1. 1/(K-1) >0

For this to be true K cannot be less than 1 and for all values that satisfy this condition, 1/K >0

(Sufficient)

2. 1/(K+1) >0

For this to be true K cannot be less than -1 and therefore K can be negative or positive for which 1/K can be greater than or less than zero.

(Not Sufficient)

Thanks for responding.

I'm still a little puzzled by statement 2. I think my thinking is really stuck in a rut.

For this to be true K cannot be less than -1. I agree.

Example if k = -3

1/(k+1) = 1/(-3+1) = 1/(-2) < 0. And so forth for all k < -1.

As far as I can see k can never be < 0 for this equation to hold true. Yet you state "... therefore K can be negative or positive...". If you could illustrate this condition with an example I would really appreciate it.

wally wrote:Hi Folks,

k is not equal to 0,1 or -1. Is 1/k >0?

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

Apparently the answer is A.

I was wondering if someone would be able to provide some insight as to how this may be.

[spoiler]In my view each statement is sufficient and hence the answer is D.
For statement 1 and 2 to hold true k > 0. Therefore 1/k > 0.[/spoiler]

IMO it is D.

In both stmts 1 and 2, K has to be +ve > 1 (coz k cannot be = 0, -1, 1).
so 1/k will be >0

Hope I dint miss anything.

please try with k = -1/2 in option 2 and you can see.. 1/K can be < 0 and with k = 1/2 . 1/k can be > 0.


Thus only A

Thanks. That's great. Just what I needed, a different perspective to shift my thinking.

Indeed if k = -1/2
1/(k+1) = (1/(1/2)) = 2 which > 0
whilst 1/k = 1(-1/2) = -2.

Therefore statement 2 is NOT SUFFICIENT.

Its good to pick nos and solve.
I'll try the normal process to explain.

Q is : is 1/k>0
If K is not equal to 0,1,-1.

For it to be true , first of all K has to >0, and K has to be a fraction.
So all we are looking for is:
Is 0<K<1

Stmt I: if 1/k-1>0
Then K>1 and it will not satisfy our condition of 0<k<1.
So 1/K is not >1
Sufficient

Stmt II:
If 1/K+1>0
We can't comment anything on K, as it can be >1,<1 or negative.
So , insufficient.


This, answer is A