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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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13 May 2015, 06:19

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.

Thanks for your nice explanations. I am not clear on one issue though. As we can multiply an inequality by a variable only if we know its sign, how can we multiply both side of an inequality by an absolute value? Would you please explain..
_________________

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.

Thanks for your nice explanations. I am not clear on one issue though. As we can multiply an inequality by a variable only if we know its sign, how can we multiply both side of an inequality by an absolute value? Would you please explain..

An absolute value of a number cannot be negative (it's 0 or positive), and since we are given that x is not 0, then |x| is positive only.
_________________

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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16 May 2015, 21:21

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.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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05 Jul 2015, 15:45

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.

I did mine a little differently.

If x ≠ 0, is |x| < 1?

Means: Is -1 < x < 1?

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

x < |x|*x If x is positive, then: \(\frac{x}{x} < |x|\) which is the same as \(1 < |x|\) This means that \(-1 > x > 1\)

If x is negative, then: \(\frac{x}{x} > |x|\) which is the same as \(1 > |x|\) This means that \(-1 < x < 1\) (we switch the direction of < to > because we divided by -1 to put |x| by itself).

These two answers are inconsistent: x is both less than 1 and greater than 1. So, insufficient.

(2) \(|x| > x\) This means that x is negative since the negative value of something is always less than the absolute value of something; whereas, a positive or 0 value of something is equal to its absolute value.

If x is equal to a negative value less than -1, then it's within the range -1 < x < 1. But, if x is equal to a large negative value, |x| can be out of range, for example, x = -1,000. Therefore, (2) is insufficient.

(1) + (2) |x| > x and |x| > 1 x is negative, and we know that x < -1. So, pick an arbitrary value less than -1, say, -5. Plug it into equation (1): \(\frac{-5}{|-5|} < -5\). Is \(-1< -5\)? False. x does not refer to values < -1.

|x| > x and |x| < 1 x is negative, and we know that x > -1. Therefore, \(-1 < x < 0\). Pick an arbitrary value between -1 and 0, non-inclusive, say, -0.5. Plug it into equation (1): \(\frac{-0.5}{|-0.5|} < -0.5\). Is \(-1 < -0.5\)? Yes, this works. Sufficient.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Jul 2015, 15:04

Here is perhaps a quicker way to solve this

Statement 1: \(\frac{x}{|x|} < x\)

This is equivalent to \(x < x |x|\) (The inequality sign does not change, because |x| is always positive.) Two case follow: a) x is positive and b) x is negative

a) when x is positive, \(\frac{x}{x}< |x|\). The inequality sign doesn't change because x is positive. So, this means, 1 < |x| b) when x is negative, \(\frac{-x}{-x}>|x|\). The inequality sign flips here because we are dividing both sides by a negative number. So, this means, 1 > |x|

From a) and b), 1 < |x|, and 1 > |x|. INSUFFICIENT.

Now, statement (2) |x| > x

This means x is negative. INSUFFICIENT.

Using statements 1 and 2, statement 2 tells us that x is negative and statement 1(b) tells us that 1 > |x| when x is negative. Hence SUFFICIENT. Answer is C

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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27 Jul 2015, 09:44

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.

Bunuel: What I was thinking is that x/|x| < x the same as x/x < |x| which then equals to 1 < |x| and as such statement 1 is suffiicient. Where am I going wrong?
_________________

"The fool didn't know it was impossible, so he did it."

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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27 Jul 2015, 10:11

samdighe 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.

Bunuel: What I was thinking is that x/|x| < x the same as x/x < |x| which then equals to 1 < |x| and as such statement 1 is suffiicient. Where am I going wrong?

x/|x| < x \(\neq\) x/x < |x| (when x<0)

You can not divide by a variable (in this case 'x') in an inequality when the sign of the variable is not clear.

Lets say

1/x < 1 and I ask is x>1 ?

If x >0 ---> in this case you can multiply both sides by x as the sign of inequality will not change ---> x>1

But if x<0, then 1/x < 1 will become x < 1 and NOT x>1. You can check this by putting x =-0.5 1/x < 1 yes but x is NOT >1 but is <1. The reason for this is that when you are given an inequality, without the knowledge the sign of a variable, you MUST not multiple or divide variables as the sign of inequality reverses when you multiply or divide by a negative number.

This is the reason why statement 1 is not sufficient.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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05 Sep 2015, 12:53

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,

Can you please explain this to me? I don't quite grasp it. Thanks in advance! You said:

(2) |x|>x. Well this basically tells that x is negative,

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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05 Sep 2015, 13:16

dlada wrote:

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,

Can you please explain this to me? I don't quite grasp it. Thanks in advance! You said:

(2) |x|>x. Well this basically tells that x is negative,

How does this tell us that x is negative?

You know that there are 2 cases for |x|:

Case 1: when x \(\geq\) 0 ---> |x| = x and

Case 2: When x<0---> |x| = -x

Now consider the following 2 cases,

Case 3: When x = 3, |x| = 3 and in this case, |x| = x but

Case 4: when x = -3 , |x| = -(-3)=3 and in this case, |x| > x. This is what is mentioned by Bunuel.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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18 Dec 2015, 11:21

1

This post received KUDOS

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,

for the first statement is it right if i solved it as shown below ?

if x>0 then |x|>1 and if x<0 then |x|< 1.. so insuff is my method right?

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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21 Feb 2016, 20:35

arjuntomar wrote:

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.

Hi,

I always thought that |-10| = 10. So x/|x| < x means that

1) if x > 0, then x / |x| < x = 1 < x, and 2) if x < 0, then -x / |x| < -x = -x/x < -x, and -1 < -x. Multiplying both the sides by -1 (sign change here) leads to 1 > x. I know i am wrong somwhere in my interpretation. Can someone throw some light please?

For the A, how did you assume that remember that x<0

Thanks

There we consider two cases: (A) when x<0 and (B) when x>0. So, A we discard the range which is not less than 0 and for B discard the range which is not more than 0.

Case 1: If x <0, x/|x| = -1 i.e. x/|x|< x can be rewritten as -1< x i.e. -1< x < 0 Giving answer to the question as YES Case 2: If x >0, x/|x| = +1 i.e. x/|x|< x can be rewritten as 1< x i.e. 1< x Giving answer to the question as NO NOT SUFFICIENT

Statement 2: |x| > x |x| can be greater than x only if x is Negative because in all other cases both will be equal i.e. x < 0 but x may be -0.5 Giving answer to the question as YES and x may be -1.5 Giving answer to the question as NO. Hence, NOT SUFFICIENT

Combining the two statements Combining -1< x < 0 and 1< x and x < 0

we get, only -1< x < 0 Giving answer to the question as YES. Hence, SUFFICIENT

Answer: option C
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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17 Aug 2016, 10:27

Bunuel wrote:

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.

I am still confused about this. Statement 1 can be re arranged to |X|>1 , so shouldn't this statement be sufficient to say that |X| is not less than 1?

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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18 Aug 2016, 08:36

drak777 wrote:

I am still confused about this. Statement 1 can be re arranged to |X|>1 , so shouldn't this statement be sufficient to say that |X| is not less than 1?

You are making the biggest mistake that you can make while dealing with variables in inequalities. Until and unless you are sure of the value of a variable, you CAN NOT cancel 'x' from both sides as the inequality sign changes for values <0. Thus , from statement 1, you WILL NOT get |x| > 1 .

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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26 Aug 2016, 10:35

udaymathapati wrote:

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?

I had a doubt , Plz tell.

x/ |x| < x, ok

then let x = -2 then value is -2/ 2 < -2 , whih is wrong , it means x can have only positve values starting from x>1 .

\(x\neq{0}\), 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?

I had a doubt , Plz tell.

x/ |x| < x, ok

then let x = -2 then value is -2/ 2 < -2 , whih is wrong , it means x can have only positve values starting from x>1 .

thus A is sufficient to answer in Yes or No .

The solution you quote clearly gives TWO ranges. Check highlighted parts above.
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