Ans: B
$\begin{array}{rcl}
\dfrac{m}{m+20} & = & \dfrac{1}{m} \\
m^2 & = & m+20 \\
m^2 – m – 20 & = & 0 \\
(m-5)(m+4) & = & 0
\end{array}$
$\begin{array}{rcl}
\dfrac{m}{m+20} & = & \dfrac{1}{m} \\
m^2 & = & m+20 \\
m^2 – m – 20 & = & 0 \\
(m-5)(m+4) & = & 0
\end{array}$
Therefore, $m=5$ or $m=-4$.
Since the number of yellow balls must be positive, then $m=5$.
