Paul has 6 boxes, each of which contains 12 trays. Paul also has 4 extra trays. If each tray can hold 8 apples, what is the largest possible number of apples that can be held by the 6 boxes and 4 extra trays?
Since each box contains 12 trays, then 6 boxes contain $12 \times 6=72$ trays.

Since Paul has 4 extra trays, then he has $72+4=76$ trays.

Since each tray can hold 8 apples, then the trays can hold $76 \times 8=608$ apples.

ANSWER: 608 apples