Bunuel wrote:

Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1

Given

is |qp + q^2| > qp ?

can be written as, |q(p + q)| > qp

Statement 1: q > 0

Plug in numbers,

q = 1, p = -1, we get LHS = 0 & RHS = -1....we get YES

q = 1, p = 0, we get LHS = 1, RHS = 0...we get YES

q = 1, p = 1, we get LHS = 2, RHS = 1...we get YES

Statement 1 is Sufficient.

Statement 2: q/p > 1, hence q & p have the same sign.

Plug in numbers

q = 2, p = 1, we get LHS = 6, RHS = 2...we get YES

q = -2, p = -1, we get LHS = 6, RHS = 2, we get YES

Statement 2 is Sufficient.

Answer D.

Thanks,

GyM