Official Solution:


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

Work on the stem before moving to the statements.

If \(x \leq 0\), then \(|x|=-x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). So, if \(x\) is 0 or negative then the value of the fraction is 1.

If \(x > 0\), then \(|x|=x\) and in this case we'd have: \(\frac{-|x| - 1}{x - 1}=\frac{-x - 1}{x - 1}\). So, in this case the value of the fraction would depend on the value of \(x\).

(1) \(\sqrt{x^6} > x^3\). Take the square root from the left hand side: \(|x^3| > x^3\). For this to hold true, \(x\) must be negative, because if \(x\) is 0 or positive, then \(|x^3|\) would be equal to \(x^3\). From above, we know that if \(x < 0\), then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.

(2) \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = -1\). Notice that if \(x\) is positive, then \(\frac{|x|}{16} + \frac{x}{8} + \frac{|x|}{4} + \frac{x}{2} = positive +positive+positive+positive=positive\), not negative number \(-1\). So, \(x\) cannot be positive. From above, we know that if \(x\) is not positive, then \(\frac{-|x| - 1}{x - 1}=\frac{-(-x) - 1}{x - 1}=\frac{x - 1}{x - 1}=1\). Sufficient.


Answer: D
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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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\(\frac{-|x| - 1}{x - 1}\)

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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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


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

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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.

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

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


(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)

=−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