If $3 \leq p \leq 10$ and $12 \leq q \leq 21$, then the difference between the largest and smallest possible values of $\frac{p}{q}$ is
(A) $\frac{29}{42}$
(B) $\frac{29}{5}$
(C) $\frac{19}{70}$
(D) $\frac{19}{12}$
(E) $\frac{19}{84}$
The maximum possible value of $\frac{p}{q}$ is when $p$ is as large as possible (that is, 10) and $q$ is as small as possible (that is, 12). Thus, the maximum possible value of $\frac{p}{q}$ is $\frac{10}{12}=\frac{5}{6}$.

The minimum possible value of $\frac{p}{q}$ is when $p$ is as small as possible (that is, 3) and $q$ is as large as possible (that is, 21). Thus, the maximum possible value of $\frac{p}{q}$ is $\frac{3}{21}=\frac{1}{7}$.

The difference between these two values is $\frac{5}{6}-\frac{1}{7}=\frac{35}{42}-\frac{6}{42}=\frac{29}{42}$.