A number is greater than five times its reciprocal by $\frac{19}{2}$. Find the number.

Let the number be $x$.According to the problem statement:
$$x-5\left(\frac{1}{x}\right)=\frac{19}{2}$$
$$x-\frac{5}{x}=\frac{19}{2}$$
$$\frac{x^2-5}{x}=\frac{19}{2}$$
$$2(x^2-5)=19x$$
$$2x^2-19x-10=0$$
Solving the quadratic equation by factoring:
$$2x^2-20x+x-10=0$$
$$2x(x-10)+1(x-10)=0$$
$$(2x+1)(x-10)=0$$
So, $x=10$ or $x=-\frac{1}{2}$.Matching with the options, the required number is $10$.