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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x

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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 09 Jun 2012, 10:10
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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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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink] New post 09 Jun 2012, 13:56
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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.



FOR THIS QUESTION: is |x|< 1

From (1) - Two cases.
Case 1 x>0,
x/|x|< x => x/x < |x| => |x|> 1 (Ans to Q stem No)

Case 2 x<0,
x/|x|< x => x/x > |x| => |x|< 1 (Ans to Q stem Yes)

Hence Insuff.

From (2) - We simply get that x < 0, Insuff.

Combined (1) and (2), we see that only x is <0
Hence case 2. Sufficient

Ans C
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink] New post 14 Jun 2012, 05:21
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Hello Bunuel, I tried to solve like this:
Basically the question asks whether -1<x<1?
Condition 1: x/|x|<x
case 1: if x>0; x/x<x ==> 1<x ==> x>1
case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right}
Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink] New post 14 Jun 2012, 05:31
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pavanpuneet wrote:
Hello Bunuel, I tried to solve like this:
Basically the question asks whether -1<x<1?
Condition 1: x/|x|<x
case 1: if x>0; x/x<x ==> 1<x ==> x>1
case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right}
Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!


You've done everything right.

Though for case 2 you could do as follows: x<0 --> \frac{x}{-x}<x --> -\frac{x}{x}<x --> x is simply reduced: -1<x. Since we consider the range x<0 then -1<x<0.

Also for (1)+(2) we have that -1<x<0. So we can answer yes to the question whether -1<x<1.

Hope it helps.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink] New post 25 Jul 2012, 04:36
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Here is a video explanation that takes a slightly different approach using graphs of basic functions:

http://www.gmatquantum.com/shared-posts ... -xx-x.html

Cheers,
Dabral
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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 24 Apr 2013, 13:39
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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
.


if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink] New post 24 Apr 2013, 15:22
This to me seems like a 700 difficulty level. I almost put C but in the end decided on B.

I honestly hate these types of questions but know they will come.

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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 24 Apr 2013, 18:25
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.


Buenel terrific explanation. This system, that you use for solving inequations, is quite fundamental and hard to be wrong. But is it really always the quickest way??
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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 25 Apr 2013, 04:14
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oyabu 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
.


if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?


When x<0, then |x|=-x, thus \frac{x}{|x|}<x becomes \frac{x}{-x}<x --> -1<x but since x<0, then -1<x<0.

Hope it's clear.
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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 25 Apr 2013, 09:17
Bunuel wrote:

When x<0, then |x|=-x, thus \frac{x}{|x|}<x becomes \frac{x}{-x}<x --> -1<x but since x<0, then -1<x<0.

Hope it's clear.


Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?
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Re: If x is not equal to 0, is |x| less than 1? [#permalink] New post 26 Apr 2013, 00:56
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oyabu wrote:
Bunuel wrote:

When x<0, then |x|=-x, thus \frac{x}{|x|}<x becomes \frac{x}{-x}<x --> -1<x but since x<0, then -1<x<0.

Hope it's clear.


Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?


First of all |-x|=|x|. Next, if x<0, then |x|=|negative|=-x=-negative=positive.

It seems that you need to go through basics: math-absolute-value-modulus-86462.html
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Re: If x is not equal to 0, is |x| less than 1?   [#permalink] 26 Apr 2013, 00:56
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