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If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0

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If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink] New post 24 Aug 2011, 10:24
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If k does not equal 0, 1 or -1, is 1/k >0

(1) 1/(k-1) > 0
(2) 1/(k+1) > 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Feb 2012, 03:08, edited 1 time in total.
Edited the question and added the OA
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Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink] New post 19 Feb 2012, 03:09
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If k#0, 1 or -1 is \frac{1}{k}> 0?

Basically the question asks: is k>0?

(1) \frac{1}{k-1}> 0 --> denominator is positive: k-1>0--> k>1, hence \frac{1}{k}>0. Sufficient.

(2) \frac{1}{k+1}> 0--> denominator is positive: k+1>0 --> k>-1, hence k can be negative as well as positive: \frac{1}{k} may or may not be >0. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink] New post 25 Aug 2013, 03:49
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SUNGMAT710 wrote:
If k <> 0, 1, or -1, is 1/k > 0 ?
(1) 1/k-1 > 0
(2) 1/1+k > 0


wats d OA?

i think both the options give you an idea that 1/k >0. since k is not taking any of the values of 0,1,-1 .

for me.. the answer is: both individually sifficcient alone.
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Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink] New post 21 Feb 2012, 04:17
+1 A

But I did the dumb ass thing of trying to plug in numbers in the fractions itself :-(

Thanks Bunuel
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Can't understand the logic behind the 1/(k+1)>0 question [#permalink] New post 16 Mar 2012, 04:30
Hi,

This might be a very simple question but i can't seem to figure out for some reason. The question states that k canonot be equal to 1, 0, or -1 and asks if k>0?
a) 1/(k+1)>0
b) 1/(k-1)>0

I'm more concerned about statement 2...since k cannot be 1, 0, or -1....in order for 1/(k-1)>0, it has to be any number greater than 1 in order for the statement to make sense. So should statement 2 be sufficient to answer the question? The answer is "A" where statement one is the only statement that's sufficient. Maybe i'm just missing something here...any help is appreciated, thanks!
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Re: Can't understand the logic behind the 1/(k+1)>0 question [#permalink] New post 16 Mar 2012, 04:41
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mprince wrote:
Hi,

This might be a very simple question but i can't seem to figure out for some reason. The question states that k canonot be equal to 1, 0, or -1 and asks if k>0?
a) 1/(k+1)>0
b) 1/(k-1)>0

I'm more concerned about statement 2...since k cannot be 1, 0, or -1....in order for 1/(k-1)>0, it has to be any number greater than 1 in order for the statement to make sense. So should statement 2 be sufficient to answer the question? The answer is "A" where statement one is the only statement that's sufficient. Maybe i'm just missing something here...any help is appreciated, thanks!


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If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink] New post 12 Dec 2012, 12:40
yogeshsheth wrote:
st 1 says 1/(k-1)>0 that means k-1>0 that means k>1 that means k is +ve and thus 1/k>0 ... suff

st2 says 1/(k+1) >0 that means k+1>0 i.e k>-1 that means k can -ve fraction or k is positive . if k is -neg fraction 1/k <0 and if k is +ve 1/k >0 so insuff

Thus the answer is A.

What is the OA?



This is lovely. When I took the GMAT Prep and encountered this q, I thought that flipping denominator would require switching the inequality too. Clearly that is not the case. This approach is so much more efficient than is plugging in numbers.
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If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink] New post 25 Aug 2013, 02:49
If k <> 0, 1, or -1, is 1/k > 0 ?
(1) 1/k-1 > 0
(2) 1/1+k > 0
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Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink] New post 25 Aug 2013, 04:32
It must be A.

Again it is not mentioned that k is an integer.

1) 1/k-1>0 => k can be any value greater than 1 to satisfy this equation hence 1/k >0; sufficient.
2) 1/1+k>0 => if k=0.5 then 1/k >0 it holds true, but if k=-0.5 this equation still holds whereas, 1/k<0; therefore not sufficient.

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Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink] New post 25 Aug 2013, 06:34
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Re: If k <> 0, 1, or -1, is 1/k > 0 ?   [#permalink] 25 Aug 2013, 06:34
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