -1<X<0; X=0; 0<X<1 and X>1 ---> they also tell you X is not equal to 0 so rule out that case.

Looking at (I) I just create a small table and solve for each potential case using smart number: 'X<-1 case' use -2; '-1<X<0 case' use -0.5; '0<X<1 case' use 0.5; and 'X>1 case' use 2; solving for each of these you will notice that 'X<-1 case' is true and '0<X<1 case' is true, since absvalue(x) < 1 can not be determined from this, (1) is insufficient

Looking at (II) it is insufficient as X could be -.5 or -2

Combining (I) and (II), you will notice that X can only be 'X<-1 case' thus we know that absvalue(x) is not < 1; thus sufficient, C.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 27 Jul 2013, 14:02
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If x is different from 0, is |x| < 1 ?
x≠0, is -1<x<1?

(1) x |x| < x
For this to hold true: 0<x<1 or x<-1

0<x<1
1/2 * |1/2| < 1/2
1/4 < 1/2 Valid

x<-1
-2 * |-2| < -2
-4 < -2 Valid

x could be between 0 and 1 or it could be less than -1.
INSUFFICIENT

(2) |x| > x
|x| is by definition ≥0 so for |x| to be greater than x, x must be negative. Depending on the value of -x, it may or may not satisfy the inequality.
INSUFFICIENT

1+2) 0<x<1 or x<-1 and x is negative. The only intersection is when x<-1 and "x is negative" meaning x<-1 and the inequality in the stem holds true.
SUFFICIENT

(C)
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Oct 2013, 01:11
i solved it in the following way.. what am i doing wrong ?

for x|x|<x

a. x>0: xx<x -> x<1 i.e. 0<x<1
b. x<0: -xx<x -> -x<1 -> x>-1 which is wrong.

ln another problem i solved x/|x| similarly, and got the correct answer .

for x/|x|< x

x>0: x/x<x -> 1<x
x<0: x/-x<x ->-1<x<0


i am confused ..
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Oct 2013, 02:44
pmsigns wrote:
i solved it in the following way.. what am i doing wrong ?

for x|x|<x

a. x>0: xx<x -> x<1 i.e. 0<x<1
b. x<0: -xx<x -> -x<1 -> x>-1 which is wrong.

i am confused ..


For (b) when x<0 you have: -x*x < x --> divie by negative x and flip the sign: -x>1 --> x<-1.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Oct 2013, 05:42
still confused ..
when -x*x<x is divided by -x, wouldn't the result be x>-1?
and wouldn't it be same as dividing by x and then by -1?
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Oct 2013, 05:45
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Oct 2013, 06:23
got it. thanks for the help.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 11 Dec 2013, 12:53
Bunuel wrote:
carcass wrote:
If x is different from 0, is |x| < 1 ?

(1) x |x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks


If x is different from 0, is |x| < 1 ?

(1) x*|x|<x --> if \(x>0\) then we can divide both parts by it and we'l get \(|x|<1\) (so we would have an YES answer. Complete range for this case would be: \(0<x<1\)) BUT if \(x<0\) then we can also divide both parts by it though this time we should switch the sign of the inequality as we are dividing by negative value: \(|x|>1\) (so we would have a NO answer. Complete range for this case would be: \(x<-1\)). Not sufficient. (Basically we should know whether \(x\) is positive or negative).

(2) |x| > x --> well this basically tells that \(x\) is negative, as if \(x\) were positive (or zero) then \(|x|\) would be equal to \(x\). Still insufficient to say whether \(|x|<1\).

(1)+(2) From (2) \(x<0\) thus from (1) \(|x|>1\) and the answer to the question is NO (solution based on both statements is \(x<-1\)). Sufficient.

Answer: C.

Similar question: if-x-is-not-equal-to-0-is-x-less-than-1-1-x-x-x-86140.html

Hope it helps.


Dear Bunuel,

I don`t understand why statement 1 is not sufficient. according to statement 1 we can infer [x]<1 and it is exactly what the question aks, so why we should test it one time with positive sign and another time with negative sign????? If we have to do so, why we don`t do that for [x]<1 in the question?????
Please clarify it...
Thanks a million...
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 12 Dec 2013, 03:04
hamidmax wrote:
Bunuel wrote:
carcass wrote:
If x is different from 0, is |x| < 1 ?

(1) x |x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks


If x is different from 0, is |x| < 1 ?

(1) x*|x|<x --> if \(x>0\) then we can divide both parts by it and we'l get \(|x|<1\) (so we would have an YES answer. Complete range for this case would be: \(0<x<1\)) BUT if \(x<0\) then we can also divide both parts by it though this time we should switch the sign of the inequality as we are dividing by negative value: \(|x|>1\) (so we would have a NO answer. Complete range for this case would be: \(x<-1\)). Not sufficient. (Basically we should know whether \(x\) is positive or negative).

(2) |x| > x --> well this basically tells that \(x\) is negative, as if \(x\) were positive (or zero) then \(|x|\) would be equal to \(x\). Still insufficient to say whether \(|x|<1\).

(1)+(2) From (2) \(x<0\) thus from (1) \(|x|>1\) and the answer to the question is NO (solution based on both statements is \(x<-1\)). Sufficient.

Answer: C.

Similar question: if-x-is-not-equal-to-0-is-x-less-than-1-1-x-x-x-86140.html

Hope it helps.


Dear Bunuel,

I don`t understand why statement 1 is not sufficient. according to statement 1 we can infer [x]<1 and it is exactly what the question aks, so why we should test it one time with positive sign and another time with negative sign????? If we have to do so, why we don`t do that for [x]<1 in the question?????
Please clarify it...
Thanks a million...


It seems that you need to brush-up your fundamentals in absolute value:
Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


As for your question: \(x*|x|<x\) does not mean that \(|x|<1\) (\(-1<x<1\)). This inequality holds for \(x<-1\) AND for \(0<x<1\).
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 23 Dec 2013, 20:13
WholeLottaLove wrote:
If x is different from 0, is |x| < 1 ?
x≠0, is -1<x<1?

(1) x |x| < x
For this to hold true: 0<x<1 or x<-1

0<x<1
1/2 * |1/2| < 1/2
1/4 < 1/2 Valid

x<-1
-2 * |-2| < -2
-4 < -2 Valid

x could be between 0 and 1 or it could be less than -1.
INSUFFICIENT

(2) |x| > x
|x| is by definition ≥0 so for |x| to be greater than x, x must be negative. Depending on the value of -x, it may or may not satisfy the inequality.
INSUFFICIENT

1+2) 0<x<1 or x<-1 and x is negative. The only intersection is when x<-1 and "x is negative" meaning x<-1 and the inequality in the stem holds true.
SUFFICIENT

(C)


I did all the same things as you up until your description in evaluating statements 1+2 together. I thought that the question stem (rephrased) was saying "is x a fraction" or "is 0<x<1". Even statements 1+2 together only tell me that x is negative, so doesn't that make the stem not true?
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 24 Dec 2013, 02:19
1
catalysis wrote:
WholeLottaLove wrote:
If x is different from 0, is |x| < 1 ?
x≠0, is -1<x<1?

(1) x |x| < x
For this to hold true: 0<x<1 or x<-1

0<x<1
1/2 * |1/2| < 1/2
1/4 < 1/2 Valid

x<-1
-2 * |-2| < -2
-4 < -2 Valid

x could be between 0 and 1 or it could be less than -1.
INSUFFICIENT

(2) |x| > x
|x| is by definition ≥0 so for |x| to be greater than x, x must be negative. Depending on the value of -x, it may or may not satisfy the inequality.
INSUFFICIENT

1+2) 0<x<1 or x<-1 and x is negative. The only intersection is when x<-1 and "x is negative" meaning x<-1 and the inequality in the stem holds true.
SUFFICIENT

(C)


I did all the same things as you up until your description in evaluating statements 1+2 together. I thought that the question stem (rephrased) was saying "is x a fraction" or "is 0<x<1". Even statements 1+2 together only tell me that x is negative, so doesn't that make the stem not true?


The question asks whether |x| < 1, which is equivalent to is -1<x<1?

Check here: if-x-is-different-from-0-is-x-126715.html#p1035973

Hope it helps.
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Jul 2014, 04:09
carcass wrote:
If x is different from 0, is |x| < 1 ?

(1) x |x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks


Maybe I am nitpicking but the above highlighted portion could have been put as x* |x| < x
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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 22 Jul 2014, 04:26
himanshujovi wrote:
carcass wrote:
If x is different from 0, is |x| < 1 ?

(1) x |x| < x

(2) |x| > x

I think in this way.

From stat 1 : |x| < 1 and so we have 2 solution INSUFF

From stat 2: the same INSUFF.

At this point I'm lost how to evaluate together ????

Thanks


Maybe I am nitpicking but the above highlighted portion could have been put as x* |x| < x

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Re: If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 23 Sep 2014, 08:04
If x is different from 0, is |x| < 1 ?

(1) x*|x| < x

(2) |x| > x
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1) x*|x| < x : on first look it seems to be sufficient, Divide by X from both side and we will get the answer, But this is the catch here.
We only know that X != 0 . X can e negative as well and if X is negative then it will reverse the sign. So we get two conditions here. X +ve and x-ve

For X +ve = divide by X = |X| <1
For X -ve = divide by -X = |X| >1
So not sufficient.

2) |x| > x : It says that X is -ve. Only in this scenerio |X| can be greater than X. This is also not sufficient.

Now Lets take ! and 2 both
We know from 2 that X is -ve so from 1 we get |X| >1. Answer is no |X| is not less than 1.
So C is sufficient
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If x is different from 0, is |x| < 1 ?  [#permalink]

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New post 09 Mar 2018, 23:42
Bunuel wrote:
carcass wrote:
I suppose that you have put in red my logic on stat 1 because not so clear.

Stat (1) : x > 1 or -x > 1.........x < -1

Stat (2) : x < x or -x > x.........x < -x

This is the logic that conduct me to say that the 2 stats are insufficient. I don't understand completely why in the 2 one |x| > x ........ x is < 0 ??'


I also read the topic that you suggested; your third reply is very helpful but the doubt remain on stat ( 2)

Thanks

OK. First of all: I marked "|x|<1" in red for (1), because you don't get that |x| < 1 from this statement. If it were so then then you'd have an YES answer right away (as the question asks exactly about the same thing "is |x|<1")

Next, "is \(|x|<1\)?" means "is \(-1<x<1\)?"

(1) x*|x|<x holds true in two cases for \(x<-1\) and \(0<x<1\):
-----(-1)----(0)----(1)----
So as you can see this one is not sufficient to answer the question.

Hope it's clear.


Hi Bunuel, can x also be x>-1 for Statement 1? Coz when I sub in the values of (-1/2) or (-1/4), x satisfies the equation. But on your number line it is in black, which indicates x>-1 does not hold true. Please kindly enlighten me.
If x is different from 0, is |x| < 1 ? &nbs [#permalink] 09 Mar 2018, 23:42

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