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Re: If x is not equal to 0, is |x| less than 1? [#permalink]
09 Jun 2012, 09:10

1

This post received KUDOS

Expert's post

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.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
14 Jun 2012, 04:21

1

This post received KUDOS

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!

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
14 Jun 2012, 04:31

1

This post received KUDOS

Expert's post

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.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
24 Apr 2013, 17:25

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?? _________________

When you feel like giving up, remember why you held on for so long in the first place.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
25 Apr 2013, 08:17

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?

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
25 Apr 2013, 23:56

1

This post received KUDOS

Expert's post

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.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
23 Jun 2013, 10:37

For number 1 why do we have two values for x<0? In other words why do we have x< 0 and -1<x<0?

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.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
11 Jul 2013, 13:43

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

|x|<1 Is -1<x<1

(1) x/|x|< x

x/|x|< x Two cases: x>0, x<0

x>0 x/|x|< x x/x<x 1<x

If x>0 and x>1 then x>1 If x>1 then |x| is NOT less than 1

x<0 x/|x|< x -x/|-x|<x -1<x

-1<x<0 if -1<x<0 then |x| IS less than 1 INSUFFICIENT

(2) |x| > x

If the absolute value of x is greater than x than x must be negative. However, |-x| could be less than 1 or greater than 1 depending on the value of x. INSUFFICIENT

1+2 x must be negative and x from #1 x is either >1 or between -1 and 0. Therefore, we know -1<x<0 which means that |x| is always less than 1. SUFFICIENT

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
06 Aug 2013, 07:42

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 bunel,

hw is that we can have two ranges x<0 & x>0. How we got this ranges?