Is ((a-k)/(b-k))>((a+k)/(b+k))? is same as
Is ((a-k)/(b-k)) -((a+k)/(b+k)) > 0 ?
Is ((a-k)/(b-k)) -((a+k)/(b+k)) > 0 ?
Simplifying the inequation

Is k(a-b)/((b+k)(b-k))>0 ?

1) a>b>k

a>b => a-b>0
b>k => b-k>0
So the question is now is k/(b+k) > 0? and We don't have sufficient information to answer the question

2)k>0

So the question is now is (a-b)/((b+k)(b-k)) > 0 ? and We don't have sufficient information to answer the question

from 1 and 2

a>b => a-b>0
b>k => b-k>0
k>0
Since k > 0 and b > k, b+k >0
So k(a-b)/((b+k)(b-k))is definitely greater than 0. IMO C

knight247 wrote:OA is C

Reducing using Dividendo Law

Is [(a - k)/ (b - k)] - 1 > [(a + k)/ (b + k)] - 1?

More simply

Is [(a - b)/ (b - k)] > [(a - b)/ (b + k)]?

I. Since a > b > k, hence a - b > 0 and hence dividing both sides of the inequality under test [(a - b)/ (b - k)] > [(a - b)/ (b + k)] by a - b won't change the inequality. Therefore the question further reduces to

Is [1/ (b - k)] > [1/ (b + k)]?

Or

Is b - k < b + k?

Or

Is -k < k?

The answer is YES if k > 0, and NO if k < 0. Insufficient

II. If k is positive and there's no information about a and b that can relate k to it, this statement alone is not sufficient.

Taken together, the question reduces to

If k > 0, is -k < k?

The answer is doubtlessly [spoiler]YES. Sufficient

Take C[/spoiler]