Marvin's trip

Marvin's average speed for the entire trip will be the total distance (in miles) divided by the total time (in hours). The total distance is $\text{10 + 2 = 12 miles}$ . The time spent for the first leg of the trip was $\dfrac{10 \text{miles}}{60 \text{mph}}=\dfrac{1}{6} \text{hours}$ and the time spent for the second leg was $\dfrac{2 \text{miles}}{30 \text{mph}}=\dfrac{1}{15} \text{hours}$ . With the $\dfrac{10}{60} \text{hour}$ stop in between, the total time spent is $\dfrac{10+4+10}{60}=\dfrac{2}{5} \text{hours}$ , so the average speed is $\dfrac{12}{\frac{2}{5}}=\boxed{30} \text{mph}$ .