smclean23 wrote:If k does not equal 0, 1, or -1, is 1/k >0?

1. 1/k-1>0
2. 1/k+1>0






Answer is A.

i go with A as well..
In Such DS question u just have to tell the answer in YES or NO

statement 1. 1/k-1>0
it can be rewritten as

1/k>1 thus we can it will surely be greater than o...sufficient

statement 2. 1/k+1>0

1/k>-1 but this tells us nothing abt whether its >0.. hence insufficient

thus A.

Hope that helps...

I still have no idea how you got this answer. Can you please rephrase or can anyone else out there help?

THANKS!

I am providing you with an explanation about condition 2, which is 1/k+1>0. If we deduct 1 on both side, we will have 1/k>-1, if 1/k =2, 1/K is greater than 0. However, you may have overlooked the numerical values between -1 and 0. What if 1/K = -1/4, will you still conclude that 1/K >0? Hope my explanation makes sense.

Since we are interested in finding out if 1/k>0, we only need to isolate 1/k on one side of the inequalities in (1) and (2).

(1) 1/k - 1 > 0

=> 1/k > 1 (adding 1 to both sides of the inequality)

anything greater than 1 will also be greater than 0.

Sufficient.

(2) 1/k + 1 > 0

=> 1/k > -1 (subtracting 1 from both sides of the inequality)

If something is greater than -1, it could be -1/2, 0 , 1, 2, ...

So we cannot tell for sure if 1/k is greater than 0 (1/k = -1/2 <0, and 1/k = 1 >0)

Not Sufficient.


Statement 1 alone is Sufficient. A

Thanks guys!

I didnt know that with an inequality you can just subtract or add a value, ie. subtracting or adding 1 from 1/k+1 or 1/k-1.

Wrongggggggg.......

everybody has read the question only wrong.....this is the problem with typing it out on a pc.....



I had this same problem....I searched it and found this thread

I KNOW HOW (A) ALONE SATISFIES THE PROBLEM.....HOWEVER, I FEEL THAT (B) ALONE WILL ALSO SATISFY THE PROBLEM....

FOR ALL VALUES OF K, WHICH ARE NEGATIVE (except -1), 1/(k+1) will be negative hence, <0, which does not satisfy the condition.....


Hence, we know that k should be Positive......

Hence, we can conclusively say that 1/k>0 by using B alone also....

so shudn the answer be "D"

Waiting for the experts to comment on this one....

Wrongggggggg.......

everybody has read the question only wrong.....this is the problem with typing it out on a pc.....

Given : k is not 1,0 or -1. It can be any other integer or fraction

Stmt II
1/k+1 > 0

k= - 1/2

1/k+1 > 0 but 1/k < 0 YES

K=5 1/K+1 > 0 and 1/k < 0 NO

Hence INSUFF

For stm I to be true k has to be positive( >1) so 1/k > 0

Hence A)

yeahh....thanks dude


i didn consider the values between 0 and -1.......when i read "not equal to 0, -1" i assumed k to be lesser than -1, which would give 1/(k+1) as always less than 0.....


stupid mistake on my part

stupid mistake on my part

Trust me most of us do! Its good to get it clarified as hopefully the next time we wont...

Good luck!

Question is 1/K > 0
now this value can be greater than 0, when and only when k>0
so if k is positive, 1/k will also be positive

St1: 1/(k -1)>0
as the whole expression is positive, and numerator (1) is positive, the denominator (k-1) will also be positive
Therefore, (k-1) > 0
k>1 SUFFICIENT


St1: 1/(k + 1)>0
as the whole expression is positive, and numerator (1) is positive, the denominator (1+k) will also be positive
(k+1)>0
k>-1
k can be both positive and negetive. INSUFFICIENT

Ans A

mental ,
Ur answer is very elegant.
Thank you.

I am still confused on this one. Why would (1) be sufficient? What if K=-2, the answer would be No. but if it equaled 2, then the answer would be yes

clsimmon wrote:I am still confused on this one. Why would (1) be sufficient? What if K=-2, the answer would be No. but if it equaled 2, then the answer would be yes

Dear friend, statement I says 1/(k-1) >0
From this we can conclude that (k-1)>0
So k>1.

Hence I is sufficient

II says 1/(k+1)>0
So k+1>0
Hence k>-1
From this we can't conclude whether 1/k is greater than 0 or not.

So A must be the answer

What i think is most imporant in this question is the fact that K is not stated to be an integer. Thus, if k>-1, k can be negative as well.

This is what always seems to get me in these kind of problems, i assume all variables are integers, even if its not stated so :\