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# Given k is a nonzero integer, is k > 0 ?

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Director
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Given k is a nonzero integer, is k > 0 ? [#permalink]

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18 Sep 2016, 23:57
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Given k is a nonzero integer, is $$k > 0$$ ?

(1) $$|k-4| =| k| +4$$

(2) $$k>k^3$$

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Director
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Re: Given k is a nonzero integer, is k > 0 ? [#permalink]

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19 Sep 2016, 02:14
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STAT1: |k-4| = |k| + 4

|k| + 4 will be a positive number and the absolute values of k and 4 will be added up.
|k-4| = |k + (-4)| this will give us the sum of absolute values of k and 4 only when k and -4 add up and this is possible only when k is also negative.
=> k < 0
=> SUFFICIENT

STAT2: k > k^3
=> possible values of k which will satisfy this statement are k<0 and k being a positive fraction. But k is a non-zero integer so k<0 is theonly solution
=> SUFFICIENT

So, answer will be D
Hope it helps!
GMATantidote wrote:
Given k is a nonzero integer, is $$k > 0$$ ?

1) $$|k-4| =| k| +4$$

2) $$k>k^3$$

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Re: Given k is a nonzero integer, is k > 0 ? [#permalink]

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19 Sep 2016, 02:49
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GMATantidote wrote:
Given k is a nonzero integer, is $$k > 0$$ ?

1) $$|k-4| =| k| +4$$

2) $$k>k^3$$

k is a non zero integer so the values it can take are ... -2, -1, 1, 2, 3, ...
Is k > 0?

1) $$|k-4| =| k| +4$$
According to the definition of absolute values,
|k - 4| = k - 4 if (k - 4) >= 0
|k - 4| = -(k - 4) = -k + 4 if (k - 4) < 0

Note that the the right hand side is |k| + 4. This can be equal to the second case only and that too when k < 0. So we know that k must be negative. We can answer the question with 'No'.
Sufficient.

2) $$k>k^3$$
On the number line, where is x greater than x^3? When either 0 < x< 1 or x < -1.
Here, since k must be an integer, it will not lie between 0 and 1. So k must be less than -1 i.e. k must be negative. We can answer the question with 'No'.
Sufficient.

P.S. - You must know the relation between x, x^2 and x^3 on the number line.
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Re: Given k is a nonzero integer, is k > 0 ? [#permalink]

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27 Jan 2018, 00:40
Statement 1: |k-4| = |k| + 4
squaring both sides, (as both LHS and RHS is positive, safe to square)
$$k^2 - 8k + 16 = k^2 + 8|k| + 16$$
=> -8k = 8|k|
=> -k = |k|
=> k <= 0 sufficient to answer our question

Statement 2: $$k > k^3$$
=> since k is an integer, k has to be negative to hold the equality true => sufficient

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Re: Given k is a nonzero integer, is k > 0 ? [#permalink]

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27 Jan 2018, 00:58
QZ wrote:
Given k is a nonzero integer, is $$k > 0$$ ?

(1) $$|k-4| =| k| +4$$

(2) $$k>k^3$$

Statemnt 1: As $$|K|$$ is always positive, so RHS will be always greater than 4. So, on LHS K has to be negative.

Statement 2: $$k(1-k)(1+k)$$ $$> 0$$. This is possible only when K is negative.

Hence, D.
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Re: Given k is a nonzero integer, is k > 0 ? [#permalink]

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28 Apr 2018, 22:21
We are required to find if k>0 or not.
St 1 : Square both sides. We get k=0 ; Sufficient
St 2 : Solve for inequality , we get k<-1. Keep in mind that k is an integer ; Sufficient

Option D it is !!
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Re: Given k is a nonzero integer, is k > 0 ?   [#permalink] 28 Apr 2018, 22:21
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# Given k is a nonzero integer, is k > 0 ?

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