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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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Manager
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If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  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]  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.

Hope it's clear.
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Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink]  25 Aug 2013, 03:49
1
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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.
D ?
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+1 Kudos if you find this answer helpful

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Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  17 Aug 2015, 07:32
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Expert's post
Turkish wrote:
Bunuel wrote:
Turkish wrote:
Hello Bunuel,

I am not getting this. why is is k>0?

Thanks

Turk

Which statement are you referring to?

In your explanation you said " Basically the question asks: is k>0? " I am not getting how the question is asking whether k>0?

The question asks whether $$\frac{1}{k}$$ is positive. The numerator, which is 1, is positive, thus in order the fraction to be positive, the denominator must also be positive.

Hope it's clear.
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Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  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]  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]  16 Mar 2012, 04:41
Expert's post
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!

Welcome to GMAT Club. Above is a solution to your problem. Please ask if anything remains unclear.

P.S. Please, DO NOT reword or shorten the questions you post, type them EXACTLY as they are presented in the original source. Thank you.
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If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  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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Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink]  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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If k does not equal 0, 1 or -1, is 1/k >0 [#permalink]  17 Aug 2015, 06:59
Bunuel wrote:
1. If k#0, 1 or -1 is $$\frac{1}{k}> 0$$?

(1) $$\frac{1}{k-1}> 0$$ --> $$k-1>0$$--> $$k>1$$, hence $$\frac{1}{k}>0$$. Sufficient.

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

Hello Bunuel,

I am not getting this. why is is k>0?

Thanks
Turk
Math Expert
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Posts: 30410
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Kudos [?]: 57463 [0], given: 8814

Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  17 Aug 2015, 07:06
Expert's post
Turkish wrote:
Bunuel wrote:
1. If k#0, 1 or -1 is $$\frac{1}{k}> 0$$?

(1) $$\frac{1}{k-1}> 0$$ --> $$k-1>0$$--> $$k>1$$, hence $$\frac{1}{k}>0$$. Sufficient.

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

Hello Bunuel,

I am not getting this. why is is k>0?

Thanks
Turk

Which statement are you referring to?
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Manager
Joined: 13 Jun 2012
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Kudos [?]: 58 [0], given: 104

Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]  17 Aug 2015, 07:28
Bunuel wrote:
Turkish wrote:
Bunuel wrote:
1. If k#0, 1 or -1 is $$\frac{1}{k}> 0$$?

(1) $$\frac{1}{k-1}> 0$$ --> $$k-1>0$$--> $$k>1$$, hence $$\frac{1}{k}>0$$. Sufficient.

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

Hello Bunuel,

I am not getting this. why is is k>0?

Thanks

Turk

Which statement are you referring to?

In your explanation you said " Basically the question asks: is k>0? " I am not getting how the question is asking whether k>0?
Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0   [#permalink] 17 Aug 2015, 07:28
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