Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?
$\textbf{(A)}\ 2 \frac{3}{4}  \qquad\textbf{(B)}\  3 \frac{3}{4} \qquad\textbf{(C)}\  4 \frac{1}{2} \qquad\textbf{(D)}\  5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}$
If Mr. Lopez chooses Route A, then he will spend $\frac{6}{30}=\frac{1}{5}$ hour, or $\frac{1}{5}\cdot60=12$ minutes.
If Mr. Lopez chooses Route B, then he will spend $\frac{9/2}{40}+\frac{1/2}{20}=\frac{11}{80}$ hour, or $\frac{11}{80}\cdot60=8\frac14$ minutes.
Therefore, Route B is quicker than Route A by $12-8\frac14=\boxed{\textbf{(B)}\  3 \frac{3}{4}}$ minutes.
~MRENTHUSIASM