dhirajx
If \(a<0\) and \(b<0\) , is \(\frac{(a-2p)}{(b-2p)}\) \(>\) \(\frac{a}{b}\) ?
(1) \(p>0\)
(2) \(a<b\)
Took time. We need to find a better way
We need to find \(\frac{(a-2p)}{(b-2p)}\) > a/b
Notice here that the question is asking if we subtract same term from Numerator and Denominator \frac{a}{b} then is the resulting fraction greater than original fraction a/b
you will have 4 case here
a>b, p>0
a>b, p<0
a<b, p<0
a<b, p>0
Clearly St 1 is not sufficient because for p>0, we can have 2 possible cases a>b and b>a and we get different result
Consider a=-3, b=-4 (a>b) and p=1 \(\frac{(-3-2)}{(-4-2)}\) =-5/-6 > -3/-4
Consider a=-4,b=-3 (b>a) and p=1 \(\frac{(-4-2)}{(-3-2)}\) or 6/5 < 4/3
Not sufficient
Similarly st 2 is not sufficient with above reasoning as p >0 or p<0
Combining 2 statements we get a<b and p>0
b=-3,a=-4 and p=1 then
Original fraction = 4/3
New fraction = \(\frac{(-4-2)}{(-3-2)}\) or 6/5 <4/3
Consider another value of p=1/2
\(\frac{(-4-1)}{(-3-1)}\)= or 5/4<4/3
Ans is C