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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?

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Math Expert V
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If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 47% (02:37) correct 53% (02:53) wrong based on 260 sessions

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GMAT CLUB'S FRESH QUESTION

If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

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GMAT 1: 720 Q51 V36 Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Bunuel wrote:

GMAT CLUB'S FRESH QUESTION

If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

The data is sufficient if we get a unique value for the expression $$\frac{-|x| - 1}{x - 1}$$
A quick info about this expression. If x is negative then the expression will have a unique value, which is 1.
If x is 0, it will have a unique value, which is 1.
If x is positive it will have multiple values. For instance, if x = 2, the value of the expression is -3. If x = 3, the value of the expression -2.

So, at some level if we can establish that x is not positive, we will have a unique value. Else we will not have a unique value.

Statement 1: $$\sqrt{x^6} > x^3$$
$$\sqrt{x^6}$$ is non negative for real x and will be |x^3|.
For instance $$\sqrt{2^6}$$ = 8 and $$\sqrt{{(-2)^6}$$ will also be 8.
But 2^3 will be 8 and (-2)^3 will be -8.
So, if we know that $$\sqrt{x^6} > x^3$$, we can infer that x is negative.

If x is negative, the value of $$\frac{-|x| - 1}{x - 1}$$ will be 1. Take any negative x and check it. Negative integer, negative non integer. It will work for all values.

So, statement 1 ALONE is sufficient.

Statement 2:$$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$
We can rewrite the expression as |x| + 2x + 4|x| + 8x = -16
5|x| + 10x = -16.
The sum of the expression is -16, which is negative.
5|x| cannot be negative. So, 10x has to be negative if the sum is -16 => x has to be negative.

If x is negative, the expression $$\frac{-|x| - 1}{x - 1}$$ has a unique value which is 1.

Statement 2 ALONE is sufficient.

Each statement is INDEPENDENTLY sufficient.
Choice D.
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If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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$$\frac{-|x| - 1}{x - 1}$$

Looking at the question, we can identify that any positive number will yield different values for the expression(such as 2, 4, 6). But any non-positive number will always yield the answer 1(such as 0, -0.5, -1, -4 ). Hence if we are able to find out that x is non-positive, we can sufficiently get the answer

Statement 1: -

$$\sqrt{x^6} > x^3$$

Therefore $$|x^3| > x^3$$

This tells us that x is negative. Hence it is SUFFICIENT

Statement 1: -

$$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

Taking LCM, we can reduce the equation to

$$\frac{|x|+2x+4|x|+8x}{16} = -1$$

5|x| +10x + 16 =0

case 1 : - Now if x is positive, then the equation above will never be 0

case 2 : - Now if x is negative, then the equation above will be 0 for a value of x

Hence, this option is also SUFFICIENT

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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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pikolo2510 wrote:
$$\frac{-|x| - 1}{x - 1}$$

Looking at the question, we can identify that any positive number will yield different values for the expression(such as 2, 4, 6). But any non-positive number will always yield the answer 1(such as 0, -0.5, -1, -4 ). Hence if we are able to find out that x is non-positive, we can sufficiently get the answer

Statement 1: -

$$\sqrt{x^6} > x^3$$

Therefore $$|x^3| > x^3$$

This tells us that x is negative. Hence it is SUFFICIENT

Statement 1: -

$$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

Taking LCM, we can reduce the equation to

$$\frac{|x|+2x+4|x|+8<}{span>/16} = -1$$

5|x| +2x + 24 =0

case 1 : - Now if x is positive, then the equation above will never be 0

case 2 : - Now if x is negative, then the equation above will be 0 for a value of x

Hence, this option is also SUFFICIENT

Hi, You missed to multiply 8 by x. This will change the shape of the equation to be

5|x| +10x =-16
BSchool Forum Moderator P
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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Mo2men wrote:
pikolo2510 wrote:
$$\frac{-|x| - 1}{x - 1}$$

Looking at the question, we can identify that any positive number will yield different values for the expression(such as 2, 4, 6). But any non-positive number will always yield the answer 1(such as 0, -0.5, -1, -4 ). Hence if we are able to find out that x is non-positive, we can sufficiently get the answer

Statement 1: -

$$\sqrt{x^6} > x^3$$

Therefore $$|x^3| > x^3$$

This tells us that x is negative. Hence it is SUFFICIENT

Statement 1: -

$$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

Taking LCM, we can reduce the equation to

$$\frac{|x|+2x+4|x|+8<}{span>/16} = -1$$

5|x| +2x + 24 =0

case 1 : - Now if x is positive, then the equation above will never be 0

case 2 : - Now if x is negative, then the equation above will be 0 for a value of x

Hence, this option is also SUFFICIENT

Hi, You missed to multiply 8 by x. This will change the shape of the equation to be

5|x| +10x =-16

Thanks for pointing that typo out. Edited my original solution _________________
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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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What is the value of $$\frac{-|x| - 1}{x - 1}$$
if x<0, = $$\frac{x-1}{x-1}$$ = 1
if x>0, $$\frac{-x-1}{x-1}$$, the value depends on value of x

Statement 1) $$\sqrt{x^6} > x^3$$
it is clear that x is negative, because if x is non negative, $$\sqrt{x^6} = x^3$$
So, x is negative and $$\frac{x-1}{x-1}$$ = 1
SUFFICIENT

Statement 2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$
Since RHS is negative, We can straight away say that x is NEGATIVE, as all terms in LHS are in addition, we cant have the additions of positive terms = negative. (There is no need to solve the equation)
So x is negative and and $$\frac{x-1}{x-1}$$ = 1
SUFFICIENT

Bunuel wrote:

GMAT CLUB'S FRESH QUESTION

If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Bunuel wrote:

GMAT CLUB'S FRESH QUESTION

If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Bunuel wrote:

GMAT CLUB'S FRESH QUESTION

If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

Statement 1) $$\sqrt{x^6} > x^3$$ implies x is negative because square root is always positive so, square root of x^6 is positive and x^3 is negative. Only in the case of x being negative, this statement holds true. So, if x is negative. -|x|=x as |x| will be -x and -(-x)=x. The expression reduces to $$\frac{x - 1}{x - 1}$$=1. Sufficient.

Statement 2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$. Simplifying, we get $$\frac{5*|x|}{16} + \frac{5*x}{8}= -1$$ or $$\frac{5*|x|}{16} + \frac{10*x}{16}= -1$$. Further simplyfying, $$5*|x| + 10*x= -16$$. Now, |x| will always be positive, so, x has to be negative to equate to -16 on RHS. Also, multiple of x is higher than of multiple of |X|, so, x has to be negative in order to total to -16. If x is positive, the expression won't hold true. Following the logic in Statement 1, if x is negative, the expression in the problem reduces to $$\frac{x - 1}{x - 1}$$=1. Sufficient.

Hence, D is the correct answer.
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Re: If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Case 1. if x<0, = x−1x−1x−1x−1 = 1
Case 2. if x>0, −x−1x−1−x−1x−1, depends on x -> Lets check now

1. It clearly shows that x < 0. So 1 is sufficient.
2. L.H.S we have mod equation. R.H.S we have -1. So it is sure that x can not be positive. It has to be negative. So no need to solve. Directly case 1. Hence Sufficient.

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If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?  [#permalink]

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Bunuel wrote:
If x ≠ 1, what is the value of $$\frac{-|x| - 1}{x - 1}$$?

(1) $$\sqrt{x^6} > x^3$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$

$$x≥0:\frac{-|x| - 1}{x - 1}=\frac{-(x)-1}{x-1}=\frac{-x-1}{x-1}=\frac{-(x+1)}{(x-1)}$$
$$x<0:\frac{-|x| - 1}{x - 1}=\frac{-(-x)-1}{x-1}=\frac{x-1}{x-1}=1$$

(1) $$\sqrt{x^6} > x^3$$ sufic.

$$\sqrt{x^6}>x^3…(\sqrt{x^2}=|x|)…\sqrt{x^6}=|x^{6/2}|=|x^3|…|x^3|>x^3…x<0$$

(2) $$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1$$ sufic.

$$\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1…|x|+2x+4|x|+8x=-16…|5x|+10x=-16$$
$$x≥0:|5x|+10x=-16…(5x)+10x=-16…15x=-16…x=-16/15=invalid…(x≥0)$$
$$x<0:|5x|+10x=-16…(-5x)+10x=-16…5x=-16…x=-16/15=valid…(x<0)$$ If x ≠ 1, what is the value of (-|x| - 1)/(x - 1) ?   [#permalink] 22 Oct 2019, 08:58
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