Bunuel wrote:

If \(a<0\) and \(b<0\) , is \(\frac{(a-2p)}{(b-2p)}\) \(>\) \(\frac{a}{b}\) ?

(1) \(p>0\)

(2) \(a<b\)

We can slightly modify (a-2p)/(b-2p) > a/b by bringing a/b to the left side. Then we get:

(a-2p)/(b-2p) - a/b > 0 OR {ab - 2pb - ab + 2pa}/{b*(b-2p)} > 0 OR {2p*(a-b)}/{b*(b-2p)} > 0

Now since b is negative, we can multiply both sides by b and change the sign of inequality. So now the question is basically asking whether:

{2p*(a-b)}/{b-2p} < 0 or not.

(1) p > 0

So 2p is positive, and denominator (b-2p) is negative (since b is already negative). But we dont know whether (a-b) in the numerator is negative or positive. Not sufficient.

(2) a < b

So (a-b) in the numerator is negative. But we dont know whether '2p' and 'b-2p' are negative or positive. Not sufficient.

Combining the statements, p is positive. This makes '2p' as positive, 'b-2p' as negative. And since a < b this makes 'a-b' as negative.

So overall {2p*(a-b)}/{b-2p} becomes positive. And this gives NO as an answer to the question asked. Sufficient.

Hence

C answer