Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10163
GPA: 3.82
Re: If k does not equal 0, 1 or -1, is 1/k > 0 (1) 1/(k-1) > 0
[#permalink]
01 Feb 2016, 05:34
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.