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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

Answer is D
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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

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

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

Answer is D

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

5|x| +10x =-16
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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

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

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

Answer is D

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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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

Answer D

Bunuel

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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Bunuel

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

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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Bunuel

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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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.

Answer : D
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Bunuel
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.

\(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1…|x|+2x+4|x|+8x=-16…|5x|+10x=-16\)
\(x≥0:|5x|+10x=-16…(5x)+10x=-16…15x=-16…x=-16/15=invalid…(x≥0)\)
\(x<0:|5x|+10x=-16…(-5x)+10x=-16…5x=-16…x=-16/15=valid…(x<0)\)

Answer (D)
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Bunuel

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

=−1

Statement 1: √x6>x3
√x6 is non negative for real x and will be |x^3|.
it only happens if x value is negative

So, we know that √x6>x3, we get x is -ve

If x is negative, the eqn will be 1

Clearly sufficient


Statement 2:|x|/16+x/8+|x|/4+x/2=−1|x|/16+x/8+|x|/4+x/2=−1
=>|x| + 2x + 4|x| + 8x = -16
5|x| + 10x = -16.

10x has to be negative if the sum is -16
=> x is negative.
Therefore the eqn equals to 1.
Sufficient

Therefore IMO D
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