If $(x+y)^{2}-(x-y)^{2}=20$, then $x y$ is equal to:
(a) 0
(b) 1
(c) 2
(d) 5
(e) 10
As $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=$ $x^{2}-2 x y+y^{2}$, we have:

$$
(x+y)^{2}-(x-y)^{2}=x^{2}+2 x y+y^{2}-x^{2}+2 x y-y^{2}=4 x y=20
$$

it follows that $x y=5$. The correct option is (d).