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(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.

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.

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

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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.

Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink]

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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.

-- Ramandeep
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--It's one thing to get defeated, but another to accept it.

(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.

(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.

Answer: A.

Hello Bunuel,

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

Thanks Turk

Which statement are you referring to?
_________________

(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.

Answer: A.

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?

(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.

A very fundamental or maybe silly question regarding such questions: when we are deducing from (1/K)>0 -> K>0.. how are we doing that, as assuming 0/0 on the RHS would be undefined.

(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.

A very fundamental or maybe silly question regarding such questions: when we are deducing from (1/K)>0 -> K>0.. how are we doing that, as assuming 0/0 on the RHS would be undefined.

Thanks

No, we are not assuming 0/0. When you are given a/b > 0 this means that the fraction a/b > 0 ---> 2 cases possible,

Case 1, either both a,b > 0 or

Case 2, both a,b < 0.

You can try the following examples to see the above 2 cases:

Case 1: a=1, b = 2, a/b = 0.5 > 0

Case 2, a=-1, b = -5 , a/b = 0.2 > 0

But if lets say you have a=-1 and b = 2 or a=2 and b= -3 , a/b <0 and NOT >0.

You can even remember this that in order for a fraction a/b to be >0 ---> both a,b MUST have the same signs. Either both of them are >0 or both of them are negative.

Coming back to the question,

When you are given 1/k > 0 ---> 1 is already >0 so based on the 'rule' above, k must also be positive in order for 1 and k to have the same 'sign'. Thus k>0.

Similarly, if 1/(k-1) > 0 ---> k-1 > 0 ---> k >1 etc.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) 1/(k-1) > 0 (2) 1/(k+1) > 0

When it comes to inequality DS questions, 2 things are important at all times. First is square. Secondly, when range of que includes range of con, the con is sufficient. Modify the original condition and the question. Multiply k^2 on the both equations and the sign of inequality doesn't change as k^2 is still a positive integer even when it's multiplied. There is 1 variable(k), which should match with the number of equations. So you need 1 equation, for 1) 1 equation, for 2) q equation, which is likely to make D the answer. For 1), multiply (k-1)^2 on the both equations, they become k-1>0, k>1. The range of que includes the range of con, which is sufficient. For 2), multiply (k-1)^2 on the both equations, they become k-1>0, k>1. The range of que doesn't include the range of con, which is not sufficient Therefore, the answer is A.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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