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

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.

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.

Hope this helps.

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.

Fun Fact => if K was an integer than then D would be the choice.

So fun, that I, in fact, committed that mistake xD

(1)\( k-1\) is positive. So k must be greater than \(1\),

Actually, \(k-1>0, \ k>1\), So;

\(\frac{1}{k} > 0\) Sufficient.

(1) \(k+1\) is positive, but k can be a negative fraction such as -0.5 is \(k+1>0\), k can be \(2\), Actually, \(k+1>0\), or\( k>-1\);

So;

\(\frac{1}{k} > 0\) can be Negative or Positive.
Insufficient.

The answer is \(A.\)

Hi Bunuel,

Please help me understand what I am assuming incorrectly here

In statement 2, we get the conclusion k> -1. However in the question statement, it mentioned that k is not -1, 0,1. Using this, we can assume that since k>-1, it is also >1 making it positive.

In my opinion, this should be marked D.

Posted from my mobile device

You are assuming that k is an integer but we are not given that. So, from (2) k can be any real number greater than -1 (including fractions and irrational numbers). So, for example, if k = -1/2 the answer is NO but if k = 10, then the answer is YES.

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